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3 years ago

A climber pulls down on a rope causing his body to rise up with the rope? Which law of motion is it?

1 answer:

4 0

This would be called the law of action-reaction. This states that every action will have an equal and opposite reaction. The action in the example is pulling down on the rope. The opposite and equal reaction is the climers body moving upward. The same law can be applied to a rocket. The action is the engines pushing down and the reaction is the rocket going up. :D

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what will be the focal lenght of a combined lens made by contact of two lenses of power +3D and -2D.

4vir4ik [10]

P_1=+3D
P_2=-2D

$\\ \bull\tt\dashrightarrow P=P_1+P_2$

$\\ \bull\tt\dashrightarrow P=+3D-2D=+1D$

Now

$\\ \bull\tt\dashrightarrow f=\dfrac{1}{P}$

$\\ \bull\tt\dashrightarrow f=\dfrac{1}{1}$

$\\ \bull\tt\dashrightarrow f=1m$

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2 years ago

What inertia is present in a stretched rubber?

iVinArrow [24]

Answer:

You input potential energy into the rubber band system when you stretched the rubber band back. Because it is an elastic system, this is kind of potential energy is specifically called elastic potential energy.

7 0

4 years ago

A balloonist drops his camera from a height of 100 m while his balloon is ascending at 5 m/s. How

ExtremeBDS [4]

20 i think is the answer

6 0

3 years ago

Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E

Kay [80]

Answer:

The true course: $40.29^\circ$ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

$V_w$ = velocity of wind = $30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h$

$V_p$ = velocity of plane in still air = $100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h$

Assume:

$V_r$ = resultant velocity of the plane

$\theta$ = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

$\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h$

Let us find the direction of this resultant velocity with respect to east direction:

$\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ$

This means the the true course of the plane is in the direction of $40.29^\circ$ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

$\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h$

Hence, the ground speed of the plane is 96.68 km/h.

5 0

3 years ago

Calculate the force at sea level that a boy of mass 50 kg exerts on a chair in which he is sitting.

kogti [31]

As per the question the mass of the boy is 50 kg.

The boy sits on a chair.

We are asked to calculate the force exerted by the boy on the chair at sea level.

The force exerted by boy on the chair while sitting on it is nothing else except the force of gravity of earth i.e the weight of the body .The direction of that force is vertically downward.

At sea level the acceleration due to gravity g = 9.8 m/s^2

Hence the weight of the boy $w = mg$ [m is the mass of the body]

we have m = 50 kg.

Hence w = 50 kg ×9.8 m/s^2

=490 N kg m/s^2

= 490 N

Here newton [N] is the unit of force.

6 0

3 years ago

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