Ask question

Ask question

All categories

3 years ago

15

A 3.0-kg mass moving in the positive x direction with a speed of 10 m/s collides with a 6.0-kg mass initially at rest. after the

collision, the speed of the 3.0-kg mass is 8.0 m/s, and its velocity vector makes an angle of 35Â° with the positive x axis. what is the magnitude of the velocity of the 6.0-kg mass after the collision?

1 answer:

OverLord2011 [107]3 years ago

6 0

Sorry i dont know the answer

You might be interested in

If the spring constant is doubled, what value does the period have for a mass on a spring? A. The period would double by square

antoniya [11.8K]

Answer:

The period would decrease by sqrt(2)

Explanation:

The restoring force is given by,

F = -kx

According to Newton's second law of motion,

ma = -kx

ma + kx = 0

The time period is given by,

T = $\frac{2\pi }{\omega }$

Where $\omega$ is the angular velocity and it is given by,

$\omega$ = $\sqrt{\frac{k}{m} }$

Now if the spring constant is doubled then,

$k_{2} = 2k$

Thus,

$T_{2}$ = $\frac{2\pi }{\sqrt{\frac{2k}{m} } }$

$\frac{T_{2} }{T} = \frac{\frac{2\pi }{\sqrt{\frac{2k}{m} } }}{\frac{2\pi }{\sqrt{\frac{k}{m} } }}$

$\frac{T_{2} }{T} = \sqrt{\frac{k}{2k} } = \sqrt{\frac{1}{2} }$

$T_{2} = \frac{T}{\sqrt{2} }$

Thus, The period would decrease by sqrt(2).

Hence, option D is correct.

3 0

2 years ago

Read 2 more answers

In 3 seconds a car moving in a straight line increases its speed from 22.4 m/s to 29.1 m/s while a truck increases its speed fro

ANTONII [103]

Is there a picture?

6 0

3 years ago

In the diagram, q1= -2.60*10^-9 C and

Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

$E_{net}=90.37\: N/c$

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

$|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}$

Where:

k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
q1 is the first charge
d1 is the distance from q1 to P

$|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}$

$|E_{1}|=80.84\: N/C$

And E2 will be:

$|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}$

$|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}$

$|E_{2}|=40.39\: N/C$

Finally, we need to use the Pythagoras theorem to find the magnitude of the net electric field.

$E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}$

$E_{net}=\sqrt{80.84^{2}+40.39^{2}}$

$E_{net}=90.37\: N/c$

I hope it helps you!

7 0

2 years ago

A ball starts from rest and rolls down a hill at a constant acceleration of 5 m/s2. If it travels for 8 m how fast is it going i

makkiz [27]

Hi there!

We can use the following kinematic equation:

$v_f^2 = v_i^2 + 2ad$

vf = final velocity (? m/s)
vi = intial velocity (0 m/s)

a = acceleration (5 m/s²)

d = displacement (8 m)

Plug in the givens and solve.

$v_f^2 = 0 + 2(5)(8)\\\\v_f = \sqrt{80} = \boxed{8.944 \frac{m}{s}}$

5 0

2 years ago

An 50kg car travels at 2m/s. What is the car's Kinetic energy? <br> 100J<br> 200J<br> 50J

Evgen [1.6K]

Answer:

The answer is 100J.

Explanation:

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. In this question, the mass is equals to 50kg and the velocity is 2m/s

Now,

25kg×4m/s^2 = 100kgm/s^2 or 100J

3 0

2 years ago