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3 years ago

13

Calculate the height from from which a body is released from rest if its velocity just before hitting the ground is30m\s

1 answer:

Kamila [148]3 years ago

5 0

Answer:

height = 45 m

Explanation:

Initial velocity = 0 m/s

Final velocity = 30 m/s

Acceleration due to gravity = +10m/s^2 ( because ball is going down )

Using the eqn of motion

${v}^{2} - {u}^{2} = 2gh$

where v= final velocity ; u = initial velocity ; g = acceleration due to gravity ; h = height

So,

${30}^{2} - {0}^{2} = 2 \times 10 \times h$

$= > 20h = 900$

$= > h = \frac{900}{20} = 45$

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Which one of the following statements is true?

Gnom [1K]

Answer:

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Explanation:

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3 years ago

Find the density of seawater at a depth where the pressure is 680 atm if the density at the surface is 1030 kg/m3. Seawater has

Pie

Answer:

$1060.41kg/m^3$

Explanation:

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3 years ago

What are the characteristics of an index fossil?

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Read 2 more answers

A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,

Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

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So we will have

$-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2$

now we know that

$v_0 = 660 m/s$

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$m = 5 \times 10^9 kg$

$r = 4 \times 10^9 m$

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now from above formula

$GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2$

now we have

$2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2$

now plug in all data

$2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2$

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3 years ago

A projectile is launched at an angle of 60° from the horizontal and at a velocity of

gayaneshka [121]

Answer:

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3 years ago