Answer:
at the highest point of the path the acceleration of ball is same as acceleration due to gravity
Explanation:
At the highest point of the path of the ball the speed of the ball becomes zero as the acceleration due to gravity will decelerate the motion of ball due to which the speed of ball will keep on decreasing and finally it comes to rest
So here we will say that at the highest point of the path the speed of the ball comes to zero
now by the force diagram we can say that net force on the ball due to gravity is given by
[tex]F_g = mg[/tex]
now the acceleration of ball is given as
[tex]a = \frac{F_g}{m}[/tex]
[tex]a = \frac{mg}{m} = g[/tex]
so at the highest point of the path the acceleration of ball is same as acceleration due to gravity
Final answer:
The magnitude of the acceleration of a tennis ball at its highest point, in an evacuated chamber and under the influence of gravity alone, is 9.8 m/s². This remains constant regardless of the ball's position in its trajectory.
Explanation:
The question asks about the magnitude of the acceleration of a tennis ball at its highest point when it is shot vertically upward in an evacuated chamber with an initial speed. This scenario is governed by the principles of classical mechanics, specifically under the domain of gravity's influence on objects in motion. Regardless of whether the object is moving up or down or is at its peak height, the acceleration due to gravity on Earth remains constant.
Acceleration at the highest point, or at any point during its motion under gravity alone (assuming no air resistance), is -9.8 m/s2. It's essential to understand that the negative sign indicates the direction of acceleration, which is towards the center of the Earth but does not affect the magnitude of acceleration. Hence, the magnitude of acceleration is 9.8 m/s2, indicating that gravity is the only force acting on the ball throughout its trajectory.
When turned on, a fan requires 5.0 seconds to get up to its final operating rotational speed of 1200 rpm. a) How large is the final operating angular speed, in rad/s? b) What was the angular acceleration of the fan, in rad/s^2?
Answer:
a)
125.6 rad/s
b)
25.12 rad/s²
Explanation:
a)
t = time required by the fan to get up to final operating speed = 5 sec
w = final operating rotational speed = 1200 rpm
we know that :
1 revolution = 2π rad
1 min = 60 sec
w = [tex]1200\frac{rev}{min}\frac{2\pi rad}{1 rev}\frac{1 min}{60 sec}[/tex]
w = [tex]\frac{1200\times 2\pi }{60}\frac{rad}{s}[/tex]
w = 125.6 rad/s
b)
w₀ = initial angular speed = 0 rad/s
α = angular acceleration
using the equation
w = w₀ + α t
125.6 = 0 + α (5)
α = 25.12 rad/s²
If the area of an iron rod is 10 cm by 0.5 cm and length is 35 cm. Find the value of resistance, if 11x10^-8 ohm.m be the resistivity of iron.
Answer:
Resistance of the iron rod, R = 0.000077 ohms
Explanation:
It is given that,
Area of iron rod, [tex]A=10\ cm\times 0.5\ cm=5\ cm^2 = 0.0005\ m^2[/tex]
Length of the rod, L = 35 cm = 0.35 m
Resistivity of Iron, [tex]\rho=11\times 10^{-8}\ \Omega-m[/tex]
We need to find the resistance of the iron rod. It is given by :
[tex]R=\rho\dfrac{L}{A}[/tex]
[tex]R=11\times 10^{-8}\times \dfrac{0.35\ m}{0.0005\ m^2}[/tex]
[tex]R=0.000077 \Omega[/tex]
So, the resistance of the rod is 0.000077 ohms. Hence, this is the required solution.
An alpha particle, which has charge 3.20 ✕ 10−19 C, is moved from point A, where the electric potential is 2.40 ✕ 103 J/C, to point B, where the electric potential is 4.95 ✕ 103 J/C. Calculate the work in electron volts done by the electric field on the alpha particle.
Answer:
5100 eV
Explanation:
q = 3.2 x 10^-19 C
Va = 2.4 x 10^3 J/C
Vb = 4.95 x 10^3 J/C
Work done = q (Vb - Va)
W = 3.2 x 10^-19 x (4.95 - 2.4 ) x 10^3 = 8.16 x 10^-16 J
A we know that 1 eV = 1.6 x 10^-19 J
So, W = (8.16 x 10^-16) / (1.6 x 10^-19) = 5100 eV
Constants Part A Two small charged spheres are 9.20 cm apart. They are moved, and the force on each of them is found to have been tripled How far apart are they now? Express your answer to three significant figures and include the appropriate units.
Answer:
5.3 cm
Explanation:
Let the charges on the spheres be q and Q.
r = 9.20 cm
F = k Q q / (9.20)^2 ..... (1)
Let the new distance be r' and force F'
F' = 3 F
F' = k Q q / r'^2
3 F = k Q q / r'^2 ...... (2)
Divide equation (1) by (2)
F / 3 F = r'^2 / 84.64
1 / 3 = r'^2 / 84.64
r' = 5.3 cm
Tsunamis are fast-moving waves often generated by underwater earthquakes. In the deep ocean their amplitude is barely noticable, but upon reaching shore, they can rise up to the astonishing height of a six-story building. One tsunami, generated off the Aleutian islands in Alaska, had a wavelength of 646 km and traveled a distance of 3410 km in 4.84 h. (a) What was the speed (in m/s) of the wave? For reference, the speed of a 747 jetliner is about 250 m/s. Find the wave's (b) frequency and (c) period.
