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1 year ago

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A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i

n the same direction at vTi 5 20.0 m/s (Fig. P9.22). The velocity of the car immediately after the collision is vCf 5 18.0 m/s to the east. (a) What is the velocity of the truck immediately after the colli

1 answer:

sukhopar [10]1 year ago

8 0

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, $v_{Ci}=25\ m/s$

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, $v_{Ti}=20\ m/s$

After the collision, velocity of the car, $v_{Cf}=18\ m/s$

Let $v$ is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

$initial\ momentum=final\ momentum$

$1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v$

$210000-21600=9000\ kg\times v$

$v=20.93\ m/s$

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

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Power is __________________. Power is __________________. the work done by a system the force required to push something the spe

DaniilM [7]

Answer:

The rate at which the energy of a system is transformed

Explanation:

Power is the rate at which energy of a system is transformed or the rate at which work is done. It is defined by Power = Workdone/time taken

Its unit is the Watt denoted by the letter W.

For example, assuming a work of 200 J is done in 10 s, then Power, P equals

P = workdone/time taken = 200 J/10 s = 20 J/s = 20 W

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1 year ago

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of

Charra [1.4K]

To solve the problem it is necessary to apply the Torque equations and their respective definitions.

The Torque is defined as,

$\tau = I \alpha$

Where,

I=Inertial Moment

$\alpha =$ Angular acceleration

Also Torque with linear equation is defined as,

$\tau = F*d$

Where,

F = Force

d= distance

Our dates are given as,

R = 30 cm = 0.3m

m = 1.5 kg

F = 20 N

r = 4.0 cm = 0.04 m

t = 4.0s

Therefore matching two equation we have that,

$d*F = I\alpha$

For a wheel the moment inertia is defined as,

I= mR2, replacing we have

$d*F= \frac{mR^2a}{R}$

$d*F= mRa$

$a = \frac{rF}{ mR}$

$a = \frac{0.04*20}{1.5*0.3}$

$a=1.77 m/s^2$

Then the velocity of the wheel is

$V = a *t \\V=1.77*4 \\V=7.11 m/s$

Therefore the correct answer is D.

4 0

1 year ago

Kenny and Candy decided to sit on a see-saw while visiting a local play park. Candy, of mass

pochemuha

Answer:

(i) 208 cm from the pivot

(ii) Move further from the pivot

Explanation:

(i) Sum of the moments about the pivot of the seesaw is zero.

∑τ = Iα

(50 kg) (10 N/kg) (2.5 m) + (60 kg) (10 N/kg) x = 0

1250 Nm + 600 N x = 0

x = -2.08 m

Kenny should sit 208 cm on the other side of the pivot.

(ii) To increase the torque, Kenny should move away from the pivot.

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1 year ago

A spaceship flies from Earth to a distant star at a constant speed. Upon arrival, a clock on board the spaceship shows a total e

m_a_m_a [10]

Answer:

35 288 mile/sec

Explanation:

This is a problem of special relativity. The clocks start when the spaceship passes Earth with a velocity v, relative to the earth. So, out and back from the earth it will take:

$10 years = \frac{2d}{v}$

If we use the Lorentz factor, then, as observed by the crew of the ship, the arrival time will be:

$0.8 = \sqrt{1-\frac{v^{2} }{c^{2} } }$

Then the amount of time wil expressed as a reciprocal of the Lorentz factor. Thus:

$0.8 = \sqrt{1 - \frac{v^{2} }{c^{2} } }$

$0.64 = 1-\frac{v^{2} }{186282^{2} }$

solving for v, gives = 35 288 miles/s

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1 year ago

The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

Margaret [11]

Let us consider two bodies having masses m and m' respectively.

Let they are separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - $F = G\frac{mm'}{r^{2} }$ where G is the gravitational force constant.

From the above we see that F ∝ mm' and $F\alpha \frac{1}{r^{2} }$

Let the orbital radius of planet A is $r_{1}$ = r and mass of planet is $m_{1}$ .

Let the mass of central star is m .

Hence the gravitational force for planet A is $f_{1} =G \frac{m_{1}*m }{r^{2} }$

For planet B the orbital radius $r_{2} =2r_{1}$ and mass $m_{2} = 3 m_{1}$

Hence the gravitational force $f_{2} =G\frac{m m_{2} }{r^{2} }$

$f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }$

$= \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }$

Hence the ratio is $\frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2} }{Gmm_{1}/r_{1} ^2 }$

$=\frac{3}{4}$ [ ans]

3 0

2 years ago

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