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Yuki888 [10]
3 years ago
6

How is the kelvin scale different from the Celsius scale?

Physics
1 answer:
k0ka [10]3 years ago
5 0
Difference is between the freezing point of water and the boiling point of water.
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. A person weighing 750 N gets on an elevator.
Kobotan [32]

 

F = 750 N  (Force)

d = 10 m  (displacement )

t = 25 s   (time)

L = ?   (Mechanical work )  =  (Energy)

P = ?   (Power)

Solve:

L = F × d = 750 × 10 = 7500 Joules

P = L / t = 7500 / 25 = 300 Watts

5 0
2 years ago
Why doesn’t a machine that increases force break the law of conservation of energy?
USPshnik [31]

Answer:

A machine in which work input equals work output. energy can be used to do work, work can be used to transfer energy. The change in the kinetic energy of an object is equal to the net work done on the object.

hope this helps

8 0
3 years ago
What is the period of a wave with a frequency of 100 Hz and a wavelength of 2.0 m
spin [16.1K]

The answer for the following answer is answered below.

  • <u><em>Therefore the time period of the wave is 0.01 seconds.</em></u>
  • <u><em>Therefore the option for the answer is "B".</em></u>

Explanation:

Frequency (f):

The number of  waves that pass a fixed place in a given amount of time.

The SI unit of frequency is Hertz (Hz)

Time period (T):

The time taken for one complete cycle of vibration to pass a given point.

The SI unit of time period is seconds (s)

Given:

frequency (f) = 100 Hz

wavelength (λ) = 2.0 m

To calculate:

Time period (T)

We know;

According to the formula;

<u>f =</u>\frac{1}{T}<u></u>

Where,

f represents the frequency

T represents the time period

from the formula;

  T = \frac{1}{f}

 T = \frac{1}{100}

  T = 0.01 seconds

<u><em>Therefore the time period of the wave is 0.01 seconds.</em></u>

5 0
2 years ago
Read 2 more answers
Suppose you wish to fabricate a uniform wire out of 1.10 g of copper. If the wire is to have a resistance R = 0.390 Ω, and if al
Citrus2011 [14]

To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:

Mass of copper wire(m) = 1.10g = 1.10*10^{-3} kg

Density (\rho)= 8.92*10^3kg/m^3

Resistively of copper (\gamma) = 1.7*10^{-8}\Omega \cdot m

Resistance (R) = 0.390\Omega

Volume is defined as,

V= lA \text{ and }\frac{m}{\rho}

lA= \frac{1.10*10^{-3}}{8.92*10^3}

lA = 1.233*10^{-7} m^3 (1)

We know that,

\frac{l}{A} = \frac{R}{\gamma}

\frac{l}{A}= \frac{0.390\Omega}{1.7*10^{-8}\Omega m}

\frac{l}{A} = 2.2941*10^7 m^{-1} (2)

Multiplying equation we have

l^2 = (1.233*10^{-7})( 2.2941*10^7)

l^2 = 2.8286m^2

l =\sqrt{2.8286m^2}

l = 1.68m

Therefore the length of the wire is 1.68m

6 0
2 years ago
Help ASAP ITS DUE IN 30 MIN
Katena32 [7]

can't read it, need larger picture

5 0
3 years ago
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