Answer: 800N
Explanation:
Given :
Mass of ball =0.8kg
Contact time = 0.05 sec
Final velocity = initial velocity = 25m/s
Magnitude of the average force exerted on the wall by the ball is can be calculated using the relation;
Force(F) = mass(m) * average acceleration(a)
a= (initial velocity(u) + final velocity(v))/t
m = 0.8kg
u = v = 25m/s
t = contact time of the ball = 0.05s
Therefore,
a = (25 + 25) ÷ 0.05 = 1000m/s^2
Therefore,
Magnitude of average force (F)
F=ma
m = mass of ball = 0.8
a = 1000m/s^2
F = 0.8 * 1000
F = 800N
The free throw line in basketball is 4.570 m (15 ft) from the basket, which is 3.050 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 7.157 m/s, releasing it at a height of 2.440 m above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket?
Answer:
[tex]\theta = 86.491^{\textdegree}[/tex]
Explanation:
The equations for the horizontal and vertical position of the ball are, respectivelly:
[tex]4.570\,m = [(7.157\,\frac{m}{s})\cdot\cos \theta]\cdot t\\3.050\,m = 2.440\,m +[(7.157\,\frac{m}{s})\cdot \sin \theta]\cdot t - \frac{1}{2}\cdot (9.807\,\frac{m}{s^{2}} )\cdot t^{2}[/tex]
By isolating each trigonometric component and summing each equation:
[tex]20.885\,m^{2} = [51.223\,\frac{m^{2}}{s^{2}}\cdot \cos^{2} \theta]\cdot t^{2}[/tex]
[tex][0.61\,m + \frac{1}{2}\cdot (9.807\,\frac{m}{s^{2}} )\cdot t^{2}]^{2} = [51.223\,\frac{m^{2}}{s^{2}}\cdot \sin^{2} \theta]\cdot t^{2}[/tex]
[tex]21.257\,m^{2} + (5.982\,\frac{m^{2}}{s^{2}})\cdot t^{2}+(24.044\,\frac{m^{2}}{s^{4}} )\cdot t^{4} = (2623.796\,\frac{m^{2}}{s^{2}})\cdot t^{2}[/tex]
[tex]21.257\,m^{2} - (2617.814\,\frac{m^{2}}{s^{2}})\cdot t^{2}+(24.044\,\frac{m^{2}}{s^{4}} )\cdot t^{4} = 0[/tex]
The positive real roots are:
[tex]t_{1} = 10.434\,s,t_{2} = 0.09\,s[/tex]
The needed angle is:
[tex]\theta = \cos^{-1} [\frac{4.570\,m}{(7.157\,\frac{m}{s} )\cdot t} ]\\\theta_{1} = 86.491^{\textdegree}\\\theta_{2} = NaN[/tex]
On your first trip to Planet X you happen to take along a 290g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You're curious about the free-fall acceleration on Planet X, where ordinary tasks seem easier than on earth, but you can't find this information in your Visitor's Guide. One night you suspend the spring from the ceiling in your room and hang the mass from it. You find that the mass stretches the spring by 21.1cm . You then pull the mass down 11.2cm and release it. With the stopwatch you find that 11 oscillations take 18.2sCan you now satisfy your curiosity?what is the new g?
Answer:
Explanation:
11 oscillations in 18.2 s
Time period is defined as the time taken to complete one oscillation.
T = 18.2 / 11 = 1.655 s
mass, m = 290 g = 0.29 kg
Δx = 21.1 cm = 0.211 m
ω = 2π / T = (2 x 3.14) / 1.655 = 3.796 rad/s
[tex]\omega =\sqrt{\frac{K}{m}}[/tex]
Where, K is the spring constant
K = ω² m = 3.796 x 3.796 x 0.29 = 4.18 N/m
Now, mg = K Δx
0.29 x g = 4.18 x 0.211
g = 3.04 m/s²
A series RLC circuit containing a resistance of 12Ω, an inductance of 0.15H and a capacitor of 100uF are connected in series across a 100V, 50Hz supply. Calculate the total circuit impedance, the circuit’s current, and the power factor.
Answer:
Impedance = 19.44ohms
Current = 5.14A
Power factor = 0.62
Explanation:
Impedance in an RLC AC circuit is defined as the total opposition to the flow of current in the resistor, inductor and capacitor.
Impedance Z = √R²+(Xl-Xc)²
Where R is the resistance = 12Ω
Inductance L = 0.15H
Capacitance C = 100uF = 100×10^-6F
Since Xl = 2πfL and Xc = 1/2πfC where f is the frequency.
Xl = 2π×50×0.15
Xl = 15πΩ
Xl = 47.12Ω
Xc = 1/2π×50×100×10^-6
Xc = 100/π Ω
Xc = 31.83Ω
Z =√12²+(47.12-31.83)²
Z = √144+233.78
Z = 19.44Ω
Impedance = 19.44ohms
To calculate the circuit current, we will use the expression V=IZ where V is the supply voltage = 100V
I = V/Z = 100/19.44
I = 5.14Amperes
To calculate the power factor,
Power factor = cos(theta) where;
theta = arctan(Xl-Xc)/R
theta = arctan(47.12-31.83)/12
theta = arctan(15.29/12)
theta = arctan1.27
theta = 51.78°
Power factor = cos51.78°
Power factor = 0.62
Answer:
The circuit impedance [tex]=19.4 \Omega[/tex]
The circuit's current [tex]=5.14 A[/tex]
Circuit Power Factor [tex]=0.62[/tex]
Explanation:
Given:
Resistance [tex]R=12 \Omega[/tex]
Inductance [tex]=0.15H[/tex]
Capacitance [tex]=100uF[/tex]
Voltage [tex]=100V[/tex]
Step 1:
To calculate the inductive reactance, [tex]$X_{L}$[/tex].
