Answer:
[tex]u_x=55.208\ m.s^{-1}[/tex]
Explanation:
Given:
horizontal distance form the point of shooting where the arrow hits ground, [tex]s=107\ ft[/tex] [tex]=32.614\ m[/tex]
angle below the horizontal form the point of release of arrow where it hits ground, [tex]\theta=3^{\circ}[/tex]
So the height above the ground from where the arrow was shot:
[tex]\tan3^{\circ}=\frac{h}{107}[/tex]
[tex]h=5.6076\ ft=1.71\ m[/tex]
Since the arrow is shot horizontally so the initial vertical component of the velocity is zero ( [tex]u_y=0[/tex] ), we've the final vertical component of the velocity as:[tex]v_y=\sqrt{2g.h}[/tex]
[tex]v_y=\sqrt{2\times 9.8\times 1.71}[/tex]
[tex]v_y=5.789\ m.s^{-1}[/tex]
Using equation of motion:
[tex]v_y=u_y+g.t[/tex]
where:
t = time taken
[tex]5.789=0+9.8\times t[/tex]
[tex]t=0.591\ s[/tex]
Now the horizontal component of speed of the arrow (which remains constant throughout the motion by the Newton's first law of motion):[tex]u_x=\frac{s}{t}[/tex]
[tex]u_x=\frac{32.614}{0.591}[/tex]
[tex]u_x=55.208\ m.s^{-1}[/tex]
g 4. A student with a mass of m rides a roller coaster with a loop with a radius of curvature of r. What is the minimum speed the rollercoaster can maintain and still make it all the way around the loop?
Answer:
minimum speed v=[tex]\sqrt({Fr}/m)}[/tex]
Explanation:
Recall the formula for centripetal force;
Centripetal force is the force that is required to keep an object moving in circular part
[tex]F=mv^{2} /r[/tex]
where;
F=centripetal force
m=mass of object
r=radius of curvature
v= minimum speed
To find minimum speed make v the subject of formula;
v=[tex]\sqrt({Fr}/m)}[/tex]
The question is incomplete.
The complete question is:
A student with a mass of m rides a roller coaster with a loop with a radius of curvature of r. What is the minimum speed the rollercoaster can maintain and still make it all the way around the loop? g= 9.8m/s
r=5.5 m= 55kg
Answer: 7.3m/s
Explanation:
Centripetal force is the force acting on a body causing circular motion towards the center which allows to keep the body on track or right path. Any combination of forces in nature can cause centripetal acceleration in various systems.
Where:
V is the tangential velocity
W is the angular velocity
M is the Mass in kg
g is the gravitational force
r is the circular radius
Apply Newton's second law of motion.
The minimum speed is also known as the critical speed is the point where the tension, frictional or normal force is zero and the only thing keeping the object in circular motion is the force of gravity.
This is explained as follows:
N + mg = mw^2r
mg = mv^2/r
Therefore,the minimum speed is:
v^2 = gr
v = square root(gr)
= sqrt(9.8N/kg)(5.5m)
=(9.8N/kg)(5.5m) ^1/2
v = 7.3m/s
A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like protons or electrons. You may have seen it used to make human hair stand on end or produce large sparks. One application of the Van de Graaff generator is to create x-rays by bombarding a hard metal target with the beam. Consider a beam of protons at 1.85 keV and a current of 5.15 mA produced by the generator.(a) What is the speed of the protons (in m/s)?(b) How many protons are produced each second?
Answer:
595391.482946 m/s
[tex]3.21875\times 10^{6}[/tex]
Explanation:
E = Energy = 1.85 keV
I = Current = 5.15 mA
e = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
t = Time taken = 1 second
m = Mass of proton = [tex]1.67\times 10^{-27}\ kg[/tex]
Velocity of proton is given by
[tex]v=\sqrt{\dfrac{2E}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.85\times 10^3\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}\\\Rightarrow v=595391.482946\ m/s[/tex]
The speed of the proton is 595391.482946 m/s
Current is given by
[tex]I=\dfrac{\Delta Q}{t}\\\Rightarrow \Delta Q=It\\\Rightarrow \Delta Q=5.15\times 10^{-3}\times (1\ sec)\\\Rightarrow Q=5.15\times 10^{-3}\ C[/tex]
Number of protons is
[tex]n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{5.15\times 10^{-3}}{1.6\times 10^{-19}}\\\Rightarrow n=3.21875\times 10^{6}\ protons[/tex]
The number of protons is [tex]3.21875\times 10^{6}[/tex]
The speed of the protons accelerated by the Van de Graaff generator is approximately 3.19 million meters per second. Additionally, the generator produces roughly 3.21 x 10^16 protons every second.
Explanation:The Van de Graaff generator is a particle accelerator that can be used to accelerate charged particles. In the given scenario, a beam of protons with an energy of 1.85 keV and a beam current of 5.15 mA is being generated.
(a) The energy of a proton (kinetic energy = 1/2 mv²) is given by the equation E = mv²/2, where m is the mass of the proton (1.67262192369 × 10⁻²⁷ kg) and v is the speed of the proton. Solving the equation v = sqrt((2*E)/m), where E is the energy in joules (1.85 keV = 1.85 * 10^-16 joules), we find that v ≈ 3.19 x 10^6 m/s. This implies that the speed of the protons is approximately 3.19 x 10^6 meters per second.
(b) The number of protons produced each second, i.e., the beam current, can be calculated using the formula I = qN/t, where I is the current, q is the charge of a proton (1.602 x 10^-19 Coulombs), N is the number of protons, and t is time. Rearranging the formula, we find that N = It/q. Substituting the values I=5.15 mA = 5.15 x 10^-3 A and t=1s, we get N ≈ 3.21 x 10^16 protons. Therefore, approximately 3.21 x 10^16 protons are produced every second under the mentioned conditions.
