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MrRa [10]
3 years ago
8

A tornado passes in front of a building, causing the pressure to drop there by 25% in 1 second. Part A If a door on the side of

the building is 8.1 feet tall and 3 feet wide, what is the net force on the closed door.
Physics
1 answer:
hram777 [196]3 years ago
8 0

Answer:

  F_net = 264, 26 pound

Explanation:

For this exercise we use Newton's second law

               F_net = F_int - F_outside

where the force can be found from the definition of pressure

              P = F / A

              F = P A

we substitute

               F_net = P_inside  A - P_outside A

               F_net = A (P_inside - P_outside)

indicate that the pressure on the outside is 25% less than the pressure on the inside

               P_outside = 0.25 P_inside

The area is

               A = L W

we substitute

              F_net = L W P_inside (1-0.25)

         

let's calculate

suppose the pressure inside is atmospheric pressure

              P_inside= P_atmospheric = 1,013 10⁵ Pa = 14.7 PSI

              F_net = 8.1 3 14.5 0.75

              F_net = 264, 26 pound

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Halley's comet orbits the sun roughly once every 76 years. It comes very close to the surface of the Sun on its closest approach
Licemer1 [7]

Answer:

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

From conservation of energy

K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

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3 years ago
If a builder of mass 75kg climbs a vertical ladder of 25m how much energy has she gained ?
erica [24]
So,

GPE (graviational potential energy) = mass x g x height

GPE is depends on where zero height is defined.  In this situation, we define h = 0 as the initial height.

GPE = 75 \ kg*9.8 \ \frac{m}{s^2}*25 \ m

GPE = 18,375 \frac{kg*m^2}{s^2}

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The builder has gained 18.375 kJ of PE.
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For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the e
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Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

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ΔUs = 0

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ΔEth = negative .

b )

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= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

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6 0
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