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finlep [7]
4 years ago
9

Astronauts on a distant planet set up a simple pendulum of length 1.20 m. The pendulum executes simple harmonic motion and makes

100 complete oscillations in 450 s. What is the magnitude of the acceleration due to gravity on this planet?
Physics
1 answer:
Ray Of Light [21]4 years ago
4 0

Answer:

Magnitude of the acceleration due to gravity on the planet = 2.34 m/s²

Explanation:

Time period of simple pendulum is given by

         T=2\pi\sqrt{\frac{l}{g}}, l is the length of pendulum, g is acceleration due to gravity value.

We can solve acceleration due to gravity as

            g=\frac{4\pi^2l}{T^2}

Here

  Length of pendulum = 1.20 m

  Pendulum executes simple harmonic motion and makes 100 complete oscillations in 450 s.

  Period, T=\frac{450}{100}=4.5s

Substituting

         g=\frac{4\pi^2\times 1.2}{4.5^2}=2.34m/s^2

Magnitude of the acceleration due to gravity on the planet = 2.34 m/s²

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Answer:

S=48.29 m

Explanation:

Given that the height of the hill h = 2.9 m

Coefficient of kinetic friction between his sled and the snow μ = 0.08

Let u be the speed of the skier at the bottom of the hill.

By applying conservation of energy at the top and bottom of the inclined plane we get.

Potential Energy=kinetic Energy

mgh = (1/2) mu²

u² = 2gh

u²=2(9.81)(2.9)

   =56.89

u=7.54 m/s

a = - f / m

a = - μ*m*g / m

a = - μg

From equation of motion

v²- u² = 2 -μ g S

v=0 m/s

-(7.54)²=-2(0.06)(9.81)S

S=48.29 m

3 0
3 years ago
3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a
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Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as \mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}.

Where, {V}_{i} is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

{V}_{i} =3 m/s, V = 0 m/s, and  ∆s = 2 m

Substitute the values in the above formula,

0=3^{2}-2 \times 2 \times a

0 = 9 - 4a

4a = 9

a=2.25 \mathrm{m} / \mathrm{s}^{2}

a=2.25 \mathrm{m} / \mathrm{s}^{2} is the acceleration.

Calculating Coefficient of friction:

\mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}

Compare the above equation

\mu \times m \times g=m \times a

Cancel "m" common term in both L.H.S and R.H.S

\text { Equation becomes, } \mu \times g=a

\text { Coefficient of friction } \mu=\frac{a}{g}

\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}

\mu=\frac{2.25}{9.8}

\mu=0.229

Hence coefficient of friction is 0.229.

calculating force:

\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

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3 years ago
A baseball player friend of yours wants to determine his pitching speed. you have him stand on a ledge and throw the ball horizo
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Answer:

The pitching speed of your friend is 33.20 m/s

Explanation:

<em>Lets explain how to solve the problem</em>

Your friend throw the ball horizontally that means the vertical initial

component of velocity is zero (u_{y}=0).

The ball is thrown from a height 4 meters above the ground.

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<u><em>Remember:</em></u> the height is negative value because its below the point of

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Then the time of the ball to land on the ground is 0.9035 seconds

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The range R=u_{x}*t, where u_{x} is the horizontal

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⇒ 30=u_{x}(0.9035) (divide both sides by 0.9035)

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<em>The pitching speed of your friend is 33.20 m/s </em>

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