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sasho [114]
2 years ago
10

An owl has a mass of 4.00 kg. It dives to catch a mouse, losing 800.00 J of its GPE. What was the starting height of the owl, in

meters?
Physics
1 answer:
vesna_86 [32]2 years ago
3 0

Answer:

height =20m

Explanation:

gpe=mgh

800=4×10×x

40x=800

x=20

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If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
yulyashka [42]

Answer:

178200 g mile pounds

Explanation:

Work= Force * Distance= Fh

F=ma=mg where m is mass and g is acceleration due to gravity

Work= 165 pounds *g* 1080 m=  178200 g mile pounds

5 0
2 years ago
A child’s toy rake is held so that its resistance length is 0.85 meters. If the mechanical advantage is 0.43, what is the effort
mart [117]

Answer:

1.28

Explanation:

7 0
1 year ago
A power plant burns 1000 kg of coal each hour and produces 500 kW of power. Calculate the overall thermal efficiency if each kg
solniwko [45]

Answer:

The overall thermal efficiency is 30%.

Explanation:

Given;

Output power = 500 kWh

input energy  per kg of coal = 6 MJ = 6 x 10⁶ J = 1.66667 kWh

1000 kg of coal will produce 1000 x 1.66667 kWh = 1666.67 kWh

Thus, total input power = 1666.67 kWh

Overall thermal efficiency = Total output power/Total input Power

Overall thermal efficiency = (500/1666.67) *100

Overall thermal efficiency = 0.29999 *100

Overall thermal efficiency = 30%

Therefore, the overall thermal efficiency is 30%.

5 0
2 years ago
A 15-cm-tall closed container holds a sample of polluted air containing many spherical particles with a diameter of 2.5 μm and a
TEA [102]

Answer:

t = 224 s

Explanation:

given,

length of container = 15 cm = 0.15 m

diameter of spherical particle = 2.5 μm

mass of particle = 1.9 x 10⁻¹⁴ kg

viscosity of air = μ = 1.18 x 10⁻⁵ kg/m.s

time taken by the particle to stop = ?

radius of particle = 2.5/2 = 1.25 μm

volume of particle = \dfrac{4}{3}\pi r^3

                              =\dfrac{4}{3}\pi (1.25 \times 10^{-6})^3

Density =\dfrac{mass}{volume}

Density =\dfrac{1.9 \times 10^{-14}}{\dfrac{4}{3}\pi (1.25 \times 10^{-6})^3}

ρ = 2322 kg/m³

terminal velocity

v_t = \dfrac{2}{9}\ \dfrac{gR^2(\rho - \rho_{air})}{\mu}

v_t = \dfrac{2}{9}\ \dfrac{9.8 \times (1.25 \times 10^{-6})^2(2322 - 1)}{1.18 \times 10^{-5}}

v_t = 6693 x 10⁻⁷ m/s

t = \dfrac{d}{v_t}

t = \dfrac{0.15}{6693 \times 10^{-7}}

t = 224 s

7 0
2 years ago
An object of mass 100 kg is initially at rest on a horizontal frictionless surface. At time t = 0, a horizontal force of 10 N is
satela [25.4K]

Answer:

(D) It is moving at a constant speed

Explanation:

Before t = 1s. Due to the force, albeit small, acting on the object, since there's no static friction stopping the object from moving, this mass object would have a constant acceleration and it's velocity would be increasing.

According to Newton's 1st law, an object will stay at a constant speed if the net force acting on it is 0. After t = 1s, horizontally speaking there's no other force exerting on the mass object. There is no friction force at play here as the surface is frictionless.

Therefore the correct statement is (D) It is moving at a constant speed

8 0
2 years ago
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