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Angelina_Jolie [31]
3 years ago
9

a student moves a box across the floor by exerting 23.3 N of force and doing 47.2 J of work on the box. How far does the student

move the box?
Physics
1 answer:
kozerog [31]3 years ago
8 0
Work = Force x Distance
47.2J = 23.3N x d
d = 47.2/23.3
d = 2.0258 m

hope this helps :P
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A 2000kg suv accelerates from rest at a rate of 3.00m/s^2. The total amount of force resisting its motion 1500N. How much force
choli [55]

The total force that the SUV exerts is:

F = 2000 kg * 3 m/s^2

F = 6000 N

 

Since a resisting force amounting to 1500 N is exerted, then the force exerted by the SUV tires is:

F tire = 6000 N – 1500 N

F tire = 4500 N

7 0
3 years ago
Which BEST describes the physical properties of the Earth’s core?
Sonja [21]

C) a solid lower part and a liquid upper part

Explanation:

The physical nature of the earth's core is made up of a solid lower part and a liquid upper part.

The core is the innermost part of the earth and it is made up of metallic minerals.

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4 0
4 years ago
Please hurry
viktelen [127]

Answer:

The velocity will be "76.8 m/s".

Explanation:

The given values are:

Acceleration,

a = 2.4 m/s²

Time,

t = 32 seconds

By equation of motion,

⇒  v=u+at

On substituting the values, we get

⇒     =0+2.4\times 32

⇒     =0+76.8

⇒     =76.8 \ m/s

6 0
3 years ago
Question 4 How much time does it take to walk 8 km north at a velocity of 3.8 km/h?​
IrinaVladis [17]

Given parameters:

Displacement = 8km

Velocity  = 3.8km/h

Unknown:

time  = ?

Solution:

Velocity is displacement divided by time.

  Velocity  = \frac{displacement}{time}  

      Displacement  = velocity x time

Input the parameters:

              8  = 3.8  x time

 Time  = \frac{8}{3.8}   = 2.1s

The time taken is 2.1s

6 0
3 years ago
A train has a length of 81.1 m and starts from rest with a constant acceleration at time t = 0 s. At this instant, a car just re
arlik [135]

Answer: a) vcar= 7 m/s ; b) a train= 0.65 m/s^2

Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.

We have for the car

distance = v car* t, considering the length of train (81.1 m) travel by the car during the first 11.6 s

the v car =  distance/time= 81.1 m/11.6s= 7 m/s

In order to calculate the acceleration we have to use the kinematic equation for the train from the rest

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so distance train= (vcar*36.35)m=421 m

the a traiin= (2* 421 m)/(36s)^2=0.65 m/s^2

4 0
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