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3 years ago

Which of the 3 states of matter is easier to increase the temperature by one degree; in a liquid, in a solid, or in a gas?

Physics

1 answer:

Mashcka [7]3 years ago

4 0

Answer:

gas

Explanation:

this is bcz they have higher kinetic energy compared to liquids and gases

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What happen when a.c is passed into the coil of dynamo?<br>

ehidna [41]

Answer:

The dynamo produces Alternating Current output, so in theory yes, it should work in reverse if Alternating Current is input.

5 0

2 years ago

A car Excel accelerates from the rest to a velocity of 5 m in four seconds what is the average acceleration

MrRa [10]

Average acceleration: velocity/time

5/4= 1.25 m/s² Acceleration

Hope I helped :)

5 0

3 years ago

An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1

goblinko [34]

Answer:

$v"_{1} = v_{1} tan$ Θ

$v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

Θ = $tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )$

Explanation:

Applying the law of conservation of momentum, we have:

Δ $p_{x = 0}$

$p_{x} = p"_{x}$

$m_{1}v_{1} = m_{2}v"_{2} cos$ Θ (Equation 1)

Δ $p_{y} = 0$

$p_{y} = p"_{y}$

$0 = m_{1} v"_{1} - m_{2} v"_{2} sin$ Θ (Equation 2)

From Equation 1:

$v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

From Equation 2:

$m_{2} v"_{2}$ sinΘ = $m_{1} v_{1}$

$v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }$

Replacing Equation 3 in Equation 4:

$v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}$

$v"_{1}=v_{1}\frac{sinΘ}{cosΘ}$

$v"_{1}=v_{1}tan$ Θ (Equation 5)

And we found Θ from the Equation 5:

tanΘ= $\frac{v"_{1}}{v_{1}}$

Θ= $tan^{-1}(\frac{v"_{1}}{v_{1}})$

7 0

3 years ago

A U-shaped tube, open to the air on both ends, contains mercury. Water is poured into the left arm until the water column is 10.

Masja [62]

Answer:

0.368 cm

Explanation:

x = distance by which the mercury rise

d = depth of the water = 10 cm = 0.10 m

ρ = density of water = 1000 kgm⁻³

ρ' = density of mercury = 13600 kgm⁻³

P₀ = atmospheric pressure

Using equilibrium of pressure on both side

P₀ + ρ g d = P₀ + ρ' g (2x)

(1000) (0.10) = (13600) (2x)

x = 0.00368 m

x = 0.368 cm

8 0

2 years ago

How is the combined gas law used to calculate changes in pressure, temperatures, and/or volume for a fixed amount of gas?

nalin [4]

Answer:

Let's start by considering the ideal gas law:

$pV=nRT$

where

p is the gas pressure

V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature

This equation can also be rewritten as

$\frac{pV}{T}=nR$

Now, if we consider a fixed amount of gas, this means that the number of moles (n) is constant. So we can rewrite the equation as

$\frac{pV}{T}=const.$

And therefore, if we consider a gas undergoing a certain transformation from 1 to 2, we can write

$\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}$

where 1 indicates the conditions of the gas at the beginning and 2 the conditions of the gas after the process. So, the change in pressure/temperature/volume of the gas can be found by using this equation.

5 0

2 years ago