Answer:
-0.912 m/s
Explanation:
When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.
[tex] (m_c + m_b)v_b + m_pv_p = 0[/tex]
where [tex]m_c = 24.6 kg, m_b = 39 kg, m_p = 5.8 kg[/tex] are the mass of the child, the boat and the package, respectively. [tex], v_p = 10m/s, v_b[/tex] are the velocity of the package and the boat after throwing.
[tex] (24.6 + 39)v_b + 5.8*10 = 0[/tex]
[tex]63.6v_b + 58 = 0[/tex]
[tex]v_b = -58/63.6 = -0.912 m/s[/tex]
Phil leans over the edge of a cliff and throws a rock upward at 5 m/s. Neglecting air resistance, two seconds later the rock's speed is A. 15 m/s. B. 5 m/s. C. 10 m/s. D. zero. E. none of the above
The final velocity of the rock is 15 m/s upwards. The rock's initial velocity is upwards at 5 m/s. The acceleration due to gravity is downwards at 9.8 m/s². Hence the correct option is A.
The initial velocity of the rock is upwards at 5 m/s. This is denoted by u.
The acceleration due to gravity is downwards at 9.8 m/s². This is denoted by a.
The time taken for the rock to reach its final velocity is 2 seconds. This is denoted by t.
The final velocity of the rock is denoted by v.
We can use the following equation to calculate the final velocity of the rock:
v = u + at
Substituting the values of u, a, and t, we get:
= 5 + 9.8 * 2
= 15 m/s
Therefore, the final velocity of the rock is 15 m/s upwards. Hence the correct option is A.
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The rock's speed two seconds after being thrown upward is zero. Gravity causes the object to slow down and eventually come to a stop before starting to fall back down.
Explanation:
The rock's speed two seconds after being thrown upward is zero (D).
When an object is thrown upwards, its speed decreases due to the force of gravity pulling it downward. Gravity causes the object to slow down and eventually come to a stop before starting to fall back down. In this case, since the rock's speed is zero after two seconds, it means it has reached its highest point and is now starting to fall back down.
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Which of the following observations represent conclusive evidence of an interaction? (Select all that apply.)
a. Change of temperature without change of velocity
b. Change of identity without change of velocity
c. Change of position without change of velocity
d. Change of shape or configuration without change of velocity
e. Change of direction without change of speed
Answer:
a,b,d and e are correct.
Explanation:
a) Change of temperature without change of velocity is a conclusive evidence of an interaction. As the temperature of a body will only change if heat energy is flows in or out of the body with surrounding. So, option a) is correct.
b) Change of the direction without change of speed is an evidence of an interaction. The direction changes when the object is under a perpendicular acceleration acceleration which clearly means the particle is interacting or external force applied. So, option b is correct.
d) Change of shape or configuration without change of velocity is conclusive evidence of an interaction. The shape can change only if external stress is applied on the particle. So, Option d is correct.
e) Change of identity without change of velocity is a conclusive evidence of an interaction. The identity of an object can only change if it interacts with surroundings. So, option e is correct
Therefore the only incorrect option is c .
In this exercise we have to use our knowledge of physics to identify the best alternative that corresponds to each of the given questions:
the only incorrect alternative is the letter C
So analyzing each alternative, we find that:
a) Change of hotness outside change of speed exist a evidence of an interplay.
As the temperature of a main part of written work will only change if heat strength exist flows fashionable or exhausted the accompanying encircling.
So, alternative A happen correct.
b) Change of the course outside change of speed happen an evidence of an interaction.
The management changes when the object happen secondary a at right angles to increasing speed acceleration which without any doubt wealth the piece exist communicate or outside force applied.
So, alternative B exist correct.
d) Change of shape or arrangement outside change of speed exist evidence of an interplay.
The shape can change only if outside stress happen used ahead of the atom.
So, alternative D exist correct.
e) Change of similarity outside change of speed happen a evidence of an interplay.
The person's individuality of an object can only change if it communicate accompanying environment.
So, alternative E exist correct
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How much heat Q H QH does a heat pump with a coefficient of performance of 2.10 2.10 deliver when supplied with 1.82 kJ 1.82 kJ of electricity?
The operating coefficient or performance coefficient of a heat pump is the ratio between the heating or cooling provided and the electricity consumed. The higher coefficients are equivalent to lower operating costs. The coefficient can be greater than 1, because it is a percentage of the output: losses, other than the thermal efficiency ratio: input energy. Mathematically can be written as,
[tex]\text{Coefficient of performance} = \frac{\text{Heat extracted}}{\text{Electrical energy supplied}}[/tex]
[tex]COP = \frac{Q}{E}\rightarrow Q =COP\times E[/tex]
Replacing,
[tex]Q = 2.1\times 1.82kJ[/tex]
[tex]Q = 3.822kJ[/tex]
Therefore the heat is 3.822kJ
Draw a circuit that contains a 5Ω resistance, a 10V independent voltage source, and a 2-AA independent current source. Connect all three elements in series.
Answer: I have attached the circuit diagram.
Explanation: NOTE that an independent voltage or current source is denoted in a circle. That is how you identify an independent source.
It simply means that the voltage and current source value does not depend on the value of voltage and current elsewhere in the circuit.
It highly utilised when using Mesh analysis on an electric circuit.
Please I have attached the circuit diagram connected in series.
Check it out!
A block lies on a plane raised an angle theta from the horizontal. Threeforces act upon the block: F_w_vec, the force of gravity;F_n_vec, the normal force;and F_f_vec, the force of friction. Thecoefficient of friction is large enough to prevent the block fromsliding .Part A)Because the block is not moving, the sum ofthe y components of the forces acting on the block must bezero. Find an expression for the sum of the y componentsof the forces acting on the block, using coordinate system b.Express your answer in terms of someor all of the variables F_n, F_f, F_w, and theta.\sum f_y=Part B)Because the block is not moving, the sum ofthe x components of the forces acting on the block must bezero. Find an expression for the sum of the x componentsof the forces acting on the block, using coordinate system b.Express your answer in terms of someor all of the variables F_n, F_f, F_w, and theta.\sum F_{x}=0Part C)To find the magnitude of the normal force,you must express F_n in terms of F_w since F_f is an unknown. Using the equations youfound in the two previous parts, find an expression for F_n involving F_w and theta but not F_f.F_n =Diagram:In the diagram, a block is placed on an inclined surface with its both x and y components
Answer:
A) N = W cos θ , B) fr = W sin θ, C) N = W cos θ
Explanation:
To find the required expressions, let's make a free body diagram of the block, see attached.
