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Alenkasestr [34]
3 years ago
14

28. Someone throws a rubber ball vertically upward from the roof of a building 8.00 m in height. The ball rises,

Physics
1 answer:
Debora [2.8K]3 years ago
6 0

Answer:

v_o= 12 m/s

Explanation:

The ball experienced a constant acceleration motion, so we need to apply the following formula:

y=y_o+v_o*t+\frac{1}{2}*a*t^2\\\\vo=\frac{y-y_o-\frac{1}{2}*a*t^2}{t}\\\\v_o=\frac{0m-8m-\frac{1}{2}*(-9.81m/s^2)*(3.00s)^2}{3.00s}\\v_o=12m/s

Note: we set the acceleration of gravity as negative because it is going down.

the ball was thrown vertically upward with an initial velocity of 12 m/s

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A proton, mass 1.67 × 10−27 kg and charge +1.6 × 10−19 c, moves in a circular orbit perpendicular to a uniform magnetic field of
stellarik [79]

Time taken by proton to complete one complete circular orbit= 7.28 x 10⁻⁸ s

Explanation:

For proton, the centripetal force required for circular motion is provided by the magnetic force,

so Fm= Fc

q v B = m v²/r

m= mass of charged particle

v= velocity

B =magnetic field

q= charge

r= radius of circular path

v= q B r/m

now v= r ω

ω= angular velocity

ω r = q B r /m

ω=q B /m

now ω= 2π/T where T =time period

so 2π/T=q B/m

T= 2 πm/q B

T= 2π (1.67 x 10⁻²⁷)/ [( 1.6 x 10⁻¹⁹)* (0.9)]

T= 7.28 x 10⁻⁸ s

6 0
3 years ago
What is the quantity of heat energy required to convert 10g cube of ice at -30oC to steam at 120oC. also draw a graph of tempera
julia-pushkina [17]

30,869.2 J

Explanation:

Given parameters:

Mass of ice cube = 10g

Initial temperature = -30°C

Final temperature = 120°C

Specific heat capacity of water = 4.2J/g°C

Specific heat capacity of ice = 2.1J/g°C

Specific heat capacity of steam  = 1.996J/g°C

Latent heat of fusion of water(l) = 334J/g

Latent heat of vaporization = 2230J/g

Unknown:

Quantity of heat required = ?

Temperature-energy graph = ?

Solution:

The temperature energy profile is attached to this solution.

q = mc∅ₓ + mlₓ + mc∅ₙ + mlₙ + mc∅ₐ

qₓ = mc∅ₓ in converting ice from -30°C to ice at 0°C

qₓ is the amount of heat supplied to the ice that changes its temperature from -30°C to that at freezing point:

qₓ  = 10 x 4.2 x (0-(-30)) = 10 x 2.1 x 30 = 630J

qₓ = mlₓ in converting ice to water

qₓ here is the latent heat used to break the ice bonds without a change in temperature:

qₓ = ml = 10 x 334 = 3340J

qₙ = mc∅ₙ is the heat from water at 0°C to boiling point.

This is the heat required to take water from freezing temperature to its boiling point

qₙ = mc∅ₙ  = 10 x 4.2 x (100 - 0) = 4200J

qₙ  = mlₙ is the heat in vaporizing water

In vaporizing water, there is no temperature change when the hydrogen bonds are broken. The heat supplied is not used to raise temperature. We use the latent heat of vaporization:

qₙ   = 10 x 2230 = 22300J

qₐ = mc∅ₐ from vapor at 100°C to 120°C

This is the heat used to raise the temperature of vapor:

qₐ = 10 x 1.996 x (120-100) = 399.2J

The total heat:

  q = qₓ + qₙ + qₐ = 630J + 3340J + 4200J + 22300J + 399.2J =30,869.2 J

Learn more:

Specific heat brainly.com/question/7210400

#learnwithBrainly

7 0
3 years ago
What are two examples of common units for each of the above measurements
kap26 [50]

Density: g/mL, kg/cubic meter  

Volume: L, teaspoon  

Mass: g, MeV/sq. C

3 0
3 years ago
If the wave represents a sound wave, explain how increasing amplitude will affect the loudness of the sound? If we decrease the
Viktor [21]

Answer:

Explanation:

Think of a sound wave like a wave on the ocean, or lake... It's not really water moving, as much as it's energy moving through the water. Ever see something floating on the water, and notice that it doesn't come in with the wave, but rides over the top and back down into the trough between them? Sound waves are very similar to that. If you looked at a subwoofer speaker being driven at say... 50 cycles a second, you'd actually be able to see the speaker cone moving back and forth. The more power you feed into the speaker, the more it moves back and forth, not more quickly, as that would be a higher frequency, but further in and further out, still at 50 cycles per second. Every time it pushed out, it's compressing the air in front of it... the compressed air moves away from the speaker's cone, but not as a breeze or wind, but as a wave through the air, similar to a wave on the ocean

More power, more amplitude, bigger "wave", louder ( to the human ear) sound.

If you had a big speaker ( subwoofer ) and ran a low frequency signal with enough power in it, you could hold a piece of paper in front of it, and see the piece of paper move in and out at exactly the same frequency as the speaker cone. The farther away from the speaker you got, the less it'd move as the energy of the sound wave dispersed through the room.

Sound is a wave

We hear because our eardrums resonates with this wave I.e. our ear drums will vibrate with the same frequency and amplitude. which is converted to an electrical signal and processed by our brain.

By increasing the amplitude our eardrums also vibrate with a higher amplitude which we experience as a louder sound.

Of course when this amplitude is too high the resulting resonance tears our eardrums so that they can't resonate with the sound wave I.e. we become deaf

6 0
2 years ago
Question 1 of 10
Bad White [126]
The answer is b maybe?
3 0
3 years ago
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