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2 years ago

What is an example of velocity?What is an example of velocity?

Physics

2 answers:

erma4kov [3.2K]2 years ago

8 0

"thirty miles per hour to the north"

givi [52]2 years ago

5 0

Speed with direction

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A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2

Pani-rosa [81]

Answer:

(A) Angular speed 40 rad/sec

Rotation = 50 rad

(b) 37812.5 J

Explanation:

We have given moment of inertia of the wheel $I=25kgm^2$

Initial angular velocity of the wheel $\omega _0=10rad/sec$

Angular acceleration $\alpha =15rad/sec^2$

(a) We know that $\omega =\omega _0+\alpha t$

We have given t = 2 sec

So $\omega =10+15\times 2=40rad/sec$

Now $\Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad$

(b) After 3 sec $\omega =10+15\times 3=55rad/sec$

We know that kinetic energy is given by $Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J$

7 0

3 years ago

A child rides a wagon down a hill. Eventually, the wagon comes to a stop. Which is most responsible for causing the wagon to sto

Dvinal [7]

Answer:

well, the hill isn't constantly going down hill, there's an ending point or goes back up hill making a v/u shape or there's nothing helping the wagon being pushed or pull currently

Explanation:

8 0

3 years ago

Will give correct answer brainliest

levacccp [35]

The potential energy= mass times gravity times height. However, we are missing height. Gravity is a constant that is 9.8 on Earth. We then solve for height by dividing 350 by 10 and then 9.8 to get about 3.5.
TLDR: 3.5

3 0

2 years ago

Read 2 more answers

Helpp pls

Kipish [7]

Answer:

The intensity of the electric field is

$|E|=10654.37 \:N/C$

Explanation:

The electric field equation is given by:

$|E|=k\frac{q}{d^{2}}$

Where:

k is the Coulomb constant
q is the charge at 0.4100 m from the balloon
d is the distance from the charge to the balloon

As we need to find the electric field at the location of the balloon, we just need the charge equal to 1.99*10⁻⁷ C.

Then, let's use the equation written above.

$|E|=(9*10^{9})\frac{1.99*10^{-7}}{0.41^{2}}$

$|E|=10654.37 \:N/C$

I hope it helps you!

5 0

3 years ago

An astronaut sitting on the launch pad on Earth's surface is 6,400 kilometers from Earth's center and weighs 400 newtons. Calcul

PIT_PIT [208]

Answer:

weight at height = 100 N .

Explanation:

The problem relates to variation of weight due to change in height .

Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .

At the surface :

Applying Newton's law of gravitation

mg₀ = G Mm / R²

At height h from centre

mg₁ = G Mm /h²

Given mg₀ = 400 N

400 = G Mm / R²

400 = G Mm / (6400 x 10³ )²

G Mm = 400 x (6400 x 10³ )²

At height h from centre

mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²

= 400 / 4

= 100 N .

weight at height = 100 N

5 0

3 years ago