Answer:
a) V = 195.70 m/s
b) f=3.02 × 10⁻⁴ Hz
c) T = 3311.25 seconds
Explanation:
Given:
Wavelength, λ = 646 Km = 646000 m
Distance traveled = 3410 Km = 3410000 m
Time = 4.84 h = 4.84 × 3600 s = 17424 seconds
a) The speed (V) of the wave is given as
V = distance / time
V = 3410000 m/ 17424 seconds
or
V = 195.70 m/s
b) The frequency (f) of the wave is given as:
f = V / λ
f= 195.70 / 646000
f=3.02 × 10⁻⁴ Hz
c) The time period (T) is given as:
T = 1/ f
T = 1/ (3.02 × 10⁻⁴) Hz
T = 3311.25 seconds
A Carnot engine's operating temperatures are 240 ∘C and 20 ∘C. The engine's power output is 910 W . Part A Calculate the rate of heat output. Express your answer using two significant figures.
Answer:1200
Explanation:
Given data
Upper Temprature[tex]\left ( T_H\right )=240^{\circ}\approx 513[/tex]
Lower Temprature [tex]\left ( T_L\right )=20^{\circ}\approx 293[/tex]
Engine power ouput[tex]\left ( W\right )=910 W[/tex]
Efficiency of carnot cycle is given by
[tex]\eta =1-\frac{T_L}{T_H}[/tex]
[tex]\eta =\frac{W_s}{Q_s}[/tex]
[tex]1-\frac{293}{513}=\frac{910}{Q_s}[/tex]
[tex]Q_s=2121.954 W[/tex]
[tex]Q_r=1211.954 W[/tex]
rounding off to two significant figures
[tex]Q_r=1200 W[/tex]
Answer:
1200
Explanation:
Determine the capacitive reactance for a 20 uF capacitor that is across a 20 volt, 60 Hz source
Answer:
Capacitive reactance is 132.6 Ω.
Explanation:
It is given that,
Capacitance, [tex]C=20\ \mu F=20\times 10^{-6}\ F=2\times 10^{-5}\ F[/tex]
Voltage source, V = 20 volt
Frequency of source, f = 60 Hz
We need to find the capacitive reactance. It is defined as the reactance for a capacitor. It is given by :
[tex]X_C=\dfrac{1}{2\pi fC}[/tex]
[tex]X_C=\dfrac{1}{2\pi \times 60\ Hz\times 2\times 10^{-5}\ F}[/tex]
[tex]X_C=132.6\ \Omega[/tex]
So, the capacitive reactance of the capacitor is 132.6 Ω. Hence, this is the required solution.
A 900 kg car initially going 15 m/s only in the x-direction runs into a stationary 1500 kg truck. After the collision the car is going 5.0 m/s at an angle of 40 degrees above the x- axis. What is the magnitude and direction of the velocity of the truck right after the collision?
Answer:
6.97 m/s, 344 degree
Explanation:
mass of car, m = 900 kg, uc = 15 m/s, vc = 5 m/s, θ = 40 degree
mass of truck, M = 1500 kg, uT = 0, vT = ?, Φ = ?
Here, vT be the velocity of truck after collision and Φ its direction above x axis.
Use conservation of momentum in X axis
900 x 15 + 1500 x 0 = 900 x 5 Cos 40 + 1500 x vT Cos Φ
13500 - 3447.2 = 1500 vT CosΦ
vT CosΦ = 6.7 .....(1)
Use conservation of momentum in y axis
0 + 0 = 900 x 5 Sin 40 + 1500 vT SinΦ
vT SinΦ = - 1.928 .....(2)
Squarring both the equations and then add
vT^2 = 6.7^2 + (-1.928)^2
vT = 6.97 m/s
Dividing equation 2 by 1
tan Φ = - 1.928 / 6.7
Φ = - 16 degree
Angle from + X axis = 360 - 16 = 344 degree
A 3.8 L volume of neon gas (Ne) is at a pressure of 6.8 atm and a temperature of 470 K. The atomic mass of neon is 20.2 g/mol, and the ideal gas constant is R=8.314 J/mol*K. The mass of neon is closest to what?
Answer:
The mass of neon is 13.534 g.
