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slava [35]
3 years ago
10

A student exerts a force of 35 newtons to move a cart loaded with lab equipment to the prep room 20 meters. If it takes the stud

ent 10 seconds to push the cart to the prep room, how much power did the student use?
a) 700 W
b) 2.5 Nm
c) 62.5 J/s
d) 70 W
Physics
1 answer:
hram777 [196]3 years ago
8 0

Answer:

option D

Explanation:

Power = work divided by time

and Work is equal to force multiplied by displacement

therefore power =\frac{35 \times 20}{10}  = 70

please mark me brainliest and 5 star

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arlik [135]

Answer: Input Resistance (Rin) = 500 ohms , overall voltage Gain (GV) = 2 volts / volts , ID = 4/9 = 0.44 A

Explanation:

5 0
3 years ago
A 1400kg car moving westward with a velocity of 15m/s collides with a utility pole and is brought to rest in 0.30sec. find the f
sergejj [24]
Force=(mass*velocity)-(mass*velocity)/time

force=0-(-15*1400)/0.30

force=70000N



8 0
3 years ago
Read 2 more answers
A horse is grazing for food. It trots rightward 15 m to eat a carrot, then walks rightward 20 m, to another carrot, then turns l
mafiozo [28]

Answer:

<h3>0.42s</h3>

Explanation:

Velocity = Displacement/time

Displacement of the horse is the distance covered in a specified direction

Total distance of the horse towards the right = 15m + 20m = 35m

Total distance of the horse towards the left = 4m

Displacement = Distance towards the right - Distance towards the left

Displacement = 35m-4m

Displacement =  31m

Time taken = 74seconds

substitute the values gotten into the velocity formula

average velocity = 31m/74s

average velocity = 0.42m/s

Hence the magnitude of its average velocity is 0.42s

6 0
4 years ago
Hi please zoom in to see it clearly, uh you don’t have to answer them all but it would be nice !!! (no links please) :D
kap26 [50]
12.could be D
13.A
14.D
15.C
16.A
3 0
3 years ago
A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a
sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0
3 years ago
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