[tex]X_{L}=2 \pi f L=2 \pi \times 50 \times 0.15=47.13 \Omega[/tex]
To calculate the Capacitive reactance,
[tex]X_{C}=\frac{1}{2 \pi f C}[/tex]
[tex]=\frac{1}{2 \pi \times 50 \times 100 \times 10^{-6}}[/tex]
[tex]=31.83 \Omega[/tex]
Step 2:
Circuit impedance,
[tex]$$Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$$[/tex]
where R is the resistance,
[tex]$$&Z=\sqrt{12^{2}+(47.13-31.83)^{2}}[/tex]
[tex]&Z=\sqrt{144+234}=19.4 \Omega\end{aligned}$$[/tex]
Step 3:
Circuits Current, I
[tex]$I=\frac{V_{S}}{Z}[/tex]
[tex]=\frac{100}{19.4}[/tex]
[tex]=5.14 \ A[/tex]
Step 4:
Voltages across the Circuit, [tex]$\mathrm{V}_{\mathrm{R}}, \mathrm{V}_{\mathrm{L}}, \mathrm{V}_{\mathrm{C}}$[/tex]
[tex]V_{R}=I \times R=5.14 \times 12=61.7$ volts[/tex]
[tex]V_{L}=I \times X_{L}=5.14 \times 47.13=242.2$ volts[/tex]
[tex]V_{C}=\ I \times X_{C}=5.14 \times 31.8=163.5$ volts[/tex]
Step 5:
Circuits Power factor
[tex]$=\frac{R}{Z}=\frac{12}{19.4}=0.619$[/tex]
Therefore,
The circuit impedance [tex]=19.4 \Omega[/tex]
The circuit's current [tex]=5.14\ A[/tex]
Power Factor [tex]=0.62[/tex]
To learn more about Circuit, refer:
https://brainly.com/question/15058220https://brainly.com/question/15170590A loop circuit has a resistance of R1 and a current of 2 A. The current is reduced to 1.5 A when an additional 1.6 Ω resistor is added in series with R1. What is the value of R1? Assume the internal resistance of the source of emf is zero. Answer in units of Ω.
Answer:
R1 = 4.8Ω
Explanation:
The loop circuit has an initial voltage of V = IR
I = 2 A , R1 = R
V = 2R1
with the current reduced to 1.5A with an additional 1.6Ω resistor
the total resistance of the circuit is 1.6 + R1
the voltage of the two scenarios has to be equal , since the same voltage flows through the circuit
therefore V = 2R1
from Ohms law V = IR
2R1= 1.5 (1.6 + R1)
2R1 = 2.4 + 1.5R1
collecting like terms
2R1 - 1.5R1 = 2.4
0.5R1 = 2.4
R1 = [tex]\frac{2.4}{0.5}[/tex]
R1 = 4.8Ω
Answer:
4.8 Ω
Explanation:
From Ohm's Law,
Using,
I = E/(R+r)................. Equation 1
E = I(R+r)................. Equation 2
Where I = current, E = emf, R = external resistance, r = internal resistance
Given: I = 2 A, R = R1, r = 0 Ω
Substitute into equation 2
E = 2(R1)
E = 2R1.
When an additional 1.6 Ω resistor is added in series,
E = 1.5(R1+1.6)
2R1 = 1.5R1+2.4
2R1-1.5R1 = 2.4
0.5R1 = 2.4
R1 = 2.4/0.5
R1 = 4.8 Ω
Show that the kinetic energy of a particle of mass m is related to the magnitude of the momentum p of that particle by KE 5 p2/2m. (Note: This expression is invalid for particles traveling at speeds near that of light.)
Answer:
Kinetic energy: [tex]E=\frac{1}{2}mv^{2}[/tex]
Momentum: p = mv
Kinetic energy in terms of momentum: [tex]E=\frac{1}{2}\frac{(mv)^{2}}{m}=\frac{p^{2}}{2m}[/tex]
Explanation:
The kinetic energy is given by this equation:
[tex]E=\frac{1}{2}mv^{2}[/tex] (1)
Now, we know that the momentum of a particle is p = m*v. This equation is true only with a classical particle, it meas particles with a speed less than the speed of light. If we had a particle traveling at speeds near that of light, the momentum would be p = γm₀v, where γ is the Lorentz factor.
So, if we see, we can rewrite the equation (1) to get this expression in terms of p.
Let's multiply and divide by mass (m) in the equation (1).
[tex]E=\frac{1}{2}\frac{m^{2}v^{2}}{m}[/tex]
[tex]E=\frac{1}{2}\frac{(mv)^{2}}{m}[/tex]
Using the p = mv here:
[tex]E=\frac{1}{2}\frac{p^{2}}{m}[/tex]
[tex]E=\frac{p^{2}}{2m}[/tex]
Therefore the kinetic energy can express in terms of momentum.
Let's see that it could not be possible using the the relativistic momentum, because it has a relativistic factor.
I hope it helps you!
