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Nutka1998 [239]
3 years ago
7

A 5.20-N force is applied to a 1.05-kg object to accelerate it rightwards across a friction-free surface. determine the accelera

tion.
Physics
2 answers:
Gnom [1K]3 years ago
7 0

Answer:

4.95 m/s/s

Explanation:

Upon neglecting air <u>resistance</u>, there are three forces acting upon the object. The up and <u>downforce</u> balance each other and the acceleration is caused by the <u>applied</u> force. The net force is 5.20 N, right (equal to the only rightward force - the applied force). So the <u>acceleration</u> of the object can be <u>computed</u> using Newton's second law.

a = Fnet / m = (5.20 N, right) / (1.05 kg) = 4.95 m/s/s, right

ZanzabumX [31]3 years ago
3 0

Explanation:

Newton's second law:

F = ma

5.20 N = (1.05 kg) a

a = 4.95 m/s²

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Answer:

an ice cream machine produced 44 ice cream per minute.

Now, after reconditioning it  speeds up to 55 ice cream per minute.

=> 55 - 44 = 11

=> 11 / 44 = 0.25

=> 0.25 * 100% = 25%.

thus it increased it production to 25%.

Explanation:

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4 years ago
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An object is moving north with an initial velocity of 14 m/s accelerates 5m/s for 20 seconds. What is the final velocity of the
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Use the kinematic equation: Vf=Vi+at

Then plug;

Vi=14 m/s

a=5 m/s²

t=20 s. Therefore;

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3 years ago
Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

3 0
3 years ago
Matter is made of small particles to small to be seen. Which of these best describe evidence of this statement? 1. Tara’s crayon
son4ous [18]

Answer:

Explanation:

I think the answer is statement no 3.

Hope it helps.

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3 years ago
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A student wearing a frictionless roller skates on a horizontal is pushed by a friend with a constant force of 55N. How far must
umka21 [38]

Answer:

6.58m

Explanation:

The kinetic energy = Workdone on the roller

Workdone = Force * distance

Given

KE = Workdone = 362J

Force = 55N

Required

Distance

Substitute into the formula;

Workdone = Force * distance

362 = 55d

d = 362/55

d = 6.58m

Hence the student must push at a distance of 6.58m

3 0
3 years ago
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