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MAVERICK [17]
3 years ago
13

2 Physics Questions WILL GIVE BRAINLIEST!!

Physics
1 answer:
In-s [12.5K]3 years ago
5 0

Answer:

a) The force of friction is 1.5 N.

b) The new force is 4/9 times as large as the original force.

Explanation:

It is given that the book is moving on the table.

∴ The friction acting on the block should be kinetic.

∴ Coeffecient of kinetic friction , k = 0.3

The force of friction = kN , where 'N' is the normal force acting on the surface where friction is acting.

In this case , N = mg , where 'm' is the mass of the book and 'g' is the acceleration due to gravity.

∴ Force of friction , f = kmg = 0.3×0.5×10 = 1.5N

b) We know that ,

Gravitational force , F ∝ m  , where 'm' is the mass of the body.

                                 F ∝ x^{2} where 'x' is the distance between the      two bodies.

Since mass of both the bodies is doubled , the force will be 4 times as large compared to the original force F from the first proportionality equation.

However , as the distance is 3 times as large , the force will be 9 times as small as compared to the original force F

∴ The new force is 4/9 times as large as the old force

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3 years ago
A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/
s344n2d4d5 [400]

Answer:

At t = (70 / 3) \; {\rm s} (approximately 23.3 \; {\rm s}.)

Explanation:

Note that the acceleration of the car between t = 0\; {\rm s} and t = 20\; {\rm s} (\Delta t = 20\; {\rm s}) is constant. Initial velocity of the car was v_{0} = 0\; {\rm m\cdot s^{-1}}, whereas v_{1} = 35\; {\rm m\cdot s^{-1}} at t = 20\; {\rm s}\!. Hence, at t = 20\; {\rm s}\!\!, this car would have travelled a distance of:

\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}.

At t = 20\; {\rm s}, the truck would have travelled a distance of x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}.

In other words, at t = 20\; {\rm s}, the truck was 400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m} ahead of the car. The velocity of the car is greater than that of the truck by 35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}. It would take another (50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s} before the car catches up with the truck.

Hence, the car would catch up with the truck at t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}.

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