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Explain whether a solenoid will attract paper clips if there is no current through its coils. Predict what will happen to the force between the solenoid and the paper clips if you increase the current in the coils or increased the number of turns of coil per un
Answer:
magnetic field strength increases.
Explanation:
No, the solenoid will not attract any magnetic substance if it does not carries a direct electric current.
The formula for the magnetic field strength of the solenoid is given as:
[tex]B=\mu.n.I[/tex]
where:
[tex]\mu=[/tex] permeability of free space
[tex]n=[/tex] no. of coil turns in the solenoid
[tex]I=[/tex] current in the coil
So, when the current & no. of coil turns in the coil are increased then the magnetic field strength also increases.
Johnson is dragging a bag on a ice surface. He pulls on the strap with a force of 112 N at an angle of 45° to the horizontal to displace it 84 m in 3.33 minutes. Determine the work done by Johnson on the bag and the power generated by Johnson
Answer with Explanation:
We are given that
Force=F=112 N
[tex]\theta=45^{\circ}[/tex]
Distance,[tex]s=84 m[/tex]
Time, t=3.33 minutes
We have to find the work done by Johnson on the bag and the power generated by Johnson.
Work done, W=[tex]Fscos\theta[/tex]
Using the formula
Work done, W=[tex]112\times 84cos45=6652.46 J[/tex]
Power, P=[tex]\frac{W}{t}=\frac{6652.46}{3.33\times 60}=33.3 watt[/tex]
Using 1 minute=60 s
Hence, the power generated by Johnson=33.3 watt
At a particular instant, a proton at the origin has velocity < 5e4, -2e4, 0> m/s. You need to calculate the magnetic field at location < 0.03, 0.05, 0 > m, due to the moving proton. What is the vector r?
Answer:
[tex]9.7\times 10^{-5} T[/tex]
Explanation:
Velocity =[tex]5\times 10^4i-2\times 10^4j[/tex]
r=[tex]0.03i+0.05j[/tex]
r=[tex]\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058[/tex]
v=[tex]\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}[/tex]
We know that
[tex]B=\frac{mv}{qr}[/tex]
Where q=[tex]1.6\times 10^{-19} C[/tex]
Mass of proton=[tex]1.67\times 10^{-27} kg[/tex]
Using the formula
[tex]B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}[/tex]
[tex]B=9.7\times 10^{-5} T[/tex]
The vector r represents the position at which the magnetic field is being calculated. The formula used to calculate the magnetic field is the Biot-Savart law. The given values can be plugged into the formula to find the magnitude and direction of the magnetic field.
Explanation:The vector r represents the position of the point where we want to calculate the magnetic field. In this case, r = < 0.03, 0.05, 0 > m.
To calculate the magnetic field at this point, we can use the Biot-Savart law. The Biot-Savart law states that the magnetic field at a point due to a moving charge is given by B = (μ₀/4π) imes ((qv)×r)/r³, where B is the magnetic field, μ₀ is the permeability of free space, q is the charge, v is the velocity of the charge, and r is the displacement vector from the charge to the point where the magnetic field is being calculated.
Plugging in the values given, we have q = +e, v = <5e4, -2e4, 0> m/s, and r = <0.03, 0.05, 0> m. The magnitude of the magnetic field can be calculated using the formula B = (μ₀/4π) imes ((qv)×r)/r³ and the direction can be determined using the right-hand rule.
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A light plane is headed due south with a speed of 200 km/h relative to still air. After 1.00 hour, the pilot notices that they have covered only 137 km and their direction is not south but 15.0∘ east of south. What is the wind velocity?
Answer: the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.
Explanation: The initial velocity of the plane is 200km/h south (supose that south is our positive x-axis here and east is the positive y-axis)
In one hour, the plane is located 137km away from the initial position, and the position in X is equal to 137km*cos(15°) = 132.33, this means that the velocity in the x axis is equal to 132.33 km/h, knowing that the initial velocity of the plane was 200km in the x-axis, this means that the velocity of the air must be:
132.33km/h - 200km/h = -67.69km/h
km and the position in "y" is equal to 137km*sin(15°) = 35.4km
This means that the velocity of the air in the y-axis is 35.4km/h
So the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.
A block with a mass of 4 kg is sliding across a horizontal surface with an initial speed of 5 m/s. Because of kinetic friction, the energy of this system will decrease linearly with the distance traveled by the block.
What is the average power (in W) supplied by the force of kinetic friction acting on the block if the block moves 4 m before coming to a stop?
Explanation:
Below is an attachment containing the solution.
The average power supplied by the force of kinetic friction acting on the block if the block moves 4 m before coming to stop is -31.25 W.
What is Power?In physics, power is the amount of energy that is transferred or transformed in a given amount of time. The International System of Units uses the watt, or one joule per second, as the unit of power. Ancient literature frequently used the term "activity" to describe power. Powers are scalar variables.
According to the question, the given values are :
Block mass, m = 4 kg,
The initial speed of the block at the horizontal surface, v₁ = 5 m/s,
The final speed of the block will be, v₂ = 0 m/s.
Distance covered, d = 4 m
Let t be the time when the box is going to stop.
d = [(v₁+v₂)/2] × t
4 = (5/2) t
t = 1.6 sec.
The change in kinetic energy of the block will be ΔE.
ΔE = mv₂²/2 - mv₁²/2
ΔE = 0 - (4)(5)²/2
ΔE = -50 J
Average power supplied by the force of kinetic friction acting on the block = P(avg)
P(avg) = ΔE / t
P(avg) = -50 / 1.6
P(avg) = -31.26 W.
