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Semenov [28]
3 years ago
8

How much force is needed to accelerate a 15kg bowling ball at 2 m/s^2

Physics
2 answers:
olga_2 [115]3 years ago
7 0
Golf no doing what I iuroeurir to do my homework and my homework
kirza4 [7]3 years ago
7 0
The answer is F=30N
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A 975-kg elevator accelerates upward at 0.754 m/s2, pulled by a cable of negligible mass. Find the tension force in the cable.
Zina [86]

To solve this problem we will apply the concepts of equilibrium and Newton's second law.

According to the description given, it is under constant ascending acceleration, and the balance of the forces corresponding to the tension of the rope and the weight of the elevator must be equal to said acceleration. So

\sum F = ma

T-mg = ma

Here,

T = Tension

m = Mass

g = Gravitational Acceleration

a = Acceleration (upward)

Rearranging to find T,

T = m(g+a)

T = (975)(9.8+0.754)

T= 10290.15N

Therefore the tension force in the cable is 10290.15N

7 0
3 years ago
Which statements correctly characterize a lunar eclipse? Select all that apply.
Zepler [3.9K]

A) it occurs when earth is between the sun and the moon

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3 years ago
Explain, briefly, what the electrical field inside a microwave oven does to heat food.
yarga [219]

Answer:

Inside the strong metal box, there is a microwave generator called a magnetron. When you start cooking, the magnetron takes electricity from the power outlet and converts it into high-powered, 12cm (4.7 inch) radio waves. Thus the microwaves pass their energy onto the molecules in the food, rapidly heating it up.May 3, 2018

5 0
3 years ago
changes to OSHA's regulations in 2013 require chemical information be provided via the _______ format. a) safety data sheets b)
Talja [164]

Don't listen to the other guy I just took the test and got it wrong because of him..

I re-took it and the correct answer is

A) Safety Data Sheets (SDS)

8 0
3 years ago
A 6.3 g bullet leaves the muzzle of a rifle with a speed of 596.2 m/s. what constant force is exerted on the bullet while it is
ivanzaharov [21]
<span>anwser will be 

F = ma

where 

F = force exerted on the bullet 
m = mass of the bullet = 5 gm (given) = 0.005 kg. 
a = acceleration of the bullet 

Substituting appropriately, 

F = 0.005a --- call this Equation 1 

Next working equation is 

Vf^2 - Vo^2 = 2as 

where 

Vf = velocity of the bullet as it leaves the muzzle = 326 m/sec (given) 
Vo = initial velocity of bullet = 0 
a = acceleration of bullet 
s = length of the rifle's barrel 

Substituting appropriately, 

326^2 - 0 = 2(a)(0.83) 

a = 64,022 m/sec^2 

the anwser will be
Substituting this into Equation 1, 

F = 0.005(64,022) 

F =320.11 Newtons 

Hope this helps. </span><span>
</span>
8 0
3 years ago
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