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horrorfan [7]
2 years ago
10

Approximately how much electrical energy does a 5-W lightbulb convert to radiant and thermal energy in one hour?​

Physics
1 answer:
k0ka [10]2 years ago
8 0

Answer:

18,000 j

Explanation:

the lightbulb dissipates 5W of power

P = ΔE / Δt

rearrange to solve for energy

ΔE = PΔt

P = 5W

t =  1 hour = 60 minutes = 3600 seconds

ΔE = 5 * 3600

ΔE = 18000 J

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777dan777 [17]
<span>Raj is trying to make a diagram to show what he has learned about nuclear fusion.
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8 0
3 years ago
I need help thanks I would appreciate your in-depth explanation thanks.
svlad2 [7]

Answer:

The coin is less dense than the water thefore it can float.

Explanation:

5 0
3 years ago
Read 2 more answers
the car turned a corner going 10m/s if it took two seconds to turn what is its acceleration is this acceleration​
exis [7]

The acceleration of this car is equal to 5 m/s^2.

<u>Given the following data:</u>

  • Initial velocity = 0 m/s (assuming it's starting from rest).
  • Final velocity = 10 m/s.
  • Time = 2 seconds.

To determine the acceleration of this car:

<h3>How to calculate acceleration.</h3>

In Science, the acceleration of an object is calculated by subtracting the initial velocity from its final velocity and dividing by the time.

Mathematically,  acceleration is given by this formula:

a = \frac{V\;-\;U}{t}

<u>Where:</u>  

  • V is the final velocity.
  • U is the initial velocity.
  • is the time measured in seconds.

Substituting the given parameters into the formula, we have;

a = \frac{10\;-\;0}{2}

Acceleration, a = 5 m/s^2

Read more on acceleration here: brainly.com/question/24728358

6 0
2 years ago
As a rock sinks deeper and deeper into water of constant density, what happens to the buoyant force on it? As a rock sinks deepe
Sidana [21]

Answer:

It remains constant

Explanation:

As we know that buoyant force on an object given as

Fb = ρ Vd g

ρ= Density of fluid

Vd=Volume displace by body

g=10 m/s²

Fb =buoyant force

So from above we can say that buoyant force does not depends on the depth. It only depends on the fluid density and volume displace by body.

So when rock gets deeper and deeper the buoyant force will remain constant.

It remains constant

3 0
3 years ago
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
Bas_tet [7]

Answer:

  W = 1,307 10⁶ J

Explanation:

Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)

              F = G m₁ M / r²

              W = ∫ F. dr

              W = G m₁ M ∫ dr / r²

we integrate

             W = G m₁ M (-1 / r)

                 

We evaluate between the limits, lower r = R_ Moon and r = ∞

           W = -G m₁ M (1 /∞ - 1 / R_moon)

            W = G m1 M / r_moon

Body weight is

             W = mg

             m = W / g

The mass is constant, so we can find it with the initial data

For the capsule

            m = 1000/32 = 165 / g_moon

            g_moom = 165 32/1000

            .g_moon = 5.28 ft / s²

I think it is easier to follow the exercise in SI system  

           W_capsule = 1000 pound (1 kg / 2.20 pounds)

           W_capsule = 454 N

           W = m_capsule g

           m_capsule = W / g

           m = 454 /9.8

           m_capsule = 46,327 kg

Let's calculate

          W = 6.67 10⁻¹¹ 46,327   7.36 10²² / 1.74 10⁶

          W = 1,307 10⁶ J

7 0
3 years ago
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