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nikdorinn [45]
3 years ago
15

There is an analogy between rotational and linear physical quantities. what rotational quantities are analogous to distance and

velocity?
Physics
2 answers:
Scrat [10]3 years ago
8 0

Answer:

Rotational units can be said to be analogous to linear physical units, for example, linear distance is analogous to angular displacement, since angular displacement is defined as the distance a body travels in a rotary motion.

On the other hand, linear velocity is analogous to angular velocity, since said angular velocity is the angular distance that a body covers per unit of time.

Explanation:

SVEN [57.7K]3 years ago
5 0

ANSWER:

Angular displacement is analogous to distance.

Angular velocity is analogous to velocity.

EXPLANATION:

FOR  ANGULAR DISPLACEMENT AND DISTANCE

Angular displacement = theta

theta = s/r

s = linear dispalcement/distance

r= radius.

so ,

linear displacement/distance  s = theta * r.

FOR  ANGULAR VELOCITY AND VELOCITY

Angular velocity = w

w = v/r

v = velocity

r = radius

so,

velocity v = w * r.

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The hypotenuse of an isosceles right triangle is 36 units. to the nearest tenth, what is the length of one of the legs
GarryVolchara [31]
The length of one of them would be 6
3 0
3 years ago
What is the strength of an electric field that will balance the weight of a 3.3 g plastic sphere that has been charged to -7.6 n
sp2606 [1]

As the plastic sphere is charged, therefore it experience an electric force when placed in an electric fields and also experiences gravitational force acts downward so the electric force must act upward.

Let  F_{E} is electric force and F_{G} is gravitational force.

If these forces are balanced, thereforeF_{E} = F_{G}

or                                                                      \left | q \right |E=mg\\\\\ E=\frac{mg}{\left | q \right |}

Given, q=-7.6 nC=-7.6\times10^{-9} C and  m=3.3 g= 3.3\times10^{-3} kg.

Substituting these values in above equation we get,

E=\frac{3.3\times10^{-3} kg\times9.8 m/s^{2} }{7.6\times10^{-9} C}\\\\E=4.2\times10^{6} N/C

Thus, the magnitude of electric field is 4.2\times10^{6} N/C.

As the charge is negative, the electric field at the location of the plastic sphere must be pointing downward.



8 0
3 years ago
In an experiment different wavelengths of light, all able to eject photoelectrons, shine on a freshly prepared (oxide-free) zinc
topjm [15]

Answer:

the energy of the photons is greater than the work function of the zinc oxide.

                     h f> = Ф

Explanation:

In this experiment on the photoelectric effect, it is explained by the Einstein relation that considers the light beam formed by discrete energy packages.

                    K_max = h f - Ф

in the exercise phase, they indicate that different wavelengths can inject electrons, so the energy of the photons is greater than the work function of the zinc oxide.

                     h f > = Ф

6 0
4 years ago
A proton moves through an electric potential created by a number of source charges. Its speed is 2.5 x 105 m/s at a point where
Kaylis [27]

Answer:

Proton’s speed, a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s

Explanation:

Given;

initial speed of proton, u = 2.5 x 10⁵ m/s

initial potential, V = 1500 V

mass of proton = 1.67 x 10⁻²⁷ kg

Work done, W = eV= ΔK.E = ¹/₂mu²

eV = ¹/₂mu² (J)

where;

e is the charge of the proton in coulombs

V is the electric potential in volts

m is the mass of the proton in kg

u is the speed of the proton in m/s

m =\frac{2eV_1}{u_1{^2}} = \frac{2eV_2}{u_2{^2}} = \frac{V_1}{u_1{^2}}} =\frac{V_2}{u_2{^2}}

\frac{V_1}{u_1{^2}}} =\frac{V_2}{u_2{^2}} = \frac{1500}{(2.5*10^5)^2} = \frac{500}{u_2{^2}} \\\\u_2{^2} =\frac{500*(2.5*10^5)^2}{1500} = 0.333*6.25*10^{10}\\\\u_2 = \sqrt{0.333*6.25*10^{10}} =1.4 *10^5 \ m/s

Therefore, proton’s speed a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s

4 0
4 years ago
The average standard rectangular building brick has a mass of 3.10 kg and dimensions of 225 m x 112 m x 75 m. The gravitational
zzz [600]

Answer:

P=3.61\times 10^{-3}\ Pa

Explanation:

Given that,

Mass of a brick, m = 3.1 kg

The dimensions of the brick 225 m x 112 m x 75 m

We need to find the maximum pressure created by the brick. We know that, the force acting per unit area is called pressure exerted. It is given by the formula as follows :

P=\dfrac{F}{A}

F = mg

A = area with minimum dimensions i.e. 112 m x 75 m

Pressure is maximum when the area is least.

So,

P=\dfrac{mg}{A}\\\\P=\dfrac{3.1\times 9.8}{112\times 75}\\\\P=3.61\times 10^{-3}\ Pa

So, the maximum pressure created by brick is 3.61\times 10^{-3}\ Pa.

5 0
3 years ago
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