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Gwar [14]
2 years ago
15

Kjebhg da sad bmm m mmm b bc f gaf see Krio ok pop it ya ta is v BSA u lie lug

Physics
1 answer:
GenaCL600 [577]2 years ago
4 0

Answer: Kjebhk das dajsd ajdn wadm awkdlwandeA fefme ef

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A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 25.5 kg.
Svetllana [295]

Answer:

72.75 kg m^2

Explanation:

initial angular velocity, ω = 35 rpm

final angular velocity, ω' =  19 rpm

mass of child, m = 15.5 kg

distance from the centre, d = 1.55 m

Let the moment of inertia of the merry go round is I.

Use the concept of conservation of angular momentum

I ω = I' ω'

where I' be the moment of inertia of merry go round and child

I x 35 = ( I + md^2) ω'

I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19

35 I = 19 I + 1164

16 I = 1164

I = 72.75 kg m^2

Thus, the moment of inertia of the merry go round is 72.75 kg m^2.

5 0
3 years ago
If a ball is rolling across a floor at a constant speed is it accelerating
KATRIN_1 [288]

You already told us that its speed is constant. That's one part of acceleration.

The other part of acceleration is the direction it's moving.

If it's rolling in a straight line, then there's no acceleration.

If it's curving left or right, then that's acceleration.

8 0
3 years ago
If someone is driving 100 miles in 60 minutes then drives 150 miles in 100 minutes west, what is his acceleration rate.
Sliva [168]

Answer:

his acceleration rate is -0.00186 m/s²

Explanation:

Given;

initial position of the car, x₀ = 100 miles = 160, 900 m ( 1 mile = 1609 m)

time of motion, t₀ = 60 minutes = 60 mins x 60 s = 3,600 s

final position of the car, x₁ = 150 miles = 241,350 m

time of motion, t₁ = 100 minutes = 100 mins x 60 s = 6,000 s

The initial velocity is calculated as;

u = 160, 900 m / 3,600 s

u = 44.694 m/s

The final velocity is calculated as;

v = 241,350 m / 6,000 s

v = 40.225 m/s

The acceleration is calculated as;

a = \frac{\Delta V}{\Delta t} = \frac{v- u}{t_1 - t_ 0} = \frac{40.225  - 44.694}{6000-3600} = -0.00186 \ m/s^2\\\\

Therefore, his acceleration rate is -0.00186 m/s²

6 0
3 years ago
Consider the following cyclic process carried out in two steps on a gas. Step 1: 44 J of heat is added to the gas, and 20. J of
notka56 [123]

Answer:37 J

Explanation:

Given

Step :1

Heat added Q=44 J

Work done=-20 J

\Delta E_1=Q+W=44-20=24 J

Step :2

Heat added Q=-61 J

work done W_2

\Delta E_2=Q+W_2

\Delta E_2=61+W_2

\Delta E_1+\Delta E_2=0

as the process is cyclic

44-20-61+W_2=0

W_2=37 J

work done in compression is 37 J

3 0
3 years ago
Objective lenses are contained in a ____________ that can be turned to put a particular objective lens in place to be used.
Masteriza [31]

Answer:

Revolving nosepiece

Explanation:

The revolving nosepiece is one of the parts of a microscope, used for holding the objective lenses. They can be turned to put a particular objective lens in place to be used in order to vary magnification.

6 0
3 years ago
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