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3 years ago

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A blue train of mass 50 kg moves at 4 m/s toward a green train of 30 kg initially at rest. The trains collide. After the collisi

on the green train moves with a speed of 3 m/s. What is the final momentum of the green train? A. 90 kgm/s B. 20 kgm/s C. 200 kgm/s D. 110 kgm/s

1 answer:

ra1l [238]3 years ago

7 0

Explanation:

Momentum = mass × speed

p = (30 kg) (3 m/s)

p = 90 kg m/s

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What will the stopping distance be for a 3,000-kg car if -3,000 N of force are applied when the car is traveling 10 m/s?

Harlamova29_29 [7]

F = force applied to stop the car = - 3000 N

m = mass of the car = 3000 kg

a = acceleration of the car = ?

v₀ = initial velocity of the car before the force is applied to stop it = 10 m/s

v = final velocity of the car when it comes to stop = 0 m/s

d = stopping distance of the car

acceleration of the car is given as

a = F/m

inserting the values

a = - 3000/3000

a = - 1 m/s²

using the kinematics equation

v² = v²₀ + 2 a d

inserting the values

0² = 10² + 2 (-1) d

0 = 100 - 2 d

2 d = 100

d = 100/2

d = 50 m

hence the correct choice is

C. 50 m

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yKpoI14uk [10]

The answer is true to your question

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4 years ago

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A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of

balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, $N=mg$

∴ $f=\mu mg=0.490\times 90\times 9.8=432.18\ N$

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

$W=-fd=-432.18d$

Now, change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J$

Therefore, from work-energy theorem,

$W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m$

Hence, the skater covers a distance of 15 m before stopping.

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3 years ago

An object accelerates from 20 m/s down to 10 m/s in 5 seconds.

DedPeter [7]

Answer:

The acceleration of the object is $-2\ \frac{m}{sec^2}$

Explanation:

we know that

The acceleration is the change in the velocity, divided by the time

Let

v -----> the change in the velocity in m/sec

t ----> the time in sec

$a=\frac{V}{t}$

we have that

$V_2=10\ m/sec\\V_1=20\ m/sec$

$V=V_2-V_1$

substitute

$V=10-20=-10\ m/sec$

$t=5\ sec$

substitute in the formula

$a=\frac{-10}{5}=-2\ \frac{m}{sec^2}$

therefore

The acceleration of the object is $-2\ \frac{m}{sec^2}$

Is negative because is a deceleration

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My name is Ann [436]

Answer:

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