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meriva
3 years ago
8

Tired of being chased by a jaguar, you set a trap. Hoping to drop it on the jaguar, you try to push a

Physics
1 answer:
Alenkasestr [34]3 years ago
6 0

Answer: acceleration = 3.27m/s^2

Explanation:

Given that the

Mass M = 44kg

Angle Ø = 15 degree

Coefficient of friction ų = 0.7

Force F = 222N

F - Fr = ma ...... (1)

Where Fr = frictional force

Fr = ųN

N = normal reaction = mg

Fr = ųmgsinØ

Fr = 0.7 × 44 × 9.81 × sin 15

Fr = 78.2N

Substitutes Fr, F and M into equation one.

222 - 78.2 = 44a

143.79 = 44a

Make a the subject of formula

a = 143.79/44

Acceleration a = 3.27 m/s^2

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3 years ago
(a) what is the acceleration of two falling sky divers (mass 132 kg including parachute) when the upward force of air resistance
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As per the question the mass of two falling sky drivers is 132 kg.

First we have to calculate their acceleration.

Whenever a body falls freely under gravity,its acceleration is acceleration due to gravity i.e g whose value is 9.8 m/s^2.

The earth pulls the object with a force equal to the weight of the body.

Hence the force gravity  F=W= mg   [ here m is mass of the body]

Here m =132 kg.

Hence force of gravity F= mg

                                        =132 kg ×9.8 m/s^2

                                        =1293.6 kg m/s^2

                                         =1293.6 N      [ here N[newton] is the unit of force.]

As per the question the air resistance is one fourth of weight of the bodies.

Hence air resistance F' =1/4 mg

                                       =\frac{1}{4} *1293.6N

                                        =323.4 N

Here F acts in vertically downward direction while F' acts in vertically upward direction.

Hence the net force acting on the particle is F-F'.

                                                 F_{net} =1293.6N -323.4N

                                                         =970.2 N

From Newton's second law of motion we know that net force is the product of mass and acceleration i.e  

                                   F_{net} =ma  [Here a is the acceleration]

                                             a =\frac{F_{net} }{m}

                                                  = \frac{970.2}{132} m/s^2

                                                  =7.35 m/s^2

In the second question it has been told that they descend with uniform speed.hence acceleration of the two bodies will be zero.

 we know that F= ma

                           =m×0

                            =0 N

Hence they will not get any force when they will descend with a uniform speed.


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