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Julli [10]
3 years ago
13

A solid cylinder of mass 20kg rolls without slipping down a 30° slope. Find the acceleration and the frictional force needed to

prevent slipping. (g=9.8m/s^2)​

Physics
1 answer:
Alex_Xolod [135]3 years ago
6 0

Answer:

a = 3.27 m/s²

F = 32.7 N

Explanation:

Draw a free body diagram.  There are three forces:

Weight force mg pulling straight down.

Normal force N pushing perpendicular to the slope.

Friction force F pushing parallel up the slope.

Sum of forces in the parallel direction:

∑F = ma

mg sin θ − F = ma

Sum of torques about the cylinder's axis:

∑τ = Iα

Fr = ½ mr²α

F = ½ mrα

Since the cylinder rolls without slipping, a = αr.  Substituting:

F = ½ ma

Two equations, two unknowns (a and F).  Substituting the second equation into the first:

mg sin θ − ½ ma = ma

Multiply both sides by 2/m:

2g sin θ − a = 2a

Solve for a:

2g sin θ = 3a

a = ⅔ g sin θ

a = ⅔ (9.8 m/s²) (sin 30°)

a = 3.27 m/s²

Solving for F:

F = ½ ma

F = ½ (20 kg) (3.27 m/s²)

F = 32.7 N

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svetoff [14.1K]

Answer:Zero

Explanation:

Given

mass of ball m=10\ kg

If the ball is thrown upward then at maximum point velocity of ball is zero because ball is no longer able to move upward

Momentum(P) of a particle is given by

P=mass\times velocity

P=10\times 0

P=0

Therefore at the highest point momentum is zero .

8 0
4 years ago
Suppose you're performing experiments in science class in which you start with 70 bacteria and the amount of bacteria triples ev
barxatty [35]
1h----------------> 70x3=210 bacteria
2h-----------------> 210*3=630 bactaeria
let be y the number of bacteria at the t=0h
it is y=70 3^0
for t= 1h
y=70*3^1=210
for t=2h
y=70*3^2=630

so we can write y=70*3^x, where x is the number of hour


5 0
4 years ago
Read 2 more answers
This physics problem is kinda complicated to my small brains...
Veronika [31]

Answer:

\huge\boxed{\sf P.E. = 14700\ Joules}

Explanation:

This work done is in the form of Potential Energy so we'll use the formula of Potential Energy.

<u>Given Data:</u>

Mass = m = 250 kg

Acceleration due to gravity = g = 9.8 m/s²

Height = h = 6 m

<u>Required:</u>

Work Done in the form of Potential Energy = P.E. = ?

<u>Formula:</u>

P.E. = mgh

<u>Solution:</u>

P.E. = (250)(9.8)(6)

P.E. = 14700 Joules

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
4 0
3 years ago
A 1.4-kg block slides freely across a rough surface such that the block slows down with an acceleration of â1.25 m/s2. what is t
Marianna [84]

Mass of the block = 1.4 kg

Weight of the block = mg = 1.4 × 9.8 = 13.72 N

Normal force from the surface (N) = 13.72 N

Acceleration = 1.25 m/s^2

Let the coefficient of kinetic friction be μ

Friction force = μN

F(net) = ma

μmg = ma

μg = a

μ = \frac{a}{g}

μ = \frac{1.25}{9.8}

μ = 0.1275

Hence, the coefficient of kinetic friction is: μ = 0.1275

6 0
4 years ago
A 7-kg bowling ball moving at 4 m/s strikes a 1 kg bowling pin. If the ball slows to 2 m/s in 0.05 s, how much force does it exe
iren [92.7K]

Answer:

280 N

Explanation:

acceleration = v2-v1 / time taken = (2-4 )/ 0.05 = -40 m/s^2  ( neg sign indicates slowing down )

force exerted = ma = 7 kg x -40 m/s^2 = - 280 N ( neg sign means opposite direction of initial velocity )

since the 7 kg ball is slowing down, the  direction of force will be opposite of the initial velocity , and it will be 280 N

7 0
3 years ago
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