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White raven [17]
3 years ago
9

If the object represented by the FBD below has a mass of 2.0 kg, what is the

Physics
1 answer:
ycow [4]3 years ago
4 0

Answer: B

Explanation:

Let us first resolve the forces.

Horizontal, the vectorial resolution is equal to zero. Since the positive direction and negative direction vector are of same magnitude. Therefore, summation of horizontal forces Fh = 0

Vertical resolution, the sum of forces will be Fv = 30 - 35 = - 5 N

Resultant force F = sqrt( Fv^2 + Fh^2)

Substitutes Fv and Fh. We get

R = sqrt ( 25) = 5N

We are given that the mass M = 2kg

From Newton second law,

F = Ma

Where a = acceleration

Substitute F and M and make a the subject of formula

5 = 2a

a = 5/2 = 2.5 m/s^2

Therefore, acceleration a of the object is 2.5 m/s^2 down.

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C humid subtropical
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Traveling waves propagate with a fixed speed usually denoted as v (but sometimes c). The waves are called __________ if their wa
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Answer:

Periodic.

Explanation:

Electromagnetic waves is a propagating medium used in all communications device to transmit data (messages) from the device of the sender to the device of the receiver.

Generally, the most commonly used electromagnetic wave technology in telecommunications is radio waves.

Radio waves can be defined as an electromagnetic wave that has its frequency ranging from 30 GHz to 300 GHz and its wavelength between 1mm and 3000m. Therefore, radio waves are a series of repetitive valleys and peaks that are typically characterized of having the longest wavelength in the electromagnetic spectrum.

Basically, as a result of radio waves having long wavelengths, they are mainly used in long-distance communications such as the carriage and transmission of data.

Generally, a fixed speed is used for the propagation of traveling waves and this speed is usually denoted with the variable "v" or sometimes "c."

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Mathematically, the period of a traveling wave is given by the formula;

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5 0
2 years ago
Review. An electron moves in a circular path perpendicular to a constant magnetic field of magnitude 1.00 mT . The angular momen
Ghella [55]

The speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

The angular momentum(L) of an electron moving in a circular path is given by the formula,

L = mvr ........(i)

We know that the radius of the path of an electron in a magnetic field is

r = mv/qB

Putting this value in equation (i),

L = mv x mv/qB

or L = (mv)^2/qB

Putting the given values in the above equation,

4 x 10^-25 = (9.1x10^-31)^2 x v^2/ 1.6 x 10^-19 x 1 x 10^-3

v comes out to be 8.88 x 10^7 m/s.

Hence, the speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

To know more about "angular momentum", refer to the following link:

brainly.com/question/15104254?referrer=searchResults

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5 0
1 year ago
Children are told to avoid standing too close to a rapidly moving train because they might get sucked under it. Is this possible
storchak [24]

Answer:

no its not like the undertow in the ocean

Explanation:

4 0
3 years ago
Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900
Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron m_{e}=9.11\times10^{-31}\ kg

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(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}

E=0.582\ Mev

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

E=0.350\ Mev

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0
3 years ago
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