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koban [17]
3 years ago
12

In an experiment, one of the forces exerted on a proton is F⃗ =−αx2i^, where α=12N/m2. What is the potential-energy function for

F⃗ ? Let U=0 when x=0.
Physics
1 answer:
Blizzard [7]3 years ago
7 0

Answer:

The potential energy is -4x^3

Explanation:

Given that,

Force F=-\alpha x^2 i

We need to calculate the potential energy

Using formula of work done

\Delta U=F(x) dx

Put the value of F

\Delta U=-\alpha x^2\ dx

On integration

U=-\alpha \dfrac{x^3}{3}+C

U=-\dfrac{\alpha x^3}{3}+C...(I)

U = 0, x = 0 so C = 0

Put the value of c and α in equation (I)

U=-\dfrac{12x^3}{3}+0

U=-4x^3

Hence, The potential energy is -4x^3

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The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont
ivann1987 [24]

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

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Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

But heat capacity of object B is twice that of object A

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

Therefore, the final temperature of both objects is 400 K.

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Thermal escape of an atmosphere is most pronounced on worlds where the gravity is low and the temperature is high.
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Answer:

3+7=          7x3=                                21➗7=          

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3+7= 10              7 8 9 10

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