Explanation:
Given that,
Volume = 3.8 L
Pressure = 6.8 atm
Temperature = 470 K
Atomic mass of neon =20.2 g/mol
Gas constant R = 8.314 = 0.082057 L atm/mol K
We need to calculate the mass of neon
Using equation of gas
[tex]PV=nRT[/tex]
[tex]6.8\times3.8=n\times0.082057\times470[/tex]
[tex]n=\dfrac{6.8\times3.8}{0.082057\times470}[/tex]
[tex]n= 0.670[/tex]
We know that,
[tex]n = \dfrac{m}{molar\ mass}[/tex]
[tex]m = n\times molar\ mass[/tex]
[tex]m=0.670\times20.2[/tex]
[tex]m=13.534\ g[/tex]
Hence, The mass of neon is 13.534 g.
First, let us consider an object launched vertically upward with an initial speed v. Neglect air resistance. Part B) At the top point of the flight, what can be said about the projectile's kinetic and potential energy?
A). Both kinetic and potential energy are at their maximum values.
B).Both kinetic and potential energy are at their minimum values.
C).Kinetic energy is at a maximum; potential energy is at a minimum.
D).Kinetic energy is at a minimum; potential energy is at a maximum.
Answer:
Option d is the correct option Kinetic energy is minimum while as potential energy is maximum
Explanation:
At the top most point of the flight since it cannot reach any further up in the vertical direction thus the potential energy at this position shall be maximum. Now since the total energy of the projectile is conserved so the remaining kinetic energy shall be minimum at that point so as the sum of the kinetic and potential energies remain constant.
If a pendulum has a period of 4 seconds at the north pole with gravity force 9.83 ms^-2 and a 3.97m pendulum length, how could the same pendulum be adjusted to have the same period as the equator with 9.78ms^-2 gravity force?
Answer:
The same pendulum could be adjusted to have the same period, in the equator must have a length of 3.949m.
Explanation:
Tnp= 4 sec
gnp= 9.83 m/sec²
Lnp= 3.97m
Tequ= 4 sec
gequ= 9.78 m/sec²
Lequ=?
Lequ= (Lnp* gequ) / gnp
Lequ= 3.949 m
A large sphere of material is placed in a dissolving liquid. (a) When the radius is 1 m, it is observed to be changing at 8 cm/hour. At that instant, what is the rate at which the sphere's volume is being dissolved with respect to time?
Final answer:
To determine the rate at which the sphere's volume is dissolving, differentiate the volume formula with respect to time and substitute the given rate of the radius change. The dissolution rate at the instant when the radius is 1 meter is 4π*0.08 m^3/hour.
Explanation:
To find the rate at which the sphere's volume is being dissolved with respect to time, we can apply the formula for the volume of a sphere, V = &frac43;πr^3, and differentiate it with respect to time (t). Since the radius (r) of the sphere is changing, we have:
dV/dt = d(&frac43;πr^3)/dt = 4πr^2*(dr/dt)
Given that the radius is shrinking at a rate of 8 cm/hour (which we need to convert to meters since our volume will be in cubic meters), the radius change in m/hour is:
dr/dt = 0.08 m/hour
At the instant when r is 1 meter, plugging the values into the equation gives us:
dV/dt = 4π(1 m)^2*(0.08 m/hour) = 4π*0.08 m^3/hour
Therefore, the rate at which the sphere's volume is being dissolved is 4π*0.08 m^3/hour.
To make a bounce pass, a player throws a 0.60-kg basketball toward the floor. The ball hits the floor with a speed of 5.4 m/s at an angle of 65° to the vertical. If the ball rebounds with the same speed and angle, what was the impulse delivered to it by the floor?
The impulse delivered to the ball by the floor after rebounding with the same speed and angle is; Δp = 2.74 N.s
We are given;
Mass of basketball; m = 0.6 kg
Speed; v = 5.4 m/s
Angle to the vertical; θ = 65°
We want to find the impulse if the ball rebounds with the same speed and angle above.
Now, as the x-component of the momentum remains constant, the impulse would be equal to the change in the y-component of the balls' momentum.
Thus;
Δp = m[(v_f * cos θ) - (v_i * cos θ)]
Earlier we saw that we were given the rebound speed to be the same with the initial speed but however, as the y-axis is pointing upwards, it means that the initial velocity will be negative.
Thus;
v_f = 5.4 m/s
v_i = -5.4 m/s
Plugging in the relevant values into the impulse equation gives;
Δp = 0.6[(5.4 * cos 65) - (-5.4 * cos 65)]
Δp = (0.6 × cos 65) × (5.4 + 5.4)
Δp = 2.74 N.s
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The impulse delivered to the ball by the floor after the rebound is 5.87 Ns.
The given parameters;
mass of the ball, m = 0.6 kgspeed of the ball, v = 5.4 m/sdirection of the ball, = 65° to the vertical.The impulse delivered to the ball is the change in the linear momentum of the ball in vertical direction.
The magnitude of the impulse delivered to the ball by the floor is calculated as;
[tex]J = \Delta P = m(v_f - v_0)\\\\J = m(vsin\theta - vsin\theta)\\\\the \ final \ speed \ occured \ in \ opposite \ direction;\\\\J = m(vsin\theta + vsin\theta)\\\\J = m(2vsin\theta)\\\\J = 2mv \times sin(\theta)\\\\J = 2\times 0.6 \times 5.4 \times sin(65)\\\\J = 5.87 \ Ns[/tex]
Thus, the impulse delivered to the ball by the floor after the rebound is 5.87 Ns.
