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1 year ago

How much heat is removed from 60 grams of steam at 100 °C to change it to 60 grams

Physics

1 answer:

Harrizon [31]1 year ago

4 0

Answer:

45200J

Explanation:

Given parameters:

Heat of vaporization of water = 2260J/g

Mass of steam = 20g

Temperature = 100°C

Unknown:

Energy released during the condensation = ?

Solution:

This change is a phase change and there is no change in temperature

To find the amount of heat released;

H = mL

m is the mass

L is the latent heat of vaporization

Insert the parameters and solve;

H = 20g x 2260J/g

H = 45200J

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n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.

9966 [12]

Answer:

a). V = 3.13*10⁶ m/s

b). T = 1.19*10^-7s

c). K.E = 2.04*10⁵

d). V = 1.02*10⁵V

Explanation:

q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

B = 1.10T

1u = 1.67 * 10^-27kg

M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

a). Centripetal force = magnetic force

Mv / r = qB

V = qBr / m

V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

V = 2.09088 * 10^-20 / 6.68 * 10^-27

V = 3.13*10⁶ m/s

b). Period of revolution.

T = 2Πr / v

T = (2*π*0.0594) / 3.13*10⁶

T = 1.19*10⁻⁷s

c). kinetic energy = ½mv²

K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²

K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

V = K / q

V = 2.04*10⁵ / (2eV).....2e-

V = 1.02*10⁵V

7 0

2 years ago

The lever shown above can be used to move the

vladimir1956 [14]

Answer:

C your answer would be C

Explanation:

It should be right

4 0

2 years ago

Power is the rate at which...........is done or the rate at which........... is converted from one form to another .

34kurt

Answer:

the answer is c I thought

8 0

2 years ago

. During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from th

mixer [17]

Answer:

106.7 N

Explanation:

We can solve the problem by using the impulse theorem, which states that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object:

$F \Delta t = m (v-u)$

where

F is the average force

$\Delta t$ is the duration of the collision

m is the mass of the ball

v is the final velocity

u is the initial velocity

In this problem:

m = 0.200 kg

u = 20.0 m/s

v = -12.0 m/s

$\Delta t = 60.0 ms = 0.06 s$

Solving for F,

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.200 kg) (-12.0 m/s-20.0 m/s)}{0.06 s}=-106.7 N$

And since we are interested in the magnitude only,

F = 106.7 N

5 0

3 years ago

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Which of these is a benefit of nuclear energy?

Mademuasel [1]

The answer i think would be D

4 0

2 years ago

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