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borishaifa [10]
2 years ago
7

a waterwheel built in Hamah, Syria, has a radius of 20 m . If the tangential velocity at the wheels edge is 7.85 m/s , what is t

he centripetal acceleration of the wheel?
Physics
1 answer:
gtnhenbr [62]2 years ago
4 0

Answer:

3.08m/s²

Explanation:

Given parameters:

Radius = 20m

Tangential velocity  = 7.85m/s

Unknown:

Centripetal acceleration  = ?

Solution:

Centripetal acceleration is the acceleration of a body along a circular path.

 it is mathematically given as;

       a  = \frac{v^{2} }{r}  

v is the tangential velocity

r is the radius

      a  = \frac{7.85^{2} }{20}   = 3.08m/s²

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hichkok12 [17]

Answer:

500ms times 2 would be when the ball reaches the max horizontal distance.

Then to find the angle, use the formula of time to reach max height t = u sin theta / g . With t being the max height time 500ms, u being 10m/s

For initial vertical velocity just use u sin theta.

5 0
2 years ago
A rock is dropped from a window 5 m above the ground. the rock hits the ground 1 s later with a speed of 10 m/s. what is the ave
Leviafan [203]
Average speed=total distance/total
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3 years ago
Only one force acts on an object, can the object have zero acceleration? can the object have zero velocity? explain.
Harlamova29_29 [7]
It can have a 0 for both of them
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3 years ago
A horizontal circular turntable rotates about its center at the uniform rate of 10.9 revolutions per minutes. Find the greatest
antiseptic1488 [7]

Answer:

6.03 m

Explanation:

First of all, let's convert the angular velocity from revolutions per minute to radians per second:

\omega = 10.9 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=1.14 rad/s

The frictional force on the block ranges from zero to a maximum value of

F=\mu mg

In order for the block to remain stuck on the turntable, the frictional force must be equal to the centripetal force, so we can write:

m\omega^2 r = \mu mg

where

m is the mass of the block

\omega is the angular velocity

r is the distance of the block from the centre

\mu = 0.8 is the coefficient of static friction

g = 9.8 m/s^2

Solving for r, we find:

r=\frac{\mu g}{\omega^2}=\frac{(0.8)(9.8)}{(1.14)^2}=6.03 m

8 0
3 years ago
Two sailboats leave a harbor in the bahamas at the same time. the first sails at 23 mph in a direction 330°. the second sails at
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First, let us calculate the total distance that each have taken after 2 hours.

Let’s say that:

A = sailboat which sails at 23 mph in a direction 330°

B = sailboat which sails at 34 mph in a direction 190°

Calculating for distances:

dA = 23 mph (2 hours) = 46 miles

dB = 34 mph (2 hours) = 68 miles

Imagining a Cartesian coordinate, the angle θ between the two sailboats is simply the difference:

θ = 330° - 190°

θ = 140°

We know that from the law of cosines:

c^2 = a^2 + b^2 – 2 a*b*cos θ

Therefore the distance between the two after 2 hours, C, is:

C^2 = 46^2 + 68^2 – 2 (46) (68) cos(140)

<span>C = 107.39 miles</span>

7 0
3 years ago
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