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schepotkina [342]
3 years ago
14

A ball whose mass is 0.3 kg hits the floor with a speed of 5 m/s and rebounds upward with a speed of 2 m/s. If the ball was in c

ontact with the floor for 1.5 ms (1.5multiply10-3 s), what was the average magnitude of the force exerted on the ball by the floor?
Physics
1 answer:
Sonbull [250]3 years ago
4 0

Answer:

1400 N

Explanation:

Change in momentum equals impulse which is a product of force and time

Change in momentum is given by m(v-u)

Equating this to impulse formula then

m(v-u)=Ft

Making F the subject of the formula then

F=\frac {m(v-u)}{t}

Take upward direction as positive then downwards is negative

Substituting m with 0.3 kg, v with 2 m/s, and u with -5 m/s and t with 0.0015 s then

F=\frac {0.3(2--5)}{0.0015}=1400N

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In a rainstorm with a strong wind, what determines the best position in which to hold an umbrella?
AVprozaik [17]

Answer:

It is best to hold the umbrella at an angle which is parallel to the direction of rain drops.

Explanation:

Velocity of vector of rain drops plays crucial role here. Direction of the wind causes the rain drop directions. Hence handle of the umbrella should be parallel to the direction of rain drops. there will be more umbrella area if we keep umbrella in the direction.

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loris [4]
The answer is d I believe
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A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling
solniwko [45]

Answer:

17.86 m

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Then, we find the height

h = ½gt²

h = ½ * 9.8 * 0.27²

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9.8H = 3.528 + 171.4952

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H = 175.02 / 9.8

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4 years ago
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Mkey [24]

Answer:

Star A would be brighter than Star B

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A hotter star will radiate more energy per second per meter square of surface area. A larger star will have a greater surface area for radiation of energy, thus increasing the luminosity. For two identical stars, the difference in apparent brightness will be dependent on their distances from Earth.

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If one star (Star B) is twice as far from the earth as the first (Star A), the brightness of Star B will be \frac{1}{2^{2} } of Star A.

Thus, Star B will appear to be a quarter of the brightness of Star A. Or, Star A will appear to be 4 times as bright as Star B.

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