In the problem of inclined plane the reference system used the x axis is parallel to the plane and the y axis is perpendicular
In this reference system the only force that we must decompose is the body weight F_w = W) for this we use trigonometry
Note that the angle of the plane is equal to the angle between the axis and the weight, therefore
sin θ = [tex]W_{x}[/tex] / W
cos θ = [tex]W_{y}[/tex] / W
[tex]W_{x}[/tex] = W sin θ
[tex]W_{y}[/tex] = W cos θ
B) Let us write Newton's second law for each axis, in this case as the block is still the acceleration is zero
X axis
fr - [tex]W_{x}[/tex] = 0
fr = W sin θ
A) Y Axis
N - [tex]W_{y}[/tex] = 0
N = W cos θ
C) N = W cos θ
Replacing an object attached to a spring with an object having 1 4 the original mass will change the frequency of oscillation of the system by a factor of?
Answer:
f = 2 f₀
Explanation:
A mass-spring system has angular velocity
w = √ k / m
The angular velocity is related to the frequency
w = 2π f
f = 1 /2π √k / m
If we change the mass for another that is ¼ of the initial mass
m = ¼ mo
We replace
f = 1 / 2π √(k 4 / mo)
f = (1 / 2π √k/m₀) √ 4
f = 2 f₀
In summary the frequency doubles from the initial frequency
The frequency of oscillation of the system by a factor of should be considered as the f = 2 f₀.
Frequency of oscillation:Since
A mass-spring system has angular velocity that applied the following equation
w = √ k / m
Also,
The angular velocity should be related to the frequency
w = 2π f
f = 1 /2π √k / m
Now
In the case when we change the mass for another that is ¼ of the initial mass
So,
m = ¼ mo
Now
f = 1 / 2π √(k 4 / mo)
f = (1 / 2π √k/m₀) √ 4
f = 2 f₀
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Which statement best describes the motion of the object whose position versus time graph is given below? (Take the positive direction to be to the right.)
Answer:
The object is moving to the right
Question:
See attachment for the complete question
Explanation:
Position vs time graph is a graphical representation of the distance covered by a particular motion plotted against the time taken.
According to the attached image;
- since the position increases with increase in time as seen on the graph, the motion is positive that means the object is moving to the right.
- the graph is not straight which means the velocity is changing (not constant speed) and also according to the shape of the curve the velocity is decreasing with time(deceleration)
The velocity is the change in position of the object with time. The given object is moving in the right direction.
Position vs time graph:It is a graphical representation of the distance covered by a moving object plotted against the time taken by the object.
In the graph, the position increases with an increase in time as seen on the graph in which the relation is positive.The graph is not straight which means the velocity of the object is changing. The shape of the velocity curve is decreasing with time which means the object is decelerating.Therefore, the given object is moving in the right direction.
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PART ONE
A student sits on a rotating stool holding two
2 kg objects. When his arms are extended
horizontally, the objects are 1 m from the axis
of rotation, and he rotates with angular speed
of 0.73 rad/sec. The moment of inertia of the
student plus the stool is 8 kg m2
and is assumed to be constant. The student then pulls
the objects horizontally to a radius 0.28 m
from the rotation axis.
Calculate the final angular speed of the
student.
Answer in units of rad/s.
(USE PICTURES)
PART TWO
Calculate the change in kinetic energy of the
system.
Answer in units of J.
Answers:
a) [tex]1.05 rad/s[/tex]
b) [tex]1.38 J[/tex]
Explanation:
a) Final angular velocity :
Before solving this part, we have to stay clear that the angular momentum [tex]L[/tex] is conserved, since we are dealing with circular motion, then:
[tex]L_{o}=L_{f}[/tex]
Hence:
[tex]I_{i} \omega_{i}=I_{f} \omega_{f}[/tex] (1)
Where:
[tex]I_{i}[/tex] is the initial moment of inertia of the system
[tex]\omega_{i}=0.73 rad/s[/tex] is the initial angular velocity
[tex]I_{f}[/tex] is the final moment of inertia of the system
[tex]\omega_{f}[/tex] is the final angular velocity
But first, we have to find [tex]I_{i}[/tex] and [tex]I_{f}[/tex]:
[tex]I_{i}=I_{s}+2mr_{i}^{2}[/tex] (2)
[tex]I_{f}=I_{s}+2mr_{f}^{2}[/tex] (3)
Where:
[tex]I_{s}=8 kgm^{2}[/tex] is the student's moment of inertia
[tex]m=2 kg[/tex] is the mass of each object
[tex]r_{i}=1 m[/tex] is the initial radius
[tex]r_{f}=0.28 m[/tex] is the final radius
Then:
[tex]I_{i}=8 kgm^{2}+2(2 kg)(1 m)^{2}=12 kgm^{2}[/tex] (4)
[tex]I_{f}=8 kgm^{2}+2(2 kg)(0.28 m)^{2}=8.31 kgm^{2}[/tex] (5)
Substituting the results of (4) and (5) in (1):
[tex](12 kgm^{2}) (0.73 rad/s)=8.31 kgm^{2}\omega_{f}[/tex] (6)
Finding [tex]\omega_{f}[/tex]:
[tex]\omega_{f}=1.05 rad/s[/tex] (7) This is the final angular speed
b) Change in kinetic energy:
The rotational kinetic energy is defined as:
[tex]K=\frac{1}{2}I \omega^{2}[/tex] (8)
And the change in kinetic energy is:
[tex]\Delta K=\frac{1}{2}I_{f} \omega_{f}^{2}-\frac{1}{2}I_{i} \omega_{i}^{2}[/tex] (9)
Since we already calculated these values, we can solve (9):
[tex]\Delta K=\frac{1}{2}(8.31 kgm^{2}) (1.05 rad/s)^{2}-\frac{1}{2}(12 kgm^{2}) (0.73 rad/s)^{2}[/tex] (10)
Finally:
[tex]\Delta K=1.38 J[/tex] This is the change in kinetic energy
To find the final angular speed of the rotating student who pulls objects closer, we apply the conservation of angular momentum, which allows us to solve for the new angular speed. To calculate the change in kinetic energy, we compare the initial and final kinetic energies of the system, considering the changes in moment of inertia and angular speed.
Explanation:The question involves the concept of conservation of angular momentum and relates to the change in angular speed when a student on a rotating stool moves two 2 kg objects closer to the axis of rotation. Initially, the objects are 1 m away from the rotation axis and the system rotates with an angular speed of 0.73 rad/sec. The moment of inertia (I) of the student plus the stool is 8 kg·m2, and the student pulls the objects to a radius of 0.28 m.
Using conservation of angular momentum, we can say that the initial angular momentum (Linitial) equals the final angular momentum (Lfinal). The formula for angular momentum is L = I·ω, where ω is the angular speed. Therefore:
Linitial = (I + 2·m·rinitial2)·ωinitial
Lfinal = (I + 2·m·rfinal2)·ωfinal
Plugging in the values:
8 kg·m2 + 2(2 kg)(1 m)2)· 0.73 rad/s = (8 kg·m2 + 2(2 kg)(0.28 m)2)·ωfinal
Solving for ωfinal, we find the final angular speed of the student.