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A child is twirling a 0.0154-kg ball on a string in a horizontal circle whose radius is 0.149 m. The ball travels once around the circle in 0.639 s. (a) Determine the centripetal force acting on the ball. (b) If the speed is doubled, by what factor does the centripetal force increase?
Answer:
The answers are: a)Fcp=0,23N b)As Fcp=0,93N, it increases 4 times when the speedis doubled
Explanation:
Let´s explain what´s the centripetal force about: It´s the force applied over an object moving on a curvilinear path. This force is directed to the rotation center.
This definition is described this way:
[tex]Fcp*r=m*V^2[/tex] where:
Fcp is the Centripetal Force
r is the horizontal circle radius
m is the ball mass
V is the tangencial speed, same as the rotation speed
w is the angular speed
Here we need to note that the information we have talks about 1cycle/0,639s (One cycle per 0,639s). We need to express this in terms of radians/seconds. To do it we define that 1cycle is equal to 2pi, so we can find the angular speed this way:
[tex]w=(1cycle/0,639s)*(2pi/1cycle)[/tex]
So the angular speed is [tex]w=9,83rad/s[/tex]
Now that we have this information, we can find the tangencial speed, which will be the relation between the angular speed and the circle radius, this way:
[tex]V=w*r[/tex] so the tangencial speed is:
[tex]V=(9,83rad/s)*(0,149m)[/tex]
[tex]V=1,5m/s[/tex]
Now we have all the information to find the Centripetal Force:
[tex]Fcp=(m*V^2)/(r)[/tex]
[tex]Fcp=((0,0154kg)*(1,5m/s)^2)/(0,149m)[/tex]
a) So the Centripetal Force is: [tex]Fcp=0,23N[/tex]
b) If the tangencial speed is doubled, its new value will be 3m/s. replacing this information we will get the new Centripetal Force is:
[tex]Fcp=((0,0154kg)*(3m/s)^2)/(0,149m)[/tex]
The Centripetal Force is: [tex]Fcp=0,93N[/tex]
Here we can see that if the speed is doubled, the Centripetal Force will increase four times.
A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.2 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 8 ft from the wall.) rad/s
Answer:0.2 rad/s
Explanation:
Given data
Velocity of the bottom point of the ladder=1.2Ft/s
Length of ladder=10ft
distance of the bottom most point of ladder from origin=8ft
From the data the angle θ with ladder makes with horizontal surface is
Cosθ=[tex]\frac{8}{10}[/tex]
θ=36.86≈37°
We have to find rate of change of θ
From figure we can say that
[tex]x^{2}[/tex]+[tex]y^{2}[/tex]=[tex]AB^{2}[/tex]
Differentiating above equation we get
[tex]\frac{dx}{dt}[/tex]=-[tex]\frac{dy}{dt}[/tex]
i.e [tex]{V_A}=-{V_B}=1.2ft/s[/tex]
[tex]{at\theta}={37}[/tex]
[tex]Y=6ft[/tex]
[tex]and\ about\ Instantaneous\ centre\ of\ rotation[/tex]
[tex]{\omega r_A}={V_A}[/tex]
[tex]{\omega=\frac{1.2}{6}[/tex]
ω=0.2rad/s
i.e.Rate of change of angle=0.2 rad/s
Find the heat that flows in 1.0 s through a lead brick 13 cm long if the temperature difference between the ends of the brick is 8.0°C. The cross-sectional area of the brick is 13 cm^2. Express your answer using two significant figures.
Final answer:
The heat flow through the lead brick can be calculated using Fourier's law of heat conduction. The thermal conductivity of lead is used along with the provided dimensions and temperature difference to find the heat flow, which is 2.2 W to two significant figures.
Explanation:
To calculate the heat flow through a lead brick, we can use Fourier's law of heat conduction, which states that the heat transfer rate (Q) through a material is proportional to the thermal conductivity of the material (k), the area (A) through which the heat is being transferred, the temperature difference (ΔT) across the material, and inversely proportional to the thickness (d) of the material.
The formula to calculate heat flow is: Q = k × A × ΔT × t / d
First, we need to convert all measurements to SI units: the length from cm to m (13 cm = 0.13 m), the area from cm² to m² (13 cm² = 0.0013 m²), and the time from seconds to hours (1.0 s). The thermal conductivity of lead is approximately 35 W/m°C. Plugging in the values: Q = 35 W/m°C × 0.0013 m² × 8.0°C × 1.0 s / 0.13 m
By simplifying the equation, we find the heat flow through the lead brick:
Q = (35 × 0.0013 × 8.0 × 1.0) / 0.13 W = 2.215 W
To two significant figures, the heat flow is 2.2 W.
