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Nuetrik [128]
3 years ago
10

A 13.6 kg block is tied at the top of an incline to a tree. If the incline is 35.5 degrees and the coefficient of friction betwe

en the sled and the incline is .45, What is the tension force between the block and the tree
Physics
1 answer:
Gre4nikov [31]3 years ago
5 0

Answer:

Explanation:

ASSUMING that block = sled AND that the rope is parallel to the slope.

The force acting parallel due to the weight is

13.6(9.81)sin35.5 = 77.475 N

The maximum friction force is

(0.45)13.6(9.81)cos35.5 = 48.877 N

If rope tension is T

77.475 - 48.877 < T < 77.475 + 48.877

            28.6 N < T < 126 N

28.6 N will occur if the block is on the verge of sliding downhill

126 N will occur if the block is on the verge of sliding uphill

Could be any value between them.

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If the accepted value of a wave is 121 m/s, who has the most accurate method of measuring the speed of a wave?
qaws [65]
The answer would be erin out of all of them thank me later :)
5 0
3 years ago
Ask Your Teacher A basketball player shoots toward a basket 5.8 m away and 3.0 m above the floor. If the ball is released 1.7 m
const2013 [10]

Answer:

The answer to your question is    vo = 5.43 m/s

Explanation:

Data

distance = d= 5.8 m

height = 3 m

height 2 = 1.7 m

angle = 60°

vo = ?

g = 9.81 m/s²

Formula

              hmax = vo²sinФ/ 2g

Solve for vo²

              vo² = 2ghmax / sinФ

Substitution

              vo² = 2(9.81)(3 - 1.7) / 0.866

Simplification

              vo² = 19.62(1.3) / 0.866

              vo² = 25.51 / 0.866

              vo² = 29.45

Result

              vo = 5.43 m/s

               

5 0
3 years ago
An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.
Georgia [21]

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

The \ weight, W  =G\dfrac{M \times m}{R^{2}} = 800 \ N

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

W_1   =G\dfrac{\dfrac{M}{2}  \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2}  \times m \times 4}{ R ^{2}} = 2 \times G \times  \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

7 0
3 years ago
Two people push on the same door from opposite sides as shown. A man pushes on a door from the left. A woman pushes on the door
AURORKA [14]

Answer:

The door will move when the forces exerted by each individual becomes unbalanced.

Explanation:

The Door will only move when the forces exerted by each of the individual is unbalanced i.e. when the Torque is unbalanced

This is because there will only be a reaction or movement of an object  when the applied force > resistant force acting on the object.

example : An object at rest will only move when the applied force is  greater than the frictional forces acting on the object at rest.

8 0
3 years ago
A heavy ball with a weight of 150 N is hung from the ceiling of a lecture hall on a 4.1-m-long rope. The ball is pulled to one s
AlladinOne [14]

Answer:

The tension in the rope is 262.88 N

Explanation:

Given:

Weight W = 150 N

Length of rope r = 4.1 m

Initial speed of ball v = 5.5 \frac{m}{s}

For finding the tension in the rope,

First find the mass of rod,

mg = 150                          ( g = 9.8 \frac{m}{s^{2} } )

  m = \frac{150}{9.8}

  m = 15.3 kg

Tension in the rope is,

  T = mg + \frac{mv^{2} }{r}

  T = 150 + \frac{15.3 \times (5.5)^{2} }{4.1}

  T = 262.88 N

Therefore, the tension in the rope is 262.88 N

7 0
3 years ago
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