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2 years ago

What happens to temperature of a substance during phase change

Physics

1 answer:

mrs_skeptik [129]2 years ago

5 0

Answer: Nothing

Explanation:

The supplied energy is used for phase transition, not molecule kinetic energy.

Example: ice cannot be heated above 0Cº when it is being melted. Heat is consumed for phase change during its transition from solid to liquid. Only after all ice has melted, the temperature will rise until the next phase transition - vapor.

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The gravitational force of a star on an orbiting planet 1 is F1. Planet 2, which is twice as massive as planet 1 and orbits at t

Andrej [43]

Answer:

ratio = 1 : 4.5

Explanation:

If m₁ is the mass of the star and m₂ the mass of the planet, the force of gravity F₁ for planet 1 is given by:

$F_1=\frac{Gm_1m_2}{r^2}$

The force F₂:

$F_2=\frac{Gm_1(2m_2)}{(3r)^2}$

The ratio:

$\frac{F_2}{F_1}=\frac{2}{9}$

8 0

3 years ago

A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,

shusha [124]

Answer:

$v=d\sqrt{\frac{k}{m}}$

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

$U_{0}+K_{0}=U_{f}+K_{f}$

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

$U_{0}=K_{f}$

The initial potential energy of the spring is given by the equation:

$U_{0}=\frac{1}{2}kd^{2}$

the Kinetic energy of the block is then given by the equation:

$K_{f}=\frac{1}{2}mv_{f}^{2}$

so we can now set them both equal to each other, so we get:

$=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}$

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

$kd^{2}=mv_{f}^{2}$

so now we can solve this for the final velocity, so we get:

$v=d\sqrt{\frac{k}{m}}$

6 0

3 years ago

Help plzzzz thank youu

dybincka [34]

The answer is B I hope this helps luv

5 0

3 years ago

Consider two wheels with fixed hubs. The hub cannot move, but the wheel can rotate about it. The hubs are fixed to a stationary

kykrilka [37]

Answer:

Magnitude of force on wheel B is 4 N

Explanation:

Given that

$I=mr^2$

For wheel A

m= 1 kg

d= 1 m,r= 0.5 m

F=1 N

We know that

T= F x r

T=1 x 0.5 N.m

T= 0.5 N.m

T= I α

Where I is the moment of inertia and α is the angular acceleration

$I_A=1 \times 0.5^2\ kg.m^2$

$I_A=0.25\ kg.m^2$

T= I α

0.5= 0.25 α

$\alpha = 2\ rad/s^2$

For Wheel B

m= 1 kg

d= 2 m,r=1 m

$I_B=1 \times 1^2\ kg.m^2$

$I_B=1 \ kg.m^2$

Given that angular acceleration is same for both the wheel

$\alpha = 2\ rad/s^2$

T= I α

T= 1 x 2

T= 2 N.m

Lets force on wheel is F then

T = F x r

2 = F x 1

So F= 2 N

Magnitude of force on wheel B is 2 N

3 0

3 years ago

A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric fiel

Salsk061 [2.6K]

Answer:

If the particle is an electron $E_y = 3.311 * 10^3 N/C$

If the particle is a proton, $E_y = 6.08 * 10^6 N/C$

Explanation:

Initial speed at the origin, $u = 3 * 10^6 m/s$

$\theta = 38^0$ to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

The particle can either be an electron or a proton:

Mass of an electron, $m_e = 9.1 * 10^{-31} kg$

Mass of a proton, $m_p = 1.67 * 10^{-27} kg$

The electric field intensity along the positive y axis $E_y$ , can be given by the formula:

$E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\$

If the particle is an electron:

$E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\$

$E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\$

$E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C$

If the particle is a proton:

$E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\$

$E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\$

$E_y = 6.08 * 10^6 N/C$

8 0

3 years ago