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OlgaM077 [116]
3 years ago
13

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.

The child grabs and clings to a bar that is 1.45 m from the center of the merry‑go‑round, causing the angular velocity of the merry‑go‑round to abruptly drop from 55.0 rpm to 17.0 rpm. What is the moment of inertia of the merry‑go‑round with respect to its central axis?
Physics
1 answer:
ozzi3 years ago
3 0

Answer:

I_1 = 32.5 kg m^2

Explanation:

Here as we know that total angular momentum is always conserved for child + round system

so here by angular momentum conservation we have

I_1 \omega_1 = (I_1 + I_2)\omega_2

here we have

I_2 = mR^2 (inertia of boy)

I_2 = (34.5)(1.45^2)

I_2 = 72.5 kg m^2

now we have

I_1(2\pi 55) = (I_1 + 72.5)(2 \pi 17)

I_1 = (I_1 + 72.5)(0.31)

I_1(1 - 0.31) = 22.4

I_1 = 32.5 kg m^2

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Answer it pls!!!!!!!!!!!
Archy [21]

Answer:

Fractional error = 0.17

Percent error = 17%

F = 112 ± 19 N

Explanation:

Plug in the values to find the force:

F = (3.5 kg) (20 m/s)² / (12.5 m) = 112 N

Find the fractional error:

ΔF/F = Δm/m + 2Δv/v + Δr/r

ΔF/F = 0.1/3.5 + 2(1/20) + 0.5/12.5

ΔF/F = 0.17

Multiply by 100% to find the percent error:

ΔF/F × 100% = 17%

Solve for the absolute error:

ΔF = 0.17 × 112 N = 19 N

Therefore, the force is:

F = 112 ± 19 N

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3 years ago
An oscillator consisting of a material point of m = 200g vibrates under the action of an elastic force according to the equation
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Answer:

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calculate the period and frequency of the oscillation

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3 years ago
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6 0
2 years ago
A ball is thrown horizontally from the top of a building 37.5 m high. The ball strikes the ground at a point 80.3 m from the bas
trasher [3.6K]

Answer:

t = 2.77 s

Explanation:

The ball in its movement describes a curved line called a semiparabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane:

Equation of movement of the ball in the X axis

X = v₀x*t   Equation  (1)

Equation of movement of the ball in the Y axis

Y = y₀+ v₀y*t -½ g*t² Equation  (2)

Where

X : horizontal position in meters (m)  

Y : vertical  position in meters (m)  

y₀ : initial vertical  position in meters (m)

v₀x :  X-initial speed in m/s  

v₀y :  Y-initial speed in m/s  

g: acceleration due to gravity in m/s²

t : time to position (X,Y)

Data

y₀ = 37.5 m

v₀y = 0

g = 9.8 m/s²

Problem development

The time the ball remains in the air is the same as the ball takes to touch the floor, that is, Y = 0

We apply the Equation (2):

Y = y₀+ (v₀y)*t - (½) g*t²

0 =  37.5 +(0)*t- (1/2)*(g)*t²

0 =  37.5 - (1/2)*(9.8)*t²

(1/2)*(9.8)*t²  = 37.5

t² = (2)(37.5)/(9.8)

t= \sqrt{\frac{2*37.5}{9.8} }

t = 2.77 s

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2 years ago
How fast does sound travel? A. 1,115 feet per second B. 1,000 feet per second C. 2,000 feet per second D. 2,115 feet per second.
lys-0071 [83]

Answer:

A

Explanation:

6 0
3 years ago
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