To practice Problem-Solving Strategy 21.1 Conservation of energy in charge interactions. An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an electric field. The particle then begins to move. Find the speed of the alpha particle after it has moved through a potential difference of −3.45×10−3 V . The charge and the mass of an alpha particle are qα = 3.20×10−19 C and mα = 6.68×10−27 kg , respectively.
Answer:
Explanation:
kinetic energy of alpha particle
= Q X V ( Q is charge on the particle and V is potential difference )
= 3.2 x 10⁻¹⁹ x 3.45 x 10⁻³
= 11.04 x 10⁻²² J
1/2 m v² = 11.04 x 10⁻²²
1/2 x 6.68 x 10⁻²⁷ x v² = 11.04 x 10⁻²²
v² = 3.305 x 10⁵
v = 5.75 x 10² m /s
Final answer:
To find the speed of the alpha particle after it has moved through a potential difference, we can use the conservation of energy. Using the given values for charge and potential difference, we can calculate the speed using the formula for change in kinetic energy. The speed of the alpha particle is approximately 1.91x10^5 m/s.
Explanation:
To find the speed of the alpha particle after it has moved through a potential difference of -3.45x10^-3 V, we can use the conservation of energy.
The potential difference is given by ΔV = qΔV, where q is the charge of the alpha particle and ΔV is the potential difference. Plugging in the values, we get ΔV = (3.20x10^-19 C)(-3.45x10^-3 V).
The change in kinetic energy is given by ΔKE = (1/2)mv^2, where m is the mass of the alpha particle and v is its velocity. Setting ΔV equal to ΔKE, we can solve for v. Plugging in the values, we get (1/2)(6.68x10^-27 kg)v^2 = (3.20x10^-19 C)(-3.45x10^-3 V).
Solving for v, we find that the speed of the alpha particle is approximately 1.91x10^5 m/s.
The electron current in a horizontal metal wire is 7.4 × 1018 electrons/s, and the electrons are moving to the left. What are the magnitude and direction of the conventional current?
Answer:
The magnitude and direction of the conventional current is then 1.186 Amps moving to the left direction
Explanation:
To answer the question, it should be noted that the direction of conventional current is in the opposite direction of the flow of electrons. Therefore, the direction of flow of conventional current will be to the right
The magnitude of the electric current is equal to the rate of flow of the electrons or the time it takes for the electrons to flow past a section of the wire. Therefore the magnitude is that of the 7.4 × 1018 electrons/s
However the unit of the electricity = ampere which is = coulombs/seconds
The 7.4 × 1018 electrons carry
7.4 × 10¹⁸×1.60217662 × 10⁻¹⁹ coulombs = 1.1856106988 coulombs
Therefore the magnitude of electric current = 1.186 coulombs/Seconds
= 1.186 Amps
Explanation:
Below is an attachment containing the solution.
Why is magnesium the limiting reactant in this experiment
Answer:
Explanation:
Magnesium is being oxidized by the oxygen in the air to magnesium oxide. This is a highly exothermic combustion reaction, giving off intense heat and light. The reaction of the combustion of magnesium in oxygen is given below: 2Mg(s) + O2(g) → 2MgO(s) The stoichiometric factor is 2 moles of magnesium are burned for every 1 mole of oxygen (2mol Mg/1mol O2). If the magnesium strip weighs 1 gram, then there is 0.04 mol of magnesium (1 gram divided by 24.3 grams/mol Mg) available in the reaction. The amount of oxygen required to completely react with the magnesium strip is:0.04 mol Mg x (1 mol O2 / 2 mol Mg) = 0.02 mol O2 x 16 g/mol O2 = 0.32 gram O2.The magnesium will burn until consumed entirely. There is much more oxygen available in the atmosphere than needed to consume the magnesium. Thus the magnesium is the limiting reactant because it determines the amount of product formed.Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and the separation of the balls, what will now be the magnitude of the attractive force on each one?A) 16F
B) 8F
C) 4F
D) F
E) F/4
Answer:
D) F
Explanation:
Let m and M be the mass of the balls A and B respectively and r be the distance between the two balls. The magnitude of attractive gravitational force experienced by the balls due to each other is given by the relation :
[tex]F=\frac{GMm}{r^{2} }[/tex] ......(1)
Now, if the masses of both the balls gets doubled as well as there separation distance also gets doubled, then let F₁ be the new gravitational force acting on them.
Since, New mass of ball A = 2M
New mass of ball b = 2m
Distance between the two balls = 2r
Substitute these values in equation (1).
[tex]F_{1} =\frac{G(2M)(2m)}{(2r)^{2} }[/tex]
[tex]F_{1} =\frac{4GMm}{4r^{2} }=\frac{GMm}{r^{2} }[/tex]
Using equation (1) in the above equation.
F₁ = F
Volume of a Cube The volume V of a cube with sides of length x in. is changing with respect to time. At a certain instant of time, the sides of the cube are 7 in. long and increasing at the rate of 0.2 in./s. How fast is the volume of the cube changing (in cu in/s) at that instant of time?
Answer:
Therefore the volume of cube is change at the 29.4 cube in./s at that instant time.
Explanation:
Formula
[tex]\frac{dx^n}{dx} =nx^{n-1}[/tex]Cube :
The volume of a cube is = [tex]side^3[/tex]
The side of length is x in.
