Forms a thin layer of gases around earth

Aye, I'm b0red so here's an easy one:

asambeis [7]

ok

1. blue red pink black

2. you can say bored look

3. why do you want to talk in brainly?

8 0

2 years ago

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Four children pull on the same stuffed toy at the same time yet there is no net force on the toy. How is this possible.

Stels [109]

Answer:

Net force is Zero.

Explanation:

If all forces that are equal and opposite are exerted on an object the resulting force will be Zero.

3 0

2 years ago

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uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago

PLEASE HELP ME SOMEONE

sasho [114]

I think its B or D, most likely D.

7 0

3 years ago

A 1100 kg sports car accelerates from 0m/s to 32m/s in 10 seconds. What is the average power of the engine?

timurjin [86]

Answer:

The average power of the engine of the sports car is 56.32 kW

Explanation:

Given;

mass of the sports car, m = 1100 kg

initial velocity of the sports car, u = 0 m/s

final velocity of the sports car, v = 32 m/s

time of motion, t = 10 s

The kinetic energy of the car is given by;

K.E = ¹/₂m(v² - u²)

K.E = ¹/₂mv²

K.E = ¹/₂ x 1100 x 32²

K.E = 563200 J

The average power of the engine of the sports car is given by;

Pavg = Energy / time

Pavg = 563200 / 10

Pavg = 56320 W

Pavg = 56.32 kW

Therefore, the average power of the engine of the sports car is 56.32 kW

8 0

3 years ago