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4 years ago

What are large glowing balls of plasma held together by their own gravity called?

Physics

1 answer:

GarryVolchara [31]4 years ago

4 0

Large glowing balls of plasma held together by their own gravity are called stars

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What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a

Daniel [21]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

v = r ω

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}$

$I = \dfrac{m(2gh - v^2)r^2}{v^2}$

$I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}$

$I =mr^2(0.287)$

I = 0.287 MR²

3 0

3 years ago

Negatively charged particles that orbit the nucleus of an atom are _________. A. electrons B. protons C. neutrons

julsineya [31]

negativily chtarge particles are electron

5 0

4 years ago

Problem: The frequency of an FM radio station is 89.3 MHz. Calculate its period. Part B: From the Library, select the general eq

vekshin1

Answer:

Time period, $T=1.11\times 10^{-8}\ s$

Explanation:

We have,

The frequency of an FM radio station is 89.3 MHz.

It is required to find the period of the wave.

The reciprocal of frequency is called time period of a wave. It can be given by :

$T=\dfrac{1}{f}\\\\T=\dfrac{1}{89.3\times 10^{6}\ Hz}\\\\T=1.11\times 10^{-8}\ s$

So, the period of the wave is $1.11\times 10^{-8}\ s$ .

5 0

3 years ago

Sound waves are also called compression waves. This means that as the wave travels through air, the ________ increases and decre

Nuetrik [128]

I believe the answer is pressure. hope this helps

3 0

3 years ago

One kind of baseball pitching machine works by rotating light and stiff rigid rod about a horizontal axis until the ball is movi

erastova [34]

Answer:

(a). The ball's centripetal acceleration is $16.17\times10^{2}\ m/s^2$

(b). The magnitude of the net force is 232.9 N.

Explanation:

Given that,

Mass of baseball = 144 g

Speed = 81 mph = 36.2 m/s

Distance = 81 cm

(a). We need top calculate the ball's centripetal acceleration just before it is released

Using formula of centripetal acceleration

$a=\dfrac{v^2}{r}$

Where, v = speed

r = radius

Put the value into the formula

$a=\dfrac{(36.2)^2}{81\times10^{-2}}$

$a=1617.82\ m/s^2$

$a=16.17\times10^{2}\ m/s^2$

(b). We need to calculate the magnitude of the net force that is acting on the ball just before it is released

Using formula of force

$F=\dfrac{mv^2}{r}$

Put the value into the formula

$F=\dfrac{144\times10^{-3}\times(36.2)^2}{81\times10^{-2}}$

$F=232.9\ N$

Hence, (a). The ball's centripetal acceleration is $16.17\times10^{2}\ m/s^2$

(b). The magnitude of the net force is 232.9 N.

4 0

3 years ago