So, the average power supplied by the force of kinetic friction on the block will be -31.26 W.
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A gas is compressed at constant temperature from a volume of 5.68 L to a volume of 2.35 L by an external pressure of 732 torr. Calculate the work done in J
Answer: The work done in J is 324
Explanation:
To calculate the amount of work done for an isothermal process is given by the equation:
[tex]W=-P\Delta V=-P(V_2-V_1)[/tex]
W = amount of work done = ?
P = pressure = 732 torr = 0.96 atm (760torr =1atm)
[tex]V_1[/tex] = initial volume = 5.68 L
[tex]V_2[/tex] = final volume = 2.35 L
Putting values in above equation, we get:
[tex]W=-0.96atm\times (2.35-5.68)L=3.20L.atm[/tex]
To convert this into joules, we use the conversion factor:
[tex]1L.atm=101.33J[/tex]
So, [tex]3.20L.atm=3.20\times 101.3=324J[/tex]
The positive sign indicates the work is done on the system
Hence, the work done for the given process is 324 J
Given values:
Pressure, [tex]P = 732 \ torr \ or \ 0.96 \ atm[/tex]Initial volume, [tex]V_1 = 5.68 \ L[/tex]Final volume, [tex]V_2 = 2.35 \ L[/tex]We know the equation,
→ [tex]W = - P\Delta V[/tex]
[tex]= -P(V_2-V_1)[/tex]
By substituting the values,
[tex]= -0.96\times (2.35-5.68)[/tex]
[tex]= 3.20 \ L.atm[/tex]
By converting it in "J",
[tex]= 3.20\times 101.3[/tex]
[tex]= 324 \ J[/tex]
Thus the above answer is right.
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fter driving a portion of the route, the taptap is fully loaded with a total of 25 people including the driver, with an average mass of 69 kg per person. In addition, there are three 15-kg goats, five 3-kg chickens, and a total of 25 kg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum amount. How much are the springs compressed
Answer:
The spring compressed is 0.44 m.
Explanation:
Given that,
Number of person = 25
Average Mass of each person [tex]m_{p}= 25\times69 = 1725\ kg[/tex]
Mass of goat [tex]m_{g}= 3\times15 = 45\ kg[/tex]
Mass of chicken [tex]m_{c}= 5\times3 = 15\ kg[/tex]
Mass of bananas = 25 kg
If the spring constant is [tex]4\times10^{4}\ N/m[/tex]
We need to calculate the total mass
[tex]M=m_{p}+m_{g}+m_{c}+m_{b}[/tex]
Put the value in the formula
[tex]M=1725+45+15+25[/tex]
[tex]M=1810\ kg[/tex]
The weight of all these things is
[tex]W=Mg[/tex]
[tex]W=1810\times9.8[/tex]
[tex]W=17738\ N[/tex]
We need to calculate the distance
Using formula of restoring force
[tex]F=kx[/tex]
[tex]x=\dfrac{F}{k}[/tex]
Put the value into the formula
[tex]x=\dfrac{17738}{4\times10^{4}}[/tex]
[tex]x=0.44\ m[/tex]
Hence, The spring compressed is 0.44 m.
In an arcade game a 0.117 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and releasing it. If the spring has a spring constant of 194 N/m and is compressed from its equilibrium position by 7 cm, find th
Find the speed with which the disk slides across the surface
Answer:
[tex]\boxed{2.85 m/s}[/tex]
Explanation:
The Potential energy of spring is transformed to kinetic energy hence
[tex]0.5kx^{2}=0.5mv^{2}\\kx^{2}=mv^{2}\\v=\sqrt{\frac {kx^{2}}{m}}[/tex]
Here k is the spring constant, x is the extension of spring, v is the velocity of disk and m is the mass of the disk.
Substituting 0.117 Kg for m, 194 N/m for k and 0.07 m for x then
[tex]v=\sqrt{\frac {194\times 0.07^{2}}{0.117}}=2.850401081 m/s\approx \boxed{2.85 m/s}[/tex]
A tennis ball is shot vertically upward from the surface of an atmosphere-free planet with an initial speed of 20.0 m/s. One second later, the ball has an instantaneous velocity in the upward direction of 15.0 m/s. A) How long does it take the ball to reach its maximum height?B) How high does the ball rise?C) What is the magnitude of the acceleration due to gravity on the surface of this planet?D) Determine the velocity of the ball when it returns to its original position. Note: assume the upward direction is positive.E) How long has the ball been in the air when it returns to its original position?
Answer:
(A) t = 4s
(B) H = 40m
(C) g = 5m/s²
(D) V = -20m/s
(E) t = 8s
The detailed solution to this problem requires the knowledge of costant linear acceleration motion.xplanation:
The detailed solution to this problem requires the knowledge of costant linear acceleration motion.
Explanation:
The full solution can be found in the attachment below.
To answer part A, we have been given some values of velocities bounding the time jnterval of 1s from which we can calculate the acceleration due to gravity and then the time.
Part B
Requires just using the acceleration due to gravity and the time taken in the equation
H = ut - 1/2gt²
Part C
Has already been calculated in part A
Part D
V = -20m/s because the tennis ball is coming down as the upward direction was assumed positive.
Part E
When the ball retirns to its original position it is the same as it never left and so H = 0m
The calculation can be found in the attachment below.
Final answer:
Ball reaches peak at 4s, rises 80m, experiences -5m/s² gravity. Returns at -20m/s and spends 8s in the air (4s up + 4s down).