For the change in kinetic energy (KE), we calculate the initial and final kinetic energies using KE = (1/2)Iω2, and then find the difference between them to get the change in kinetic energy.
A railroad diesel engine weighs 4 times as much as a flatcar. If the engine coasts
at 5 km/h into a flatcar that is initially at rest, how fast do the two coast after
they couple together?
Answer:
[tex]\displaystyle v'=4\ km/h[/tex]
Explanation:
Linear Momentum
The total linear momentum of an isolated system (with no external forces) is conserved regardless of the internal mutual interactions of its parts. Recall the momentum is obtained by multiplying the speed by the object's mass. We'll refer to the railroad diesel engine as mass 2 and the flatcar as mass 1.
We have the following data
[tex]\displaystyle m_2=4\ m_1[/tex]
[tex]\displaystyle v_1=0[/tex]
[tex]\displaystyle v_2=5\ km/h[/tex]
Since both objects remain coupled after their encounter, the final speed is common to both:
[tex]\displaystyle v_1'=v_2'=v'[/tex]
Let's sketch the principle of conservation of linear momentum as follows
[tex]m_1v_1+m_2v_2=m_1v_1'+m_2v_2'[/tex]
Using the mentioned conditions for the speeds
[tex]\displaystyle m_1\ v_1+m_2\ v_2=(m_1+m_2)\ v'[/tex]
Solving for v'
[tex]\displaystyle v'=\frac{m_1\ v_1+m_2\ v_2}{m_1+m_2}[/tex]
[tex]\displaystyle v'=\frac{m_1.0+4m_1\ v_2}{m_1+4m_1}[/tex]
[tex]\displaystyle v'=\frac{4\cancel{m_1} \ v_2}{5 \cancel{m_1}}[/tex]
[tex]\displaystyle v'=\frac{4}{5}v_2[/tex]
[tex]\displaystyle v'=\frac{4}{5}\ 5km/h[/tex]
[tex]\boxed{\displaystyle v'=4\ km/h}[/tex]
To find the speed at which the two vehicles coast after coupling together, we can apply the principle of conservation of momentum. The railroad diesel engine weighs 4 times as much as the flatcar, so we can use this information to calculate the final velocity of the coupled vehicles.
Explanation:To solve this problem, we can apply the principle of conservation of momentum. The momentum before the coupling is equal to the momentum after the coupling.
Let's assume the mass of the flatcar is m kg. According to the given information, the mass of the railroad diesel engine is 4 times the mass of the flatcar, so the mass of the engine is 4m kg.
Before the coupling, the momentum of the engine is (4m)(5 km/h) = 20m km/h, and the momentum of the flatcar is (m)(0) = 0.
After the coupling, the combined mass is (4m+m) = 5m kg, and their common velocity is v km/h. Using the principle of conservation of momentum, we can write the equation:
(20m) + (0) = (5m)(v)
Simplifying the equation, we have:
20m = 5mv
Dividing both sides by 5m, we get:
4 = v
Therefore, the two coast after they couple together at a speed of 4 km/h.
A horizontal 745 N merry-go-round of radius
1.45 m is started from rest by a constant
horizontal force of 56.3 N applied tangentially
to the merry-go-round.
Find the kinetic energy of the merry-goround after 3.62 s. The acceleration of gravity
is 9.8 m/s^2
Assume the merry-go-round is a solid cylinder.
Answer in units of J.
Answer:
The kinetic energy of the merry-goround after 3.62 s is 544J
Explanation:
Given :
Weight w = 745 N
Radius r = 1.45 m
Force = 56.3 N
To Find:
The kinetic energy of the merry-go round after 3.62 = ?
Solution:
Step 1: Finding the Mass of merry-go-round
[tex]m = \frac{ weight}{g}[/tex]
[tex]m = \frac{745}{9.81 }[/tex]
m = 76.02 kg
Step 2: Finding the Moment of Inertia of solid cylinder
Moment of Inertia of solid cylinder I =[tex]0.5 \times m \times r^2[/tex]
Substituting the values
Moment of Inertia of solid cylinder I
=>[tex]0.5 \times 76.02 \times (1.45)^2[/tex]
=> [tex]0.5 \times 76.02\times 2.1025[/tex]
=> [tex]79.91 kg.m^2[/tex]
Step 3: Finding the Torque applied T
Torque applied T = [tex]F \times r[/tex]
Substituting the values
T = [tex]56.3 \times 1.45[/tex]
T = 81.635 N.m
Step 4: Finding the Angular acceleration
Angular acceleration ,[tex]\alpha = \frac{Torque}{Inertia}[/tex]
Substituting the values,
[tex]\alpha = \frac{81.635}{79.91}[/tex]
[tex]\alpha = 1.021 rad/s^2[/tex]
Step 4: Finding the Final angular velocity
Final angular velocity ,[tex]\omega = \alpha \times t[/tex]
Substituting the values,
[tex]\omega = 1.021 \times 3.62[/tex]
[tex]\omega = 3.69 rad/s[/tex]
Now KE (100% rotational) after 3.62s is:
KE = [tex]0.5 \times I \times \omega^2[/tex]
KE =[tex]0.5 \times 79.91 \times 3.69^2[/tex]
KE = 544J
two 10-kilogram boxes are connected by a massless string that passes over a massless frictionless. What is tension in the string
Answer:
The tension in the string is 87.72 N.
Explanation:
Given that,
Mass of two each boxes = 10 kg
Suppose the inclined plane makes an angle is 60° with the horizontal and the coefficient of kinetic friction between the box and plane is 0.15.
We need to calculate the acceleration
Using balance equation
[tex]mg-T=ma[/tex]
Put the value into the formula
[tex]10g-T=10a[/tex]
[tex]98-T=10a[/tex]....(I)
Again using balance equation
[tex]T-mg\sin\theta-F_{f}=ma[/tex]
[tex]T-mg\sin\theta-\mu mg=ma[/tex]
Put the value into the formula
[tex]T-10g\sin60-\mu10\cos60=10a[/tex]
[tex]T-10\times9.8\dfrac{\sqrt{3}}{2}-0.15\times10\times9.8\cos\times\dfrac{1}{2}=10a[/tex]
[tex]T-84.8+7.35=10a[/tex]
[tex]T-77.45=10a[/tex]...(II)
From equation (I) and (II)
[tex]98-77.45=20a[/tex]
[tex]a=\dfrac{98-77.45}{20}[/tex]
[tex]a=1.0275\ m/s^2[/tex]
Put the value of acceleration in equation (I)
[tex]98-T=10\times1.028[/tex]
[tex]-T=10\times1.028-98[/tex]
[tex]T=87.72\ N[/tex]
Hence, The tension in the string is 87.72 N.
