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gregori [183]
2 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]2 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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The energy stored in the membrane is 6.44\cdot 10^{-14} J

Explanation:

The capacitance of a parallel-plate capacitor is given by

C=\frac{k\epsilon_0 A}{d}

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A is the area of the plates

d is the separation between the plates

For the membrane in this problem, we have

k = 4.6

A=4.50\cdot 10^{-9} m^2

d=8.1\cdot 10^{-9} m

Substituting, we find its capacitance:

C=\frac{(4.6)(8.85\cdot 10^{-12})(4.50\cdot 10^{-9})}{8.1\cdot 10^{-9}}=2.26\cdot 10^{-11} F

Now we can find the energy stored: for a capacitor, it is given by

U=\frac{1}{2}CV^2

where

C=2.26\cdot 10^{-11} F is the capacitance

V=7.55\cdot 10^{-2} V is the potential difference

Substituting,

U=\frac{1}{2}(2.26\cdot 10^{-11} F)(7.55\cdot 10^{-2})^2=6.44\cdot 10^{-14} J

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A 1400 kg car traveling at 17.0 m/s to the south collides with a 4700 kg truck that is at rest. The car and truck stick together
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Answer:

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KE after crash = 182,702.62 J

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We are given;

m1 = 1400 kg

m2 = 4700 kg

u1 = 17 m/s

u2 = 0 m/s

Using formula for inelastic collision, we have;

m1•u1 + m2•u2 = (m1 + m2)v

Where v is final velocity after collision.

Plugging in the relevant values;

(1400 × 17) + (4700 × 0) = (1400 + 1700)v

23800 = 3100v

v = 23800/3100

v = 7.677 m/s

Kinetic energy before crash = ½ × 1400 × 17² = 202300 J

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Answer:

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Substitute speed and time

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Therefore, the average speed for the remaining time of the journey is equal to 93 km/h

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