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PIT_PIT [208]
3 years ago
6

How does an increase in thermal energy affect the state of matter?

Physics
1 answer:
shusha [124]3 years ago
8 0

Answer:

When thermal energy is added to a substance, its temperature increases, which can change its state from solid to liquid (melting), liquid to gas (vaporization), or solid to gas (sublimation). ... When the pressure exerted on a substance increases, it can cause the substance to condense.

Explanation:

Hope it will help use

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HELPPPPPPPPP Which option is a series of movement exercises for physical and mental health and is based on Hindu philosophy? O p
nika2105 [10]

Answer:

The correct answer is option ( D ) Yoga

Explanation:

hope it helps you dear :)

have a nice day

7 0
3 years ago
A circular flat coil that has N turns, encloses an area A, and carries a current i, has its central axis parallel to a uniform m
gogolik [260]

Answer:

A. Zero

Explanation:

The force on a coil of N turns, enclosing an area, A and carrying a current I in the presence of a magnetic field B, is :

F = N * I * A * B * sinθ

Where θ is the angle between the normal of the enclosed area and the magnetic field.

Since the normal of the area is parallel to the magnetic field, θ = 0

Hence:

F = NIABsin0

F = 0 or Zero

3 0
2 years ago
Are small molecular units joined together in large molecules
inysia [295]
<span>Polymers are small molecular units joined together in large molecules.

hope this helps!</span>
7 0
3 years ago
Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit
Ede4ka [16]

Answer:

386.2^{\circ}F

Explanation:

We are given that

P_1=200lbf/in^2

P_2=60lbf/in^2

v_1=200ft/s

v_2=1700ft/s

T_1=500^{\circ}F

Q=0

C_p=1BTU/lb^{\circ}F

We have to find the exit temperature.

By steady energy flow equation

h_1+v^2_1+Q=h_2+v^2_2

C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}

1BTU/lb=25037ft^2/s^2

Substitute the values

1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}

500+1.598=T_2+115.4

T_2=500+1.598-115.4

T_2=386.2^{\circ}F

7 0
3 years ago
a projectile is launched at an angle of 30 degrees and lands later at the same level. if it's initial speed is 50 m/s, solve for
Mrrafil [7]
using \: the \: formula \\ t = \frac{2u \sin( \alpha ) }{g} where \: u = initial \: speed \: \\ \alpha = angle \: of \: projection \\ g = acceleration \: due \: to \: gravity \\ \frac{2 \times 50 \times \sin(30) }{10} \\ \frac{100 \times 0.5}{10} = \frac{50}{10} = 5seconds

Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres
3 0
2 years ago
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