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Molodets [167]
2 years ago
10

Find the momentum of a body of mass 5kg moving with velocity of 4.5m/s​

Physics
1 answer:
Mnenie [13.5K]2 years ago
7 0

Momentum= mass × velocity

momentum= P

P= m×v

P= 5×4.5

P=22.5

I hope this helps.

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A descending elevator of mass 1,000 kg is uniformly decelerated to rest over a distance of 8 m by a cable in which the tension i
Stolb23 [73]

The speed  V_{i} of the elevator at the beginning of the 8 m descent is nearly 4 m/s. Hence, option A is the correct answer.

We are given that-

the mass of the elevator (m) = 1000 kg ;

the distance the elevator decelerated to be y = 8m ;

the tension is T = 11000 N;

let us determine the acceleration 'a' by using Newton's second law of motion.

∑Fy = ma

W - T = ma

(1000kg x 9.8 m/s² ) - 11000N = 1000 kg x a

9800 - 11000 = 1000

a = - 1.2 m/s²

Using the equation of kinematics to determine the initial velocity.

V_{f} ² = V_{i}² + 2ay

V_{i} = √ ( 2 x 1.2m/s² x 8 m )

V_{i} = √19.2 m²/s²

V_{i} = 4.38 m/s   ≈ 4 m/s

Hence, the initial velocity of the elevator is 4m/s.

Read more about the Equation of kinematics:

brainly.com/question/12351668

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8 0
1 year ago
Why do we learn useless nonsense school, why don't we learn what we will actually use in life
laila [671]
I totally agree but, in my opinion its because of the government and what the state has control over. Teacher have little control over it.
4 0
3 years ago
Your friend says an appliance uses energy. How would you correct his statement?
madreJ [45]

Answer:

Not all appliances run on energy. Some of them run on gas. Some both. It just depends on the age of the appliance, the make of the appliance, and the company who made it.

3 0
3 years ago
A rocket car on a horizontal rail has an initial mass of 2500 kg and an additional fuel mass of 1000 kg. At time t0 the rocket m
slamgirl [31]

Answer: Acceleration of the car at time = 10 sec is 108 m/s^{2} and velocity of the car at time t = 10 sec is 918.34 m/s.

Explanation:

The expression used will be as follows.

M\frac{dv}{dt} = u\frac{dM}{dt}

\int_{t_{o}}^{t_{f}} \frac{dv}{dt} dt = u\int_{t_{o}}^{t_{f}} \frac{1}{M} \frac{dM}{dt} dt

       = u\int_{M_{o}}^{M_{f}} \frac{dM}{M}

v_{f} - v_{o} = u ln \frac{M_{f}}{M_{o}}

v_{o} = 0

As, v_{f} = u ln (\frac{M_{f}}{M_{o}})

u = -2900 m/s

M_{f} = M_{o} - m \times t_{f}

           = 2500 kg + 1000 kg - 95 kg \times t_{f}s

           = (3500 - 95t_{f})s

v_{f} = -2900 ln(\frac{3500 - 95 t_{f}}{3500}) m/s

Also, we know that

     a = \frac{dv_{f}}{dt_{f}} = \frac{u}{M} \frac{dM}{dt}

        = \frac{u}{3500 - 95 t} \times (-95) m/s^{2}

        = \frac{95 \times 2900}{3500 - 95t} m/s^{2}

At t = 10 sec,

v_{f} = 918.34 m/s

and,   a = 108 m/s^{2}

3 0
3 years ago
The number of wave cycles that pass in one second is called the _______ of the wave.
postnew [5]

That is the frequency.

6 0
3 years ago
Read 2 more answers
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