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3 years ago

The big bang theory of the information and expansion of the universe is supported by the observed

Physics

1 answer:

dusya [7]3 years ago

6 0

By Hubble theory in which universe is expanding,

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What three things do cells / organisms have to do to maintain homeostasis? ____ from food, get rid of ____, and _____(mitosis/me

user100 [1]

1) use energy from food
2) get rid of wastes
3) maintain

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3 years ago

Why is this event important to include in the biography?

nadya68 [22]

Answer:

D: It shows that Frida Kahlo used art to cope with her pain.

Explanation:

Within the text given it shows her emotions being lonely, immobile and in pain. But it all shows her asking her father for art which states that art is her sort of relief and happy place.

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3 years ago

Read 2 more answers

A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal di

Lyrx [107]

Answer:

Explanation:

From the information given:

The angular speed for the block $\omega = 50 \ rad/s$

Disk radius (r) = 0.2 m

The block Initial velocity is:

$v = r \omega \\ \\ v = (0.2 \times 50) \\ \\ v= 10 \ m/s$

Change in the block's angular speed is:

$\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s$

However, on the disk, moment of inertIa is:

$I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2$

The time t = 10s

∴

Frictional torques by the wall on the disk is:

$T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10}) \\ \\ =0.6 \ N.m$

Finally, the frictional force is calculated as:

$F = \dfrac{T}r{}$

$F= \dfrac{0.6}{0.2} \\ \\ F = 3N$

8 0

3 years ago

Select all the correct ANSWERS.

MissTica

The ground exerts an equal force on the golf ball.

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4 years ago

Calculate the magnitude of the linear momentum for the following cases. (a) a proton with mass 1.67 10-27 kg, moving with a spee

abruzzese [7]

Answer:

Explanation:

Mass of proton :

$m_P=1.67\times 10^-^2^7\:kg\\$

Speed of Proton:

$v_P=4.85\times 10^6$

Linear Momentum of a particle having mass (m) and velocity (v) :

$-> p =m->v\:\:\: (1)$

Magnitude of momentum :

$p=mv\:\:\: (2)$

Frome equation (2), magnitude of linear momentum of the proton :

$p_P=m_P\:v_P\\\\p_P=1.67\times 10^-^2^7 \:kg\times4.85\times 10^6\:ms^-^1\\\\p_P= 8.0995\times 10^-^2^1\:kgms^-^1$

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3 years ago