The tension in the string connecting the two 10-kilogram boxes is 98 N.
Explanation:The tension in the string connecting the two 10-kilogram boxes can be determined using Newton's second law and the equilibrium equation for the x-direction. Based on these equations, the tension in the shorter string is twice the tension in the longer string. Therefore, the tension in the string connecting the two boxes would be:
Tension = weight of the hanging mass
So, the tension in the string would be equal to the weight of each 10-kilogram box which is:
Tension = (10 kg) × (9.8 m/s²) = 98 N
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The small piston of a hydraulic lift has a crosssectional area of 2.23 cm2 and the large piston 297 cm2 . What force must be applied to the small piston for the lift to raise a load of 2.4 kN? (In service stations, this force is usually exerted by compressed air.) Answer in units of N.
Answer:
The force that must be applied to the small piston = 18.02 N.
Explanation:
Hydraulic Press: This is a device that produce a very large force to compress something e.g printing press.
From pascal's principle,
In an hydraulic press,
f/a = F/A......................... Equation 1
Where f = force applied to the small piston, a = area of the small piston, F = force required to lift the load/ force produced at the large piston. A = Area of the large piston.
Making f the subject of the equation above,
f = F×a/A.............................. Equation 2
Given: F = 2.4 kN = 2400 N, a = 2.23 cm², A = 297 cm².
Substituting into equation 2
f = 2400(2.23)/297
f = 18.02 N.
Thus the force that must be applied to the small piston = 18.02 N.
A spacecraft is in a 400-km by 600-km LEO. How long (in minutes) does it take to coast from the perigee to the apogee?
To solve this problem we will use Kepler's third law for which the period is defined as
[tex]T= 2\pi \sqrt{\frac{a^3}{GM}}[/tex]
The perigee altitude is the shortest distance between Earth's surface and Satellite.
The average distance of the perigee and apogee of a satellite can be defined as
[tex]a = R+\frac{r_1+r_2}{2}[/tex]
Here,
R = Radius of Earth
[tex]r_1[/tex]= Lower orbit
[tex]r_2[/tex]= Higher orbit
Replacing we have,
[tex]a = 6378.1+\frac{400+600}{2}= 6878.1km[/tex]
Time Taken to fly from perigee to apogee equals to half of orbital speed is
[tex]t = \frac{T}{2}[/tex]
[tex]t = \pi \sqrt{\frac{a^3}{GM}}[/tex]
Replacing,
[tex]t =\pi \sqrt{\frac{(6878.1*10^3)^3}{(6.67*10^{-11})(6*10^{24})}}[/tex]
[tex]t = 2832.79s[/tex]
[tex]t = \text{47min 12.8s}[/tex]
Therefore will take around to 47 min and 12.8s to coast from perigee to the apogee.
Final answer:
The task requires understanding of orbital mechanics and Kepler's laws to calculate the coasting time from perigee to apogee in a specific orbit, but the information provided is not sufficient for a direct calculation without additional details on velocity or using specific orbital mechanics equations.
Explanation:
The question asks about the time it takes for a spacecraft to coast from the perigee (the point in the orbit closest to Earth) to the apogee (the point in the orbit farthest from Earth) in a low Earth orbit (LEO) with altitudes of 400 km and 600 km respectively. To solve this, we would typically use Kepler's laws and orbital mechanics equations. However, the information provided does not directly enable the calculation of the coasting time without additional steps or assumptions about the spacecraft's orbit, such as its velocity, the mass of the Earth, and orbital mechanics principles that allow for the calculation of orbital period. An approximate method to find the orbital period (the total time for one complete orbit) involves using the semi-major axis of the orbit, which can be derived from the given perigee and apogee. However, without the specific velocities or applying advanced physics formulas, we can't provide an accurate answer.
Two identical cylindrical vessels with their bases at the same level each contain a liquid of density 1.23 g/cm3. The area of each base is 3.89 cm2, but in one vessel the liquid height is 0.993 m and in the other it is 1.76 m. Find the work done by the gravitational force in equalizing the levels when the two vessels are connected.
Explanation:
Work done by gravity is given by the formula,
W = [tex]\rho A (h_{1} - h)g (h - h_{2})[/tex] ......... (1)
It is known that when levels are same then height of the liquid is as follows.
h = [tex]\frac{h_{1} + h_{2}}{2}[/tex] ......... (2)
Putting value of equation (2) in equation (1) the overall formula will be as follows.
W = [tex]\frac{1}{4} \rho gA(h_{1} - h_{2})^{2})[/tex]
= [tex]\frac{1}{4} \times 1.23 g/cm^{3} \times 9.80 m/s^{2} \times 3.89 \times 10^{-4} m^{2}(1.76 m - 0.993 m)^{2})[/tex]
= 0.689 J
Thus, we can conclude that the work done by the gravitational force in equalizing the levels when the two vessels are connected is 0.689 J.
What is the gage pressure at the bottom of the lake?
What is the absolute pressure at the bottom of the lake?
Answer:
[tex]P_g=\rho_w.g.h[/tex] (gauge pressure)
[tex]P_{abs}=P_{atm}+\rho_w.g.h[/tex] (absolute pressure)
Explanation:
The gauge pressure is the pressure measured with reference to the atmospheric pressure where as the absolute pressure is the pressure measured including the atmospheric pressure taking absolute vacuum as the reference.
So when gauge measuring the pressure at the bottom of a lake the pressure only due to the water column above that point is measured which can be calculated as:
[tex]P_g=\rho_w.g.h[/tex]
where:
[tex]\rho_w=[/tex] density of water
[tex]g=[/tex] acceleration due to gravity
[tex]h=[/tex] height of the water column
Since the atmospheric pressure at the surface of the sea level is defined as the atmospheric pressure.
So, the absolute pressure:
[tex]P_{abs}=P_{atm}+\rho_w.g.h[/tex]
where:
[tex]P_{atm}=[/tex] the atmospheric pressure at the sea level
Therefore, the gage pressure at the bottom of the lake is [tex]\( 294300 \, \text{Pa} \)[/tex], and the absolute pressure at the bottom of the lake is [tex]\( 395625 \, \text{Pa} \).[/tex]
To solve the problem, we can use the formula for gage pressure in a fluid, [tex]( P_g = \rho g h \))[/tex], where ρ is the density of the fluid g is the acceleration due to gravity, and h is the depth below the surface of the fluid.
