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shutvik [7]
3 years ago
6

A 3.0 kg mass is released from rest at point A. The mass slides along the curved surface to point B in 6.0 seconds. Point B is 2

.0 m lower than A and the mass is moving at 2.0 m/s at point B. Find the average power exerted by kinetic friction on the mass during this segment. a) 0 W b) -3 W c) -6 W d) -9 W e) -18 W
Physics
1 answer:
timurjin [86]3 years ago
5 0

Answer:

option d) -9 J

Explanation:

Given:

Mass, m = 3.0 kg

time, t = 6.0 seconds

Velocity of mass, v = 2.0 m/s

height, h = 2 m

Now, using the concept of work-Energy theorem

we have

Net work done = change in kinetic energy

or

Work done by gravity + work done by the friction = Final kinetic energy - Initial kinetic energy

mgh +W_f = \frac{1}{2}mv^2-0

on substituting the values in the above equation, we get

3 × 9.8 × 2 + W_f = \frac{1}{2}\times 3\times2^2

or

58.8 + W_f = 6

or

W_f = -52.8 J

here negative sign depicts that the work is done against the motion of the mass

also,

Power = (Work done)/time

or

Power = -52.8/6 = -8.8 W ≈ 9 J

Hence, option d) -9 J is correct

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In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.2 106 m/s.
Murrr4er [49]

The central force acting on the electron as it revolves in a circular orbit is 9.52 \times 10^{-8} \ N.

The given parameters;

  • <em>speed of electron, v = 2.2 x 10⁶ m/s</em>
  • <em>radius of the circle, r = 4.63 x 10⁻¹¹ m</em>

<em />

The central force acting on the electron as it revolves in a circular orbit is calculated as follows;

F = \frac{M_e v^2}{r} \\\\

where;

M_e is mass of electron = 9.11 x 10⁻³¹ kg

F = \frac{(9.11 \times 10^{-31}) \times(2.2\times 10^6)^2 }{4.63 \times 10^{-11}} \\\\F = 9.52 \times 10^{-8} \ N

Thus, the central force acting on the electron as it revolves in a circular orbit is 9.52 \times 10^{-8} \ N.

Learn more about centripetal force here:brainly.com/question/20905151

8 0
2 years ago
0/2 File Limit
slamgirl [31]

Answer:

Speed at which it will reach the ground is given as

v_f = 46.8 m/s

Total time for which it will remain in air is given as

t = 6.3 s

Explanation:

As we know that the object is projected upwards with speed

v_i = 15 m/s

g = - 9.81 m/s^2

now when it will reach the ground then we have

y = v_y t + \frac{1}{2} at^2

so we have

-100 = 15 t - \frac{1}{2}(-9.81) t^2

4.905 t^2 - 15 t - 100 = 0

so we have

t = 6.3 s

Now speed of the object when it reaches the ground is given as

v_f = v_i + at

v_f = -15 + (9.81)(6.3)

v_f = 46.8 m/s

8 0
3 years ago
Explain the following observations:
Deffense [45]

Answer and Explanation:

a. An oxygen-filled balloon is not able to float in the air, because the oxygen inside the balloon is of the same density, that is, the same "weight" as the oxygen outside the balloon and present in the atmosphere. The balloon can only float if the gas inside it is less dense than atmospheric oxygen. Helium gas is less dense than atmospheric gas, so if a balloon is filled with helium gas, that balloon will be able to float because of the difference in density.

b. The ship is able to float in the water because its steel construction is hollow and full of air. This makes the average density of this ship less than the density of water, which makes the ship lighter than water and for this reason, this ship is able to float. In addition, the ship is partially immersed, allowing the weight of the ship on the water to counteract the buoyant force that the water promotes on the ship. Weight and buoyant are two opposing forces that keep the ship afloat.

6 0
2 years ago
A zero-order reaction has a constant rate of 2.30×10−4 M/s. If after 80.0 seconds the concentration has dropped to 1.50×10−2 M,
dolphi86 [110]

Answer:

Initial concentration of the reactant = 3.34 × 10^(-2)M

Explanation:

Rate of reaction = 2.30×10−4 M/s,

Time of reaction = 80s

Final concentration = 1.50×10−2 M

Initial concentration = Rate of reaction × Time of reaction + Final concentration

= 2.30×10−4 M/s × 80s + 1.50×10−2 M = 3.34 × 10^(-2)M

Initial concentration = 3.34 × 10^(-2)M

6 0
2 years ago
An object of mass 20 kg is raised vertically through a distance of 8 m above ground level. If g = 10 m/s2, what is the gravitati
IgorC [24]

2400joules

Explanation:

P.E

m= 20kg h=8m g=10m

P.E= 20×8×10

=2400joules

6 0
2 years ago
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