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Andreyy89
2 years ago
15

Jay pushes the box with a force of 25 Newtons. There are 5 Newtons of friction present. In what direction will the box accelerat

e?
To the right (same direction as Fp)

Downward (same direction as Fg)

To the left (same direction as Ff)

Upward (same direction as FN)
Physics
1 answer:
yKpoI14uk [10]2 years ago
8 0

Answer: To the right

Explanation:

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The rate of change of velocity or speed is known as acceleration. If a car increases it speed form 10 to 20m/s in 2 seconds, the
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Answer:

A) 5 m/s/s

Explanation:

<u>Given the following data;</u>

Initial velocity = 10m/s²

Final velocity = 20m/s²

Time, t = 2 seconds.

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

Acceleration, a = \frac{final \; velocity  -  initial \; velocity}{time}

Substituting into the equation, we have;

Acceleration, a = \frac{20 - 10}{2}

Acceleration, a = \frac{10}{2}

<em>Acceleration, a = 5m/s²</em>

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3 years ago
In what ways does a river carry its load?
MArishka [77]
It carry its loads from the sand and outher partials in it
7 0
3 years ago
HELP I WILL GIVE BRAINLIEST!
Ann [662]

Answer:

Answer:B

Explanation:

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3 years ago
A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 34.5 kg.
rodikova [14]

Answer:

the moment of inertia of the merry go round is 38.04 kg.m²

Explanation:

We are given;

Initial angular velocity; ω_1 = 37 rpm

Final angular velocity; ω_2 = 19 rpm

mass of child; m = 15.5 kg

distance from the centre; r = 1.55 m

Now, let the moment of inertia of the merry go round be I.

Using the principle of conservation of angular momentum, we have;

I_1 = I_2

Thus,

Iω_1 = I'ω_2

where I' is the moment of inertia of the merry go round and child which is given as I' = mr²

Thus,

I x 37 = ( I + mr²)19

37I = ( I + (15.5 x 1.55²))19

37I = 19I + 684.7125

37I - 19 I = 684.7125

18I = 684.7125

I = 684.7125/18

I = 38.04 kg.m²

Thus, the moment of inertia of the merry go round is 38.04 kg.m²

7 0
4 years ago
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