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OLEGan [10]
3 years ago
10

a 12000.0-kg car is traveling at 19 m/s. the driver suddenly slams on the brakes and skids to a stop. The coefficient of kinetic

friction between the tires and the road is 0.80. for what length of time is the car skidding?
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
6 0
What I don’t get it ?
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Which of the following is NOT usually published with a scientific report?
dusya [7]

C cost (I need points)

5 0
3 years ago
I'll mark brainliest
irga5000 [103]

Answer:

A) 35 ft

B) 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

Explanation:

A) Total distance covered by the dog = 20 + 15

                                  = 35 ft

B) Since the other distance covered by the dog before chewing the stick, after the retrieval, was in an opposite direction to the initial direction, then;

total displacement of the dog = 20 - 15

                                  = 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick + distance covered before the dog starts chewing the stick

But, displacement involves a specified direction. The distance covered before the dog starts chewing the stick was in an opposite direction to the initial direction.

Thus,

Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

7 0
3 years ago
You exert of force of 75 newtons on a rock. You push and you push, but you can't budge it. You are exhausted! How much work did
sergiy2304 [10]
You performed 0 work for the fact that work means the distance of movement made on an object not the amount of force it is exposed to. 0 work because it didn't move
7 0
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20 Points available for physics help
Sonja [21]
The second one is correct not sure about the first one sorry
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A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force of friction on the cart
Ber [7]
Newton's second law states that the resultant of the forces applied to an object is equal to the product between the object's mass and its acceleration:
\sum F = ma
where in our problem, m is the mass the (child+cart) and a is the acceleration of the system.

We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force F_f acting in the opposite direction. So Newton's second law can be rewritten as
F-F_a = ma
or
F=ma + F_f

since the frictional force is 15 N and we want to achieve an acceleration of a=1.50 m/s^2, we can substitute these values to find what is the force the man needs:
F=(30 kg)(1.5 m/s^2)+15 N=60 N
8 0
3 years ago
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