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maxonik [38]
3 years ago
9

Can anyone help in physics plz

Physics
1 answer:
Brrunno [24]3 years ago
3 0

Answer:

The answer is B. 8000.

Explanation:

She gained 10,000J , but when she reached the bottom she was at 2,000J. All you need to do is 10,000-2,000 and you get 8,000. So the answer is 8,000J.

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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
4 years ago
you slide across home plate during baseball game. If you have a mass of 82 kg, and the coefficient of kinetic friction between y
Tju [1.3M]

m = mass of the person = 82 kg

g = acceleration due to gravity acting on the person = 9.8 m/s²

F = normal force by the surface on the person

f = kinetic frictional force acting on the person by the surface

μ = Coefficient of kinetic friction = 0.45

The normal force by the surface in upward direction balances the weight of the person in down direction , hence

F = mg                                          eq-1

kinetic frictional force on the person acting is given as

f = μ F

using eq-1

f = μ mg

inserting the values

f = (0.45) (82) (9.8)

f = 361.6 N

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Answer:

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Answer:

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