All categories

2 years ago

Temperature

Physics

1 answer:

Orlov [11]2 years ago

8 0

Answer:

Particles spread apart, temperatue increases, pressure increases

Explanation:

You might be interested in

How much kinetic energy does a proton gain if it is accelerated, with no friction, through a potential difference of 1.00 V? The

ra1l [238]

Answer:

If energy is conserved, then the sum of the potential energy and the kinetic energy is a constant.

Assuming the proton starts from rest, so it's kineitc energy is zero, but it has a potential energy, PE equal to:

PE = qV

where q =1.6 x 10^-19 C

and V = 1.00 V

Assuming the proton no longer experiences the potential energy and it is all converted to kinetic energy then:

PE* = 0,

KE* = 1/(2mv^2)

Now since

PE + KE = Total energy =PE* + KE*

Therefore,

qV + 0 = 0 + 1/2mv^2

KE = qV = 1.6 10^-19 J

4 0

3 years ago

A painter on a ladder, painting the ceiling of a room comments. “It is hotter up here by the ceiling than it is down on the floo

ira [324]

Answer:

1. Hot air is less dense and has moved upward allowing cool air to move downward which is less dens

(Convection current)

2. The ceiling has transferred heat to him by radiation.

Explanation:

During the day when air is heated as a result of the ceiling transferring heat to it, it becomes less dens and it gains energy, which make it lighter in weight than cooler air, this hot air moves upward allowing cooler air to move downward.

4 0

3 years ago

I shared a picture of the problem. It’s a basic Physics question and an Algebra question.

julia-pushkina [17]

Hence the expression of ω in terms of m and k is

$\omega = \sqrt{\frac{k}{m}$

Given the expressions;

$T_s = 2 \pi \sqrt{\frac{m}{k} } \ and \ T_s = \frac{2 \pi}{\omega}$

Equating both expressions we will have;

$2 \pi \sqrt{\frac{m}{k} } = \frac{2 \pi}{\omega}$

Divide both equations by 2π

$\frac{2 \pi\sqrt{\frac{m}{2 \pi} } }{2 \pi}=\frac{\frac{2 \pi}{\omega} }{2\pi}\\\sqrt{\frac{m}{2 \pi} } = \frac{1}{\omega}\\$

Square both sides

$(\sqrt{\frac{m}{k} } )^2 = (\frac{1}{\omega} )^2\\\frac{m}{k} = \frac{1}{\omega ^2} \\\omega ^2 = \frac{k}{m}$

Take the square root of both sides

$\sqrt{\omega ^2} =\sqrt{\frac{k}{m} } \\\omega = \sqrt{\frac{k}{m}$

Hence the expression of ω in terms of m and k is

$\omega = \sqrt{\frac{k}{m}$

3 0

3 years ago

An astronaut spends 3 months in space and then returns to Earth. When she returns to Earth, her muscles have weakened. This is b

Pepsi [2]

False. there's less gravitational force in space than on earth

6 0

3 years ago

Read 2 more answers

An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of 25.0 with the horizontal. The coeffici

uranmaximum [27]

Answer:

V = 10.88 m/s

Explanation:

V_i =initial velocity = 0m/s

a= acceleration= gsinθ- $\mu_k$ cosθ

putting values we get

a= 9.8sin25-0.2cos25= 2.4 m/s^2

v_f= final velocity and d= displacement along the inclined plane = 10.4 m

using the equation

$v^2_f=v^2_i-2as$

$v^2_f=0^2-2(2.4)(10.4)$

v_f= 7.04 m/s

let the speed just before she lands be "V"

using conservation of energy

KE + PE at the edge of cliff = KE at bottom of cliff

(0.5) m V_f^2 + mgh = (0.5) m V^2

V^2 = V_f^2 + 2gh

V^2 = 7.04^2 + 2 x 9.8 x 3.5

V = 10.88 m/s

6 0

3 years ago