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Amanda [17]
3 years ago
15

1. Определите, используя график, удельную теплоемкость жидкости, если на плавление и нагревание до 65 градусов 1 кг вещества был

о затрачено 200 кДжэнергии. Удельная теплота плавления вещества 95 кДж\кг.
Physics
1 answer:
Sholpan [36]3 years ago
6 0

translate it to english

Explanation:

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Which factors affect gravitational potential force but not elastic potential energy
Inga [223]

Answer:

The mass of the object involved and the value of the gravitational acceleration

Explanation:

- Gravitational potential energy is defined as the energy possessed by an object in a gravitational field due to its position with respect to the ground:

U=mgh

where m is the mass of the object, g is the gravitational acceleration and h is the heigth of the object with respect to the ground.

- Elastic potential energy is defined as the energy possessed by an elastic object and it is given as:

U=\frac{1}{2}kx^2

where k is the spring constant of the elastic object, while x is the compression/stretching of the spring with respect to the equilibrium position.

As we can see from the equations, both types of energy depends on the relative position of the object/end of the spring with respect to a certain reference position (h in the first formula, x in the second formula), but gravitational potential energy also depends on m (the mass) and g (the gravitational acceleration) while the elastic energy does not.

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Smooth surfaces produce a __________________ reflection.
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What is the genetic makeup of the offspring of asexual reproduction
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3 years ago
One hundred jumping beans are placed along the center line of a gymnasium floor at six-inch intervals. Twelve hours later, the d
stira [4]

Answer:

a) Diffusion  coefficient, D = 1.5 in/hr

b) Mean jump frequency, f = 0.0833 Hz

Explanation:

a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:

^{2} = 2Dt..........(1)

Where <r> = mean displacement

D = Diffusion coefficient

t = time = 12 hrs

sum of the squares of the distance divided by 100 is 36 in2.

<r>²= 36 in²

Substituting these values into equation (1) above

36 = 2 * D *12\\36 = 24 D\\D = 36/24\\D = 1.5 in/hr

b) Mean jumping distance, <r> = 0.1 inches

Applying equation (1) again

Where D = 1.5 in/hr

^{2} = 2Dt

0.1^{2}  = 2 * 1.5t\\0.01 = 3t\\t = 0.01/3\\t = 0.0033 hrs\\t = 0.0033 * 3600\\t = 12 seconds

The mean jump frequency, f = 1/t

f = 1/12

f = 0.0833 Hz

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