Explanation:
Here are the answers to your questions about the tennis ball on the atmosphere-free planet:
A) Time to reach maximum height: Between the beginning and one second later, the ball's velocity decreases by 5.0 m/s. Since acceleration is constant (no atmosphere), this decrease is due to gravity acting against the initial velocity. We can use the equation:
v = u + at
where:
v = final velocity (15.0 m/s)
u = initial velocity (20.0 m/s)
a = acceleration due to gravity (unknown, negative direction)
t = time (1 second)
Solving for a:
a = (v - u) / t = (15.0 m/s - 20.0 m/s) / 1 s = -5.0 m/s²
Now, to find the time to reach maximum height, we know that at the peak, the velocity is 0 m/s. We can again use the same equation:
0 = u + at
Solving for t:
t = -u / a = -20.0 m/s / (-5.0 m/s²) = 4.0 seconds (positive result since time cannot be negative)
Therefore, it takes 4.0 seconds for the ball to reach its maximum height.
B) Maximum height:
We can use the kinematic equation to find the maximum height (h):
h = ut + 1/2 × at²
Substituting the values:
h = 20.0 m/s × 4.0 s + 1/2 × (-5.0 m/s²) × (4.0 s)² = 40.0 m + 40.0 m = 80.0 meters
C) Acceleration due to gravity:
We already calculated the acceleration due to gravity in part A: -5.0 m/s² (the negative sign indicates downward direction).
D) Velocity upon returning to the original position:
Since the motion is symmetrical (same initial and final positions), the velocity when the ball returns will be the negative of its initial velocity: -20.0 m/s.
E) Time spent in the air:
The total time in the air is the sum of the time to reach the maximum height and the time to fall back to the original position. Since the motion is symmetrical, the time to fall back is also 4.0 seconds. Therefore, the total time in the air is:
4.0 seconds (upward) + 4.0 seconds (downward) = 8.0 seconds
A 14-kg block rests on a level frictionless surface and is attached by a light string to a 5.0-kg hanging mass where the string passes over a massless frictionless pulley. If g=9.8 m/s2 what is the tension in the connecting string?
Answer:
Explanation:
Let T be the tension in the string .
For hanging mass
net force = m₁g - T
m₁g - T = m₁a
For mass placed on horizontal plane
T = m₂a
m₁g - m₂a = m₁a
m₁g = a ( m₂ +m₁)
a = m₁g / ( m₂ +m₁)
= 5 x 9.8 / 19
= 2.579 m /s²
m₁g - T = m₁a
T = m₁g - m₁a
= m₁ ( g - a )
= 5 ( 9.8 - 2.579)
= 36.10 N
how is Newton’s three laws of motion are used in testing the safety of our automobiles.
Answer
Newton's third law is used for the impact test on the vehicle.
In the frontal collision of the vehicle, the impact of the vehicle is made against the concrete wall then the impact on the passengers and driver is calculated.
In the impact test time of the opening of the airbag is also calculated.
Similar tests also take place when the impact is sideways.
ou are unloading a refrigerator from a delivery van. The ramp on the van is 5.0 m long, and its top end is 1.4 m above the ground. As the refrigerator moves down the ramp, you are on the down side of the ramp trying to slow the motion by pushing horizontally against the refrigerator with a force of 370 N . Part A How much work do you do on the refrigerator during its trip down the ramp
Answer:
Explanation:
component of force along the ramp
= 370 cos θ where θ is the slope of the ramp with respect to ground.
sinθ = 1.4 / 5
θ = 16 degree
370 cos16
= 355.67 N
Work done
= component of force along the ramp x length of the ramp
= 355.67 x 5
= 1778.35 J
The resistivity of a metallic, single-walled carbon nanotube is 2.30 ✕ 10−8 Ω · m. The electron number density is 6.60 ✕ 1028 m−3. What is the mean free time for the electrons flowing in a current along the carbon nanotube?
Answer:
The mean free time is given as z [tex]= 2.3*10^{-14}sec[/tex]
Explanation:
Generally the formula for resistivity
[tex]\rho = \frac{m_e}{e^2 n_ez}[/tex]
Where
[tex]\rho[/tex] is he resistivity of metal
[tex]m_e[/tex] is the mass of the electron
[tex]e[/tex] is the charge of the electron
[tex]n_e[/tex] is electron density
[tex]z[/tex] is the mean free time
Now making z the subject of the formula
=> [tex]z = \frac{m_e}{e^2n_e\rho}[/tex]
Substituting the given values
[tex]z = \frac{9*10^{-31}}{(1.6*10^{-19}(16.1*10^{28}(2.5*10^{-8})))}[/tex]
[tex]= 2.3*10^{-14}sec[/tex]
The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s).
Find the speed of the passengers when the Ferris wheel is rotating at this rate.
Answer:
V = 5.24m/s
Explanation:
See attachment below.
An airplane is flying at 150 ft/s at an altitude of 2000 ft in a direction that will take it directly over an observer at ground level. Find the rate of change of the angle of elevation between the observer and the plane when the plane is directly over a point on the ground that is 2000 ft from the observer.
Answer:
Explanation:
Let x be the horizontal distance of airplane . angle of elevation of airplane from observer = θ , altitude of airplane = 2000 ft ,
from the construction θ = 45 degree. , x = 2000 ft .
Tanθ = 2000 / x
sec²θ dθ / dt = (2000 / x²) dx / dt
dθ / dt = 2000 /(sec²θ x²) x dx / dt
dx / dt = 150 ft /s
dθ / dt = 2000x 150 /(sec²θ x²)
= 300000 / sec²45 x 2000²
= .15 degree/ s
The rate of change of the angle of elevation when an airplane (traveling at 150 ft/s) is over a point 2000 ft from the observer is approximately -2.15° per second.
