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3 years 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]3 years 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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4. Calculate the kinetic energy of a 4.7 kg object moving at a speed of 7 m/s. SHOW YOUR WORK

SashulF [63]

Answer:

$\boxed {\boxed {\sf 115.15 \ J}}$

Explanation:

Kinetic energy is the energy an object possesses due to motion. It is calculated with the following formula.

$E_K= \frac{1}{2} mv^2$

The mass of the object is 4.7 kilograms. The velocity of the object is 7 meters per second.

m= 4.7 kg
v= 7 m/s

Substitute the values into the formula.

$E_K= \frac{1}{2} (4.7 \ kg)(7 \ m/s)^2$

Solve the exponent.

(7 m/s)²= 7 m/s * 7 m/s = 49 m²/s²

$E_K= \frac{1}{2} (4.7 \ kg)(49 \ m^2/s^2)$

Multiply the numbers together.

$E_K = 2.35 \ kg * 49 \ m^2/s^2$

$E_K= 115.15 \ kg*m^2/s^2$

Convert the units. 1 kilogram square meter per square second is equal to 1 Joule.

$E_K= 115.15 \ J$

The object has <u>115.15 Joules</u> of kinetic energy.

3 0

2 years ago

Read 2 more answers

a block is pushed up a frictionless 40 incline. if the initial velocity after the push is 5.00 m/s. how far along the incline do

Lana71 [14]

Answer:

d=1.982m, t=1.019s

Explanation:

There are different approaches we can take to solve this problem. You could either solve this by using conservation of energy or by taking a kinematic approach. I'll solve this by using kinematics. So, the very first thing we need to do in order to solve this is do a drawing of the situation so we can analyze it better. (See attached picture).

So, since we are talking about an inclined plane, we can see that the force of gravity is being split into an x and y components if we incline the axis of coordinates. Taking this into account we can see that:

$\sum F_{x}=ma_{x}$

Since there is no friction in our system, then the only force acting upon the box is the force of gravity, or weight. Since we are taking the upwards direction as the positive direction of movement, we can say that the force of gravity is excerting a negative influence on our box, so this acceleration will be negative, so our sum of forces will now look like this:

$-mg sin(40^{o})=ma$

we can cancel the masses out so we can see that:

a=-g sin(40°)

$a=-9.81m/s^{2} sin(40^{o})$

We have now enough information to solve our problem.

we can take the following equation to find the distance the block travels up the incline:

$x=\frac{V_{f}^{2}-V_{0}^{2}}{2a}$

we know the final velocity must be zero, so we can use the provided data to solve our formula:

$x=\frac{(0)^{2}-(5m/s)^{2}}{2(-9.81m/s^{2})sin 40^{o}}$

which yields:

x=1.982m

In order to find the time it takes for the block to return to its original position we can use the following formula:

$x=V_{0}t+\frac{1}{2}at^{2}$

since x=0 is the starting point we can use that to solve our equation:

$0=5t+\frac{1}{2}(-9.81sin 40^{o})t^{2}$

which simplifies to:

$0=5t-4.905t^{2}$

which can now be solved for t

t(5-4.905t)=0

t=0 and 5-4.905t=0

t=0 and $t=\frac{5}{4.905}=1.019$

so the time it takes the block to return to its original position is

t= 1.019s

6 0

3 years ago

Consider a positive charge Q and a point B twice as far away from Q as point A. What is the ratio of the electric field strength

Vikentia [17]

Answer:

$\frac{E_{A}}{E_{B}}=4$

Explanation:

The electric field is defined as the electric force per unit of charge, this is:

$E=\frac{F}{q}$ .

The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as

$F=\frac{kQq}{r^{2}}$ .

By substitution we get that

$E=\frac{kQq}{qr^{2}}\\\\E=\frac{kQ}{r^{2}}$

Now, letting $E_{A}$ be the electric field at point A, letting $E_{B}$ be the electric field at point B, and letting R be the distance from the charge to A:

$E_{A}=\frac{kQ}{R^{2}}\\\\E_{B}=\frac{kQ}{(2R)^{2}}$ .

The ration of the electric fields is

$\frac{E_{A}}{E_{B}}=\frac{\frac{kQ}{R^{2}}}{\frac{kQ}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(4)R^{2}}}\\\\\\\frac{E_{A}}{E_{B}}=\frac{1}{\frac{1}{(4)}}\\\\\frac{E_{A}}{E_{B}}=4$

This means that at half the distance, the electric field is four times stronger.

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3 years ago

To be skeptical when you hear new information means that you should _____.

adelina 88 [10]

Answer:

C.

Explanation:

I TOOK THE TEST

3 0

3 years ago

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How much kinetic energy does a care 1000 kg car moving at 7 m/s have?

aalyn [17]

Hey there!

<span>How much kinetic energy does a care 1000 kg car moving at 7 m/s have?

Answer:
</span><span>A. 24500 J

Hope this helps
Have a great day (:
</span>

5 0

3 years ago