Then volume of the cube is (V) = [tex]x^3[/tex]
∴ V = [tex]x^3[/tex]
Differentiate with respect to t
[tex]\frac{d}{dt}(V)=\frac{d}{dt} (x^3)[/tex]
[tex]\Rightarrow \frac{dV}{dt} =3x^2\frac{dx}{dt}[/tex]....(1)
Given that the side of the cube is increasing at the rate of 0.2 in/s.
i.e [tex]\frac{dx}{dt} = 0.2[/tex] in/s.
And the sides of the cube are 7 in i.e x= 7 in
Putting [tex]\frac{dx}{dt} = 0.2[/tex] and x= 7 in equation (1)
[tex]\therefore \frac{dV}{dt} =3 \times 7^2 \times 0.2[/tex] cube in./s
=29.4 cube in./s
Therefore the volume of cube is change at the 29.4 cube in./s at that instant time.
To find the rate at which the volume of the cube is changing, differentiate the volume formula with respect to time. For a cube with sides of 7 in length growing at 0.2 in/s, the rate of volume change is 6.3 in³/s.
The volume of a cube is determined by the formula V = x³, where x is the length of the side of the cube.
Given that the sides of the cube are 7 in long and increasing at 0.2 in/s, we can calculate how fast the volume is changing by differentiating the volume formula with respect to time. By taking the derivative of V = x³, we find that the rate of change of the volume at that instant of time is 6.3 in³/s.
Chang drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 12 hours. When Chang drove home, there was no traffic and the trip only took 8 hours. If his average rate was 20 miles per hour faster on the trip home, how far away does Chang live from the mountains?
Answer:
Chang live 480 miles from the mountains
Explanation:
Constant Speed Motion
An object is said to have constant speed if it takes the same time t to travel the same distances x. The speed is calculated as
[tex]\displaystyle v=\frac{x}{t}[/tex]
From that equation, we can solve for x
[tex]x=v\cdot t[/tex]
and for t
[tex]\displaystyle t=\frac{x}{v}[/tex]
Let's assume the distance from the mountains and Chang's house is x. We know he took t1=12 hours in heavy traffic at an average speed v1, thus we can set
[tex]x=v_1\cdot t_1[/tex]
In his way back home Chang took t2=8 hours at a speed v2, thus:
[tex]x=v_2\cdot t_2[/tex]
Since the distance is the same
[tex]v_1\cdot t_1=v_2\cdot t_2[/tex]
The speed back is 20 mph more than the speed to the mountain:
[tex]v_2=v_1+20[/tex]
Replacing in the above equation
[tex]v_1\cdot t_1=(v_1+20)\cdot t_2[/tex]
[tex]v_1\cdot t_1=v_1\cdot t_2+20\cdot t_2[/tex]
[tex]v_1\cdot 12=v_1\cdot 8+20\cdot 8[/tex]
Solving for v1
[tex]v_1\cdot 4=160[/tex]
[tex]v_1=40\ mph[/tex]
Now we can compute the value of x
[tex]x=40\cdot 12[/tex]
[tex]\boxed{x=480 \ miles}[/tex]
Chang lives 480 miles from the mountains
Suppose there was a star with a parallax angle of 1 arcsecond. How far away would it be?
Answer: 3.26 light years
Explanation:
Each star has a parallax of one arcsecond at a distance of one parsec, which is equivalent to 3.26 light years.
so the parallax of 1 arcsecond will be at a distance of 1/1 × 3.26 light years
A star with a Parallax Angle of 1 arcsecond is 1 parsec or about 3.26 light-years away. This method of determining star distance is fundamental in astronomy.
The distance of a star from us can be determined using its parallax angle.
The unit of measurement is the parsec, which stands for 'parallax-second'.
If a star has a parallax angle of one arcsecond (which is essentially a measurement of the angular shift in the star's position due to the Earth's orbit around the Sun), it's defined to be 1 parsec away from us.
A parallax angle of 1 arcsecond, therefore, means that the star is 1 parsec away.
Because 1 parsec equals 3.26 light-years, such a star would actually be about 3.26 light-years distant from Earth.
This method of calculating the distance of stars using their parallax angles was revolutionized by the Hipparcos spacecraft and is critical in the field of astronomy.
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What should be the angle of incidence for sunlight on a plane mirror so that the rescue pilot sees the reflected light?
Answer:
The incidence angle is 27°
Explanation:
As the complete question is not given, the complete question is given here
The angle between the Sun and a rescue aircraft is 54 degrees. What should be the angle of incidence for sunlight on a plane mirror so that the rescue pilot sees the reflected light?
From the question, the total angle between the Sun and the aircraft is 54 degrees which is the sum of the incidence and the reflected angle so
[tex]\theta_{total}=\theta_{incidence}+\theta_{reflection}=54[/tex]
Also from the law of reflection
[tex]\theta_{incidence}=\theta_{reflection}[/tex]
So now the equation becomes
[tex]\theta_{total}=\theta_{incidence}+\theta_{reflection}=54\\\theta_{total}=\theta_{incidence}+\theta_{incidence}=54\\2\theta_{incidence}=54\\\theta_{incidence}=27\\[/tex]
So the incidence angle is 27°
The angle of incidence for sunlight on a plane mirror should be set so that the angle of reflection directs the light towards the rescue pilot. The mirror's orientation is crucial, and light reflects at the same angle relative to the normal. Practical adjustments have to be made based on the positions of the sun and pilot.
Explanation:For sunlight to be reflected from a plane mirror to a rescue pilot, the angle of incidence should be such that the angle of reflection directs the light towards the pilot. Since the angle of reflection is equal to the angle of incidence, the mirror must be tilted to reflect the light into the pilot's eyes.
