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Thepotemich [5.8K]

3 years ago

15

A 480 kg car moving at 14.4 m/s hits from behind another car moving at 13.3 m/s in the same direction. If the second car has a m

ass of 570 kg and a new speed of 17.9 m/s, what is the velocity of the first car after the collision?

1 answer:

Nostrana [21]3 years ago

5 0

Answer:

Velocity of the first car after the collision, $v_1=8.93\ m/s$

Explanation:

It is given that,

Mass of the car, $m_1 = 480\ kg$

Initial speed of the car, $u_1 = 14.4\ m/s$

Mass of another car, $m_2 = 570\ kg$

Initial speed of the second car, $u_2 = 13.3\ m/s$

New speed of the second car, $v_2 = 17.9\ m/s$

Let $v_1$ is the final speed of the first car after the collision. The total momentum of the system remains conserved, Using the conservation of momentum to find it as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$m_1u_1+m_2u_2-m_2v_2=m_1v_1$

$480\times 14.4+570\times 13.3-570\times 17.9=480v_1$

$v_1=8.93\ m/s$

So, the velocity of the first car after the collision is 8.93 m/s. Hence, this is the required solution.

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Please tell me if this is Newton's first law, second, or third law of motion. There can be more than one of the same answer very

AfilCa [17]

Answer: Hope This Helps!

Explanation:

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2: Newton's First Law of Motion is defined as "An object at rest and an object in motion will stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force."In soccer, when the soccer ball is in the soccer field and it is not moving, that means that it is at rest and there is no force acting upon it. When there is a person that is ready to play soccer and wants to kick the ball and play, then the unbalanced force would be the power from the person's foot.

3: Newtons third law can explain, as the cannonball is pushed forwards by the expanding high-pressure gases created by the exploding gunpowder, it pushes back on these gases. The gases push back on the cannon itself, causing it to roll backwards. Alternative answer: the cannon pushes forward on the cannonball. the reaction force is the cannonball pushing backwards on the cannon.

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3 years ago

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

3 years ago

Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s

frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

y₁ = v₀₁ t + ½ g t²

y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

t ’= (t-t₀)

y₂ = v₀₂ t ’+ ½ g t’²

y₂ = 0 + ½ g (t-t₀)²

y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

S = y₁ -y₂

S = ½ g t²– ½ g (t-t₀)²

S = ½ g [t² - (t²- 2 t to + to²)]

S = ½ g (2 t t₀ - t₀²)

S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to. The rock y has not left and the distance increases

For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂

3 0

3 years ago

A scientist testing to see how a cat “always” lands on their feet drops a cat upside down out of a third story window at a heigh

makkiz [27]

<u>Answer: </u>

Cat has 2.02 seconds to right itself.

<u>Explanation: </u>

Initial height of cat from ground = 20 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of cat in vertical direction = 0 m/s, acceleration = acceleration due to gravity = 9.8 $m/s^2$ , we need to calculate time when s = 20 meter.

Substituting

$20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds$

So, cat has 2.02 seconds to right itself.

7 0

2 years ago

Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of

FrozenT [24]

Gravity obeys the inverse square law. At 6400 km above the center of the Earth (Earth's surface) you weigh x. Twice that reduces your weight to 1/4th. Four times that height reduces your weight to 1/16th. 4 times 6400 km is 25,600 km. But that is above the center of the earth, and the question requests the height above the surface, so we deduct 6400 km to arrive at our final answer: 19,200 km.

Incidentally, it doesn't exactly work the opposite way. At the center of the Earth the mass would be equally distributed around you, and you would therefore be weightless.

6 0

3 years ago