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Effectus [21]
3 years ago
6

Will every solid with the same dimensions have the same density?Explain your answer.

Physics
1 answer:
Stella [2.4K]3 years ago
6 0
No. Density is a property of a given material, regardless of how much or how little you have. Density is mass divided by volume
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E) Thermal energy is released during
vichka [17]

Answer:

e) True, f) False

Explanation:

e) Let consider a close system, that is, a system with no mass interactions with surroundings. Then, we get the following expression by the First Law of Thermodynamics:

Q_{net,in} - W_{net, out} = \Delta U (1)

Where:

Q_{net, in} - Net input heat, measured in joules.

W_{net, out} - Net output work, measured in joules.

\Delta U - Change in thermal energy, measured in joules.

Please notice that work comprises all kind of work (i.e. mechanical, electric, magnetic), whereas heat comprises all heat interactions including chemical and radioactive phenomena.

If thermal energy is released, then \Delta U < 0, which is caused by three scenarios:

(i) Q_{net,in} < 0, W_{net, out} < 0, |Q_{net,in}|>|W_{net,out}|

(ii) Q_{net, in} > 0, W_{net,out} > 0, |Q_{net,in}|

(iii) Q_{net,in}< 0, W_{net, out}>0

In the case Q_{net,in} > 0, W_{net, out}, the thermal energy of the system is increased. Therefore, thermal energy is released during some energy conversions. Answer: True

f) A liquid solidifies when temperature goes below point of fusion, meaning a realease of heat with no work interactions. That is:

Q_{net, in} = \Delta U, Q_{net, in} < 0 (2)

If Q_{net, in} < 0, then  \Delta U < 0. Then, if a liquid absorbs heat energy, then thermal energy is increase and the liquid does not solidifies. Answer: False.

7 0
3 years ago
Why are temperatures of the currents generally colder at the poles than the equator?
patriot [66]

Answer:

the answer is A

Explanation:

6 0
3 years ago
The position of a particle moving along the x axis varies in time according to the expression x = 4t 2, where x is in meters and
Pie

Answer:

a) X = 17.64 m

b) X = 17.64 + 4∆t^2 + 16.8∆t

c) Velocity = lim(∆t→0)⁡〖∆X/∆t〗 = 16.8 m/s

Explanation:

a) The position at t = 2.10s is:  

                 X = 4t^2

                 X = 4(2.10)^2

                 X = 17.64 m

b) The position at t = 2.10 + ∆t s  will be:

                 X = 4(2.10 + ∆t)^2

                 X = 17.64 + 4∆t^2 + 16.8∆t  m

c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,

                 ∆X= 4∆t^2 + 16.8∆t

Divide by ∆t on both sides:  

                ∆X/∆t =  4∆t + 16.8

 Taking the limit as ∆t approaches to zero we get:  

              Velocity =lim(∆t→0)⁡〖∆X/∆t〗 = 4(0) + 16.8

              Velocity = 16.8 m/s

 

8 0
3 years ago
Read 2 more answers
Scientists have concluded that Earth is at risk of future impacts with meteoroids. What are some criteria that engineers conside
Aleksandr [31]
Answer:
Explain step by step
Explanation:
Collisions with asteroids, comets and other stuff from space have been responsible for huge landmarks in our planet’s history: global shifts in climate, the creation of our moon, the reshuffling of our deepest geology, and the extinction of species.

Asteroid threats pop up in the news every now and then, but the buzz tends to fizzle away as the projectiles pass us by. Other times, as with the 2013 Chelyabinsk meteor in Russia, we don’t know they’re here until they’re here.
Perhaps most useful to remember is that when near-Earth objects (including asteroids, comets and meteoroids) enter the atmosphere, they’re called meteors; and if there’s anything left when they hit the ground, the resulting object is called a meteorite. We tend to focus on asteroids when talking about potential collisions, because they’re more likely to hit us than other stuff like comets, but still big enough to pose a threat.
5 0
2 years ago
A frictionless spring with a 3-kg mass can be held stretched 0.8 meters beyond its natural length by a force of 40 newtons. If t
Free_Kalibri [48]

Answer:

Explanation:

mass m = 3 kg

spring constant be k

k x .8 = 40 N

k = 40 / .8 = 50 N /m

angular frequency ω = √ ( k / m )

= √ ( 50 / 3 )

= 4.08 rad /s

Let amplitude of oscillation be A .

1/2 k A² = 1/2 m v²

50 A² = 3 x 1²

A = .245 m = 24.5 cm

For displacement , the equation of SHM is

x = A sinωt

= 24.5 sin4.08 t

x = 24.5 sin4.08 t

Here, angle 4.08 t is in radians .

3 0
3 years ago
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