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3 years ago

How do I find frictional force with force applied?

Physics

1 answer:

lions [1.4K]3 years ago

8 0

Answer: Choose the normal force acting between the object and the ground. Let's assume a normal force of 250 N.

Determine the friction coefficient.

Multiply these values by each other: 250 N * 0.13 = 32.5 N .

You just found the force of friction!

Explanation:

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Calculate the change in velocity of a 0.070kg tennis ball hit by Serena with a force of 140 N over 0.020 s

Brilliant_brown [7]

V=at and a=F/m

140/.070 = 2000m/s^2

2000*.020 = 40m/s

The ball’s velocity increased by 40m/s.

6 0

3 years ago

if vector u has lenght 70 and direction 40 degrees, and vector v has length 85 and direction 335 degrees what is the length and

Anastaziya [24]

Answer:

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

Explanation:

||u|| = 70

θ = 40°

$\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62$

$\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99$

||v|| = 85

θ = 335°

$\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03$

$\vec{v}_y=||v||sin\theta \\\Rightarrow \vec{v}_y=85sin335=-35.92$

Resultant

$R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15$

$\theta=tan^{-1}\frac{R_y}{R_x}\\\Rightarrow \theta=tan^{-1}\frac{70sin40+85sin335}{70cos40+85cos335}\\\Rightarrow \theta=tan^{-1}0.069=3.97^{\circ}$

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

4 0

3 years ago

How many air molecules are in a 13.0×12.0×10.0 ft room (28.2 L=1 ft3)? Assume atmospheric pressure of 1.00 atm, a room temperatu

aksik [14]

Answer:

1963.93 Moles

Explanation:

-We know the standard conversion ratio for the volume of a mole is $1 Mole=22.4L$

Given volume of rooms as $13.0ft\times12.0ft\times10ft=1560 ft^3$

Convert the volume into liters: $1560\times28.2l=43992l$

#From our conversion ratio above, we get the volume of air molecules in moles as:

$V_m=\frac{43992L}{22.4L}\times 1\ mole\\\\=1963.93\ Moles$

Hence, the volume of air molecules is 1963.93 Moles

4 0

3 years ago

A crate is being lifted into a truck. If it is moved with a 2470 N force and 4850 J of work is done, then how far is the crate b

olchik [2.2K]

Answer:1.96m

Explanation:

Work=4850J

Force=2470N

Distance =work ➗ force

Distance =4850 ➗ 2470

Distance =1.96m

7 0

3 years ago

what is the force of friction between an 80kg box and the ground on Earth if the coefficient is 0.2?

Reil [10]

Answer:

160N

Explanation: When 80kg mass is one group . It's reaction force acting on a ground.

Weight of the object = 80*10

= 800 N

Here we are given cofficient of static friction its 0.2. It should be smaller than 1

Friction force = Reaction * Friction Cofficient

Reaction = 800N ( Considering Vertical Equilibrium )

F = 800* 0.2

F = 160N

3 0

3 years ago

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