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Tema [17]
3 years ago
9

How can doing work on an object increase its kinetic energy?

Physics
1 answer:
Lyrx [107]3 years ago
6 0

Answer:If the object's speed increases.

Explanation:

If the object's speed increases, then its kinetic energy will increase. If the kinetic energy increases, the change in kinetic energy will be positive.

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28. In the diagram above which symbol represents the light bulb ?
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28. It's B
29. It's A
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4 years ago
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A vertical spring stretches 9.6 cm when a 1.3 kg block is hung from its end. (a) Calculate the spring constant. This block is th
tatiyna

Answer:

F = - K x

a) K = 1.3 kg * 9.8 m/s^2 / .096 m = 133 kg/sec^2

b)  ω = (K/m)^1/2     angular frequency of SHM

ω = (133 / 1.3)^1/2 = 10.1 / sec

f = 2 π ω = 6.28 * 10.1 / sec = 63.5 / sec

P = 1/f = .0157 sec

5 0
3 years ago
Using poiseulli assumption derive an expression for the rate of flow of a liquid through a horizontal tube in terms of a, l, p a
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3 years ago
As a pendulum swings from its highest to lowest position, what happens to its kinetic and potential energy?
DIA [1.3K]

The potential energy decreases while the kinetic energy increases.



7 0
3 years ago
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Electric charge is distributed over the disk x2 + y2 ≤ 4 so that the charge density at (x, y) is rho(x, y) = 4x + 4y + 4x2 + 4y2
maw [93]

Answer:

Q=185.84C

Explanation:

We have to take into account the integral

Q=\int \rho dV

In this case we have a superficial density in coordinate system.

Hence, we have for R: x2 + y2 ≤ 4

Q=\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\rho dydx

but, for symmetry:

Q=4\int_0^2\int_0^{\sqrt{4-x^2}}\rho dydx\\\\Q=4\int_0^2\int_0^{\sqrt{4-x^2}}(4x+4y+4x^2+4y^2) dydx\\\\Q=4\int_0^{2}[4x\sqrt{4-x^2}+2(4-x^2)+4x^2\sqrt{4-x^2}+\frac{4}{3}(4-x^2)^{3/2}]dx\\\\Q=4[46.46]=185.84C

HOPE THIS HELPS!!

8 0
3 years ago
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