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LiRa [457]
3 years ago
11

A 30 kg rock sits motionless on the ground. What is the normal force on the rock?

Physics
1 answer:
Anna [14]3 years ago
5 0

Answer:30 kg

Explanation:

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Jacki evaluated the expression below. 2 cubed (3 minus 1) + 4 (8 minus 12) = 2 cubed (2) + 4 (4) = 8 (2) + 16 = 16 + 16 = 32. Wh
Naddika [18.5K]

Answer:

Her error was that she did not subtract 12 from 8 correctly

Explanation:

Jackie did 8-12 instead of 12-8

8 0
3 years ago
In the photoelectric effect, the greater the frequency of the illuminating light, the greater the:_______
TEA [102]

Answer:

B. Maximum velocity of ejected electrons.

Explanation:

The ejection of electrons form a metal surface when the metal surface is exposed to a monochromatic electromagnetic wave of sufficiently short wavelength or higher frequency (or equivalently, above a threshold frequency),  which leads to the enough energy of the wave to incident and get absorbed to the exposed surface emits electrons. This phenomenon is known as the photoelectric effect or photo-emission.

The minimum amount of energy required by a metal surface to eject an electron from its surface is called work function of metal surface.

The electrons thus emitted are called photo-electrons.

The current produced as a result is called photo electricity.

Energy of photon is given by:

E=h.\nu

where:

h = Planck's constant

\nu= frequency of the incident radiation.

8 0
3 years ago
A truck tire rotates at an initial angular speed of 21.5 rad/s. The driver steadily accelerates, and after 3.50 s the tire's ang
erma4kov [3.2K]

Given:

initial angular speed, \omega _{i} = 21.5 rad/s

final angular speed, \omega _{f} = 28.0 rad/s

time, t = 3.50 s

Solution:

Angular acceleration can be defined as the time rate of change of angular velocity and is given by:

\alpha = \frac{\omega_{f} - \omega _{i}}{t}

Now, putting the given values in the above formula:

\alpha = \frac{28.0 - 21.5}{3.50}

\alpha = 1.86 m/s^{2}

Therefore, angular acceleration is:

\alpha = 1.86 m/s^{2}

5 0
3 years ago
Calculate the radiative and collisional energy losses (in keV/micron) for a 1.9 MeV electron in lead and determine the rad./coll
Andrew [12]

Answer:

Explanation:

During an energy transfer, the collision loss for an electron can be determined by using the formula:

Q = \dfrac{4mME }{(m+M)^2}

However; from the total stopping power & power loss of the electron;

\dfrac{radiational \ energy \ loss}{colisional \ energy \ loss } = \dfrac{ZE}{800}

where;

Z = atomic no. for lead = 82

E = 1.9 MeV

∴

radiational energy loss = collisional energy loss  =\dfrac{82 \times 1.9}{800}

= 0.19475

b)

Normally, the traditional lead shielding in its pure shape contains high brittleness. However, the functionality of this carbon group chemical element is useful for protection because it has an excessive density.

Initially, the conventional lead protection however reduces the mild clarity at the same moment as plexiglass is useful for light transmittance and readability.

Moreover, the traditional lead with its high density and thickness reduces observation features, in the meantime, the plexiglass is a whole lot higher than the stated.

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4 0
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How does changing the wavelength alter the frequency of the wave? & How does altering frequency alter pitch?
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for a given type of wave in a given medium a larger frequency means a smaller wavelength

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