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laila [671]
3 years ago
9

A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo

ving to the left at 1.0 m/s. What is the total kinetic energy after the collision?
Physics
1 answer:
Zinaida [17]3 years ago
3 0

Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2

E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

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Arlecino [84]

Answer:

6.5

Explanation:

Because 1.5+5=6.5

7 0
2 years ago
A ball initially at rest rolls down a hill with an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how far will it move?
goldfiish [28.3K]

<u>Answer:</u>

  Ball will move 92.8125 meter along the cliff in 7.5 seconds.

<u>Explanation:</u>

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 case initial velocity = 0 m/s, acceleration = 3.3 m/s^2, we need to calculate displacement when time = 7.5 seconds.

Substituting

  s=0*7.5+\frac{1}{2} *3.3*7.5^2\\ \\ =92.8125 meter

  So ball will move 92.8125 meter along the cliff in 7.5 seconds.

5 0
3 years ago
A string under a tension of 68 N is used to whirl a rock in a horizontal circle of radius 3.7 m at a speed of 16.53 m/s. The str
alekssr [168]

Answer:

F = 5253.7 N

Explanation:

As we know that tension force in the string will be equal to the centripetal force on the string

so we will have

T = \frac{mv^2}{L}

now we have

68 = \frac{m(16.53^2)}{3.7}

now we have

68 = 73.8 m

m = 0.92 kg

now when string length is 0.896 m and its speed is 71.5 m/s then we will have

F = \frac{mv^2}{r}

F = \frac{0.92(71.5^2)}{0.896}

F = 5253.7 N

8 0
3 years ago
Read 2 more answers
The space shuttle orbits 310 km above the surface of the Earth.
MAVERICK [17]

Answer:

44.7 N

Explanation:

The gravitational force between the objects is given by:

F=G\frac{mM}{r^2}

where

G is the gravitational constant

m and M are the masses of the two objects

r is the distance between the centres of the two objects

In this problem, we have:

m=5.0 kg is the mass of the sphere

M=5.98\cdot 10^{24} kg is the Earth's mass

R=6370 km is the Earth's radius, while h=310 km is the altitude of the sphere, so the distance of the sphere from Earth's centre is

r=6370 km+310 km=6680 km=6.68\cdot 10^6 m

Substituting into the equation, we find

F=(6.67\cdot 10^{-11})\frac{(5.0 kg)(5.98\cdot 10^{24} kg)}{(6.68\cdot 10^6 m)^2}=44.7 N

8 0
3 years ago
Which star makes up Orion's knee to your right, or the end of his tunic?
lord [1]

Answer:

Betelgeuse and Rigel

Explanation:

100% correct

7 0
3 years ago
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