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Harlamova29_29 [7]
3 years ago
9

A 3 kg stone is dropped from a height of 4 m what is its momentum as it strikes the ground

Physics
1 answer:
Vitek1552 [10]3 years ago
7 0

Okay first you have to recognize that the maximum Gravitational potential energy will equal the maximum kinetic energy. The maximum GPE will equal when the stone is at its highest point and the max KE will be right before the stone hits the ground. So

GPE max = KE max

mgh = 1/2mv^2

Mass cancels as it is on both sides

gh = 1/2v^2

Multiply by 2

2gh = v^2

Square root

v = √2gh

Now plug in

v = √2(9.8)(4)

v = 8.85 m/s

Now use the mass to calculate the momentum just before it hits the ground as this is the speed right before it hits the ground

p = mv

p = (3)(8.85)

p = 26.55 kgm/s

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Water is returned from earth’s surface to the atmosphere by
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Answer:

Evaporation

Explanation:

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3 years ago
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dem82 [27]
Answer: C. 1.64 x 10-3 m/s2
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leva [86]

Answer:

Option C

Maximum potential energy is at point R.

Explanation:

Potential energy is a product of mass, acceleration due to gravity and height ie

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3 years ago
An organ pipe open at both ends has a length of 0.80 m. If the velocity of sound in air is 340 m/s, what is the frequency of the
bazaltina [42]

Answer:

the frequency of the second harmonic of the pipe is 425 Hz

Explanation:

Given;

length of the open pipe, L = 0.8 m

velocity of sound, v = 340 m/s

The wavelength of the second harmonic is calculated as follows;

L = A ---> N   +  N--->N   +   N--->A

where;

L is the length of the pipe in the second harmonic

A represents antinode of the wave

N represents the node of the wave

L = \frac{\lambda}{4} + \frac{\lambda}{2} + \frac{\lambda}{4} \\\\L = \lambda

The frequency is calculated as follows;

F_1 = \frac{V}{\lambda} = \frac{340}{0.8} = 425 \ Hz

Therefore, the frequency of the second harmonic of the pipe is 425 Hz.

5 0
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Force F=2.0N i - 3.0N k acts on a pebble with position vectorr=0.50m j - 2.0m k relative to the origin. In unit vector notation,
klasskru [66]

Answer with Explanation:

We are given that

Force acts on a pebble=2\hat{i}-3\hat{k} N

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Torque=r\times F

Substitute the values then we get

Torque= (0.5j-2k)\times (2i-3k)

Torque=-k-1.5i-4jN-m

By using i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0

b.r=2i-3k

r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k

Torque about point (2,0,-3)

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6 0
3 years ago
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