Given that the density of the lake water is [tex]\( 1000 \, \text{kg/m}^3 \)[/tex], the acceleration due to gravity is [tex]\( 9.81 \, \text{m/s}^2 \)[/tex], and the depth of the lake is [tex]\( 30 \, \text{m} \),[/tex] we can calculate the gage pressure:
[tex]\[ P_g = (1000 \, \text{kg/m}^3) \times (9.81 \, \text{m/s}^2) \times (30 \, \text{m}) = 294300 \, \text{Pa} \][/tex]
To find the absolute pressure[tex](\( P_{abs} \)),[/tex] we add the atmospheric pressure [tex](\( P_{atm} = 101325 \, \text{Pa} \)):[/tex]
[tex]\[ P_{abs} = P_g + P_{atm} = 294300 \, \text{Pa} + 101325 \, \text{Pa} = 395625 \, \text{Pa} \][/tex]
The maximum safe air pressure of a tire is typically written on the tire itself. The label on a tire indicates that the maximum pressure is 32 psi (gage).
Express this maximum pressure in kPa.
Answer:
220.508 kPa
Explanation:
32 psi = 32 pound force per square inch = 32 lbf/in2 = 32 * (4.448 N/lbf) * (12 in/ft * 3.28 ft/m)^2
[tex] = 32*4.448*(12*3.28)^2 = 220508 N/m^2[/tex] or 220508 Pa or 220.508 kPa
So the maximum safe air pressure of a tire is 220.508 kPa
The correct conversion of 32 psi (gage) to kPa is approximately 220.6 kPa.
To convert from psi (pounds per square inch) to kPa (kilopascals), one can use the conversion factor of 1 psi being equal to 6.89476 kPa. Therefore, to convert 32 psi to kPa, we multiply 32 by 6.89476:
[tex]\[ 32 \, \text{psi} \times 6.89476 \, \frac{\text{kPa}}{\text{psi}} = 220.632 \, \text{kPa} \][/tex]
Rounding to a reasonable number of significant figures, we get approximately 220.6 kPa. This is the maximum pressure in kPa that the tire can safely withstand as indicated by the label.
An airplane is flying in a horizontal circle at a speed of 104 m/s. The 85.0 kg pilot does not want the centripetal acceleration to exceed 6.35 times free-fall acceleration.
(a) Find the minimum radius of the plane’s path.
(b) At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth?
Answer:
Part A:
[tex]r\geq 173.806 m[/tex]
Part B:
[tex]F=5289.5757 N[/tex]
Explanation:
Part A:
Airplane is moving in circular path so acceleration in circular path is given by:
[tex]a_c=\frac{V^2}{r}[/tex]
Where:
V is the velocity of object in circular path
r is the radius of circular path
In order to find radius r. Above equation will become:
[tex]r=\frac{V^2}{a_c}[/tex]
V=104 m/s
a_c=6.35*g=6.35*9.8=62.23 m/s^2
[tex]r\geq \frac{104^2}{62.23}\\ r\geq 173.806 m[/tex]
Part B:
Force on a object moving in a circular path is:
[tex]F=\frac{mV^2}{r}\\[/tex]
r=173.806 m
m=85 kg
V=104 m/s
[tex]F=\frac{85*(104)^2}{173.806} \\F=5289.5757 N[/tex]
Given:
Velocity, [tex]V = 104 \ m/s[/tex]Mass, [tex]m = 85 \ kg[/tex](a)
We know,
→ [tex]a_c = mg[/tex]
[tex]= 6.35\times g[/tex]
[tex]= 6.35\times 9.8[/tex]
[tex]= 62.23 \ m/s^2[/tex]
Now,
Acceleration in circular path will be:
→ [tex]a_c = \frac{V^2}{r}[/tex]
or,
→ [tex]r = \frac{V^2}{a_c}[/tex]
By substituting the values, we get
[tex]= \frac{(104)^2}{62.23}[/tex]
[tex]= 173.806 \ m[/tex] (radius)
(b)
The force on object will be:
→ [tex]F = \frac{mV^2}{r}[/tex]
[tex]= \frac{85\times (104)^2}{173.806}[/tex]
[tex]= 5289.58 \ N[/tex]
Thus the approach above is correct.
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Consider a positive charge Q and a point B twice as far away from Q as point A. What is the ratio of the electric field strength at point A to the electric field strength at point B
Answer:
[tex]\frac{E_{A}}{E_{B}}=4[/tex]
Explanation:
The electric field is defined as the electric force per unit of charge, this is:
[tex]E=\frac{F}{q}[/tex].
The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as
[tex]F=\frac{kQq}{r^{2}}[/tex].
By substitution we get that
[tex]E=\frac{kQq}{qr^{2}}\\\\E=\frac{kQ}{r^{2}}[/tex]
Now, letting [tex]E_{A}[/tex] be the electric field at point A, letting [tex]E_{B}[/tex] be the electric field at point B, and letting R be the distance from the charge to A:
[tex]E_{A}=\frac{kQ}{R^{2}}\\\\E_{B}=\frac{kQ}{(2R)^{2}}[/tex].
The ration of the electric fields is
[tex]\frac{E_{A}}{E_{B}}=\frac{\frac{kQ}{R^{2}}}{\frac{kQ}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(4)R^{2}}}\\\\\\\frac{E_{A}}{E_{B}}=\frac{1}{\frac{1}{(4)}}\\\\\frac{E_{A}}{E_{B}}=4[/tex]
This means that at half the distance, the electric field is four times stronger.
The ratio of the electric field strength at point A to the electric field strength at point B is 4.
Given to us
A positive charge Q and a point B is twice as far away from Q as point A.
What is Coulomb's Law?According to Coulomb's law, the electric force between two electrical charges particles is inversely proportional to the square of the distance between them and it is directly proportional to the product of their charges.
[tex]F =\dfrac{k\ Q_1 Q_2}{r^2}[/tex]
We know that an electric field is written as,
[tex]F= EQ[/tex]
Merging the two-equation we get
[tex]E = \dfrac{k Q}{r^2}[/tex]
What is the ratio of the electric field?As we can write
[tex]\dfrac{E_A}{E_B} = \dfrac{\dfrac{k Q}{R^2}}{\dfrac{k Q}{(2R)^2}}[/tex]
[tex]\dfrac{E_A}{E_B} = 4[/tex]
Hence, the ratio of the electric field strength at point A to the electric field strength at point B is 4.
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particle’s position along the x-axis is described by the function x(t) = A t + B t2,
where t is in seconds, x is in meters, and the constants A and B are given below.
Randomized Variables
A = -4.1 m/s
B = 5.4 m/s2Part (a) Enter an expression, in terms of A, B, and t, for the velocity of the particle as a function of time. Part (b) At what time, in seconds, is the particle’s velocity zero?
a) The expression in terms of A , B and t for the velocity of the particle as a function of time is v ( t ) = A + 2Bt.
b) The particle's velocity is zero at time t = 0.379 seconds.
