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3 years ago

A stone is dropped from a certain height, distance covered by it in one second is:

Physics

1 answer:

Alinara [238K]3 years ago

5 0

Answer:

d = 4.9 m

Explanation:

It is mentioned that a stone is dropped from a certain height. It is required to find the distance covered by it in one second.

The initial speed of the stone is equal to 0 as it was at rest. Let d is the distance covered by the stone.

Using second equation of motion :

$d=ut+\dfrac{1}{2}at^2$

Put u = 0 and a = g

$d=\dfrac{1}{2}gt^2\\\\d=\dfrac{1}{2}\times 9.8\times 1^2\\\\d=4.9\ m$

So, the distance covered by it in one second is 4.9 m.

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A 1000-kilogram car traveling with a velocity of 20. meters per second decelerates uniformly at -5.0 meters per second2 until it

Elis [28]

Answer:

-20000 kgm/s

Explanation:

Impulse: This can be defined as the product of the mass of a body and its change in velocity. The S.I unit of impulse is kgm/s.

Mathematically, impulse can be expressed as

I = m(v-u).............. Equation 1.

Where I = impulse applied to the car to bring it to rest, m = mass of the car, u = initial velocity of the car, v = final velocity of the car.

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2 years ago

A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res

Natalija [7]

Answer:

The time constant is $\tau = 17.5 \ s$

Explanation:

From the question we are told that

The spring constant is $k = 11.5 \ N/m$

The mass of the ball is $m_b = 490 \ g = 0.49 \ kg$

The amplitude of the oscillation t the beginning is $x = 6.70 cm = 0.067 \ m$

The amplitude after time t is $x_t = 2.20 cm = 0.022 \ m$

The number of oscillation is $N = 30$

Generally the time taken to attain the second amplitude is mathematically represented as

$t = N * T$ Here T is the period of oscillation

$t = N * 2\pi \sqrt{\frac{m}{k} }$

=> $t = 30 * 2 * 3.142 * \sqrt{\frac{ 0.490}{11.5} }$

=> $t = 38.88 \ s$

Generally the amplitude at time t is mathematically represented as

$x(t) = x e^{-\frac{at}{2m} }$

Here a is the damping constant so

at $t = 38.88 \ s$ , $x_t = 2.20 cm = 0.022 \ m$

$0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }$

=> $0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }$

taking natural log of both sides

=> $ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }$

=> $a = 0.028$

Generally the time constant is mathematically represented as

$\tau = \frac{m}{a}$

=> $\tau = \frac{0.490}{ 0.028}$

=> $\tau = 17.5 \ s$

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2 years ago

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Answer:

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2 years ago