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yarga [219]
3 years ago
11

What type of acceleration does an object moving in a circular path with constant speed experience?

Physics
2 answers:
Marysya12 [62]3 years ago
7 0

An object moving in a circular path has centripetal acceleration. <em>(A)</em>

AnnyKZ [126]3 years ago
5 0
An object moving in a circular path with constant speed experiences centripetal acceleration.

Centripetal Acceleration- “the rate of change of tangential velocity”

“The direction of the centripetal acceleration is always inwards along the radius vector of circular motion.”
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Consider a block of mass m at rest on an inclined plane of angle theta. An acrobat of mass m_A is standing on the top corner of
shtirl [24]

Answer:

μ = tan θ

Explanation:

For this exercise let's use the translational equilibrium condition.

Let's set a datum with the x axis parallel to the plane and the y axis perpendicular to the plane.

Let's break down the weight of the block

         sin θ = Wₓ / W

         cos θ = W_y / W

         Wₓ = W sin θ

         W_y = W cos θ

The acrobat is vertically so his weight decomposition is

         sin θ = = wₐₓ / wₐ

         cos θ = wₐ_y / wₐ

         wₐₓ = wₐ  sin θ

         wₐ_y = wₐ cos θ

let's write the equilibrium equations

Y axis  

     N- W_y - wₐ_y = 0

     N = W cos θ + wₐ cos θ

X axis

        Wₓ + wₐ_x - fr = 0

         fr = W sin θ + wₐ sin θ

the friction force has the formula

         fr = μ N

         fr = μ (W cos θ + wₐ cos θ)

we substitute

         μ (Mg cos θ + mg cos θ) = Mgsin θ + mg sin θ

         μ = \frac{(M +m) \ sin \  \theta  }{(M +m) \ cos  \ \theta }

 

         μ = tan θ

this is the minimum value of the coefficient of static friction for which the system is in equilibrium.

8 0
3 years ago
A person has a gravitational force (weight) of 600 N on Earth. Suppose the mass of the Earth is double and the radius shrinks to
Alenkasestr [34]

Explanation:

that's impossible,the radius of the earth can't decrease when the mass doubles!

4 0
3 years ago
One billiard ball is shot east at 2.2m/s. A second, identical billiard ball is shot west at 1.2m/s. The balls have a glancing co
Kruka [31]

Answer:

v_{1}=1.886 \frac{m}{s}

β= 57.99 south of east

Explanation:

v_{1}=2.2 \frac{m}{s} \\v_{2}=1.2 \frac{m}{s} \\m_{1}=m_{2}=m\\v_{fx}=1.6 \frac{m}{s} \\v_{fy}=?

Velocity in axis x the two balls come one from east and west

m_{1}*v_{1x}+m_{2}*v_{2x}=m_{1}*v_{fx1}+m_{2}*v_{fx2}\\m*(v_{1x}+v_{2x})=m*(v_{fx1}+v_{fx2})\\v_{fx2}=0\\v_{1x}+v_{2}=v_{f1}+0\\v_{fx1}=2.2 \frac{m}{s}+(1.2\frac{m}{s})\\  v_{fx1}=1 \frac{m}{s} \\

Velocity in axis y initial is zero so:

v_{y1}+v_{y2}=v_{y1f}+v_{y2f}\\v_{y1}=0\\v_{y2}=0\\v_{y1f}+v_{y2f}=0\\v_{y1f}=-v_{y2f}\\v_{y2f}=1.6\frac{m}{s}

v=\sqrt{v_{1fx}^{2}+v_{1fy}^{2}}\\ v=\sqrt{1^{2}+1.6^{2}}\\v=1.886 \frac{m}{s}

Angle is find using:

tan(β)=\frac{v_{fy}}{v_{fx}}

\beta =tan^{-1}*\frac{1.6}{1}=57.99

5 0
4 years ago
Please help me!!!!!!!!!!!!!!!​
qaws [65]

Answer:

3

Explanation:

i did it

3 0
3 years ago
What is the largest component of M1?
mariarad [96]

The largest component of M1 is- Currency.

8 0
3 years ago
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