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2 years ago

Describe how work done is related to a change in volume of a fluid. 100 points

2 answers:

8 0

Just as we say that work force acting over a distance,for fluids,we can say that work is the pressure acting through the change in volume. for pressure- volume work, pressure is analogous to force, and the traditional defination of work.

Nookie1986 [14]2 years ago

4 0

Answer: work is the pressure acting through the change in volume

Explanation:

In the same way that work is defined as force operating over a distance, work is the pressure acting through the change in volume. Pressure is comparable to force in pressure–volume work, while volume is analogous to distance in the classic definition of work.

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) a 1.0 kilogram laboratory cart moving with a velocity of 0.50 meter per second due east collides with and sticks to a similar

ZanzabumX [31]

Momentum would be the same before and after the collision
Before the collision:
Momentum of the single cart: 1 * 0.50 = 0.50
After the collision
velocity = 0.25m / s
1 * 0.25 + 1 * 0.25 =
0.25 * (1 + 1) =
0.25 * 2 =
0.50
Now new momentum will be 0.5
answer
the same before and after the collision

4 0

4 years ago

Read 2 more answers

A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the

zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

$E= \frac{kqr}{R^3}$

The potential at a distance can be represented as:

V(r) - V(0) = $-\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

V(r) = $-[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

= 0.56 × 10⁻² m

R = 1.29 cm

= 1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

$V(r)$ $= -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}$

$V(r)$ = -2.15 × 10⁻⁴ V

The difference between the radial distance and center can be expressed as:

V(r) - V(0) = $-\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]^R$

V(r) = $-\frac{qR^2}{8 \pi E_0R^3 }$

V(r) = $-\frac{q}{8 \pi E_0R }$

V(r) $= -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}$

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0

3 years ago

A person strikes a ball with a bat. The temperature of the ball increases by 0.06ᵒC. What accounts for the increase?

marusya05 [52]

The increase in the average kinetic energy of the ball causes the increase in the temperature of the ball.

Kinetic energy of a particle is directly proportional to its temperature.

A ball initially at rest acquires kinetic energy when an external force is applied to it. As the person strikes the ball with a bat, the ball gains momentum which increases its kinetic energy of the ball.

Temperature on the other hand, is the measure of the average kinetic energy of a particle. Consequently, as the kinetic energy of the ball increases, the temperature of the ball increases as well.

Thus, we can conclude that the increase in the average kinetic energy of the ball causes the increase in the temperature of the ball.

Learn more here: brainly.com/question/18833622

6 0

3 years ago

One wire carries a current of I1= 2 Amps, and the other carries a parallel current (same direction, wires are side by side) of I

Yuki888 [10]

Answer: $F=3\times 10^{-5} N$