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3 years ago

If a marble is released from a height of 10 meters, how long would it take for it to hit the ground?

Physics

2 answers:

azamat3 years ago

8 0

Time to reach ground = √2h/g = √2*10/9.8 = √20/9.8 = 1.43 second

Hope it helps!

Dvinal [7]3 years ago

3 0

It would take approximately 1.43 seconds

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A man wishes to lift a stone weighing 1440 N, using a first-class lever that measures 5 meters. What force should it perform if

Crazy boy [7]

Answer:

3360 N

Explanation:

In a first-class lever, the effort force and load force are on opposite sides of the fulcrum.

The lever is 5 m long. The load force is 1.50 m from the fulcrum, so the effort force must be 3.50 m from the fulcrum.

The torques are equal:

Fr = Fr

(1440 N) (3.5 m) = F (1.5 m)

F = 3360 N

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3 years ago

True or False: A change of 1 pH unit represents a tenfold change in the<br> acidity of the solution.

alisha [4.7K]

Answer:

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3 years ago

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Monica [59]

You are dealing with pulleys?

can be done with addition of the two equations to eliminate T.

$(2^{*}) \quad {} -m_1 g \cos \theta + T = m_1 a _2$
+
$(4^{*}) \quad 3m_1 g - T = 3m_1 a_2$
=

$(-m_1 g \cos\theta + 3m_1 g) + (T - T) = m_1 a _2 + 3m_1 a_2 \implies \\ \\
(-m_1 g \cos\theta + 3m_1 g) = 4m_1 a_2 \implies \\ \\
m_1(- g \cos\theta + 3g)= 4m_1 a_2$

we can cancel m₁ by dividing both sides by it, assuming mass is not zero

$(- g \cos\theta + 3g)= 4a_2 \implies \\ \\
a_2 = \dfrac{- g \cos\theta + 3g}{4} \\ \\
a_2 = \dfrac{- 9.80 \cos 60 + 3(9.80)}{4} \\ \\
a_2 = 6.125 \text{m/s}^2
$

a₂ = 6.125 m/s² ( do significant digits if you need to)

6 0

3 years ago

A pendulum is made of a small sphere of mass 0.250 kg attached to a lightweight string 1.20 m in length. As the pendulum swings

olchik [2.2K]

Answer:

0.833 N

Explanation:

Formula for Kinetic Energy $E_k = \frac{mv^2}{2}$

Formula for Potential Energy $E_p = mgy$

First we need to find the vertical distance between the maximum-angle position and the pendulum lowest point:

Using the swinging point as the reference, the vertical distance from the maximum-angle (34 degree) position to the swinging point is:

$L * cos(34^o) = 1.2cos(34^o) = 1.2*0.83 = 0.995 \approx 1 m$

At the lowest position, pendulum is at string length to the swinging point, which is 1.2 m. Therefore, the vertical distance between the maximum-angle position and the pendulum lowest point would be

y = 1.2 - 1 = 0.2 m.

As the pendulum is traveling from the maximum-angle position to the lowest point position, its potential energy would be converted to the kinetic energy.

By law of energy conservation:

$E_k = E_p$