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2 years ago

Can an object have both kinetic energy and gravitational potential energy? Explain.

Physics

1 answer:

marin [14]2 years ago

8 0

Energy flows with kinetic energy

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An optical disk drive in a computer can spin a disk at up to 10,000 rpm. If a particular disk is spun at 7570 rpm while it is be

anastassius [24]

Answer:

The magnitude of the average angular acceleration is calculated as $1822.36\ rad/s^{2}$

Explanation:

Maximum speed that can be attained by the disk, $N_{m}$ = 10,000 rpm

Speed of spinning of the disk, N = 7570 rpm

Time taken to come to rest, t = 0.435 s

Now,

The initial angular velocity is given by:

$\omega = \frac{2\pi N}{60} = 792.73\ rads^{-1}$

Final angular velocity, $\omega' = 0\ rads^{- 1}$

The average angular acceleration of the disk can be computed by using the kinematic eqn:

$\omega' = \omega + \alpha t$

$0 = 792.73 + 0.435\alpha$

$\alpha = - 1822.36\ rads^{- 2}$

5 0

3 years ago

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What is the magnitude (in N/C) and direction of an electric field that exerts a 3.50 ✕ 10−5 N upward force on a −1.55 µC charge?

IRISSAK [1]

Answer:

The magnitude of electric field is 22.58 N/C

Solution:

Given:

Force exerted in upward direction, $\vec{F_{up}} = 3.50\times 10^{- 5} N$

Charge, Q = $- 1.55\micro C = - 1.55\times 10^{- 6} C$

Now, we know by Coulomb's law,

$F_{e} = \frac{1}{4\pi\epsilon_{o}\frac{Qq}{R^{2}}$

Also,

Electric field, $E = \frac{1}{4\pi\epsilon_{o}\frac{q}{R^{2}}$

Thus from these two relations, we can deduce:

F = QE

Therefore, in the question:

$\vec{E} = \frac{\vec F_{up}}{Q}$

$\vec{E} = \frac{3.50\times 10^{- 5}}{- 1.55\times 10^{- 6}}$

$\vec{E} = - 22.58 N/C$

Here, the negative side is indicative of the Electric field acting in the opposite direction, i.e., downward direction.

The magnitude of the electric field is:

$|\vec{E}| = 22.58\ N/C$

6 0

2 years ago

ANSWER ASAP

shtirl [24]

5.6 g/ml. That is the density.

5 0

3 years ago

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A man has a mass of 85kg, what is his weight

Ostrovityanka [42]

The answer is he weighs 187.39 LBS/Pounds

5 0

3 years ago

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A hockey puck with mass 0.3 kg is shot across an ice-covered pond. Before the hockey puck was hit, the puck was at rest. After t

JulsSmile [24]

Answer:

The net friction force is 8.01 N

Explanation:

Net friction force = mass of hockey puck × acceleration

From the equations of motion

v^2 = u^2 + 2as

v = 40 m/s

u = 0 m/s (puck was initially at rest)

s = 30 m

40^2 = 0^2 + 2×a×30

60a = 1600

a = 1600/60 = 26.7 m/s^2

The acceleration of the puck is 26.7 m/s^2

Net friction force = 0.3 × 26.7 = 8.01 N

3 0

3 years ago

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