Given data:
a)
The position function is represented as [tex]x ( t ) = At + Bt^2[/tex].
Velocity is the derivative of position with respect to time. To find the velocity function, we need to differentiate the position function.
So, [tex]v(t)=\frac{d(x(t))}{dt}[/tex].
[tex]v(t)=\frac{d(At+Bt^2)}{dt}[/tex]
[tex]v(t)=A+2Bt[/tex].
So, the expression is [tex]v(t)=A+2Bt[/tex].
Now, the value of A = -4.1 m/s and B = 5.4 m/s².
Substituting the values in the expression:
v(t) = -4.1 + 10.8t
b)
The velocity of the particle is 0.
So, v(t) = 0
-4.1 + 10.8t = 0
t = 0.379 seconds
Hence, the time is 0.379 seconds where the particle velocity is 0.
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The velocity of the particle as a function of time is obtained by differentiating the position function, yielding v(t) = A + 2Bt. Setting this equal to zero and solving for t will give the time at which the particle's velocity is zero, which for given values of A and B, is about 0.38 s
Explanation:The motion of a particle is described by the function x(t) = A t + B t2. To find the velocity of the particle as a function of time, we will differentiate the function x(t) with respect to time (t). This is because velocity is the rate of change of the position of an object.
The derivative of x(t) yields: v(t) = A + 2Bt.
To find the time at which the particle's velocity is zero, simply set the velocity function v(t) equal to zero and solve for t: 0 = A + 2Bt.
After doing the math with the provided values for A and B, we get t = -A / (2B) = -(-4.1 m/s) / (2*(5.4 m/s2)) = 0.38 s.
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A 100.0 gram sample of Polonium-210 is contained for 552 days. How many half-lives occur during this period of time, if the half-life is 138 days?
Answer:
4 half-lives will occur during this period of time.
Explanation:
Formula used :
[tex]a_o=a\times e^{-\lambda t}[/tex]
[tex]\lambda =\frac{0.693}{t_{1/2}}[/tex]
[tex]a=\frac{a_o}{2^n}[/tex]
where,
a = amount of reactant left after n-half lives and time t
[tex]a_o[/tex] = Initial amount of the reactant.
[tex]\lambda = [/tex] decay constant
[tex]t_{1\2}[/tex] = half life of an isotope
n = number of half lives
We have :
[tex]a_o=100.0 g[/tex]
a = ?
t = 552 days
[tex]t_{1/2}=138 days[/tex]
[tex]a=100.0 g\times e^{-\frac{0.693}{138}\times 552}[/tex]
[tex]a=6.254 g[/tex]
[tex]6.254 g=\frac{100.0 g}{2^n}[/tex]
[tex]2^n=\frac{100.0 g}{6.254 g}[/tex]
n = 4
4 half-lives will occur during this period of time.
By dividing the total time of 552 days by the half-life of Polonium-210 of 138 days, we find that 4 half-lives occur in this period.
Explanation:The number of half-lives that occur during a given period of time can be found by dividing the total amount of time by the period of one half-life. Given that we are asked to find out the number of half-lives for a 552-day period and the half-life of Polonium-210 is 138 days, we can use this approach. So, the calculation would be 552 days ÷ 138 days/half-life = 4 half-lives. Therefore, 4 half-lives of Polonium-210 would occur in a 552-day period.
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We are usually not aware of the electric force acting between two everyday objects because
a.) most everyday objects have as many plus charges as minus charges
b.) the electric force is invisible
c.) the electric force is one of the weakest forces in nature
d.) the electric force is due to microscopic-sized particles such as electrons and protons
Answer:We are usually not aware of the electric force acting between two everyday objects because most everyday objects have as many plus charges as minus charges. Option A
Explanation:An electric force is exerted between any two charged objects( either positive or negative). Objects with the same charge will repel each other, and objects with opposite charge will attract each other. The strength of the electric force between any two charged objects depends on the amount of charge that each object contains and on the distance between the two charges. Electric charges are generated all around us due to different surfaces bearing different types of charges. We are usually not aware of it because the quantity of positive charges equals the number of negative charges.
The electric force between everyday objects is typically not noticeable because these objects are generally electrically neutral, having an equal number of positive and negative charges cancelling out any major electrostatic force.
Explanation:We are usually not aware of the electric force acting between two everyday objects because most everyday objects have as many plus charges as minus charges, which results in them being electrically neutral. This neutrality means that there are no noticeable electrostatic forces being exerted between such objects. In contrast, if everyday objects were not electrically neutral, we would experience significant electrostatic interactions on a daily basis. Furthermore, the electric force is actually one of the four fundamental forces in nature and is much stronger than gravity (considering two charged particles like protons). This force can be felt in the form of both attractive and repulsive forces depending on the nature of the charges involved. It is this force that governs several interactions at the atomic level including the bond between atoms in a molecule and the forces that we perceive as contact forces, like friction.
Moreover, Coulomb's law describes how the electric force varies as the product of the charges and inversely with the square of the distance between them. The fact that electric force can be both attractive and repulsive, whereas gravity is always attractive, can lead to forces cancelling out when objects have both positive and negative charges evenly distributed, as is the case with most macroscopic objects we encounter.
Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin. 2) What is the distance in meters?
Answer:
The distance is 1.69 m.
Explanation:
Given that,
First charge [tex]q_{1}= 3.8\times10^{-6}\ C[/tex]
Second charge [tex]q_{2}=3.2\times10^{-6}\ C[/tex]
Distance = 3.25 m
We need to calculate the distance
Using formula of electric field
[tex]E_{1}=E_{2}[/tex]
[tex]\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}[/tex]
[tex]\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}[/tex]
[tex]\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}[/tex]
[tex]x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}[/tex]
Put the value into the formula
[tex]x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}[/tex]
[tex]x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}[/tex]
[tex]x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}[/tex]
[tex]x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}[/tex]
[tex]x=1.69\ m[/tex]
Hence, The distance is 1.69 m.
If a proton and an electron are released when they are 5.50×10−10 mm apart (typical atomic distances), find the initial acceleration of each of them. Express your answer in meters per second squared.
Final answer:
The initial acceleration of a proton and an electron released 5.50×10−10 mm apart is calculated using Coulomb's law and Newton's second law. The proton's acceleration is approximately 4.91×10¹⁴ m/s², and the electron's acceleration is approximately 8.99×10¹⁶ m/s².
