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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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Resistance in wires causes thermal energy. The correct answer between all the choices given is the third choice or letter C. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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Determine the maximum theoretical speed that may be achieved over a distance of 66 m by a car starting from rest, knowing that t
My name is Ann [436]

Answer:

v=32.49 m/s

Explanation:

Given that

Distance ,d= 66 m

Initial speed of the car ,u = 0 m/s

Coefficient of friction ,μ = 0.8

Lets take the total mass of the car = m

The acceleration of the car is given as

a = μ g                               ( g= 10 m/s² )

Now by putting the values in the above equation we get

a= 0.8 x 10 m/s²

a= 8 m/s²

We know that ,final speed is given as

v²= u ²+ 2 a d

Now putting the value

v²=0² + 2 x 8 x 66

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3 0
3 years ago
An astronomer discovers a massive galaxy which has four nuclei. What is a likely explanation for a galaxy having more than one n
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6 0
2 years ago
A mass of M-kg rests on a frictionless ramp inclined at 30°. A string with a linear mass density of μ=0.025" kg/m" is attached t
I am Lyosha [343]

Answer:

44.3 m/s

Explanation:

a) Draw a free body diagram of the mass M.  There are three forces:

Weight force mg pulling down,

Normal force N pushing perpendicular to the ramp,

and tension force T pulling parallel up the ramp.

Sum of forces in the parallel direction:

∑F = ma

T − Mg sin 30° = 0

T = Mg sin 30°

T = Mg / 2

Draw a free body diagram of the hanging mass m.  There are two forces:

Weight force mg pulling down,

and tension force T pulling up.

Sum of forces in the vertical direction:

∑F = ma

T − mg = 0

T = mg

Substitute:

mg = Mg / 2

m = M / 2

M = 2m

b) Velocity of a standing wave in a string is:

v = √(T / μ)

T = mg, and m = 5 kg, so T = (5 kg) (9.8 m/s²) = 49 N.  Therefore:

v = √(49 N / 0.025 kg/m)

v = 44.3 m/s

7 0
3 years ago
An automobile starter motor has an equivalent resistance of 0.055 Ï and is supplied by a 12.0 v battery which has a 0.0305 Ï int
LiRa [457]
12 V is the f.e.m. \epsilon of the battery. The potential difference that is applied to the motor is actually the fem minus the voltage drop on the internal resistance r:
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this is equal to the voltage drop on the resistance of the motor R:
RI
so we can write:
\epsilon - Ir = RI
and using r=0.0305~\Omega and R=0.055~\Omega we can find the current I:
I= \frac{\epsilon}{R+r}=140~A
8 0
3 years ago
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