Ask question

Ask question

All categories

2 years ago

6

A 2200 kg car moving east at 10.3 m/s collides with a 3160 kg car moving east. The carsstick together and move east as a unit af

ter the collision at a velocity of 5.32 m/s. a) What is the velocity of the 3160 kg car before the collision?

1 answer:

IgorLugansk [536]2 years ago

8 0

Answer:

u = 1.77 m/s

Explanation:

Conservation of momentum

2250(10.3) + 3160u = (2250 + 3160)(5.32)

u = 1.77 m/s

You might be interested in

Formulating a Hypothesis: Part I Since the investigative question has two variables, you need to focus on each one separately. T

Westkost [7]

Answer : The kinetic energy depends directly on the mass of a particle.

Explanation :

We know that the kinetic energy of any particle is given by :

$KE=\dfrac{1}{2}mv^2$

Where,

m is the mass of an object.

v is the velocity with which it is moving

Kinetic energy is due to the motion of the particle.

So, the kinetic energy of a particle is directly proportional to its mass.

Hence, the conclusion of the question is if the mass of a particle is increases then its kinetic energy also increase.

3 0

3 years ago

Read 2 more answers

Please help.................

Goryan [66]

Answer:

I'dont know sooooooooooooooooooooooorrrrrrrrrrrrrrrrry

7 0

2 years ago

Read 2 more answers

Reflected light from a thin film of oil gives constructive interference for light with a wavelength inside the film of λfilm. By

My name is Ann [436]

Answer:D. λfilm/4

Explanation: Destructive interference is a type of wave interference which means the coming together or over-lapping of two opposing waves creating No effect or the Cancellation of the wave impact. An example of destructive wave is when Noise cancel the effect of sound from a head phone.

The film thickness will need to be increased by λfilm/4 for it to be able to give a destructive interference.

3 0

3 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$

Generally from the law energy conservation we have that

$T__{E}} = KE_p$

So

$2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2$

=> $5.4564 *10^{8} = v^2$

=> $v = \sqrt{5.4564 *10^{8}}$

=> $v = 2.3359 *10^{4} \ m/s$

4 0

2 years ago

n a level football field a football is projected from ground level. It has speed 9.0 m/sm/s when it is at its maximum height. It

Pani-rosa [81]

Answer:

T = 0.225 s

Explanation:

The speed of a projectile at the highest point of its motion is the horizontal speed of the projectile. Considering the horizontal motion with negligible air resistance, we can use the following formula:

$v_x = RT\\\\T = \frac{v_x}{R}$

where,

T = Total time of ball in air = ?

R = Horizontal distance covered = 40 m

$v_x$ = horizontal speed = 9 m/s

Therefore,

$T = \frac{9\ m/s}{40\ m}$

<u>T = 0.225 s</u>

4 0

2 years ago