If the specific gravity of copper is 8.91, it weighs:
A. 55.6 lb/ft^3
B. 238.8 lb/ft^3
C. 23.88 lb/ft^3
D. 556 lb/ft^3
Answer:
Option D is the correct answer.
Explanation:
The specific gravity of copper is 8.91.
We have
[tex]\frac{\texttt{Density of copper}}{\texttt{Density of water}}=8.91\\\\\texttt{Density of copper}=8.91\times 1000kg/m^3=8910kg/m^3[/tex]
We also have
1 kg = 2.205 lb
1 m = 3.28 ft
Substituting
[tex]\texttt{Density of copper}=8910kg/m^3=\frac{8910\times 2.205}{3.28^3}=556.76lb/ft^3[/tex]
Option D is the correct answer.
Answer:
Option D is the correct answer.
The _____ is used to express absolute temperatures in the English system of measurement. A. Fahrenheit scale B. Kelvin scale C. Celsius scale D. Rankine scale
Answer:
B). Kelvin Scale
Explanation:
Kelvin scale is the absolute scale which is used to express temperature in English System
We have different temperature scales
1) Fahrenheit Scale
Generally use in British system of units
2) Celcius Scale
It is used to given temperature of different scales and its relation with kelvin
[tex]^0 C = K - 273[/tex]
3) Rankine Scale
It is used in thermodynamic scales with large temperature range
A 22g bullet traveling 210 m/s penetrates a 2.0kg block of wood and emerges going 150m/s. If the block were stationary on a frictionless plane before the collision, what is the velocity of the block after the bullet passes through? What is the kinetic energy loss of the system?
Answer:
speed of block is 0.665 m/s
Kinetic energy loss is 237.16 J
Explanation:
Here we can use momentum conservation as there is no external force on the system in horizontal direction
so here we will have
[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]
[tex]0.022(210) + 2(0) = 0.022(150) + 2 v[/tex]
[tex]4.62 = 3.3 = 2 v[/tex]
[tex]v = \frac{4.63 - 3.3}{2}[/tex]
[tex]v = 0.665 m/s[/tex]
Now kinetic energy loss in the system is given as
[tex]Loss = KE_i - KE_f[/tex]
[tex]Loss = \frac{1}{2}(0.022)(210^2) - (\frac{1}{2}(0.022)(150^2) + \frac{1}{2}(2)(0.665)^2)[/tex]
[tex]Loss = 237.16 J[/tex]
The block's final velocity after the bullet passes through it is about 1.68 m/s. The kinetic energy loss of the system is about 602 Joules. These conclusions are drawn using the conservation of momentum and energy.
Explanation:This problem can be approached using the principles of conservation of momentum and energy. Let's first focus on momentum. Before the bullet hits, the momentum of the system is only contributed by the bullet. After the bullet passes through the block, both the bullet and the block contribute to the total momentum, which should be the same as before. This gives us the equation: 0.022 kg * 210 m/s (bullet's initial momentum) = 0.022 kg * 150 m/s (bullet's final momentum) + 2.0 kg * v (final momentum of the block). Solving this for the block's final velocity, v, will yield about 1.68 m/s.
Next, let's examine the kinetic energy loss. Before the collision, kinetic energy is only contributed by the bullet: KE_initial = 0.5 * 0.022 kg * (210 m/s)². After, it comes from both the bullet and the block: KE_final = 0.5 * 0.022 kg * (150 m/s)² + 0.5 * 2.0 kg * (1.68 m/s)². The kinetic energy loss is KE_initial - KE_final, which should be about 602 Joules. It's worth mentioning that this energy loss is transferred to other forms like heat and sound, illustrating the law of conservation of energy.
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A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 110,000 kg of scrap metal into it. (a) What is the final velocity of the loaded freight car? (b) How much kinetic energy is lost?
Explanation:
The mass of freight car, m₁ = 30,000 kg
Velocity of freight car, u₁ = 0.85 m/s
Mass of hopper, m₂ = 110,000 kg
(a) Let v is the final velocity of the loaded freight car. Initial momentum of the car before the dump, [tex]p_i=30000\ kg\times 0.850\ m/s=25500\ kg-m/s[/tex]
Final momentum, [tex]p_f=(30000\ kg+110000\ kg)v=140000\ v[/tex]
According to the conservation of momentum,
initial momentum = final momentum
25500 kg-m/s = 140000 v
v = 0.182 m/s
So, the final velocity of the loaded freight car is 0.182 m/s.
(b) Initial kinetic energy, [tex]k_i=\dfrac{1}{2}\times 30000\ kg\times (0.850\ m/s)^2=10837.5\ J[/tex]
Final kinetic energy, [tex]k_f=\dfrac{1}{2}(m_1+m_2)v^2[/tex]
[tex]k_f=\dfrac{1}{2}\times (30000\ kg+110000\ kg)\times (0.182\ m/s)^2=2318.68\ J[/tex]
So, loss in kinetic energy, [tex]\Delta k=k_f-k_i[/tex]
[tex]\Delta k=2318.68\ J-10837.5\ J=-8518.82\ J[/tex]
So, 8518.82 J of kinetic energy is lost after collision. Hence, this is the required solution.