According to the law of reflection, light incident on a mirror will reflect off at the same angle relative to the normal (an imaginary line perpendicular to the surface of the mirror). Hence, for the rescue pilot to see the reflected light, the angle of incidence must be adjusted accordingly based on the position of the sun and the pilot's location in the sky.
It is important to note that if we wish to maximize the reflection towards the pilot, utilizing the mirror's orientation is key. At very high angles of incidence, approaching 90 degrees, almost all the light is reflected, according to physical principles. However, such angles may not be practical when trying to target a specific viewer such as a pilot.
A cyclist is coasting at 13 m/s when she starts down a 460 m long slope that is 30 m high. The cyclist and her bicycle have a combined mass of70 kg. A steady 12 N drag force due to air resistance acts on her as she coasts all the way to the bottom
What is her speed at the bottom of the slope?
Express your answer to two significant figures and include the appropriate units.
.
Answer:
Explanation:
Given that
Initial velocity u=13m/s
Length of slope
L=460m
Height of slope =30m
Mass of cyclist and bike =70kg
Drag force, fictional force=12 N
Final velocity?
Because the system is not isolated, there is some workdone by the drag force.
Therefore,
∆E=W
K.E(f) - K.E(i) + P.E(f) - P.E(i)=W
½mVf² - ½mVi² + mgy(f) - mgy(i)=W
Note, y(f) = 0, the cyclists is already on the floor
½mVf² -½mVi² - mgy(i) = -Fd × d
½×70×Vf² - ½×70×13²-70×9.81×30=-12×450
35Vf²- 5915 - 20601=-5400
35Vf²=-5400+5915+20601
35Vf²=21116
Vf²=21116/35
Vf²=603.314
Vf=√603.314
Vf=24.56m/s
The final velocity is 24.46m/s at the bottom of the track.
You have been hired to help improve the material movement system at a manufacturing plant. Boxes containing 16 kg of tomato sauce in glass jars must slide from rest down a frictionless roller ramp to the loading dock, but they must not accelerate at a rate that exceeds 2.6 m/s2 because of safety concerns.a. What is the maximum angle of inclination of the ramp?b. If the vertical distance the ramp must span is 1.4 m, with what speed will the boxes exit the bottom of the ramp?c. What is the normal force on a box as it moves down the ramp?
a) [tex]15.4^{\circ}[/tex]
b) 5.2 m/s
c) 151.2 N
Explanation:
a)
When the box is on the frictionless ramp, there is only one force acting in the direction along the ramp: the component of the forc of gravity parallel to the ramp, which is given by
[tex]mg sin \theta[/tex]
where
m =16 kg is the mass of the box
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
[tex]\theta[/tex] is the angle of the ramp
According to Newton's second law of motion, the net force on the box is equal to the product of mass and acceleration, so:
[tex]F=ma\\mgsin \theta = ma[/tex]
where a is the acceleration.
From the equation above we get
[tex]a=g sin \theta[/tex]
And we are told that the acceleration must not exceed
[tex]a=2.6 m/s^2[/tex]
Substituting this value and solving for [tex]\theta[/tex], we find the maximum angle of the ramp:
[tex]\theta=sin^{-1}(\frac{a}{g})=sin^{-1}(\frac{2.6}{9.8})=15.4^{\circ}[/tex]
b)
Here we are told that the vertical distance of the ramp is
[tex]h=1.4 m[/tex]
Since there are no frictional forces acting on the box, the total mechanical energy of the box is conserved: this means that the initial gravitational potential energy of the box at the top must be equal to the kinetic energy of the box at the bottom of the ramp.
So we have:
[tex]GPE=KE\\mgh=\frac{1}{2}mv^2[/tex]
where:
m = 16 kg is the mass of the box
[tex]g=9.8 m/s^2[/tex]
h = 1.4 m height of the ramp
v = final speed of the box at the bottom of the ramp
Solving for v,
[tex]v=\sqrt{2gh}=\sqrt{2(9.8)(1.4)}=5.2 m/s[/tex]
c)
There are two forces acting on the box in the direction perpendicular to the ramp:
- The normal force, N, upward
- The component of the weight perpendicular to the ramp, downward, of magnitude
[tex]mg cos \theta[/tex]
Since the box is in equilibrium along the perpendicular direction, the net force is zero, so we can write:
[tex]N-mg cos \theta[/tex]
and by substituting:
m = 16 kg
[tex]g=9.8 m/s^2[/tex]
[tex]\theta=15.4^{\circ}[/tex]
We can find the normal force:
[tex]N=mg cos \theta=(16)(9.8)cos(15.4^{\circ})=151.2 N[/tex]
The maximum angle of inclination of the ramp and the final velocity of the boxes are calculated based on the given acceleration and distance. The normal force on a box moving down the ramp remains unaffected by the angle of inclination in this case as the ramp is devoid of friction. It's an application of physics concepts in real-world situations.
Explanation:This problem is a practical application of the concepts of physics, specifically mechanics. Let's break it down.
a. The maximum angle of inclination of the ramp can be found by utilizing the relationship between the acceleration, the gravitational constant, and the angle of inclination. The formula is as follow:
sin(θ) = acceleration / g
Substitute the given acceleration (2.6 m/s²) and the gravitational constant g (9.8 m/s²) and solve for θ.
b. To find the final velocity of the boxes, we can apply the equations of motion. Using the formula v² = u² + 2gs (where u is the initial velocity, g is the gravitational constant, s is the distance) and substituting the given values (u=0, g=2.6 m/s², s=1.4m), we find the final velocity.
c. The normal force on a box moving down the ramp can be found from the formula: Normal force = mg cos (θ) (where m is the mass, g is gravitational constant and θ is the angle of inclination). Here, θ does not affect the normal force because the box is moving down a frictionless ramp.
