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romanna [79]
3 years ago
10

A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to acc

elerate across the ice. If drag and the friction from the ice apply a force of 5 N on the boy, what is the magnitude of the net force acting on him?
Physics
1 answer:
Bezzdna [24]3 years ago
7 0

Answer:

70 N

Explanation:

Draw a free body diagram of the boy.  There are four forces:

Weight force mg pulling down,

A 300 N normal force pushing up,

A 75 N applied force pulling right,

and a 5 N friction force pushing left.

The boy's acceleration in the y direction is 0, so the net force in the y direction is 0.

The net force in the x direction is 75 N − 5 N = 70 N.

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Can you sent a picture of the answers
5 0
3 years ago
An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in
Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

T {V}^{\gamma - 1} = \text{constant}

\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}

Given:-

{T}_{1} = 275 \; K  

{T}_{2} = T \left( \text{say} \right)

{V}_{1}  = V

{V}_{2} = 6V

\gamma = \cfrac{5}{3} \;    (the gas is monoatomic)

\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}

 

\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}  

T  =  275 \times 0.30

T  =  83 K.

3 0
3 years ago
Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the
balandron [24]

Answer:

a) Δx = 11.6 m

b) t = 3.9 s

Explanation:

a)

  • Since the snowmobile is moving at constant speed, and the drive force is 195 N, this means that thereis another force equal and opposite acting on it, according to Newton's 2nd Law, due to there is no acceleration present in the horizontal direction .
  • This force is just the force of kinetic friction, and is equal to -195 N (assuming the positive direction as the direction of the movement).
  • Once the drive force is shut off, the only force acting on the snowmobile remains the friction force.
  • According Newton's 2nd Law, this force is causing a negative acceleration (actually slowing down the snowmobile) that can be found as follows:

       a = \frac{F_{fr} }{m} = \frac{-195N}{128kg} = -1.5 m/s2 (1)

  • Assuming the friction force keeps constant, we can use the following kinematic equation in order to find the distance traveled under this acceleration before coming to an stop, as follows:

       v_{f} ^{2}  -v_{o} ^{2} = 2* a* \Delta x (2)

  • Taking into account that vf=0, replacing by the given (v₀) and a from (1), we can solve for Δx, as follows:

       \Delta x =- \frac{v_{o}^{2}}{2*a} =- \frac{(5.90m/s)^{2}}{2*(-1.5m/s2)} = 11.6 m (3)

b)

  • We can find the time needed to come to an stop, applying the definition of acceleration, as follows:

       v_{f} = v_{o} + a*\Delta t (4)

  • Since we have already said that the snowmobile comes to an stop, this means that vf = 0.
  • Replacing a and v₀ as we did in (3), we can solve for Δt as follows:

       \Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s   (5)

7 0
2 years ago
if the efficiency of an electric furnace is 96%, then 96% of the input energy is transformed into thermal energy. what is the ot
Nonamiya [84]
It is wasted, most likely as light, in this case, or it is lost during the transport of electricity.
5 0
3 years ago
A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s.
wlad13 [49]

Answer:

12m/s

Explanation:

v_f=v_o+at

Let's call the velocity that the car maintains for 10 seconds v_f_1, and the final velocity v_f_2.

v_f_1=0+(4)(5)=20m/s \\\\v_f_2=20+(-2)(4)=12m/s

Hope this helps!

5 0
3 years ago
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