Using the conservation of momentum, we find the final velocity of the loaded freight car to be 0.182 m/s. By comparing the initial and final kinetic energies, we find that 8,459.89 Joules of kinetic energy is lost during the process.
Explanation:This problem is a classic example of conservation of momentum, and to a lesser extent, conservation of kinetic energy. Firstly, we calculate the initial momentum of the system. Momentum is mass multiplied by velocity, so the initial momentum of the freight car is 30,000 kg * 0.850 m/s = 25,500 kg*m/s. After the scrap metal is dumped into the freight car, the total mass of the system is now 30,000 kg + 110,000 kg = 140,000 kg. Because momentum must be conserved, we calculate the final velocity of the freight car as the total momentum divided by the total mass, 25,500 kg*m/s / 140,000 kg = 0.182 m/s.
Secondly, before the scrap metal is added, the initial kinetic energy of the freight car is 0.5 * 30,000 kg * (0.850 m/s)² = 10,781.25 Joules. After the metal is added, the final kinetic energy is 0.5 * 140,000 kg * (0.182 m/s)² = 2,321.36 Joules. The change in kinetic energy can then be obtained by subtracting the final kinetic energy from the initial one, which gives us 8,459.89 Joules as the amount of kinetic energy that is lost.
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The Eiffel tower is made of iron and is about 300 m tall. Due to thermal expansion, its height changes between winter and summer. Estimate the height change the tower undergoes if the difference in temperature between the seasons is 23º C. Enter your answer in meters but do not include units. Provide your answer to the hundredths place.
Answer:
Change in length is 8.3 cm
Explanation:
Coefficient of thermal expansion for iron is given as
[tex]\alpha = 12 \times 10^{-6} per ^0C[/tex]
now as we know that the change in the length due to thermal expansion depends of change in temperature and initial length of the object
so here we will have
[tex]\Delta L = L_o \alpha \Delta T[/tex]
here we know that
[tex]L_o = 300 m[/tex]
[tex]\Delta T = 23^o C[/tex]
now we will have
[tex]\Delta L = (300)(12 \times 10^{-6})(23)[/tex]
[tex]\Delta L = 0.083 m[/tex]
so change in length is approx 8.3 cm
A 63.9-kg wrecking ball hangs from a uniform, heavy-duty chain having a mass of 20.5 kg . (Use 9.80 m/s2 for the gravitational acceleration at the earth's surface.)
Find the maximum tension in the chain.
Find the minimum tension in the chain.
What is the tension at a point three-fourths of the way up from the bottom of the chain?
Answer:
a) Tmax=827.12N
b) Tmin = 626.22N
c) 776.895 N
Explanation:
Given:
Mass of wrecking ball M1=63.9 Kg
Mass of the chain M2=20.5 Kg
acceleration due to gravity, g=9.8m/s²
Now,
(a)The Maximum Tension generated in the chain,
Tmax=(M1+M2)×g)
Tmax=(M1+M2)×(9.8 m/s²)
Tmax=(63.9+ 20.5)×(9.8 m/s²)
Tmax=827.12N
(b) The Minimum Tension Tmin will be due to the weight of the wrecking ball only
Mathematically,
Tmin=weight of the wrecking ball
Tmin = 63.9kg×9.8m/s²
Tmin = 626.22N
(c)Now. the tension at 3/4 from the bottom of the chain
In this part we have to use only 75% of the chain i.e the weight acting below the point of consideration
thus, the tension will be produced by the weight of the 3/4 part of the chain and the wrecking ball
Therefore, the weight of the 3/4 part of the chain = [tex]\frac{3}{4}\times 20.5\times 9.8 N[/tex]
= 150.675 N
Hence, the tension at a point 3/4 of the way up from the bottom of the chain will be = 150.675 N + (63.9×9.8) N = 776.895 N
In a two-slit experiment, monochromatic coherent light of wavelength 500 nm passes through a pair of slits separated by 1.30 x 10^-5 m. At what angle away from the centerline does the first bright fringe occur? O 1.56° O 2.20° O 3.85° O 2.73° O 4.40°
Answer:
2.20°
Explanation:
λ = wavelength of the coherent light = 500 x 10⁻⁹ m
d = slits separation = 1.30 x 10⁻⁵ m
n = order of the fringe = 1
θ = angle made by the first bright fringe with the center line = ?
For first bright fringe, Using the equation
d Sinθ = n λ
(1.30 x 10⁻⁵) Sinθ = (1) (500 x 10⁻⁹)
[tex]Sin\theta =\frac{500\times 10^{-9}}{1.30\times 10^{-5}}[/tex]
Sinθ = 0.0385
θ = 2.20°
Final answer:
To find the angle of the first bright fringe in a double-slit experiment, the formula for constructive interference is used. By substituting the provided measurements into the formula and calculating the inverse sine, the angle is found to be approximately 2.20°.
