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TEA [102]
3 years ago
12

For phyics again , pls !!!

Physics
1 answer:
CaHeK987 [17]3 years ago
3 0

Answer: same

Explanation: They both weigh a kilogram and there is no friction

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Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after
s2008m [1.1K]

Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?

a-velocity

b-mass

c-momentum

d-direction

Answer:

b. Mass

Explanation:

This question has to do with the principle of the law of conservation of momentum which states that the momentum of a system remains constant if no external force is acting on it.

As the question states, two objects collide with each other and eventually bounce apart, so their momentum may not be conserved but the mass of the objects is constant for each non-relativistic motion. Because of this, the mass of each object prior to the collision would be the same as the mass after the collision.

Therefore, the correct answer is B. Mass.

6 0
3 years ago
Read 2 more answers
Which energy does a car travelling 30 m/ph as it slows have:
Paladinen [302]

Answer:

c) kinetic energy

Explanation:

6 0
3 years ago
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What is the resistance (R) when voltage is 179V and current is 5 Amps?
Evgesh-ka [11]

Answer:

R = 35.8 Ω

Explanation:

Recall Ohm's Law:

V = I * R

then R = V / I

in our case:

R = 179 V / 5 A = 35.8 Ω

3 0
3 years ago
A heat engine performs (245 + A) J of work in each cycle while also delivering (142 + B) J of heat to the cold reservoir. Find t
Ganezh [65]

Answer:

The value is \eta  =  54.4 \%

Explanation:

From the question we are told that

    The work input is  W = ( 245 + A ) \  J

     The heat delivered is Q =  (142 + B) \  J

      The value of A is  A =  14

        The value of B  is  B  = 72

Generally the efficiency of the heat engine is mathematically represented as

          \eta  =  \frac{W}{Q_t}

Here  Q_t is the total out energy produce by the heat engine and this is mathematically represented as

           Q_t= Q + W

=>         Q_t=  245 + A + 142 + B

=>         Q_t=  390 + A+B

So

               \eta  =  \frac{245 + A }{390 + A+ B}

=>          \eta  =  0.544

=>          \eta  =  0.544 *100

=>          \eta  =  54.4 \%

5 0
3 years ago
An object with a mass of m = 3.85 kg is suspended at rest between the ceiling and the floor by two thin vertical ropes.
Aleksandr-060686 [28]

The tension in the upper rope is determined as 50.53 N.

<h3>Tension in the upper rope</h3>

The tension in the upper rope is calculated as follows;

T(u) = T(d)+ mg

where;

  • T(u) is tension in upper rope
  • T(d) is tension in lower rope

T(u) = 12.8 N + 3.85(9.8)

T(u) = 50.53 N

Thus, the tension in the upper rope is determined as 50.53 N.

Learn more about tension here: brainly.com/question/918617

#SPJ1

6 0
2 years ago
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