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2 years ago

A person throws a 0.21-kg ball straight up into the air. It reaches a height of

Physics

1 answer:

Brrunno [24]2 years ago

7 0

Answer:

Explanation:

As it begins to fall

F = ma a = 9.81

F = .21 * 9.81 = 2.06 N

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A 2.0 kilogram ball rolling along a flat surface starts up a hill. If the ball reaches a height of 0.63 meters, what was its ini

Agata [3.3K]

The potential energy as it reaches a height of 0.63 meters is mgh
Since total mechanical energy is conserved, Set that equal to the initial kinetic energy which is 0.5mv^2

2*9.8*0.63 = 0.5*2*v^2
12.348 = v^2
3.51 = v

That's it's initial speed.

5 0

3 years ago

The time constant of a simple RL circuit is defined as _______. We say that R is the resistance of the circuit and L is the indu

Gennadij [26K]

Answer:

The correct answer will be " $\tau =\frac{L}{R}$ ".

Explanation:

The time it would take again for current or electricity flows throughout the circuit including its LR modules can be connected its full steady-state condition is equal to approximately 5 $\tau$ as well as five-time constants.

It would be calculated in seconds by:

⇒ $\tau=\frac{L}{R}$

, where

R seems to be the resistor function in ohms.
L seems to be the inductor function in Henries.

5 0

3 years ago

Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 NN at

dexar [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

The amplitude of the lateral force is $F = 25 \ N$

The frequency is $f = 1 \ Hz$

The mass of the bridge per unit length is $\mu = 2000 \ kg /m$

The length of the central span is $d = 144 m$

The oscillation amplitude of the section considered at the time considered is $A = 75 \ mm = 0.075 \ m$

The time taken for the undriven oscillation to decay to $\frac{1}{e}$ of its original value is t = 6T

Generally the mass of the section considered is mathematically represented as

$m = \mu * d$

=> $m = 2000 * 144$

=> $m = 288000 \ kg$

Generally the oscillation amplitude of the section after a time period t is mathematically represented as

$A(t) = A_o e^{-\frac{bt}{2m} }$

Here b is the damping constant and the $A_o$ is the amplitude of the section when it was undriven

So from the question

$\frac{A_o}{e} = A_o e^{-\frac{b6T}{2m} }$

=> $\frac{1}{e} =e^{-\frac{b6T}{2m} }$

=> $e^{-1} =e^{-\frac{b6T}{2m} }$

=> $-\frac{3T b}{m} = -1$

=> $b = \frac{m}{3T}$

Generally the amplitude of the section considered is mathematically represented as

$A = \frac{n * F }{ b * 2 \pi }$

=> $A = \frac{n * F }{ \frac{m}{3T} * 2 \pi }$

=> $n = A * \frac{m}{3} * \frac{2\pi}{25}$

=> $n = 0.075 * \frac{288000}{3} * \frac{2* 3.142 }{25}$

=> $n = 1810 \ people$

3 0

3 years ago

How can you determine the diameter of water

katrin [286]

Answer:

To test your hydraulic skills, your Boss has requested you calculate the difference in water surface elevation between two reservoirs that are connected.

Explanation:

8 0

3 years ago

Which of the following would probably not work with circuits on a daily basis

Anni [7]

You need to attach the answer choices

8 0

3 years ago