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mr_godi [17]
3 years ago
5

How many grandfather clocks could you power with the same amount of power as is used by a single light bulb?

Physics
1 answer:
rosijanka [135]3 years ago
3 0

Yes we can run almost three lakhs of the grandfather clocks to run a single light bulb .

<u>Explanation:</u>

  • Yes we can run almost three lakhs of the grandfather clocks to run a single light bulb .
  • In simple words, power can be defined as the work done by the time and the SI unit of the power is Joules per second.
  • For example in the case of the light bulb, the Incandescent light bulb takes 60 watts per hour.
  • For example in the case of the clocks, the modern clocks use between 1 and 2 watts power.

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A 79.7 kg base runner begins his slide into second base while moving at a speed of 4.77 m/s. The coefficient of friction between
SashulF [63]

To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:

\Delta W = \Delta KE

\Delta W = \frac{1}{2} mv^2

Here,

m = mass

v = Velocity

Our values are given as,

m = 79.7kg

v = 4.77m/s

Replacing,

\Delta W = \frac{1}{2} (79.7kg)(4.77m/s)^2

\Delta W = 907J

Therefore the mechanical energy lost due to friction acting on the runner is 907J

6 0
3 years ago
A force of 10 N pressing on an area of 2m2. Calculate the pressure please answer fast it’s a test rn
musickatia [10]

Answer:

pressure=force/area

=10/2

5

Explanation:

3 0
3 years ago
Read 2 more answers
A motorbike travels 100m in 2.5 seconds. What is its speed?
lora16 [44]

Answer:

The motorbike is traveling at 40 m/s

Explanation:

100m over 2.5 seconds or 100/2.5 is 40 m/s

4 0
3 years ago
An object with a mass of 78kg is lifted through a height of 6 meters. How much work is done?
g100num [7]
Work = force x distance. 

force = mass x acceleration 

work = mass x acceleration x diastance 

use acceleration of gravity in this problem 

W (J) = m (kg) x a (m/s/s) x d (m) 
W = 78 x 9.8 x 6 
W = 4586.4

3 0
3 years ago
A block of ice with mass 2.00 kg slides 0.890m down an inclined plane that slopes downward at an angle of 28.3 degrees below the
Butoxors [25]

Answer:  The final speed, ignoring the effects of friction, will be 2.88 m/s.

Explanation:

The problem can be solved using two different physical principles. If we resort to Newton's second Law, we can say that, neglecting friction (which is reasonable for a ice block) the only force acting on the block  in the direction of the movement along the plane, is the component of the weight that is parallel to the slope, i.e.

Fp = mg sin 28.3º = ma ⇒ a=g sin 28.3º

Now, as we know that g = constant, we can use the following kinematic equation:

vf² - v₀² = 2 a x

if the block starts from rest, this means that v₀ = 0.

Replacing by the values in the equation, and solving for vf, we get;

vf = vf = \sqrt{2. 9.8.0.89. sin 28.3}  = 2.88 m/s

The other approach is using the conservation of energy principle:

When the block starts, it has some potential energy = mgh

This height h, can be expressed in terms of x (the length travelled by the block downward the plane) and the angle that forms with the horizontal, as follows:

h = x sin 28.3 (applying sin definition) ⇒ U = mg x sin 28.3

At the the end of the slide, the potential energy has been converted to kinetic energy, so we can write the following equation:

m. g. x. sin 28.3º = 1/2 m vf²

Simplifying, replacing by the values and solving for vf, we arrive to the same result as above.

5 0
4 years ago
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