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10. A satellites is in a circular orbit around the earth at a height of 360 km above the earth’s surface. What is its time period? What is its orbital speed?
Answer:
Orbital speed=8102.39m/s
Time period=2935.98seconds
Explanation:
For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)
V2R+h=g(R2(R+h)2)
V=√g(R2R+h)
V= sqrt(9.8 × (6371000)^2/(6371000+360000)
V= sqrt(9.8× (4.059×10^13/6731000)
V=sqrt(65648789.18)
V= 8102.39m/s
Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)
T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)
T=sqrt(3.40×10^21)/ (3.99×10^14)
T= sqrt(0.862×10^7)
T= 2935.98seconds
How high does a rocket have to go above the Earth’s surface so that its weight is reduced to 58.8 % of its weight at the Earth’s surface? The radius of the Earth is 6380 km and the universal gravitational constant is 6.67 × 10−11 N · m 2 /kg2 . Answer in units of km
Final answer:
To determine the height a rocket needs to go above the Earth's surface so that its weight is reduced to 58.8% of its weight at the Earth's surface, we can use the concept of gravitational force.
Explanation:
To determine the height a rocket needs to go above the Earth's surface so that its weight is reduced to 58.8% of its weight at the Earth's surface, we can use the concept of gravitational force. The force of gravity between two objects is given by the equation: F = (G * m1 * m2) / r^2, where G is the universal gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between the objects. We can set up a ratio of the weight of the rocket at a certain height to its weight at the Earth's surface:
Weight at height / Weight at surface = (G * m1 * m2) / (r + h)^2 / (G * m1 * m2) / r^2
By substituting the given values, such as the radius of the Earth and the weight reduction percentage, we can solve for h, which represents the height the rocket needs to reach. The final answer should be in units of kilometers, which can be obtained by converting the radius of the Earth from meters to kilometers.
A box of mass m = 17.5 kg is pulled up a ramp that is inclined at an angle θ = 23.0 ∘ angle with respect to the horizontal. The coefficient of kinetic friction between the box and the ramp is μ k = 0.295 , and the rope pulling the box is parallel to the ramp. If the box accelerates up the ramp at a rate of a = 2.29 m/s 2 , calculate the tension F T in the rope. Use g = 9.81 m/s 2 for the acceleration due to gravity.
Answer:
T = 153.77 [N]
Explanation:
To solve this type of problems, we must make a free body diagram, with the forces acting on the box. Then performing a sum of forces on the Y axis equal to zero we can find the value of the normal force. After finding the friction force, we performed a sum of forces equal to the product of mass by acceleration (newton's second law). We can find the T-Force value.
To calculate the tension in the rope, we can use Newton's second law. Using the given values of mass, angle, coefficient of kinetic friction, and acceleration, we can set up an equation and solve for FT. The tension in the rope is approximately 173.5 N.
Explanation:To calculate the tension in the rope, we can use Newton's second law. The only external forces acting on the mass are its weight and the tension in the rope. Using the given information, we can set up the equation FT - mgsinθ - μkmgcosθ = ma, where FT is the tension, m is the mass, θ is the angle, μk is the coefficient of kinetic friction, a is the acceleration, and g is the acceleration due to gravity.
Substituting the given values, we have FT - (17.5 kg)(9.81 m/s2)sin(23.0°) - (0.295)(17.5 kg)(9.81 m/s2)cos(23.0°) = (17.5 kg)(2.29 m/s2).
Solving for FT, we find that the tension in the rope is approximately 173.5 N.
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A wheel of radius 0.4 m rotates with a constant angular velocity of 50 rad/s. What are the magnitudes of the tangential velocity
Answer:
Therefore the magnitude of tangential velocity is 20 m/s.
Explanation:
Tangential velocity:The tangential velocity is the straight line velocity of at any point of rotating object.
It is denoted by [tex]v_t[/tex]
[tex]v_t= \omega r[/tex]
ω= angular velocity
r = radius of rotating object.
Angular velocity: Angular velocity is ratio of angle to time.
Here ω= 50 rad/s and r = 0.4 m
Tangential velocity=(50 ×0.4)m/s
=20 m/s
Therefore the magnitude of tangential velocity is 20 m/s.
an object of unknown mass oscillates on the end of a spring wit period 8 s. A 10kg object is attached to the first object, changing te period to 12 s. What is the mass of the first object
Answer:
Explanation:
Let the first object have a mass of M
And a period of T1=8sec
The second object has a mass 10kg and a period of T2=12 sec
It is know that,
The period of a spring-mass system is proportional to the square root of the mass and inversely proportional to the square root of the spring constant.
T=2π√(m/k)
Then the constant in this equation is the spring constant (k) and 2π, which does not change for the same material.
