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swat32
3 years ago
15

A ball on the end of a string is rotating with constant speed in a horizontal plane. When the ball is moving North it is located

to the East of the pivot point.
At that time the angular velocity of the ball points in what direction?

A. up

B. down

C. East

D. West

E. North
Physics
1 answer:
joja [24]3 years ago
4 0

Answer:

Option A

Up

Explanation:

we can determine the direction of the angular velocity of a rotaing body by using the right hand rule.

The right hand rule says that if you hold the axis with your right hand and rotate the fingers in the direction of motion of the rotating body then your thumb will point the direction of the angular velocity.

Following this, curving the fingers in such a way that they depict motion from the east to north, we can see that our thumb will point upwards. This makes the direction of the angular velocity at that point in time to be up

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Answer:

With air resistance, acceleration throughout a fall gets less than gravity (g) because air resistance affects the movement of the falling object by slowing it down. How much it slows the object down depends on the surface area of the object and its speed

Explanation:

PLZ MARK AS THE BRAINLIEST

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4 0
3 years ago
The diameter of an automobile tire is closest to
slamgirl [31]

Answer:

(1)  10^−2 m

Explanation:

The diameter of the tire of an automobile is generally expressed in centimetres; we can say that the diameter of a tire is generally about

d = 20 cm (20 centimetres)

Now we have to verify which option is closest to this value. To do that, we have to keep in mind the equivalence between metres and centimetres; in fact, we have:

1 cm = 10^{-2} m

This means that we can rewrite the diameter of the tire of a car as

d=20 cm = 20\cdot 10^{-2} m

By comparing it with the given options, we see that the closest option is

(1)  10^−2 m

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6 0
4 years ago
A proton and an alpha particle (q = +2e, m = 4 u) are fired directly toward each other from far away, each with an initial speed
stich3 [128]

Answer:

Distance of closest approach, r=1.91\times 10^{-14}\ m

Explanation:

It is given that,

Charge on proton, q_p=e

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Mass of proton, m_p=1.67\times 10^{-27}\ kg

Mass of alpha particle, m_a=4m_p=6.68\times 10^{-27}\ kg

The distance of closest approach for two charged particle is given by :

r=\dfrac{k2e^2(m_p+m_a)}{2m_am_pv_p^2}

r=\dfrac{9\times 10^9\times 2(1.6\times 10^{-19})^2(1.67\times 10^{-27}+6.68\times 10^{-27})}{2\times 6.68\times 10^{-27}\times 1.67\times 10^{-27}(0.01\times 3\times 10^8)^2}

r=1.91\times 10^{-14}\ m

So, their distance of closest approach, as measured between their centers 1.91\times 10^{-14}\ m. Hence, this is the required solution.

4 0
3 years ago
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3 0
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