Explanation:
The question refers to a Young's double-slit experiment, where monochromatic light of a known wavelength is used to produce an interference pattern on a screen. To find the angle at which the first bright fringe occurs, we can use the formula for constructive interference in a double-slit setup, which is given by:
d sin(θ) = mλ
Where d is the separation between the slits, θ is the angle of the fringe from the centerline, m is the order number of the fringe (m=1 for the first bright fringe), and λ is the wavelength of the light.
Plugging in the values provided:
d = 1.30 x 10-5 m
λ = 500 x 10-9 m
m = 1
Now we solve for θ:
sin(θ) = mλ / d
sin(θ) = (1)(500 x 10-9 m) / (1.30 x 10-5 m)
sin(θ) ≈ 0.038462
Using a calculator to find the inverse sine, we get:
θ ≈ 2.20°
A dentist's drill starts from rest. After 3.50 s of constant angular acceleration, it turns at a rate of 2.49 104 rev/min. (a) Find the drill's angular acceleration. Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. rad/s2 (along the axis of rotation) (b) Determine the angle through which the drill rotates during this period. rad
Answer:
Part A)
[tex]\alpha = 745 rad/s^2[/tex]
Part B)
[tex]\theta = 4563.1 rad[/tex]
Explanation:
Drill starts from rest so its initial angular speed will be
[tex]\omega_i = 0[/tex]
now after 3.50 s the final angular speed is given as
[tex]f = 2.49 \times 10^4 rev/min[/tex]
[tex]f = {2.49 \times 10^4}{60} = 415 rev/s[/tex]
so final angular speed is given as
[tex]\omega = 2\pi f[/tex]
[tex]\omega_f = 2607.5 rad/s[/tex]
now we have angular acceleration given as
[tex]\alpha = \frac{\omega_f - \omega_i}{\Delta t}[/tex]
[tex]\alpha = \frac{2607.5 - 0}{3.50}[/tex]
[tex]\alpha = 745 rad/s^2[/tex]
Part b)
The angle through which it is rotated is given by the formula
[tex]\theta = \frac{(\omega_f + \omega_i)}{2}\delta t[/tex]
now we have
[tex]\theta = \frac{(2607.5 + 0)}{2}(3.50)[/tex]
[tex]\theta = 4563.1 rad[/tex]
If given both the speed of light in a material and an incident angle. How can you find the refracted angle?
Answer:
[tex]r = Sin^{-1}\left ( \frac{v Sini}{c} \right )[/tex]
Explanation:
Let the speed of light in vacuum is c and the speed of light in medium is v. Let the angle of incidence is i.
By using the definition of refractive index
refractive index of the medium is given by
n = speed of light in vacuum / speed of light in medium
n = c / v ..... (1)
Use Snell's law
n = Sin i / Sin r
Where, r be the angle of refraction
From equation (1)
c / v = Sin i / Sin r
Sin r = v Sin i / c
[tex]r = Sin^{-1}\left ( \frac{v Sini}{c} \right )[/tex]
A ladder that is 4.6 m long is leaning against a wall at an angle of 66.0º with respect to the ground. If the base of the ladder is moved 0.31 m away from the wall, how far will the top of the ladder go down?
Answer:
The top of the ladder go down by 0.15 m.
Explanation:
Here we have right angled triangle.
Hypotenuse = 4.6 m
Bottom angle = 66º
Length from ladder bottom to wall bottom = 4.6 cos66 = 1.87 m
Length from ladder top to wall bottom = 4.6 sin66 = 4.20 m
New length from ladder bottom to wall bottom = 1.87 + 0.31 = 2.18 m
By Pythagoras theorem
New length from ladder top to wall bottom is given by
[tex]\sqrt{4.6^2-2.18^2}=4.05m[/tex]
Distance the top of the ladder go down = 4.20 - 4.05 = 0.15 m
A person exerts a force of 50 N on the end of a door 71 cm wide. (a)What is the magnitude of the torque if the force is exerted perpendicular to the door? (b) What is the magnitude of the torque if the force is exerted at a 45 degree angle to the face of the door?
Answer:
a) 35.5
b)25.1
Explanation:
Torque=distance* (perpendicular component of the force to the distance)
a) force=50 N distance=0,71m(we need the distance in meters to have the same units with the Newton)
T=50*0.75
b) only one component of the force makes torque, so we multiply all the 50N with sin(45) to find the perpendicular component to the distance
T=50*sin(45)*0.71
Two spherical objects with a mass of 6.22 kg each are placed at a distance of 1.02 m apart. How many electrons need to leave each object so that the net force between them becomes zero?
