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nexus9112 [7]
3 years ago
10

a uniform disc and hollow right circular cone have the same formula for their moment of inertia when rotating about the central

axis why is it so?
Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
6 0

Answer:

This is as a result that about the central axis a collapsed hollow cone is equivalent to a uniform disc

Explanation:

The integration of the differential mass of the hollow right circular cone yields

I=\int\limits   dmr^2 = \int\limits^a_b {\frac{2Mxr^2}{R^2 +H^2} } \, dx  = \frac{2MR^2dx}{(R^2 +H^2)^2} \frac{(R^2 +H^2)^2}{4} = \frac{1}{2}MR^2

and for a uniform disc

I = 1/2πρtr⁴ = 1/2Mr².

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Is it point<br> A,B,C or D
e-lub [12.9K]

the answer is (c) i think

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2 years ago
F an object has a mass of 200 kg and a weight of 1000 N, what is g?
jeka94

Answer:

g = 5 m/s square

Explanation:

Weight(W), Mass(m), Gravity(g)

W = mg

1,000N = 200g

g = 1000/200

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7 0
2 years ago
How many minutes would it take a light wave to travel from the planet Venus to Earth? (Average distance from Venus to Earth = 28
Gnesinka [82]

Answer:

t=2.51min

Explanation:

The time taken by the light to travel a given distance is defined as:

t=\frac{d}{c}

Here c is obviously the speed of light. Now we convert the average distance form Venus to Earth to meters:

28*10^{6}mi*\frac{1609.34}{1mi}=4.51*10^{10}m

Finally, we calculate the minutes taken by the light to travel from Venus to Earth:

t=\frac{4.51*10^{10}m}{3*10^8\frac{m}{s}}\\t=150.33s*\frac{1min}{60s}=2.51min

6 0
3 years ago
A 200-kg object and a 500-kg object are separated by 4.00 m. (a) find the net gravitational force exerted by these objects on a
pantera1 [17]
Check the attached file for the solution for this problem.

5 0
3 years ago
A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help"
Digiron [165]

Answer: a) The cliff is 532.05m high

b) Her speed just before hitting the ground is 102.12 m/s

Explanation: To solve This, I'll use a sketch diagram, attached to this solution,

In 3seconds, the teacher heard the echo of her initial scream back. We can obtain the distance the teacher had fallen at the end of 3 seconds using the equations of motion,

Y1 = ut + 0.5g(t^2)

Since she's falling under the influence of gravity, her initial velocity, u = 0m/s, g = 9.8m/s2, t = 3s

Y1, distance she fell through in 3 seconds = 0.5×9.8(3^2) = 44.1m

Let the total height of the cliff be (44.1 + x); where is the remaining height of cliff that the teacher will fall through.

Using the equations of motion again, we can obtain distance travelled by the sound waves in 3s. sound waves travel with a constant speed of 340m/s, no acceleration,

Y2 = ut + 0.5g(t^2) where g = 0, u = 340m/s, t = 3seconds

Y2 = 340 × 3 = 1020m

But in 3 secs, the sound waves would have travelled through the total height of the cliff (44.1 + x) and back to the teacher's current height, x. That is, 1020 = 44.1 + x + x

x = 487.95m

So, total height of cliff = 44.1 + 487.95 = 532.05m

b) the speed of the teacher just before she hits the ground.

Using the equations of motion again,

(V^2) = (U^2) + 2gs

Where v is the final velocity to be calculated

U is the initial velocity = 0m/s

g is acceleration due to gravity = 9.8m/s2

S is the total height she fell through, that is, the height of the cliff = 532.05m

(V^2) = 0 + 2×9.8×532.05 = 10428.18

V = √(10428.18) = 102.12m/s

QED!

4 0
3 years ago
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