Then, make k subject of formulas
T²=4π²(m/k)
T²k=4π²m
Then, k/4π²=m/T²
So the k is directly proportional to m and inversely proportional to T²
M1/T1²=M2/T2²
Since, M1 is unknown, M2=10kg, T1=8sec and T2=12
Then,
M1/T1²=M2/T2²
M1/8²=10/12²
M1/64=0.06944
M1=0.06944×64
M1=4.444kg
The mass of the first object is 4.44kg
Narrow belts of high-speed winds that blow in the upper troposphere and lower stratosphere are known as
Explanation:
Jet streams are the winds that flow in the upper levels of the atmosphere.These are relatively narrow bands of strong wind.These high-speed winds blow from west to east in jet streams.Sometimes, the flow of wind shifts from north to south.Jet streams follow the boundaries between hot and cold air.Carbon dioxide enters an adiabatic compressor at 100 kPa and 300K at a rate of 0.5 kg/s and leaves at 600 kPa and 450K. Neglecting kinetic energy changes, determine a) the volume flow rate of the carbon dioxide at the compressor inlet (Ans. around 0.3 m3/s) and b) the power input to the compressor (Ans. around 70 kW).
Explanation:
Below is an attachment containing the solution.
Answer:
The answers to the question are
a) The volume flow rate of the carbon dioxide at the compressor inlet is 0.2834 m³/s ≈ 0.3 m³/s
b) The power input to the compressor is 73.35 kW ≈ 70 kW
Explanation:
We note the following
Mass flow rate = 0.5 kg/s
Inlet pressure = 100 pKa
Outlet pressure = 600 kPa
Inlet temperature = 300 K
Outlet temperature = 450 K
Molar mass of CO₂ = 44.01 g/mol
R Universal Gas Constant = 8.314 4621. J K−1 mol−1
a) Number of moles = [tex]\frac{Mass}{Molar.Mass}[/tex] = [tex]\frac{500g}{44.01g}[/tex] = 11.361 moles
P·V= n·R·T ∴ V = [tex]\frac{n*R*T}{P}[/tex] = [tex]\frac{11.361*8.3145*300}{ 100 }[/tex] = 0.2834 m³
Therefore the volume flow rate = 0.2834 m³/s ≈ 0.3 m³/s
b) Cp at 300 K = 0.846 kJ/(kg K)
Cp at 600 K = 0.978 kJ/(kg K)
Cv = 0.657
K = 1.289
While the power input to the compressor can be calculated by
m'×Cp×(T₂-T₁)
Where m' = mass flow rate = 0.5 kg/s
Therefore power = 0.5 kg/s×0.978 kJ/(kg K)×(450 K - 300 K)
= 73.35 kJ/s = 73.35 kW ≈ 70 kW
Most of the visible light we see coming from the sun originates from the
Answer:
The answer for this is photosphere.
Explanation:
Most of the visible light we see coming from the sun originates from photosphere.The Photosphere is 300km dense and the temperature at the bottom of the Photosphere is 6400K and the top of the Photosphere is 4600K respectively.
Following are the feature of photosphere that is given below.
Limb Darkening: The edges are darker than the centre part of the sun.Sunspots: The size of the sunspots is similar to the size of the Earth.
Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height of 30.0 m. During the collision at the bottom of the elevator shaft, a 86.0 kg passenger is stopped in 5.00 ms. (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision
Explanation:
Below is an attachment containing the solution.
A car accelerates uniformly from rest and reaches a speed of 21.5 m/s in 11.4 s. The diameter of a tire is 66.5 cm. Find the number of revolutions the tire makes during this motion, assuming no slipping. Answer in units of rev.
Answer:
57.39 rev
Explanation:
From circular motion,
s = rθ................... Equation 1
Where s = distance, r = radius, θ = angular distance.
make θ the subject of the equation
θ = s/r............... Equation 2
Where can look for s using any of the equation of motion
s = (v+u)t/2............ Equation 3
Where v and u = Final and initial velocity respectively, t= time.
Given: v = 21.5 m/s, u = 0 m/s (at rest), t = 11.4 s
Substitute into equation 3
s = (21.5+0)11.4/2
s = 122.55 m.
given: r = 66.5/2 = 33.25 cm = 0.3325 m
Substitute into equation 2
θ = 122.55/0.3325
θ = 368.57 rad
θ = (360.57×0.159155) rev
θ = 57.39 rev
Answer:
58.6886 revolutions
Explanation:
First we need to know the total distance travelled by the car, and we can do that using Torricelli formula:
V2= Vo2 + 2aDS
V = 21.5
Vo = 0
a = 21.5/11.4 = 1.886
(21.5)^2 = 2*1.886*DS
DS = 462.25/3.772 = 122.5477 m
For each revolution of the tire, the car moves the circunference of the tire, which is pi*d = 3.14*66.5 = 208.81 cm = 2.0881 m
So, to know the number of revolutions, we divide the total travel distance by the circunference of the tire:
122.5477/2.0881 = 58.6886
Is it easier to balance a long rod with a mass attached to it when the mass is closer to your hand or when the mass is farther away?
Answer:
Yes, It is easier to balance a long rod with a mass attached to it when the mass is farther from your hand.
Explanation:
There it is illustrated how we can hold a long rod while the hand is further away from the surface.
Rotational inertia depends on whether the mass is closer to or further to the rotation point.
The more the mass is, the greater the acceleration of the rotation.
The explanation for this is that the mass variance is directly proportional to the roational inertia height.
When the mass attached to a long rod is closer to your hand, the rod is easier to balance. This is because the center of gravity and mass concentration is closer to the axis of rotation, which reduces the force needed to maintain balance.
