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makkiz [27]
3 years ago
5

HELP PLEASE!!

Physics
1 answer:
aliya0001 [1]3 years ago
3 0

Answer:

m=41 kg

v=0,02 ms

R=2,1 m

F-?

a=v²/R

a=(0,02) ²/2,1=0,0002

F=m*a

F=41*0,0002=0,0082 H

F=0,0082 H

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In the United States, household electric power is provided at a frequency of 60 HzHz, so electromagnetic radiation at that frequ
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Answer:

the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

Explanation:

Given the data in the question;

To determine the maximum intensity of an electromagnetic wave, we use the formula;

I = \frac{1}{2}ε₀cE_{max²

where ε₀ is permittivity of free space ( 8.85 × 10⁻¹² C²/N.m² )

c is the speed of light ( 3 × 10⁸ m/s )

E_{max is the maximum magnitude of the electric field

first we calculate the maximum magnitude of the electric field ( E_{max  )

E_{max = 350/f kV/m

given that frequency of 60 Hz, we substitute

E_{max = 350/60 kV/m

E_{max = 5.83333 kV/m

E_{max = 5.83333 kV/m × ( \frac{1000 V/m}{1 kV/m} )

E_{max = 5833.33 N/C

so we substitute all our values into the formula for  intensity of an electromagnetic wave;

I = \frac{1}{2}ε₀cE_{max²

I = \frac{1}{2} × ( 8.85 × 10⁻¹² C²/N.m² ) × ( 3 × 10⁸ m/s ) × ( 5833.33 N/C )²

I = 45 × 10³ W/m²

I = 45 × 10³ W/m² × ( \frac{1 kW/m^2}{10^3W/m^2} )

I = 45 kW/m²

Therefore, the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

7 0
3 years ago
As a car accelerates its tires increase their angular velocity from 114 rad/s to 131 rad/s in 0.43 seconds. What is the angular
e-lub [12.9K]
I think it's 39.53

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5 0
3 years ago
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Hello!

We can use Faraday's Law of Electromagnetic Induction to solve.

\epsilon = -N \frac{d\Phi_B}{dt}

ε = Induced emf (4.08 V)
N = Number of loops (?)

\Phi_B = Magnetic Flux (Wb)
t = time (s)

**Note: The negative sign can be disregarded for this situation. The sign simply shows how the induced emf OPPOSES the current.

Now, we know that \frac{d\Phi_B}{dt} is analogous to the change in magnetic flux over change in time, or \frac{\Delta \Phi_B}{\Delta t}, so:
\epsilon = N \frac{\Delta \Phi_B}{\Delta t}\\\\\epsilon = N \frac{\Phi_{Bf} - \Phi_{Bi}}{\Delta t}

Rearrange the equation to solve for 'N'.

N = \frac{\epsilon}{ \frac{\Phi_{Bf} - \Phi_{Bi}}{\Delta t}}

Plug in the given values to solve.

N = \frac{4.08}{ \frac{9.44*10^{-5} - 2.57*10^{-5}}{0.0154}}  = 914.585 = \boxed{915 \text{ coils}}

**Rounding up because we cannot have a part of a loop.

4 0
2 years ago
An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference
kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

K=qV          (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J

The kinetic energy of the particle is also:

K=\frac{1}{2}mv^2       (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

|F_B|=qvBsin(\theta)

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The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

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the radius of the trajectory of the electron is 10.1 cm

6 0
4 years ago
which of the following are examples of circular motion? A. Roller skating down a hill. B. A race car going around a rounded curv
Katena32 [7]

Answer:

C. Earth going around the sun.

Explanation:

Circular motion should have a center to repeat its motion.

6 0
3 years ago
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