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Sav [38]
3 years ago
14

With what magnitude of force does a ball of mass 0.75 kilograms need to be hit so that it accelerates at the rate of 25 meters/s

econd2? Assume that the ball undergoes motion along a straight line.
Physics
2 answers:
SSSSS [86.1K]3 years ago
8 0

Answer:

Assume that the ball undergoes motion along a straight line. ... F = m A Force = (mass) x (acceleration) The question tells you the mass and the acceleration. All YOU have to do is take the numbers and pluggum into Newton's 2nd law. F = m A = (0.75 kg) (25 m/s²) = (0.75 x 25) kg-m/s² = 18.75 Newtons .

Explanation:

i looked it up ok

Zielflug [23.3K]3 years ago
8 0

<u>Answer: </u>

A ball of mass 0.75 kilograms needs to be hit so that it accelerates at the rate of 25 meters/second^2 assuming that the ball undergoes motion along a straight line and the mount of Force required is 18.75 newton.

<u>Explanation:</u>

According to Newton`s 2nd  law F=Ma,

where M mass of the object, a- Acceleration due to gravity.

Given-

Mass (m) = 0.75 kg

Acceleration (a) = 25 m/s^2

We know That, Force = mass × acceleration

Substituting the given values , we get

  F = 0.75 × 25

   F= 18.75 N  

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Katyanochek1 [597]

Answer:

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Explanation:

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8 0
3 years ago
Read 2 more answers
How do you do this question?
umka2103 [35]

Explanation:

The moment of inertia of each disk is:

Idisk = 1/2 MR²

Using parallel axis theorem, the moment of inertia of each rod is:

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The total moment of inertia is:

I = 2Idisk + 5Irod

I = 2 (1/2 MR²) + 5 [1/2 mr² + m (R − r)²]

I = MR² + 5/2 mr² + 5m (R − r)²

Plugging in values:

I = (125 g) (5 cm)² + 5/2 (250 g) (1 cm)² + 5 (250 g) (5 cm − 1 cm)²

I = 23,750 g cm²

7 0
3 years ago
A(n) 96.1 g ball is dropped from a height of 59.1 cm above a spring of negligible mass.The ball compresses the spring to a maxim
Serhud [2]

Answer:

Explanation:

Mass of ball Is m=96.1g=0.0961kg

Height above spring is 59.1cm

L=0.591m

Extension of the spring is 4.75403cm

e=0.0475403m

Then the distance the ball traveled is H=L+e

H=0.591+0.0475403

H=0.6385403m

Then, the potential energy of the ball is given as

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P.E=0.602J

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Then, the P.E is transferred to the work done by the spring

Then, Work done by spring is given as

W=½ke²

W=P.E=½×k×0.0475403²

0.602=½×k×0.0475403²

k=0.602×2/0.0475403²

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The spring constant is 532.72 N/m

4 0
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Hello!

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According to their number of protons, these are Aluminum isotopes: ²⁷Al (stable), and ²⁶Al (radioactive), respectively. 

Have a nice day!

7 0
3 years ago
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\huge \tt \underline \red{answer}

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4 0
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