Explanation:This problem falls under the department of mathematics known as trigonometry specifically, its applications in real-world scenarios. In this case, we're looking to find the rate of change of the angle of elevation when the airplane is at a certain point.
When the airplane is directly over a point 2000 ft from the observer, it forms a right triangle with the observers' location and the point over which it is flying. The hypotenuse of this triangle is the line between observer and airplane.
The angle of elevation, θ, from the observer to the plane changes over time as the airplane moves overhead so its rate of change(dθ/dt) is what we need to find. We use the trigonometric relation tangent (tan), which in this case equals to the altitude (opposite side, 2000 ft) over the horizontal distance between the observer and the plane (adjacent side, 2000 ft).
The relation is tan θ = opposite/adjacent = 2000/2000 = 1, thus θ = 45 degrees. With the plane's speed (150 ft/s), this changes the horizontal distance over time, so we differentiate tan θ giving us sec² θ*dθ/dt = -150/2000² = -0.0375 radians/sec, by applying the chain rule and remembering sec² θ is 1/cos² θ.
In degrees per second, this is approximately -2.15°/s, so that is the rate of change of the angle of elevation when the airplane is directly over a point 2000 ft from the observer.
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(1 point) Find a linearly independent set of vectors that spans the same subspace of R3R3 as that spanned by the vectors ⎡⎣⎢3−1−2⎤⎦⎥, ⎡⎣⎢−923⎤⎦⎥, ⎡⎣⎢−30−1⎤⎦⎥. [3−1−2], [−923], [−30−1]. A linearly independent spanning set for the subspace is:
Answer:
Explanation:
[tex]A=\left[\begin{array}{ccc}3&-9&-3\\-1&2&0\\-2&3&-1\end{array}\right] \\\\R_2\rightarrow 3R_2+R_1,R_3\rightarrow 3R_3+2R_1\\\\=\left[\begin{array}{ccc}3&-9&-3\\0&-3&-3\\0&-9&-9\end{array}\right] \\\\R_3\rightarrow 3R_3-9R_2\\\\=\left[\begin{array}{ccc}3&-9&-3\\0&-3&-3\\0&0&0\end{array}\right][/tex]
This is the row echelon form of A. This means that only two of the vectors in our set are linearly independent. In other words, the first two vectors alone will span the same subspace of [tex]R^4[/tex] as all three vectors.
Therefore, the linearly independent spanning set for the subspace is
[tex]\left[\begin{array}{ccc}3\\-1\\-2\end{array}\right] \left[\begin{array}{ccc}-9\\2\\3\end{array}\right] \left[\begin{array}{ccc}3\\0\\-1\end{array}\right][/tex]
A bat strikes a 0.145 kgkg baseball. Just before impact, the ball is traveling horizontally to the right at 50.0 m/sm/s; when it leaves the bat, the ball is traveling to the left at an angle of 30∘30∘ above horizontal with a speed of 38.0 m/sm/s. The ball and bat are in contact for 1.75 msms. find the horizontal and vertical components of the average force on the ball.
Answer with Explanation:
We are given that
Mass of ball=m=0.145 kg
Initially horizontal velocity of ball,ux=50 m/s
[tex]\theta=30^{\circ}[/tex]
Final velocity of ball,v=38m/s
Time ,t=1.75 ms=[tex]1.75\times 10^{-3} s[/tex]
[tex]1 ms=10^{-3} s[/tex]
Horizontal component of average force, [tex]F_x=\frac{m(vcos\theta-u_x)}{t}[/tex]
Using the formula
Horizontal component of average force, [tex]F_x=\frac{0.145(-38cos30-50)}{1.75\times 10^{-3}}=-6.9\times 10^3[/tex]N
Vertical component of average force, [tex]F_y=\frac{m(vsin\theta-u_y)}{t}[/tex]
Vertical component of average force,[tex]F_y=\frac{0.145(38sin30-0}{1.75\times 10^{-3}}=1.6\times 10^3 N[/tex]
During a walk on the Moon, an astronaut accidentally drops his camera over a 20.0-m cliff. It leaves his hands with zero speed, and after 2.00 s it attains a velocity of 3.40 m/s downward. How far has the camera fallen after 9.10 s
Answer:
Δy=70.3885 downwards
Explanation:
First we get the acceleration on moon by substituting the values we have(v=0m/s at t=0s and v=-3.40m/s at t=2.0 s) into equation of simple motion:
[tex]v_{f}-v_{i}=at\\-3.40m/s-0=(2s)a\\a=-1.7m/s^{2}[/tex]
As we know that
Δy=vit+(1/2)at²
Substitute the values of acceleration and time t=9.10s
So
Δy=vit+(1/2)at²
[tex]=(0)(9.10s)+(1/2)(-1.7m/s^{2})(9.10s)^{2}\\ =-70.3885m[/tex]
Δy=70.3885 downwards
The camera has fallen approximately 402.5 meters after 9.10 seconds.
Explanation:To find the distance the camera has fallen after 9.10 s, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance fallen
u = initial velocity (0 m/s)
t = time (9.10 s)
a = acceleration (due to gravity, -9.8 m/s^2)
Plugging the values into the equation, we get:
s = 0 + (1/2)(-9.8)(9.10)^2
Simplifying, we find that the camera has fallen approximately 402.5 meters after 9.10 seconds.
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Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls. Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k. Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward. Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle. Use g for the magnitude of the acceleration due to gravity.
(a) How far below the bridge will Kate eventually be hanging, once she stops oscillating and comes finally to rest? Assume that she doesn't touch the water.
(b) If Kate just touches the surface of the river on her first downward trip (i.e. before the first bounce), what is the spring constant k?
