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melisa1 [442]
2 years ago
15

Derive an expression for the block's centripetal acceleration ac in terms of m , θ , and physical constants, as appropriate

Physics
1 answer:
maxonik [38]2 years ago
7 0

The expression for the block's centripetal acceleration is derived as ω²r or v²/r.

<h3>What is centripetal acceleration?</h3>

The centripetal acceleration of an object is the inward or radial acceleration of an object moving in a circular path.

The expression for the block's centripetal acceleration is derived as follows;

ω = dθ/dt

where;

  • ω is the angular speed
  • θ is the angular displacement
  • t is the time of motion

ac = ω²r

where;

  • r is the radius of the circular path

Also, ω = v/r

ac = (v/r)²r

ac = v²/r

Thus, the expression for the block's centripetal acceleration is derived as ω²r or v²/r.

Learn more about centripetal acceleration here: brainly.com/question/79801

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Answer:

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Explanation:

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3 0
4 years ago
A car is travelling at a speed of 30 m/s. It decelerates at a rate of 12 m/s^2. Calculate the time taken for the car to come to
user100 [1]

Answer:

The answer to your question is: t = 2.5 s

Explanation:

Data

vo = 30 m/s

a = -12 m/s2

t = ?

vf = 0 m/s

Formula

vf = vo + at

Substitution

                              0 m/s = 30 + (-12)t

Solve it for t           -30 = -12t

                              t = -30 / -12

                              t =  30/12  = 15/6 = 5/2

                              t = 2.5 s                              

8 0
3 years ago
Read 2 more answers
a parent swings a 18.5 kg child in a circle of radius 1.05m, making 5 revolutions in 13.4s. what centripetal acceleration does t
Zolol [24]

Answer: 0.146 m/s^{2}

Explanation:

The <u>centripetal acceleration</u> a_{c} of an object moving in a uniform circular path is given by the following equation:

a_{c}=\frac{V^{2}}{r}  (1)

Where:

V is the tangential velocity

r=1.05 m is the radius of the circle

On the other hand, the tangential velocity  is expressed as:

V=\omega r (2)

Where \omega is the angular velocity, which can be found knowing the child makes 5 revolutions in 13.4s:

\omega=\frac{5 rev}{13.4 s}=0.37 rev/s (3)

Substituting (3) in (2):

V=(0.37 rev/s)(1.05 m) (4)

V=0.39 m/s (5)

Substituting (5) in (1):

a_{c}=\frac{(0.39 m/s)^{2}}{1.05 m}  (6)

Finally:

a_{c}=0.146 m/s^{2}  

6 0
3 years ago
Read 2 more answers
You have just landed on Planet X. You release a 100-g ball from rest from a height of 10.0 m and measure that it takes 3.40 s to
Nana76 [90]

Answer:

w = 0.173 N

Explanation:

The weigh of any object is computed by multiplying its mass to the acceleration of gravity, so we need to find the gravity on that planet in order to compute the weigh we want.

The ball has a mass of 0.1 kg and its released from a height of 10 m, therefore it is in a free fall motion with gravity acting as a constant acceleration on the body, we can use the equations for free fall movement in order to determine the value for this acceleration:

y(t) = v_0 * t + y_0 - 0.5 * g * t^2

y(t) is the position in the end of the movement, when t = 3.4 s, so y(t) = 0 m.

v_0 is the initial velocity, in this case v_0 = 0 m/s.

y_0 is the initial position of the ball, in this case it is 10 m.

g is the gravity that we want to know.

Applying these values in the equation we have:

0 = 0*(3.4) + 10 - 0.5*g*(3.4)^2

0 = 10 - 0.5*11.56*g

0 = 10 -5.78*g

5.78*g = 10

g = 1.73 m/s^2

Then we can use this value to find out the weigh of the ball in that planet:

w = g*m = 0.1*1.73 = 0.173 N

6 0
4 years ago
If f(x) = -x2+x-1 then value of f(f(2)) is
qaws [65]

Answer:

<u>-13</u>

Explanation:

<u>Function</u>

  • f(x) = -x² + x - 1

<u>Solving</u>

  • Substitute 2 in place of x
  • f(2) = -(2)² + 2 - 1
  • f(2) = -4 + 2 - 1
  • f(2) = -3
  • f(f(2)) = f(-3)
  • f(-3) = -(-3)² - 3 - 1
  • f(-3) = -9 - 3 - 1
  • f(f(2)) = <u>-13</u>
6 0
2 years ago
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