Will the 79 kg skier in the figure below slide down if f the coefficient of static friction is 0.25?

Will The 79 Kg Skier In The Figure Below Slide Down If F The Coefficient Of Static Friction Is 0.25?

Answers

Answer 1

Answer:

Man will not slide down

Explanation:

Given:

Coefficient of static friction = 0.25

Angle = 13°

Computation:

Man will slide down if

tan13° > Coefficient of static friction

Tan 13 = 0.23

So,

0.23 < 0.25

So,

Man will not slide down


Related Questions

distace x time graph will be ................ if the body is in ununiform motion​

Answers

A non-straight ... possibly a zig-zag, wiggly, curvy, or snaking ... line. But always rising.

Answer:

Uniform: Straight line; Non uniform: Curved.

Explanation:

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Which of the following factors can affect the strength of electric or magnetic fields? (YOU MAY SELECT MORE THAN ONE ANSWER)

Strength of the magnet/electric charge

The material that is creating the field

The mass of the object creating the field

The distance from the source of the field

Answers

All of the above .........


C
D
7
The sun is the original source of
energy for many of our energy
resources
Which energy resource does not
originate from the sun? *
(1 Point)
.
A. Geothermal
B. Hydroelectric
C. Waves
D. Win

Answers

Answer:

geothermal

Explanation:

geothermal energy is the heat energy obtained from within the Earth. Hence not derived from Sun's energy.

Suppose that 2 J of work is needed to stretch a spring from its natural length of 24 cm to a length of 42 cm. (a) How much work is needed to stretch the spring from 32 cm to 34 cm

Answers

Answer:

Workdone = 0.025 Joules

Explanation:

Given the following data;

Workdone = 2J

Extension = 42 - 24 = 18 cm to meters = 18/100 = 0.18m

The workdone to stretch a string is given by the formula;

Workdone = ½ke²

Where;

k is the constant of elasticity.

e is the extension of the string.

We would solve for string constant, k;

2 = ½*k*0.18²

2 = ½*k*0.0324

Cross-multiplying, we have;

4 = 0.0324k

k = 4/0.0324

k = 123.46 N/m

a. To find the workdone when e = 32, 34.

Extension = 34 - 32 = 2 to meters = 2/100 = 0.02m

Workdone = ½*123.46*0.02²

Workdone = 61.73 * 0.0004

Workdone = 0.025 Joules

Therefore, the amount of work (in J) needed to stretch the spring from 32 cm to 34 cm is 0.67.

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