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3 years ago

A physicist drops astone off the top of cliff over looking a lake. She hears the splash 4 second after releasing the stone?

Physics

1 answer:

Angelina_Jolie [31]3 years ago

6 0

Answer:

Height of cliff(h) = 80 meter (Approx)

Explanation:

Given:

Time taken t = 4 sec

Gravitational acceleration = 10 m/s (Approx)

Find:

Height of cliff(h)

Computation:

h = ut + 1/2(g)(t)²

Height of cliff(h) = (0)(4) + 1/2(10)(4)²

Height of cliff(h) = 80 meter (Approx)

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A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, caus

babunello [35]

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

$I = m\cdot (\vec{v}_{2} - \vec{v_{1}})$ (1)

Where:

$I$ - Impulse, in kilogram-meters per second.

$m$ - Mass, in kilograms.

$\vec{v_{1}}$ - Initial velocity of the hockey park, in meters per second.

$\vec{v_{2}}$ - Final velocity of the hockey park, in meters per second.

If we know that $m = 0.2\,kg$ , $\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right]$ and $\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right]$ , then the impulse applied by the stick to the park is approximately:

$I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]$

$I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]$

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.

8 0

3 years ago

The rate of cooling determines ....... and ......

PilotLPTM [1.2K]

Answer:

freezing point and melting point

7 0

2 years ago

If the angular frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum acceleration

Nataly_w [17]

Answer:

When we double the angular velocity the maximum acceleration $(a_{max})$ will changes by a factor of 4.

Explanation:

Given the angular frequency $(\omega)$ of the simple harmonic oscillator is doubled.

We need to find the change in the maximum acceleration of the oscillator.

$a_{max}=A\omega^2$

Now, according to the problem, the angular frequency $(\omega)$ got doubled.

Let us plug $\omega=2\times \omega$ . Then the maximum acceleration will be $a_{max'}$

$a_{max}=A\omega^2$

$a_{max'}=A(2\times \omega)^2\\a_{max'}=A\times 4\omega\\a_{max'}=4A\omega$

$a_{max'}=4a_{max}$

We can see, when we double the angular velocity the maximum acceleration will changes by a factor of 4.

6 0

3 years ago

Due to efficiency considerations related to its bow wake, the supersonic transport aircraft must maintain a cruising speed that

givi [52]

Answer:

decreases.

Explanation:

When the aircraft is flies from the warm air into the colder air then its speed will be decreases.

as we know that

we know mach number is constant

so that here Mach number M is expressed as

M = $\frac{u}{v}$ .............................1

here u is Local flow velocity with respect to the boundarie and v is the speed of sound in the medium

If the aircraft flies from hot air to cold air, the speed of sound in the medium will decrease. But the Mach number remains constant. Therefore, the local flow velocity relative to the boundaries also decreases.

7 0

3 years ago

What are possible formulas for impulse? Check all that apply. J = Fdeltat J = StartFraction force over change in time EndFractio

Alex

<u>The possible formulas for impulse are as follows:</u>

J = FΔt

J = mΔv

J = Δp

Answer: Option A, E and F

<u>Explanation:</u>

The quantity which explains the consequences of a overall force acting on an object (moving force) is known as impulse. It is symbolised as J. When the average overall force acting on an object than such products are formed and in given duration than the start fraction force over change in time end fraction J = FΔt.

The impulse-momentum theorem explains that the variation in momentum of an object is same as the impulse applied to it: J = Δp J = mΔv if mass is constant J = m dv + v dm if mass changes. Logically, the impulse-momentum theorem is equivalent to Newton second laws of motion which is also called as force law.

6 0

3 years ago