A 25 N force at 60° is required to set a crate into motion on a floor. What is the value of the static friction?

1 answer:

3 0

The normal force of the force given is calculated through the equation,

Fn = F(sin θ)
where Fn is the normal force, F is the force, and θ is the angle.

Fn = (25 N)(sin 60°) = 21.65 N

The x-component of the force applied is,
Fx = (25 N)(cos 60°) = 12.5 N

The value of the coefficient of static friction is calculated through the equation,
F = μFn
μ = Fx / Fn = 12.5 N / 21.65 N = 0.577

You might be interested in

What is the speaker’s power output if the sound intensity level is 102 dBdB at a distance of 25 mm ? Express your answer to two

lesya [120]

Answer:

Power = 124.50 W

Explanation:

Given that:

The Sound intensity of a speaker output is 102 dB

and the distance r = 25 m

For the intensity of sound,

$\beta (dB)= 10 \ log_{10 } (\dfrac{I}{I_o})$

where;

the threshold of hearing $I_o = 10^{-12} (W/m^2)$

$\dfrac{102 }{10}= log_{10}( \dfrac{I}{10^{-12}})$

$10^{10.2} = \dfrac{I}{10^{-12}}$

$I = 10^{10.2} \times 10^{-12}$

I = 0.01585 W/m²

If we recall, we know remember that ;

Power = Intensity × A rea

Power = 0.01585 W/m² × 4 × 3.142 × (25 m)²

Power = 124.50 W

7 0

2 years ago

Increase in Space Suit Pressure 0.0/3.0 points (graded) If the pressure in a space suit increases, how will each of the followin

Art [367]

Answer:

Flexibility Increases

Pre-breathe time decreases

Mass of suit decreases.

Explanation:

Spacesuits are designed for space shuttles when a person goes to explore the galaxy. The spacesuits shuttle era are pressurized at 4.3 pounds per inch. The gas in the suit is 100% of oxygen and there is more oxygen to breathe when the altitude of 10,000 is reached. This will decrease the breathing time and mass of suit.

3 0

3 years ago

Chi walked 100 meters in 5 minutes. She took another 15 minutes to walk another 400 meters to reach her house. What is her avera

Law Incorporation [45]

Average speed is the ratio of total distance moved by Chi in total time interval

So here we will have

Total distance = 100 m + 400 m

$d = 500 m = 0.5 km$