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zalisa [80]
3 years ago
10

A brick is released with no initial speed from the roof of a building and strikes the ground in 1.80 s , encountering no appreci

able air drag. how tall, in meters, is the building?
Physics
1 answer:
Mars2501 [29]3 years ago
8 0
<span>Kinematics is used in this problem. The mass does not matter here because the question is mass independent. vi = 0 vf = x d = ? d = vi + 1/2 a t^2 d = 0 + 1/2 (9.8) (1.8)^2 d = 15.9 m (counting sig figs)</span>
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A nova occurs when
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C. hydrogen accreted onto a white dwarf from a close companion rapidly fuses to helium, releasing a large amount of energy.

The accreted material, composed mainly of hydrogen, is compacted on the surface of the white dwarf due to the intense gravitational force on that place. As material accumulates, The white dwarf becomes increasingly hot, until it reaches the critical temperature for ignition of nuclear fusion.

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3 years ago
If a force of 25 N is applied to an object with a mass of 8 kg, the object will accelerate at
Blizzard [7]
3.13 m/s2
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the formula for acceleration is as follows:
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so 25/8 = 3.13
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Engineers at the Space Centre must determine the net force needed for a rockets engine to achieve an acceleration of 70 m/s2. As
Troyanec [42]

Answer:

3,150,000N

Explanation:

According to Newton's second law;

F = mass * acceleration

Given

Mass = 45000kg

acceleration = 70m/s^2

Substitute

F = 45000 * 70

F = 3,150,000N

Hence the force required to be produced by the rocket engines is 3,150,000N

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3 years ago
Which of the following are examples of energy being conserved? Choose all that apply A light bulb constantly changes electrical
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A plant collects sunlight to form glucose, and your friend proposes an idea for a fan. Conserved = saving
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It has been suggested, and not facetiously, that life might have originated on Mars and been carried to Earth when a meteor hit
damaskus [11]

Answer:

a=3125000 m/s^2\\a=3.125*10^6 m/s^2

Acceleration, in m/s, of such a rock fragment = 3.125*10^6m/s^2

Explanation:

According to Newton's Third Equation of motion

V_f^2-V_i^2=2as

Where:

V_f is the final velocity

V_i is the initial velocity

a is the acceleration

s is the distance

In our case:

V_f=V_{escape},  V_i=0,s=4 m

So Equation will become:

V_{escape}^2-V_i^2=2as\\V_{escape}^2-0=2as\\V_{escape}^2=2as\\a=\frac{V_{escape}^2}{2s}\\a=\frac{(5*10^3m)^2}{2*4}\\a=3125000 m/s^2\\a=3.125*10^6 m/s^2

Acceleration, in m/s, of such a rock fragment = 3.125*10^6m/s^2

5 0
3 years ago
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