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Andreas93 [3]
3 years ago
12

A light bulb is left on for an hour and has a power of 0.1KW. How much did it cost?

Physics
1 answer:
KengaRu [80]3 years ago
3 0

(0.1 KW) x (1 hour) = 1.0 kWh of energy

Cost =
     <em>(1.0) x (cost of 1 kWh from that utility company in that neighborhood).</em>


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The slope of a velocity time graph will give___
nadya68 [22]

Answer:

<h3>THE ANSWER IS <em>(</em><em> </em><em>acceleration of the object</em><em> </em><em>)</em></h3>
6 0
3 years ago
A car accelerates at a constant rate from 12 m/s to 27 m/s while it travels 125 m. How long does it take to achieve this speed?
MArishka [77]

Answer:

6.41 s

Explanation:

Under constant acceleration we know that

average velocity × time taken = displacement

s=\frac{u+v}{2}t

125=\frac{27+12}{2}t

    t = 6.41 s

The proof of used equation is given in the attachment.

3 0
3 years ago
How was the scientific revolution important to the enlightenment
Margaret [11]

Answer:  The scientific revolution laid the foundations for the Age of Enlightenment, which centered on reason as the primary source of authority and legitimacy, and emphasized the importance of the scientific method.

Explanation:

7 0
3 years ago
Ask Your Teacher Suppose the roller coaster below(h1 = 36 m, h2 = 13 m, h3 = 30) passes point A with a speed of 1.00 m/s. If the
Oliga [24]

Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = (\frac{1}{5} × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

v₂² = 216.82

v₂  =  14.72 m/s

The roller coaster will reach point B with a speed of 14.72 m/s

8 0
3 years ago
A sample of an unknown substance has a mass of 0. 158 kg. If 2,510. 0 J of heat is required to heat the substance from 32. 0°C
guapka [62]

The specific heat of the unknown substance with a mass of 0.158kg is 0.5478 J/g°C

HOW TO CALCULATE SPECIFIC HEAT CAPACITY:

The specific heat capacity of a substance can be calculated using the following formula:

Q = m × c × ∆T

Where;

  • Q = quantity of heat absorbed (J)
  • c = specific heat capacity (4.18 J/g°C)
  • m = mass of substance
  • ∆T = change in temperature (°C)

According to this question, 2,510.0 J of heat is required to heat the 0.158kg substance from 32.0°C to 61.0°C. The specific heat capacity can be calculated:

2510 = 158 × c × (61°C - 32°C)

2510 = 4582c

c = 2510 ÷ 4582

c = 0.5478 J/g°C

Therefore, the specific heat capacity of the unknown substance that has a mass of 0.158 kg is 0.5478 J/g°C.

Learn more about specific heat capacity at: brainly.com/question/2530523

4 0
3 years ago
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