Explanation:
To find the initial acceleration of a proton and an electron when they are released 5.50×10−10 mm apart, we use Coulomb's law to calculate the force and then apply Newton's second law to find the acceleration. First, convert the distance to meters, so 5.50×10−10 mm is 5.50×10−12 m. Coulomb's law gives us the force as F = k×|q1×q2|/r2, where k is Coulomb's constant (9.00×109 N×m2/C2), q1 and q2 are the charges of the proton and electron (1.60×10−19 C), and r is the separation distance. Substituting the values gives F = 8.20×10−10 N. Then, use Newton's second law, a = F/m, where m is the mass of the particle. For the proton, m = 1.67×10−27 kg, and for the electron, m = 9.11×10−31 kg. Therefore, the initial acceleration of the proton is ap ≈ 4.91×1014 m/s2 and the initial acceleration of the electron is ae ≈ 8.99×1016 m/s2.
The initial acceleration of the proton is approximately [tex]\(4.56 \times 10^{19} \, \text{m/s}^2\)[/tex] and the initial acceleration of the electron is approximately [tex]\(8.35 \times 10^{21} \, \text{m/s}^2\)[/tex].
To calculate the initial acceleration of the proton and electron when they are 5.50×10^(-10) mm apart, we can use Coulomb's law, which states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's law is:
[tex]\[ F = \frac{{k \cdot |q_1 \cdot q_2|}}{{r^2}} \][/tex]
Where:
- F is the force between the charges,
- k is Coulomb's constant [tex](\(8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2\))[/tex],
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the magnitudes of the charges (in this case, the charge of a proton and an electron, which are [tex]\(1.6 \times 10^{-19} \, \text{C}\) and \(-1.6 \times 10^{-19} \, \text{C}\)[/tex], respectively),
- r is the distance between the charges.
First, we need to convert the distance from milli meters to meters:
[tex]\[ r = 5.50 \times 10^{-10} \, \text{mm} = 5.50 \times 10^{-13} \, \text{m} \][/tex]
Now, we can calculate the force between the charges using Coulomb's law:
[tex]\[ F = \frac{{8.9875 \times 10^9 \times |(1.6 \times 10^{-19}) \times (-1.6 \times 10^{-19})|}}{{(5.50 \times 10^{-13})^2}} \][/tex]
[tex]\[ F = \frac{{8.9875 \times 10^9 \times 2.56 \times 10^{-38}}}{{3.025 \times 10^{-25}}} \][/tex]
[tex]\[ F \approx 7.61 \times 10^{-8} \, \text{N} \][/tex]
Since force F = ma, we can rearrange to find acceleration [tex]\( a = \frac{F}{m} \)[/tex]. For both the proton and electron, we use their respective masses, which are approximately [tex]\(1.67 \times 10^{-27} \, \text{kg}\) and \(9.11 \times 10^{-31} \, \text{kg}\)[/tex], respectively.
For the proton:
[tex]\[ a_p = \frac{7.61 \times 10^{-8}}{1.67 \times 10^{-27}} \][/tex]
[tex]\[ a_p \approx 4.56 \times 10^{19} \, \text{m/s}^2 \][/tex]
For the electron:
[tex]\[ a_e = \frac{7.61 \times 10^{-8}}{9.11 \times 10^{-31}} \][/tex]
[tex]\[ a_e \approx 8.35 \times 10^{21} \, \text{m/s}^2 \][/tex]
So, the initial acceleration of the proton is approximately [tex]\(4.56 \times 10^{19} \, \text{m/s}^2\)[/tex] and the initial acceleration of the electron is approximately [tex]\(8.35 \times 10^{21} \, \text{m/s}^2\)[/tex].
These values represent the accelerations they experience due to the electrostatic force when they are [tex]5.50\times10^{-10}[/tex] mm apart.
Complete question :- Determine the initial acceleration of a proton and an electron when they are initially [tex]5.50\times10^{-10}[/tex] mm apart, a typical atomic distance. Provide the answer in meters per second squared.
Charged beads are placed at the corners of a square in the various configurations shown in
(Figure 1) . Each red bead carries a charge +q, and the blue bead carries a charge −q.
Rank the configurations according to the magnitude of the electric field at the center of the square, largest magnitude first.
According to the principles of physics, the magnitude of the electric field at the center of the square depends on the charge and location of the beads. Oppositely charged beads cancel out, while same-charged beads add up, resulting in a larger field.
Explanation:The configurations need to be analyzed based on the direction of the electric field each bead creates at the center of the square. As known, the electric field generated by a positive charge points away from it, while for a negative charge it points towards it. In configurations where positive and negative charges are opposite each other, their fields at the center will cancel each other out, leaving a smaller net field. In a scenario where all beads are positively charged or all are negatively charged, the electric field at the center will be maximized due to the fields all adding up. Field magnitude is always proportional to the amount of charge - in this case +q or −q.
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A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 21.5 m/s , and the distance between them is 52.0 m . After t1 = 4.00 s , the motorcycle starts to accelerate at a rate of 6.00 m/s2 . The motorcycle catches up with the car at some time t2. How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words, find t2-t1? How far does the motorcycle travel from the moment it starts to accelerate (at time t1) until it catches up with the car (at time t2 )?
Answer:
a) t = 4.16 s
b) x = 141.51 m
Explanation:
Given
v = 21.5 m/s
x0 = 52.0 m
a = 6.0 m/s²
a) Motorcycle
x = v0*t + (a*t²/2)
x = 21.5t + (6*t²/2)
x = 21.5t + 3t² (I)
Car
x = x0 + v0*t
x = 52 + 21.5t (II)
then we can apply I = II
21.5t + 3t² = 52 + 21.5t
⇒ 3t² = 52
⇒ t = 4.16 s
b) We can use I or II, then
x = 52 + 21.5*(4.16)
⇒ x = 141.51 m
To find the time it takes for the motorcycle to catch up with the car, we can use the equation of motion and the given information. The motorcycle starts accelerating at t1, so it covers a distance of 0 m. The distance covered by the car until the motorcycle catches up is equal to the initial distance between them, which is 52.0 m.
Explanation:To find the time it takes for the motorcycle to catch up with the car, we first need to find the acceleration of the car. Since it is traveling at a constant speed, its acceleration is zero. The motorcycle's acceleration is given as 6.00 m/s². We can use the equation of motion, s = ut + 0.5at², to find the distance each vehicle covers from time t1 to t2. The motorcycle will cover a distance of 0 m, as it starts accelerating at t1. The car will cover a distance of (21.5 m/s)(t2 - t1). Since the motorcycle catches up with the car, the distance covered by the car is equal to the initial distance between them, which is 52.0 m.
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If you see Sagittarius high in your night sky on June 20 and today is your birthday, what is your zodiac constellation?
