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3 years ago

True or false A temperature change and a change of state can both occur at the same time as a substance gains or loses energy.

Physics

1 answer:

jeka57 [31]3 years ago

5 0

Answer:

False

Explanation:

When ice melts, temperature always 0°C. Only when we have water the temperature begins to grow.

The change of state always goes under constant temperature.

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static electricity and friction

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3 years ago

What is the magnitude of the electric force of attraction between two point charges + 2.8 mc and -1.2 mc if the distance

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2 years ago

A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t

julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

$F_k = F_{W,E}$

$kx_1 = mg$

The extension of the spring due to the weight of the object on Moon is a value of $x_2$ , then

$kx_2 = mg_m$

Recall that gravity on the moon is a sixth of Earth's gravity.

$kx_2 = m\frac{g}{6}$

$kx_2 = \frac{1}{6} mg$

$kx_2 = \frac{1}{6} kx_1$

$x_2 = \frac{1}{6} x_1$

We have that the displacement at the earth was $x_1 = 0.3m$ , then

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3 years ago

In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur

Margaret [11]

Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

$3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)$

First, we isolate C from equation (2):

$2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)$

using this value of C from equation (3) in equation (1):

$3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}$

Put this value in equation (3), we get:

$C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C = \frac{17}{7}\\$

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3 years ago

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2 years ago