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dezoksy [38]
2 years ago
15

Please help with this problem. Thank you.​

Physics
1 answer:
4vir4ik [10]2 years ago
4 0

Answer:

Explanation:

F = mω²R

F = 15(2π/8.5)²(7.8)

F = 63.93044788...

F = 63.9 N

answer a) is the closest. No idea how they got a value that low unless they used a poor approximation for π.

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If two balls collide with each other, they will move apart at the same speed if
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A.
if you have seen a newton's cradle this will make sense.

in order for both of them to travel at the same speed, the balls need to have the same mass and the speed to begin with tocontinue to travel at the same speed because mass can affect the impact of the force on the balls by each other, causing each ball to have different speeds.
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3 years ago
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Two wires are made of the same material and have the same length but different radii. They are joined end-to- end and a potentia
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2 years ago
Two trains A and B of length 400 m each are moving on two parallel tracks with
Vinvika [58]

<u>Answer:</u>

<em>The initial distance between the trains is 1450 m. </em>

<u>Explanation:</u>

In the question two trains are of equal length 400 m and moves at a uniform speed of 72 km/h. train A is moving ahead of train B. If the train B has to overtake train A it should accelerate.

Train B’s acceleration  is 1m/s^2   and it accelerated for 50 seconds.

<em>a=1 m/s^2</em>

<em>t=50 s </em>

<em>initial speed u=72km/h </em>

<em>we have to convert this speed into m/s  </em>

<em>u=72 \times 5/18=20 m/s</em>

<em>Distance covered in accelerating phase  S=ut+1/2  at^2  </em>

<em>=20 \times 50+1/2 \times 1 \times 50^2</em>

<em>=1000+1250=2250 m </em>

If  a train is just behind another, the distance covered by the train located behind during overtaking phase will be equal to the sum of the lengths of the trains.

<em>Here length of train A+length of train B=400+400=800 m</em>

<em>Hence the initial distance between the trains = 2250-800=1450 m</em>

6 0
2 years ago
A ball is thrown at 60 degrees and lands in 18.5 seconds what is the velocity of the ball at the start
n200080 [17]

Answer:

the initial velocity of the ball is 104.67 m/s.

Explanation:

Given;

angle of projection, θ = 60⁰

time of flight, T = 18.5 s

let the initial velocity of the ball, = u

The time of flight is given as;

T = \frac{2u\times sin(\theta)}{g} \\\\2u\times sin(\theta) = Tg\\\\u = \frac{Tg}{2\times sin(\theta)} \\\\u = \frac{18.5 \times 9.8}{2\times sin(60^0)} \\\\u = 104.67 \ m/s

Therefore, the initial velocity of the ball is 104.67 m/s.

3 0
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antiseptic1488 [7]

Answer:

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Explanation:

your welcome :)

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