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3 years ago

What is the input of energy for a device that provides 50 units of useful energy and has an efficiency of 0.55?

1 answer:

3 0

If that's the case, then

      50 units = 0.55 x the input energy

Divide each side by 0.55 :

   50 units/0.55 = the input energy =

   <span> 90 and 10/11 units</span>

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Which statement is true?

vladimir2022 [97]

<span>a.current varies throughout a parallel circuit.

Hope this helps!</span>

8 0

4 years ago

The Sun delivers an average power of 1.575 W/m2 to the top of Neptune's atmosphere. Find the magnitudes of max and max for the e

FrozenT [24]

Answer:

$1.1486813808\times 10^{-7}\ T$

34.46 V/m

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

c = Speed of light = $3\times 10^8\ m/s$

I = Intensity = 1.575 W/m²

The maximum magnetic field intensity is given by

$B_m=\sqrt{\dfrac{2\mu_0I}{c}}\\\Rightarrow B_m=\sqrt{\dfrac{2\times 4\pi \times 10^{-7}\times 1.575}{3\times 10^8}}\\\Rightarrow B_m=1.1486813808\times 10^{-7}\ T$

The magnetic field intensity is $1.1486813808\times 10^{-7}\ T$

The maximum electric field intensity is given by

$E_m=B_m\times c\\\Rightarrow E_m=1.1486813808\times 10^{-7}\times 3\times 10^8\\\Rightarrow E_m=34.46\ V/m$

The electric field intensity is 34.46 V/m

8 0

3 years ago

If the depth of water in a well is 10m, what is the pressure exerted by it the bottom of the well ? ( Use g = 10 m/s2)

Tasya [4]

Answer:

The precise answer depends on the density and therefore the temperature of the water, but we can obtain a reasonable approximation by assuming that the density of the water is 1000 kilograms per cubic meter (kg/m³).

Since the depth of the water in the well is 10 m, the volume of water directly above an area A of a square meters (m²) at the bottom of the well is 10×a m³.

Since the density of the water is 1,000 kg/m³, the mass of water directly above area A is (1,000 kg/m³) × (10×a m³) = (1000×10×a kg) = 10,000×a kg.

Since g = 9.8 m/s², the force of gravity acting on the water directly above area A is (9.8 m/s²) × (10,000×a kg) = 9.8×10,000×a N (newtons) = 98,000×a N.

So the pressure of water acting on area A is (98,000×a N)/(a m²) = (98,000×a)/a N/m² = 98,000 pascals (pa). And since A could be any given area at the bottom of the well, this is the pressure at any point at the bottom of the well.

So the pressure at the bottom of the well is 98,000 pascals (or 98,000/101,325 standard atmospheres = 560/579 atmospheres ~ 0.967 standard atmospheres).

Please comment below if you have any questions.

7 0

3 years ago

Initially, a particle is moving at 5.25 m/s at an angle of 35.5Â° above the horizontal. Three seconds later, its velocity is 6.0

ivolga24 [154]

Answer:

a =( -0.32 i ^ - 2,697 j ^) m/s²

Explanation:

This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.

Break down the speeds in two moments

initial

v₀ₓ = v₀ cos θ

v₀ₓ = 5.25 cos 35.5

v₀ₓ = 4.27 m / s

$v_{oy}$ = v₀ sin θ

$v_{oy}$ = 5.25 sin35.5

$v_{oy}$ = 3.05 m / s

Final

vₓ = 6.03 cos (-56.7)

vₓ = 3.31 m / s

$v_{y}$ = v₀ sin θ

$v_{y}$ = 6.03 sin (-56.7)

$v_{y}$ = -5.04 m / s

Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order

a = ( $v_{f}$ - v₀) /t

aₓ = (3.31 -4.27)/3

aₓ = -0.32 m/s²

$a_{y}$ = (-5.04-3.05)/3

$a_{y}$ = -2.697 m/s²

6 0

4 years ago

Whitch two options are forms of kinetic energy?

valentina_108 [34]

Answer:the witch has nothing to do with the problem

Explanation:

7 0

3 years ago