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Pavel [41]
3 years ago
14

When an object is moving with uniform circular motion, the centripetal acceleration of the object a. is circular. b. is perpendi

cular to the plane of motion. c. is zero. d. is directed toward the center of motion. Please select the best answer from the choices provided A B C D
Physics
1 answer:
never [62]3 years ago
5 0
Hi Pupil Here's Your answer :::


➡➡➡➡➡➡➡➡➡➡➡➡➡


When an object is moving with uniform circular motion, the centripetal acceleration of the object \textbf{is\:Zero}.


Because the body is moving with a uniform velocity and hence, there might not be any acceleration due to it.


⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅



Hope this helps .......
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gravity

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A rollercoaster accelerates from 10 m/s to 100 m/s2 for 25 seconds. What is the acceleration?
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A roller coasters accelerates from an initial velocity of of 6.0 m/s to a final velocity of 70 m/s over 4 seconds. What's the acceleration? Q. Acceleration only takes place when things speed up. Q. A drag racer accelerated from 0 m/s to 200 m/s in 5 s.

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2 years ago
A glass rod and a steel rod are of equal length at 0C. At 100C they differ in length by
NeX [460]

The given lengths at 0 °C are 2.5 m

Let l₀ be the given lengths of the glass and steel rods at 0 °C. Let l and l' be the lengths of the glass and steel rods at 100 °C respectively.

From our expression for linear expansivity,

l = l₀ + l₀αΔθ where α = linear expansivity of glass = 0.000008/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

Also,

l' = l₀ + l₀α'Δθ where α' = linear expansivity of steel = 0.000012/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

Since the difference in their lengths at 100 °C = 0.001 m, we have that

l - l' = l₀ + l₀αΔθ - (l₀ + l₀α'Δθ)

l - l' = l₀ + l₀αΔθ - l₀ - l₀α'Δθ)

l - l' = l₀αΔθ - l₀α'Δθ

l - l' = l₀(α- α')Δθ

Making l₀ subject of the formula, we have

l₀ = (l - l')/[(α- α')Δθ]

Substituting the values of the variables into the equation, we have

l₀ = (l - l')/[(α- α')Δθ]

l₀ = 0.001 m/[(0.000008/°C - 0.000012/°C)100 °C.]

l₀ = 0.001 m/[(-0.000004/°C)100 °C.]

l₀ = 0.001 m/-0.0004

l₀ = -2.5 m

Neglecting the negative sign,

l₀ = 2.5 m

So, the given lengths at 0 °C are 2.5 m

Learn more about linear expansion here:

brainly.com/question/14089545

6 0
2 years ago
A skier starts from rest at the top of a hill that is inclined 10.5° with respect to the horizontal. The hillside is 200 m long,
Svetlanka [38]

Answer:

d) 289.31 m

Explanation:

Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m

Energy used by friction = μmgcos 10.5 x 200 = .075 x m x 9.8 x cos 10.5 x200 = 144.54 m .

Energy used by friction on plain surface = μmg x d.( dis distance covered on plain ) =.075x m x 9.8 xd = .735 m d

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4 0
3 years ago
How do I calculate the tension in the horizontal string?
matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

T_1\cos 30\degree-m\cdot g=0

Solve for T₁,

T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N

Now, we use the second equation to find the tension in the horizontal string,

T_2-T_1\sin 30\degree=0

Solve for T₂,

T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

8 0
1 year ago
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