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2 years ago

How does electricity turn into light in a common lightbulb?

Physics

2 answers:

Pachacha [2.7K]2 years ago

8 0

Answer:

i think its B

Explanation:

im pretty sure.

Travka [436]2 years ago

3 0

It’s b garunteed I promise

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PLEASE HELPPPP!!! Density is the amount of mass per unit volume. True or False

Alexxandr [17]

Answer:

true :D

Explanation:

4 0

2 years ago

Consider a situation where a constant force of 25 N acts on an object having a mass of 2 kg for 3 seconds. What is the work done

blagie [28]

Answer:

Work done W =1406.25 J

Explanation:

Work done on a body can be calculated using newton's 2nd laws:

F=ma

$\Rightarrow a=\frac{F}{m}$

Hence acceleration of the block is given by:

$\Rightarrow a=\frac{25}{2}=12.5m/s^2$

Displacement of the object is given by:

$\Rightarrow S=ut+\frac{1}{2}at^2$

Substitute the values

$\Rightarrow S=0*3+\frac{1}{2}(12.5)3^2$

$\Rightarrow S=56.25 m$

Now work done is given by:

W=F.S

W = 25×56.25

W =1406.25 J

3 0

2 years ago

Cloudy nights are usually warm because

almond37 [142]

Answer:

When you have a cloudy night, the clouds prevent heat from escaping through the atmosphere and into space.

The clouds act like a blanket and trap the heat it, and that is why every time you have a cloudy night, it is always warmer than a clear night.

5 0

2 years ago

A force of 44 N will stretch a rubber band 88 cm (0.080.08 m). Assuming that Hooke's law applies, how far will aa 11-N forc

Setler79 [48]

Answer:

The rubber band will be stretched 0.02 m.

The work done in stretching is 0.11 J.

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = 0.02 m

The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.

potential energy stored = $\frac{1}{2}ke^{2}$

==> $\frac{1}{2}* 550* 0.02^{2}$ = 0.11 J

3 0

2 years ago

Please help

inysia [295]

Explanation:

Start with what you know and list your knowns and unknowns

F = ma

F= 3N

m = 6kg

a =?

3N = 6kg x a

solve for a

3N / 6kg = a

3 0

3 years ago