Final answer:
To determine how far below the bridge Kate will eventually be hanging, we need to consider the forces acting on her. When she stops oscillating and comes to rest, the gravitational force pulling her downwards will be balanced by the spring force exerted by the bungee cord. To determine the spring constant k, we need to use the equation Fs = -kx and substitute the values of the gravitational force and the displacement x.
Explanation:
(a) To determine how far below the bridge Kate will eventually be hanging, we need to consider the forces acting on her. When she stops oscillating and comes to rest, the gravitational force pulling her downwards will be balanced by the spring force exerted by the bungee cord. At this point, her net force will be zero. The gravitational force can be calculated as mg, where m is her mass and g is the acceleration due to gravity. The spring force can be calculated using Hooke's Law: Fs = -kx, where k is the spring constant and x is the displacement of the cord from its equilibrium position. Equating the gravitational force and the spring force and solving for x will give us the distance below the bridge where Kate will be hanging.
(b) To determine the spring constant k, we need to use the equation Fs = -kx and substitute the values of the gravitational force and the displacement x. Solving for k will give us the spring constant of the bungee cord.
A 20 centimeter long copper wire sensor is designed to alert if the temperature elevates beyond a safe region causing the wire to make contact with the wall, 1.5 mm away at 25 C. What temperature (in K) will the copper wire make contact with the wall, setting the sensor off? The thermal coefficient of expansion, α, for copper (Cu) is found to be 16.6 ppm/K (16.6x10-6 /K).
Answer:
The temperature is 749.8 K
Explanation:
Final temperature (T2) = (distance apart/thermal coefficient of expansion×length) + initial temperature
distance apart = 1.5 mm = 1.5/1000 = 0.0015 m
thermal coefficient of expansion for copper = 16.6×10^-6/K
Length of copper = 20 cm = 20/100 = 0.2 m
Initial temperature = 25 °C = 25 + 273 = 298 K
T2 = (0.0015/16.6×10^-6×0.2) + 298 = 451.8 + 298 = 749.8 K
A water molecule perpendicular to an electric field has 1.40×10−21 J more potential energy than a water molecule aligned with the field. The dipole moment of a water molecule is 6.2×10−30Cm.
What is the strength of the electric field?
To solve this problem we will apply the concepts related to the potential energy in the molecules and obtain its electric field through the relationship given by the dipole moment. The change in potential energy from one state to another is given by,
[tex]U_2-U_1 = 1.4*10^{-21} J[/tex]
From this difference we can identify that [tex]U_1[/tex] is equivalent to the potential energy when it is perpendicular to the electric field. At the same time, the potential energy [tex]U_2[/tex] would be equivalent when it is aligned with the electric field.
From there the relationship between energy, the dipole moment and the electric field would be subject to
[tex]U_2-U_1 = pE[/tex]
Here,
[tex]p = \text{Dipole moment} = 6.2*10^{-30} C \cdot m[/tex]
Rearranging to find the electric field,
[tex]E = \frac{(U_2-U_1)}{p}[/tex]
[tex]E = \frac{(1.4*10^{-21})}{(6.2*10^{-30})}[/tex]
[tex]E =2.25*10^8 N/C[/tex]
Therefore the electric field is [tex]2.25*10^8N/C[/tex]
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.30 A. When they are wired in parallel with the same battery, the current is 1.6 A. What are the values of the two resistors?
Note: I understand other people have asked the same concept with different numbers, and their questions have been answered, however I do not understand why you need to multiple R1 by R2 when the resistors are in parallel, and furthermore, why are you supposed to multiply the "in parallel" equation by (R1+R2; the in series resistors) to find R1*R2?
Answer:
[tex]30\Omega, 10\Omega[/tex]
Explanation:
Let two resistors R1 and R2 are wired in series.
Potential difference, V=12 V
Current=I=0.3 A
We have to find the value of two resistors.
When two resistors are connected in series
[tex]R=R_1+R_2[/tex]
[tex]V=IR=I(R_1+R_2)[/tex]
Substitute the values
[tex]12=0.3(R_1+R_2)[/tex]
[tex]R_1+R_2=\frac{12}{0.3}=40[/tex]
[tex]R_1+R_2=40[/tex]..(1)
In parallel
[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]
[tex]\frac{1}{R}=\frac{R_2+R_1}{R_1R_2}[/tex]
[tex]R=\frac{R_1R_2}{R_1+R_2}[/tex]
Current in parallel, I=1.6 A
[tex]V=IR[/tex]
[tex]V=1.6(\frac{R_1R_2}{R_1+R_2})[/tex]
[tex]\frac{12}{1.6}=\frac{R_1R_2}{40}[/tex]
[tex]R_1R_2=\frac{12\times 40}{1.6}=300[/tex]
[tex]R_1-R_2=\sqrt{(R_1+R_2)^2-4R_1R_2}[/tex]
[tex]R_1-R_2=\sqrt{(40)^2-4(300)}=20[/tex]....(2)
Adding equation (1) and (2)
[tex]2R_1=60[/tex]
[tex]R_1=\frac{60}{2}=30\Omega[/tex]
Substitute the value in equation (1)
[tex]30+R_2=40[/tex]
[tex]R_2=40-30=10\Omega[/tex]
Final answer:
To find the values of two resistors based on their behavior in series and parallel circuits, one must calculate the total equivalent resistances for each configuration with Ohm's law and then solve the equations relating the individual resistances.
Explanation:
When looking to determine the values of two resistors R1 and R2 based on current measurements in both series and parallel configurations, the approach involves some electrical principles and algebra. Let's break down the steps needed to solve for the resistors' values.
