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2 years ago

What is the angular momentum at a radius of 2 m with an object of 5 kg at a velocity of 20 m/s?

Physics

2 answers:

aev [14]2 years ago

5 0

The angular momentum is 200 kg m^2 s^-1

<h3>what is angular momentum?</h3>

Angular momentum is the product of linear momentum and the perpendicular distance. Linear momentum is the product of mass and velocity, where radius is the perpendicular distance

Angular momentum = mass * velocity * radius

Angular momentum = 5 * 2 * 20

Angular momentum = 200 kg m^2 s^-1

read more about angular momentum here: brainly.com/question/4126751

ahrayia [7]2 years ago

3 0

Answer:

200

Explanation:

angular momentum=mvr

=2×5×20

= 200kgm2/s

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A car travels a distance of 320 km in 4 hours. What is your average speed in meters per second?

Andreas93 [3]

Answer:

22.2 m/s

Explanation:

First, we need to convert km to m by multiplying by 1000. This means that the car traveled 320 000 meters.

Next, we convert hours to minutes by multiplying by 3600 (the number of seconds in an hour). This means that overall, the car traveled 320 000 m in 14 400 seconds.

The average speed can be found by using the equation $\frac{distance}{time}$ . After substitution, this gives the fraction $\frac{320 000}{14 400}$ , which reduces to 22 $\frac{2}{9}$ m/s, or about 22.2 m/s.

4 0

4 years ago

. An unbalanced force of 500 N is applied to a 75 kg object. What is the acceleration of the object?

Hoochie [10]

The acceleration of the object is $6.7 m/s^2$

Explanation:

We can solve the problem by using Newton's second law, which states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:

$F=ma$

where

F is the net force

m is the mass of the object

a is its acceleration

For the object in this problem,

F = 500 N is the applied force

m = 75 kg is the force

Solving the equation for a, we find the acceleration:

$a=\frac{F}{m}=\frac{500}{75}=6.7 m/s^2$

Learn more about Newton's second law:

brainly.com/question/3820012

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5 0

3 years ago

1. A car starting from rest accelerates uniformly at 3

amid [387]

The car's final speed in m/s after the acceleration is 30.

<u />

<u>Given the following data:</u>

Initial velocity, U = 0 m/s (since the cars starts from rest)

Time, t = 10 seconds

Acceleration, a = 3 meter per seconds square.

To find the car's final speed in m/s after the acceleration, we would use the first equation of motion;

Mathematically, the first equation of motion is given by the formula;

$V = U + at\\\\V = 0 + 3(10)$

<em>Final speed, V </em><em>=</em><em> 30 m/s</em>

Therefore, the car's final speed in m/s after the acceleration is 30.

Read more: brainly.com/question/8898885

5 0

3 years ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago

Juan and Kuri are on a carousel. Juan is closer to the center of the carousel than Kuri. Which statement describes their tangent

Licemer1 [7]

Answer:

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

Explanation:

Since Juan is closer to the center and Kuri is away from the center so we can say that Juan will move smaller distance in one complete revolution

As we know that the distance moved in one revolution is given as

$d = 2\pi r$

also the time period of revolution for both will remain same as they move with the time period of carousel

Now we can say that the speed is given as

$v = \frac{2\pi r}{T}$

so Juan will have less tangential speed. so correct answer will be

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

6 0

3 years ago

Read 2 more answers