Answer:
Gemini
Explanation:
Zodiac Constellation or Sun sign is the constellation in which the Sun resides on the date of birth of a person. Throughout the year the Sun crosses across 12 constellations thus there are 12 Sun-signs. Though astronomically the Sun crosses across 13 constellations so there should be 13 zodiacs but most of the astrologers do not accept this. On the date of June 20, the Sagittarius which is a summer constellation and a zodiac can be seen high up in the sky in the night time. At this time the Sun will be in a constellation which is almost opposite to Sagittarius in the celestial sphere. That constellation is Gemini. Thus for a person born on 20 June, zodiac will be Gemini.
Your zodiac constellation according to modern Western astrology would be Gemini, not Sagittarius, if you see Sagittarius high in the sky around your birthday on June 20. This is due to the precession of constellations, which has caused a shift of about 1/12th of the zodiac, or roughly the width of one sign, since the astrological signs were first designated over 2000 years ago.
If you see Sagittarius high in your night sky on June 20 and today is your birthday, which would mean your birthday is also around June 20, your zodiac constellation is not Sagittarius but the one that precedes it. Due to precession, the sun signs are about a month behind the real constellations. Since Sagittarius is the zodiac for those born around December 22 to January 19, if we go back one sign from Sagittarius, we find Scorpio, which is for those born around October 23 to November 21. However, due to precession, you would actually be a Gemini by the sun sign astrology, which is designated for those born around May 21 to June 20. Another important aspect to consider here is the difference between astronomy and astrology: while astronomy is a scientific study of celestial bodies and the universe, astrology is a belief system that suggests a relationship between the positions of celestial bodies and events on Earth. Even though astrology assigns a sign to the dates when it was first set up, the actual constellations have since shifted.
If the torque required to loosen a nut that is holding a flat tire in place on a car has a magnitude of 50 N·m, what minimum force must be exerted by the mechanic at the end of a 22 cm-long wrench to loosen the nut? Answer in units of N.
Answer:
227.27 N
Explanation:
Torque: This can be defined as the force that tend to cause an object to twist or rotate. the torque of a body is also called moment of a body. The S.I unit of torque is N.m.
Mathematically
T = F×d ....................... Equation 1
Making F the subject of the equation,
F = T/d ..................... Equation 2
T = torque required to loosen the nut, F = force applied , d = distance between the force and the center of the nut.
Given: T = 50 N.m, d = 22 cm = (22/100) = 0.22 m.
F = 50/0.22
F = 227.27 N.
Thus the minimum force required to loosen the nut = 227.27 N
How many electrons must be removed from an electrically neutral silver dollar to give it a charge of +3.4 C?
Answer:
There are [tex]2.12\times 10^{19}\ electrons[/tex] that must be removes from an electrically neutral silver dollar.
Explanation:
In this case, we need to find the number of electrons must be removes from an electrically neutral silver dollar to give it a charge of +3.4 C. Let n number of electrons removed. It is a case of quantization of electric charge. It is given by :
q = ne
[tex]n=\dfrac{q}{e}[/tex]
e is the charge on an electron
[tex]n=\dfrac{3.4\ C}{1.6\times 10^{-19}\ C}[/tex]
[tex]n=2.12\times 10^{19}\ electrons[/tex]
So, there are [tex]2.12\times 10^{19}\ electrons[/tex] that must be removes from an electrically neutral silver dollar.
An object at the surface of the Earth (thus, a distance R from the center of the Earth) weighs 180 N. Its weight at a distance 3R from the center of the Earth is:
Answer: 20N
Explanation:
According to Newton's law of gravitation which states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers
F = Gm1m2/r^2
Where
F = force between the masses
G universal gravitational constant
m1 and m2 = mass of the two particles
r = distance between the centre of the two mass
Therefore, weigh of an object on earth is inversely proportional to the square of its distance from the centre of the earth
W₁/W₂ = r₂²/r₁²
W₂ = W₁r₁²/r₂²
W₁ = 180 N
r₁ = R
r₂ = 3R
W₂ = (180 × R²)/(3R)²
W₂ = 180R²/9R²
W₂ = 180/9
W₂ = 20N
its weight at 3R is 20N
Final answer:
The weight of an object decreases with the square of the distance from the center of the mass it is orbiting. At 3R, the distance from the Earth's center, the gravitational force would be 1/9th of what it is on Earth's surface. Hence, an object that weighs 180 N on Earth would weigh 20 N at a distance of 3R from Earth's center.
Explanation:
To determine the weight of an object at a distance of 3R from the center of Earth, we need to understand how gravitational force decreases with distance. The gravitational force and, therefore, gravitational acceleration (g) at a distance r from a mass M is given by Newton's law of universal gravitation:
F = G(Mm/r²)
where G is the gravitational constant, M is the mass of Earth, m is the mass of the object, and r is the distance to the center of Earth. The weight (W) of an object is the product of its mass m and the gravitational acceleration g, so:
W = mg
The gravitational acceleration (g) on Earth's surface is 9.80 m/s². We can use the formula:
g' = GM/(3R)²
to find the gravitational acceleration at a distance of 3R. Since the mass of the object cancels out, we get:
g' = g/R²
And we know the object's weight on Earth's surface (W = 180 N), so:
W' = (W/R²) x 1/9
Substituting the values, we have:
W' = 180 N / 9 = 20 N
Therefore, at a distance of 3R from the center of the Earth, the object would weigh 20 N.
CH===The radius of a sphere is increasing at a rate of 3 mm/s. How fast is the volume increasing when the diameter is 40 mm? Evaluate your answer numerically. (Round the answer to the nearest whole number.)'
Answer:
15086 mm³/s
Explanation:
Given
dr / dt = 3 mm/s
v = 4/3 π r³
dv / dt = (4 / 3) π 3r² dr/dt
= 4πr² x 3
= 12πr²
= 12 x 22/7 x 20²
= 15086 mm³/s
The volume of a sphere with a radius increasing at 3 mm/s will be increasing at a rate of approximately 15072 mm³/s when the diameter is 40 mm. This is calculated using the derivative of the sphere's volume formula.
Given that the radius of a sphere is increasing at a rate of 3 mm/s, we need to find how fast the volume is increasing when the diameter is 40 mm. First, we note that the diameter is 40 mm, so the radius is 20 mm.
The formula for the volume of a sphere is:
V = (4/3)πr³
To find the rate at which the volume is increasing, we take the derivative of the volume with respect to time (t):
dV/dt = 4πr²(dr/dt)
We know from the problem that dr/dt (the rate at which the radius is increasing) is 3 mm/s, and r (the radius at the given moment) is 20 mm. Plugging these values into the derivative formula, we get:
dV/dt = 4π(20)²(3)
dV/dt = 4π(400)(3) = 4800π mm³/s
Evaluating this value numerically and rounding to the nearest whole number, we get:
dV/dt ≈ 15072 mm³/s