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3 years ago

A fox with a thick fur would have a survival advantage over other foxes if

1 answer:

8 0

This question is poorly stated, but I assume you mean what conditions are needed. It would have to be cold outside, correct?

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When it gets that hot the evergreens can explode?

Ghella [55]

If you mean the tree, evergreen trees can exploded if theres extreme stress on the trunk

3 0

3 years ago

_____ is the transfer of thermal energy between two bodies which are at different temperatures. The si unit for this is the joul

ss7ja [257]

Answer:

Heat

Explanation:

The thermal energy is usually defined as a heat energy that occurs due to the increase in temperature. Due to this increasing temperature, the atoms, as well as the molecules, start to migrate at a much faster rate and rapid collisions occur with one another.

This temperature change that is involved in this process can be measured.

Thus, in thermal energy, heat (heat energy) is transferred from one body to another at different distinct temperatures.

The SI unit of heat is Joule.

6 0

3 years ago

Suppose you apply a force of 75 N to a 25-kg object. What will the acceleration of the object be?

Serjik [45]

Answer:

I belive it would be ture

Explanation:

It's been a while since I learned this but I think that is right.

6 0

3 years ago

Read 2 more answers

A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.

LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

$\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}$

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

$W = \frac{1}{2}(50)(5.61^2) = 786.8025 J$

Now, we can define work:

$\large\boxed{W = Fdcos\theta}}$

Now, plug in the values:

$786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta$

Solve for theta:

$cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}$

4 0

3 years ago

Heat transfers energy from a hot object to a cold object. Both objects are isolated from their surroundings. The change in entro

aniked [119]

To develop this problem we will start from the definition of entropy as a function of total heat, temperature. This definition is mathematically described as

$S = \frac{Q}{T}$

Here,

Q = Total Heat

T = Temperature

The total change of entropy from a cold object to a hot object is given by the relationship,

$\Delta S = \frac{Q}{T_{cold}}-\frac{Q}{T_{hot}}$

From this relationship we can realize that the change in entropy by the second law of thermodynamics will be positive. Therefore the temperature in the hot body will be higher than that of the cold body, this implies that this term will be smaller than the first, and in other words it would imply that the magnitude of the entropy 'of the hot body' will always be less than the entropy 'cold body'

Change in entropy $\Delta S_{hot}$ is smaller than $\Delta S_{cold}$

Therefore the correct answer is C. Will always have a smaller magnitude than the change in entropy of the cold object

5 0

3 years ago