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2 years ago

7

Goldie Doldfish, a speed swimmer, loves to rece around the parks pond, which is 0.5 miles around. If she can swim around the tra

ck in 2 hours, whar is her average speed?

1 answer:

rosijanka [135]2 years ago

8 0

Answer:

Her average speed is 0.25 miles / hour.

Explanation:

I did 0.5 divided by 2 = 0.25 miles / hour.

<em>Hope this helps! :)</em>

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The empire state building is 1,450 feet tall. King Kong weighs 20 tons and he climbs to the very top. If he jumps off the top, w

lozanna [386]

Answer:velocity= $93.12m/s$

Explanation:

We shall use conservation of energy to solve this problem

we have

Potential energy of King Kong at top of building = His kinetic energy at the bottom

Potential energy of an object = $m\times g\times h$ = 20000 kg

Where m is mass of an object(20000 kg)

g is accleration due to gravity = 9.81 $m/s^{2}$

h is the height above the surface of earth = 1450 feet = 441.96 meters

Applying values we get

$(P.E)_{TOP}=(20000\times 9.81\times 441.96) Joules\\\\(P.E)_{TOP}= 86712.552KiloJoules$ ......................(i)

Now Kinetic energy is given by

$(K.E)_{Bottom}=\frac{1}{2}mv^{2}$ .......................(ii)

Equating i and ii we get

86712.552 = $\frac{1}{2}mv^{2}$

$v=\sqrt{\frac{2E}{m}}$

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$v=\sqrt{\frac{2\times 86712.552\times 10^{3}}{20\times 10^{3}}}\\\\v= 93.12m/s$

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2 years ago

What is statement can most likey be useing in the shock waves of atmoic bomb

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Answer:

The initial energy emission occurs by 80% or more in the form of gamma rays but these are quickly absorbed and dispersed mostly by air in little more than a microsecond, converting gamma radiation into thermal radiation (thermal pulse ) and kinetic energy (shock wave) which are actually the two dominant effects in the initial moments of the explosion. The rest of the energy is released in the form of delayed radiation (fallout or fallout) and is not always counted when measuring the performance of the explosion.

Explanation:

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2 years ago

A 4-kg hammer is lifted to a height of 10 m and dropped from rest. What was the velocity (in m/s) of the hammer when it was at a

ipn [44]

Answer:

v = 10.84 m/s

Explanation:

using the equation of motion:

v^2 = (v0)^2 + 2×a(r - r0)

<em>due to the hammer starting from rest, vo = 0 m/s and a = g , g is the gravitational acceleration.</em>

v^2 = 2×g(r - r0)

v = \sqrt{2×(-9.8)×(4 - 10)}

= 10.84 m/s

therefore, the velocity at r = 4 meters is 10.84 m/s

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2 years ago

The actual depth of a shallow pool 1.00 m deep is not the same as the apparent depth seen when you look straight down at the poo

DedPeter [7]

Answer:

d' = 75.1 cm

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It is given that,

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or

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