Determination of resistors in series
Firstly, we calculate the equivalent resistance for the series circuit using Ohm's law (V = I × R), where V is the voltage, I is the current, and R is the resistance. With a 12 V battery and a current of 0.30 A, the equivalent resistance, Req-series, is 12 V / 0.30 A = 40 Ω.
Determination of resistors in parallel
For the parallel configuration, the current through the battery increases to 1.6 A with the same voltage of 12 V. Again, using Ohm's law, we find the equivalent parallel resistance, Req-parallel, which is 12 V / 1.6 A = 7.5 Ω.
In parallel circuits, the reciprocal of the total resistance is the sum of the reciprocals of each individual resistance. This can be expressed as 1/Req-parallel = 1/R1 + 1/R2. To find R1 and R2, we need to relate the resistances in parallel to the sum of their series counterpart (R1 + R2 = Req-series).
Finding Individual Resistances
Given the equations R1 + R2 = 40 Ω and 1/R1 + 1/R2 = 1/7.5 Ω, we can simplify the latter to R1×R2/(R1 + R2) = 7.5 Ω. Substituting R1 + R2 = 40 Ω into this equation, we finally solve for the individual resistance values, R1 and R2.
A 37.5 kg box initially at rest is pushed 4.05 m along a rough, horizontal floor with a constant applied horizontal force of 150 N. If the coefficient of friction between box and floor is 0.300, find the following.(a) the work done by the applied force
J
(b) the increase in internal energy in the box-floor system due to friction
J
(c) the work done by the normal force
J
(d) the work done by the gravitational force
J
(e) the change in kinetic energy of the box
J
(f) the final speed of the box
m/s
Answer:
a) 607.5 J
b) 160.531875 J
c) 0 J
d) 0 J
e) 2.925 m\s
Explanation:
The given data :-
Mass of the box ( m ) = 37.5 kg.Displacement made by box ( x ) = 4.05 m.Horizontal force ( F ) = 150 N.The co-efficient of friction between box and floor ( μ ) = 0.3Gravitational force ( N ) = m × g = 37.5 × 9.81 = 367.875Solution:-
a) The work done by applied force ( W )
W = force applied × displacement = 150 × 4.05 = 607.5 J
b) The increase in internal energy in the box-floor system due to friction.
Frictional force ( f ) = μ × N = 0.3 × 367.875 = 110.3625 N
Change in internal energy = change in kinetic energy.
ΔU = ( K.E )₂ - ( K.E )₁
Since the initial velocity is zero so the ( K.E )₁ = 0
ΔU = ( K.E )₂ = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J
c) The work done by the normal force .
Displacement of box vertically = 0
W = force applied × displacement = 367.875 × 0 = 0 J
d) The work done by the gravitational force.
Displacement of box vertically = 0
W = force applied × displacement = 367.875 × 0 = 0 J
e) The change in kinetic energy of the box
( K.E )₂ - ( K.E )₁ = ( K.E )₂ - 0 = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J
f) The final speed of the box
( K.E )₂ = 160.531875 J = 0.5 × 37.5 × v²
v² = 8.56
v = 2.925 m\s.
A 30.0-kg box is being pulled across a carpeted floor by a horizontal force of 230 N , against a friction force of 210 N . What is the acceleration of the box? How far would the box move in 3 s , if it starts from rest?
Answer:
0.67 m/s² or 2/3 m/s²
3 m
Explanation:
Using
F-F' = ma............ Equation 1
Where F = Horizontal force applied to the box, F' = Frictional force, m = mass of the box, a = acceleration of the box.
make a the subject of the equation
a = (F-F')/m............ Equation 2
Given: F = 230 N, F' = 210 N, m = 30 kg.
substitute into equation 2
a = (230-210)/30
a = 20/30
a = 0.67 m/s² or 2/3 m/s²
The acceleration of the box = 2/3 m/s²
Using,
s = ut+1/2at²............ Equation 3
Where u = initial velocity, t = time, a = acceleration, s = distance.
Given: u = 0 m/s (from rest), t = 3 s, a = 2/3 m/s²
substitute into equation 3
s = 0(3)+1/2(2/3)(3²)
s = 3 m.
Hence the box moves 3 m
The acceleration of the 30.0-kg box is approximately 0.67 m/s². If it starts from rest, it would move around 3.02 meters in 3 seconds.
Explanation:In order to solve the problem, we'll apply the principle of Newton's second law, which states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
First, we identify the net force on the box. The box is being pulled by a force of 230 N, and there is friction acting against this pull with a force of 210 N. The net force (F) is therefore 230 N (pulling force) - 210 N (friction) = 20 N.
To find the acceleration (a), we use the formula:
a = F/m
Substituting the given values, we get:
a = 20 N / 30.0 kg = 0.67 m/s².
Now, the second part of the problem asks for the distance the box would move in 3 seconds if it starts from rest. We use the formula for distance (d) traveled under constant acceleration:
d = 0.5 * a * t²
Substituting the calculated acceleration and time (3 seconds), we find:
d = 0.5 * 0.67 m/s² * (3 s)² = 3.02 m.
So, under the given conditions, the box's acceleration is 0.67 m/s², and it would move approximately 3.02 meters in 3 seconds.
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A "mechanical wave" occurs in a physical medium because of some restoring force acting on the medium, and can be described by an amplitude, an oscillation frequency, and an amplitude.
You can easily decrease the wavelength of a mechanical wave by half by doubling the frequency ONLY if which of the following is true? Select all that apply
(A) The ratio between the restoring force and the density of the medium remain constant
(B) The amplitude is doubled
(C) The amplitude is reduced by 1/2
(D) The same medium is used at the same temperature
Answer:
(A) The ratio between the restoring force and the density of the medium remain constant
Explanation:
The ratio between the restoring force and the density of the medium is equal to the square of the velocity of the wave.
