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solniwko [45]
3 years ago
15

A 5 kg ball of clay is moving with a speed of 25 m/s directly toward a 10 kg ball of clay which is at rest. The two clay balls c

ollide and stick together. What is their velocity immediately after the collision?​
Physics
1 answer:
Dafna11 [192]3 years ago
7 0

Answer:

8.3m/s

Explanation:

Given parameters:

mass of clay ball  = 5kg

Speed of clay ball  = 25m/s

mass of clay ball at rest  = 10kg

speed of clay ball at rest  = 0m/s

Unknown:

Velocity after collision  = ?

Solution:

 Since the balls stick together, this is an inelastic collision:

   m1v1 + m2v2  = v(m1 + m2)

  5(25) + 10(0)  =  v (5 + 10)

         125 = 15v

           v  = 8.3m/s

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Can a body possess velocity at the same time in horizontal and vertical directions?​
iogann1982 [59]

Answer:

Yes

Explanation:

A body can possess velocity at the same time in horizontal and vertical direction

For example

A projectile

5 0
3 years ago
after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF
Snezhnost [94]

Answer:

The time is 110.16\times10^{-3}\ sec

Explanation:

Given that,

Capacitor = 120 μF

Voltage = 150 V

Resistance = 1.8 kΩ

Current = 50 mA

We need to calculate the discharge current

Using formula of discharge current

i_{0}=\dfrac{V_{0}}{R}

Put the value into the formula

i_{0}=\dfrac{150}{1.8\times10^{3}}

i_{0}=83.3\times10^{-3}\ A

We need to calculate the time

Using formula of current

i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}

Put the value into the formula

50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}

\dfrac{50}{83.3}=e^{\frac{-t}{RC}}

\dfrac{-t}{RC}=ln(0.600)

t=0.51\times1.8\times10^{3}\times120\times10^{-6}

t=110.16\times10^{-3}\ sec

Hence, The time is 110.16\times10^{-3}\ sec

4 0
3 years ago
Plzzzz helpppppppo Near the end of the life cycle of a star why does the star lose its size
gavmur [86]

Answer:

Because the energy is waning

Explanation: hope this helps

8 0
3 years ago
A moving rope (parallel to the slope) is used to pull skiers up the mountain. If the slope of the hill is 37" and friction is ne
Charra [1.4K]

Since rope is parallel to the inclined plane so here we can say that net force parallel to the person which is pulling upwards must counterbalance the component of weight of the person.

Now here we will do the components of the weight of the person

given that weight of the person = 500 N

now its components are

W_x = 500 cos37

W_y = 500 sin37

now here as we can say that one of the component is balanced here by the normal force perpendicular to plane

while the other component of the weight is balanced by the force applied on the rope

So here the force applied on the rope will be given as

F = W_y = 500 sin37

F = 300 N

so it apply 300 N force along the inclined plane

5 0
3 years ago
In order to prevent injury in a car crash, it is recommended that you _______. Increase the initial velocity of the collision.
Vlad1618 [11]
Increase the change in momentum of the collision
5 0
3 years ago
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