Explanation:In the context of balancing a long rod with a mass attached to it, it's crucial to consider the concept of the center of gravity, and moment of inertia. The center of gravity is the point at which the weight of an object is concentrated. When the mass is closer to your hand, the center of gravity is also closer, making the rod easier to balance as less torque-force is required.
This is analogous to two people carrying a load – whoever is closer to the center of gravity carries more of the weight, making it easier for them. Furthermore, the moment of inertia, a measure of an object's resistance to changes to its rotation, also plays a role. Objects with their mass concentrated closer to the axis of rotation have a lower moment of inertia and are thus easier to rotate or balance. This is reflective of the rod's behavior – when the mass is closer to your hand (essentially the axis of rotation), balancing it becomes easier.
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What is the term that describes how a plant responds to gravity
Answer:
The term is geotropism (also known as gravitropism)
A boulder is raised above the ground, so that its potential energy relative to the ground is 200 J. Then it is dropped. Estimate what its kinetic energy will be just before hitting the ground.
Answer:
200 J
Explanation:
In this problem, I assume there is no air resistance, so the system is isolated (=no external forces).
For an isolated system, the total mechanical energy is constant, and it is given by:
[tex]E=KE+PE[/tex]
where
KE is the kinetic energy
PE is the potential energy
The kinetic energy is the energy due to the motion of the object, while the potential energy is the energy due to the position of the object relative to the ground.
At the beginning, when the boulder is raised above the ground, its height above the ground is maximum, while its speed is zero; it means that all its mechanical energy is just potential energy, and it is:
[tex]E=PE_{max}=200 J[/tex]
As the boulder falls down, its altitude decreases, so its potential energy decreases, while the speed increases, and the kinetic energy increases. Therefore, potential energy is converted into kinetic energy.
Eventually, just before the boulder hits the ground, the height of the object is zero, and the speed is maximum; this means that all the energy has now converted into kinetic energy, and we have
[tex]E=KE_{max}=200 J[/tex]
Therefore, the kinetic energy just before hitting the ground is 200 J.
Which vessels have a tunica media with relatively more smooth muscle than elastic tissue, and an elastic membrane on each face of the tunica media?
Answer:
Muscular Arteries
Explanation:
Muscular arteries continue from elastic arteries and control the distribution of blood throughout the body.
Muscular arteries are the vessels that have more smooth muscle than elastic tissue in their tunica media, along with an elastic membrane on each face of the tunica media. They are found farther away from the heart and have a significant role in vasoconstriction due to their increased amount of smooth muscle. They have an internal and external elastic membrane.
Explanation:The vessels that have a tunica media with more smooth muscle than elastic tissue and an elastic membrane on each face of the tunica media are known as muscular arteries. These arteries exist farther from the heart and due to their increased amount of smooth muscle, they play a significant role in vasoconstriction.
In these arteries, the percentage of elastic fibers decreases, while the presence of smooth muscle increases. This results to the artery having a thick tunica media. It's important to note that the diameter of muscular arteries can range from 0.1mm to 10mm.
Additionally, muscular arteries possess an internal elastic membrane (also known as the internal elastic lamina) at the boundary with the tunica media, as well as an external elastic membrane in larger vessels. This gives these arteries increased structure while allowing them the ability to stretch. Due to the decreased blood pressure, muscles arteries can accommodate, elasticity is less crucial in these types of vessels.
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The area of each plate of a parallel plate capacitor is 0.021 m2. The plates are 2.75 mm apart with a dielectric material (κ = 3.0) between them. The maximum possible electric field between the plates is 3.25 ✕ 105 V/m.What is the maximum energy that can be stored in the capacitor?
Explanation:
The given data is as follows.
Dielectric constant, K = 3.0
Area of the plates (A) = 0.021 [tex]m^{2}[/tex]
Distance between plates (d) = [tex]2.75 \times 10^{-3} m[/tex]
Maximum electric field (E) = [tex]3.25 \times 10^{5} V/m[/tex]
Now, we will calculate the capacitance as follows.
C = [tex]\frac{k \epsilon_{o} \times A}{d}[/tex]
= [tex]\frac{3.0 \times 8.85 \times 10^{-12} \times 0.021}{2.75 \times 10^{-3}}[/tex]
= [tex]\frac{0.55755 \times 10^{-12}}{2.75 \times 10^{-3}}[/tex]
= [tex]0.203 \times 10^{-9}[/tex] F
Formula to calculate electric charge is as follows.
E = [tex]\frac{\sigma}{k \epsilon_{o}}[/tex]
or, Q = [tex]E \times k \times \epsilon_{o}A[/tex] (as [tex]\frac{\sigma}{\epsilon_{o}} = \frac{Q}{A}[/tex])
= [tex]3.25 \times 10^{5} \times 3.0 \times 8.85 \times 10^{-12} \times 0.021[/tex]
= [tex]181.2 \times 10^{-9} C[/tex]
Formula to calculate the energy is as follows.
U = [tex]\frac{1 \times Q^{2}}{2 \times C}[/tex]
= [tex]\frac{(181.2 \times 10^{-9} C)^{2}}{2 \times 1.6691 \times 10^{-9}}[/tex]
= [tex]\frac{32833.44 \times 10^{-18}}{3.3382 \times 10^{-9}}[/tex]
= [tex]9835.67 \times 10^{-9}[/tex]
or, = [tex]98.35 \times 10^{-7} J[/tex]
Thus, we can conclude that the maximum energy that can be stored in the capacitor is [tex]98.35 \times 10^{-7} J[/tex].