[tex]v = \sqrt{\frac{F}{\mu}}[/tex]
The general formula that relates the displacement and velocity (x = vt) can be written in wave mechanics such that
[tex]v = \lambda f[/tex]
where f is the frequency, λ is the wavelength, and v is the velocity of the wave.
According to this equation, in order to halve the wavelength by doubling the frequency, the velocity should be constant. Therefore, the correct answer is (A).
The wavelength of a mechanical wave can be decreased by half by doubling the frequency ONLY if; Choice (A) The ratio between the restoring force and the density of the medium remain constant and Choice D: The same medium is used at the same temperature.
Discussion:
The speed of the mechanical wave is dependent on the ratio of the restoring force and the density of the medium.
Additionally, when the same medium is used at the same temperature; the density of the medium remains constant.
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Ballistic pendulum conservation of momentum/mechanical energy?
"Mechanical energy and momentum are conserved only when certain conditions rae met. Explain why the collision between the ball and the pendulum conserves momentum but not mechanical energy. Similarly, explain why the motion of the pendulum during its swing conserves mechanical energy but (apparently) not momentum."
I understand why during the collision kinetic energy is lost and momentum is conserved, but why would mechanical energy be conserving during the swing and not momentum?
Answer:
In the explanation the answers for the two questions are analyzed.
Explanation:
1. When a collision occurs, it can be said that the total energy is not conserved, but the momentum is conserved. Kinetic energy is converted to heat in the case of an inelastic collision, therefore the sum of all energies is the same before and after the collision. Regarding the net moment before the collision is equal to the net moment after said collision.
2. Regarding the impulse, this is not conserved in the case of an oscillating pendulum because the resulting force acting on the pendulum is not equal to zero. While the total energy would be equal to the sum of the potential energy plus the kinetic energy.
Air enters a 16-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. Air is heated as it flows, and it leaves the pipe at 180 kPa and 43°C. The gas constant of air is 0.287 kPa·m3/kg·K. Whats the volumetric flow rate of the inlet/outlet, mass flow rate and velocity & volume flow rate at the exit?
Explanation:
(a) We will determine the mass flow rate as follows.
m = [tex]\rho_{1} V_{1}[/tex]
= [tex]\frac{P_{1}}{RT_{1}}A_{1}v_{1}[/tex]
= [tex]\frac{P_{1}}{RT_{1}} \times \frac{D^{2}}{4} \pi v_{1}[/tex]
Putting the given values into the above formula as follows.
m = [tex]\frac{P_{1}}{RT_{1}} \times \frac{D^{2}}{4} \pi v_{1}[/tex]
= [tex]\frac{200}{0.287 \times 293 K} \times \frac{(0.16)^{2}}{4} \pi \times 5[/tex]
= 0.239 kg/s
Hence, the mass flow rate of the inlet/outlet is 0.239 kg/s.
(b) Now, we will determine the final volume rate as follows.
[tex]V_{2} = \frac{m}{\rho_{2}}[/tex]
= [tex]\frac{RT_{2}m}{P_{2}}[/tex]
= [tex]\frac{0.287 \times 313 \times 0.239}{180}[/tex]
= 0.119 [tex]m^{3}/s[/tex]
And, the final velocity will be determined as follows.
[tex]v_{2} = \frac{V_{2}}{A}[/tex]
= [tex]\frac{4V_{2}}{D^{2} \times \pi}[/tex]
= [tex]\frac{4 \times 0.119}{(0.16)^{2} \times \pi}[/tex]
= 5.92 m/s
Therefore, the volumetric flow rate is 0.119 [tex]m^{3}/s[/tex] and velocity rate is 5.92 m/s.
The power rating of a 400-Ω resistor is 0.800 W.(a) What is the maximum voltage that can be applied across this resistor without damaging it? Use three significant figures in your answer.
Answer:
[tex]V=17.9\ Volt[/tex]
Explanation:
Joule's Law in Electricity
The Joule's law allows us to calculate the power dissipated in a resistor of resistance R through which goes a current I.
[tex]P=I^2R[/tex]
The relation between the voltage and the current is given by Ohm's law:
[tex]V=RI[/tex]
Solving for I and replacing int the first equation
[tex]\displaystyle P=\frac{V^2}{R}[/tex]
Solving for V
[tex]V=\sqrt{PR}[/tex]
[tex]V=\sqrt{0.8\cdot 400}=17.9[/tex]
[tex]\boxed{V=17.9\ Volt}[/tex]
Maximum voltage for given power rating, applied across this resistor without damaging it is 17.9 volts.
What is the Ohm's law?Ohm's law states that for a flowing current the potential difference of the circuit is directly proportional to the current flowing in it. Thus,
[tex]V\propto I[/tex]
Here, (V) is the potential difference and (I) is the current.
It can be written as,
[tex]V=IR[/tex]
Here, (R) is the resistance of the circuit.
Given information-
The value of resistance is 400-Ω.
The value of power rating is 0.800 W.
By the Joule's law, the power of a circuit is equal to the product of the square of the current flowing in it and the resistance. It can be given as,
[tex]P=I^2\times R\\I=\sqrt{\dfrac{P}{R}}[/tex]
Put this value of current in ohm's law as,
[tex]V=\sqrt{\dfrac{P}{R}}\times R\\V=\sqrt{PR}[/tex]
Put the value of power and current in the above formula,
[tex]V=\sqrt{0.800\times 400}\\V=17.9\rm Volts[/tex]
'
Thus the maximum voltage that can be applied across this resistor without damaging it is 17.9 volts.
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