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Leviafan [203]
3 years ago
15

6. traction a. friction between a tire and the road.b. pushes a moving object out of a curve and into a straight linec. the abil

ity of a car to hold a straight lined. a road higher in the center than on either edge 7. kinetic energy a.a force that keeps objects moving in a straightb. reduces the effects of inertiac. energy of motiond. a road higher in the center than on either edge 8. force of impact a. a force that keeps objects moving in a straight line.b. a road higher in the center than on either edgec. the force with which a moving vehicle hits another objectd. perception distance, reaction distance, and braking distance
Physics
1 answer:
Gennadij [26K]3 years ago
4 0

Answer:

6. a. friction between a tire and the road

7. c. energy of motion

8. c. the force with which a moving vehicle hits another object

Explanation:

6. As a car moves along the road, the tires push back against the ground. As tires push back against the ground, the road exerts and opposing force to the motion of the tires. This opposing force is the friction between the tires and the road. <u>This opposing force between the tires and the rad is called traction.</u>

So, the answer is a

7. As an object moves, it has energy. <u>This energy due to its motion is called kinetic energy.</u>

So, the answer is c

8. When a moving vehicle hits another object, it exerts a force on the object. The process of the vehicle hitting the other object is called impact and the<u> force exerted on the object is called the force of impact. </u>

So, the answer is c.

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Question 1.

  • mass = 4500 kg
  • potential energy (p.e) = 67500 J

now, we know :

=》

p.e =  mgh

=》

67500 = 4500 \times 10 \times h

=》

67500 = 45000 \times h

=》

h =  \dfrac{67500}{45000}

=》

h = 1.5 \: m

note : if we take acceleration due to gravity as 9.8, then height = 1.53 m

Question 2.

  • mass = 4500 kg
  • kinetic energy = 63000 j

we know,

=》

k.e =  \dfrac{1}{2} mv {}^{2}

=》

63000 =  \dfrac{1}{2}  \times 4500 \times  {v}^{2}

=》

{v}^{2}  =  \dfrac{63000 \times 2}{4500}

=》

{v}^{2}  = 28

=》

v =  \sqrt{28}

=》

v = 2 \sqrt{7} \:  \:  ms {}^{ - 1}

or

=》

5.29 \:  \: ms {}^{ - 1}

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3 years ago
How much energy has 4×10^10m^3 of water collected in a reservoir at a hight of 100 m from the power house ?What kind of energy i
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Answer:

PE = 3.92x10^16J

potential energy

Explanation:

PE = m*g*h

mass of water = 1000kg/m³

(4*10^10m³)*1000kg = 4*10^13kg

PE = (4*10^13kg)*(9.81m/s²)*(100m)

PE = 3.92x10^16J

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Although there are different ways to approach a scientific investigation, all scientific investigations begin with some sort of
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A bullet with a mass of 0.02 kg is fired horizontally into a block of wood hanging on a string. The bullet sricks in the wood an
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u= 20.09 m/s

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