Answer:
The number of electrons need to leave each object is [tex]3.35\times10^{9}[/tex]
Explanation:
Given that,
Mass of object = 6.22 kg
Distance = 1.02 m
We need to calculate the number of electron
Using formula of electric force
[tex]F_{e}=\dfrac{k(q)^2}{r^2}[/tex]....(I)
We know that,
[tex]q = Ne[/tex]
Put the value of q in equation (I)
[tex]F_{e}=\dfrac{k(Ne)^2}{r^2}[/tex].....(II)
Using gravitational force
[tex]F_{G}=\dfrac{Gm^2}{r^2}[/tex].....(III)
Equating equation (II) and (III)
[tex]F_{e}=F_{G}[/tex]
[tex]\dfrac{k(Ne)^2}{r^2}=\dfrac{Gm^2}{r^2}[/tex]
[tex]N=\sqrt{\dfrac{G}{k}}\times\dfrac{m}{e}[/tex]....(IV)
Put the value in the equation(IV)
[tex]N=\sqrt{\dfrac{6.67\times10^{-11}}{9\times10^{9}}}\times\dfrac{6.22}{1.6\times10^{-19}}[/tex]
[tex]N=3.35\times10^{9}[/tex]
Hence, The number of electrons need to leave each object is [tex]3.35\times10^{9}[/tex]
A tennis player tosses a tennis ball straight up and then catches it after 2.00 s at the same height as the point of release. (a) What is the acceleration of the ball while it is in flight? (b) What is the velocity of the ball when it reaches its maximum height? Find (c) the initial velocity of the ball and (d) the maximum height it reaches
Answer:
Part a)
a = -9.81 m/s/s
Part b)
v = 0
Part c)
v = 9.81 m/s
Part d)
[tex]H = 4.905 m[/tex]
Explanation:
Part a)
During the motion of ball it will have only gravitational force on the ball
so here the acceleration of the ball is only due to gravity
so it is given as
[tex]a = g = 9.81 m/s^2[/tex]
Part b)
As we know that ball is moving against the gravity
so here the velocity of ball will keep on decreasing as the ball moves upwards
so at the highest point of the motion of the ball the speed of ball reduce to zero
[tex]v_f = 0[/tex]
Part c)
We know that the total time taken by the ball to come back to the initial position is T = 2 s
so in this time displacement of the ball will be zero
[tex]\Delta y = 0 = v_y t + \frac{1}{2} at^2[/tex]
[tex]0 = v_y (2) - \frac{1}{2}(9.81)(2^2)[/tex]
[tex]v_y = 9.81 m/s[/tex]
Part d)
at the maximum height position we know that the final speed will be zero
so we will have
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
here we have
[tex]0 - (9.81^2) = 2(-9.81)H[/tex]
[tex]H = 4.905 m[/tex]
The acceleration of the ball is -9.8 m/s², the velocity of the ball at max height is 0 m/s, the initial velocity is 9.8 m/s and the max height is 4.9 m.
Explanation:The questions refer to the motion of a tennis ball under gravity. (a) The acceleration of the ball while it is in flight is -9.8 m/s², due to Earth's gravity. This will be the case throughout the ball's flight, regardless of whether it's moving up or down. (b) The velocity of the ball when it reaches its maximum height is 0 m/s, as the ball stops moving vertically for an instant before it starts coming down. (c) We can calculate the initial velocity using the equation v = u + at. Here, v is final velocity, u is initial velocity, a is acceleration and t is time. As v = 0 m/s at max height, a = -9.8 m/s² and t = 1 s (time taken to reach max height, half of total time), we find u to be 9.8 m/s. (d) The maximum height can be calculated using the equation h = ut + 0.5at². Here, h is height, u is initial velocity, t is time and a is acceleration. Subsituting u = 9.8 m/s, a = -9.8 m/s², t = 1 s, we get h to be 4.9 m.
Learn more about Motion under gravity here:https://brainly.com/question/28554700
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A bullet of mass 12 g strikes a ballistic pendulum of mass 2.2 kg. The center of mass of the pendulum rises a vertical distance of 10 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.
Answer:
The bullet's initial speed 258.06 m/s.
Explanation:
It is given that,
It is given that,
Mass of the bullet, m₁ = 12 g = 0.012 kg
Mass of the pendulum, m₂ = 2.2 kg
The center of mass of the pendulum rises a vertical distance i.e. h = 10 cm = 0.1 m
We need to find the initial speed of the bullet. Here, the kinetic energy of the bullet gets converted to potential energy of the system as :
[tex]\dfrac{1}{2}(m_1+m_2)V^2=(m_1+m_2)gh[/tex]
[tex]V=\sqrt{2gh}[/tex]....................(1)
Where
V is the speed of bullet and pendulum at the time of collision.
Let v be the initial speed of the bullet. Using conservation of momentum as :
[tex]m_1v=(m_1+m_2)V[/tex]
[tex]V=(\dfrac{m_1+m_2}{m_1})\sqrt{2gh}[/tex]
[tex]V=(\dfrac{0.012\ kg+2.2\ kg}{0.012 kg})\sqrt{2\times 9.8\ m.s^2\times 0.1\ m}[/tex]
V = 258.06 m/s
So, the initial speed of the bullet is 258.06 m/s. Hence, this is the required solution.