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gregori [183]
3 years ago
15

A ball with an initial velocity of 5 m/s rolls off a horizontal table with a height of 1.0

Physics
1 answer:
Dmitry [639]3 years ago
8 0

Answer:

Approximately 0.45\; \rm s, assuming that air resistance is negligible and that g = 9.81\; \rm m\cdot s^{-2}.

Explanation:

The ball starts to fall the moment it rolls of the edge of the table.

  • Let h_{0} denote the initially height of this ball.
  • Let v_{0} denote the initial vertical velocity of this ball.

Assume that gravity is the only force acting on the ball during its fall (that is, there's no air resistance to slow the ball down.) The vertical acceleration of this ball during the fall would be constantly equal to (-g) (negative because the ball is accelerating downwards.)

The following SUVAT equation would give the height h of this ball at time t:

\displaystyle h = \frac{1}{2}\, (-g)\cdot t^{2} + v_{0} \cdot t + h_{0}.

Since the table is horizontal, the vertical velocity of this ball would be 0 the moment it rolls of the edge. In other words: v_{0} = 0\; \rm m\cdot s^{-1}.

The initial height of this ball when it rolls of the table is h_{0} = 1.0\; \rm m, same as the height of the table.

Hence, the height h of this ball at time t would be:

\begin{aligned}h &= \frac{1}{2}\, (-g)\cdot t^{2} + v_{0} \cdot t + h_{0} \\ &= \frac{1}{2}\, (-g) \cdot t^{2} + h_{0}\end{aligned}.

At t = 0\; \rm s, the height of this ball would be 1.0\; \rm m. The ball would be on the ground by the time h = 0\; \rm m, Set the right-hand side of this equation to 0 and solve for the time t at which the ball is on the ground:

\displaystyle \frac{1}{2}\, (-g) \cdot t^{2} + h_{0} = 0.

\displaystyle \frac{1}{2}\, (-g) \cdot t^{2} = - h_{0}.

\displaystyle t = \sqrt{\frac{2\, h_{0}}{g}}.

Substitute in the values h_{0} = 1.0\; \rm m and g = 9.81\; \rm m\cdot s^{-2}:

\begin{aligned} t &= \sqrt{\frac{2\, h_{0}}{g}} \\ &= \sqrt{\frac{1.0\; \rm m}{9.81\; \rm m\cdot s^{-2}}} \\ &\approx0.45\; \rm s\end{aligned}.

In other words, the ball would be in the air for approximately 0.45\; \rm s. (The initial horizontal velocity of this ball does not affect the duration of this fall.)

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3 years ago
You are making a telephone out of two aluminum cans and some string. You can choose between two types of string: a 2-m length of
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Answer:

C)You should use the thin cooking twine.

Explanation:

A)You can choose either because they are the same length and will produce the same wave speed.

B)You should use the heavy rope.

C)You should use the thin cooking twine.

The speed of wave in a string is given by the following formula:

|v| = \sqrt{\frac{F_T}{u} }

Where |v| = speed of wave, F_T = tension in the string, and μ = mass per length of the string.

<em>Even though the two strings have the same length, the μ (mass/length) for the heavy rope will be more than the that of a thin rope. Consequently, the </em>F_T<em>:μ for the thin rope will be higher than that of the heavy rope and as such, gives a bigger |</em>v<em>|. </em>

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The correct option is C.

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3 years ago
What is the necessary conditions for the production of sound?
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Answer: Something that's vibrating, and you also need medium for those vibrations to start in.

I hope this helped!

<!> Brainliest is appreciated! <!>

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Sanjay is making plans for his Saturday afternoon. He can choose a health-enhancing activity like lifting weights, or he can cho
Bezzdna [24]

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A 0.9 µF capacitor is charged to a potential difference of 10.0 V. The wires connecting the capacitor to the battery are then di
jeyben [28]

Answer:

3.6μF

Explanation:

The charge on the capacitor is defined by the formula

q = CV

because the charge will be conserved

q₁ = C₁V₂

q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor  and the voltage drop across the two capacitor will be the same

q = q₁ + q₂ = C₁V₂ + C₂V₂

CV = CV₂ + C₂V₂

CV - CV₂ = C₂V₂

C ( V - V₂) = C₂V₂

C ( V/ V₂ - V₂ /V₂) = C₂

C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF

7 0
3 years ago
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