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Anna11 [10]
3 years ago
8

Which of the following are not examples of circular motion? Check all that apply

Physics
1 answer:
CaHeK987 [17]3 years ago
5 0

B.riding on a ferris wheel

E.a satellite orbiting the Earth

Explanation:

A person riding on a ferris wheel and a satellite orbiting the earth both depicts circular motion.

Circular motion is the motion of a body about a fixed axis or center.

  • The motion path of the body forms a circle around the center.
  • In circular motion, the travel path is along a circle.
  • An orbiting satellite traces out its orbit round the earth.
  • At certain times, it passes through that point over and over again.
  • This is similar to that of a ferris wheel

Other options gives other types of motion.

learn more:

circular motion brainly.com/question/2562955

#learnwithBrainly

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What is the weight (in newtons) of a bowling ball that has a mass of 3 kilograms?
ipn [44]
It will be <span>29.419950086 Newtons.</span>
8 0
3 years ago
A solenoid has a length , a radius , and turns. The solenoid has a net resistance . A circular loop with radius is placed around
geniusboy [140]

This question is incomplete, the complete question is;

A solenoid has a length 11.34 cm , a radius 1.85 cm , and 1627 turns. The solenoid has a net resistance of 144.9 Ω . A circular loop with radius of 3.77 cm is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance of 1651.6 Ω. At a time of 0 s , the solenoid is connected to a battery that supplies a potential 34.95 V. At a time 2.58 μs , what current flows through the outer loop?

Answer:

the current flows through the outer loop is 1.3 × 10⁻⁵ A

Explanation:

Given the data in the question;

Length l = 11.34 cm = 0.1134 m

radius a = 1.85 cm = 0.0185 m

turns N = 1627

Net resistance R_{sol = 144.9 Ω

radius b = 3.77 cm = 0.0377 m

R_o = 1651.6 Ω

ε = 34.95 V

t = 2.58 μs = 2.58 × 10⁻⁶ s

Now, Inductance; L = μ₀N²πa² / l

so

L = [ ( 4π × 10⁻⁷ ) × ( 1627 )² × π( 0.0185 )² ] / 0.1134

L = 0.003576665 / 0.1134

L = 0.03154

Now,

ε = d∅/dt = \frac{d}{dt}( BA ) =  \frac{d}{dt}[ (μ₀In)πa² ]

so

ε = μ₀n \frac{dI}{dt}( πa² )

ε = [ μ₀Nπa² / l ] \frac{dI}{dt}

ε = [ μ₀Nπa² / l ] [ (ε/L)e^( -t/R_{sol) ]

I = ε/R_o = [ μ₀Nπa² / R_ol ] [ (ε/L)e^( -t/R_{sol) ]

so we substitute in our values;

I = [ (( 4π × 10⁻⁷ ) × 1627 × π(0.0185)²) / (1651.6 ×0.1134) ] [ ( 34.95 / 0.03154)e^( -2.58 × 10⁻⁶ / 144.9 ) ]

I = [ 2.198319 × 10⁻⁶ / 187.29144 ] [ 1108.116677 × e^( -1.7805 × 10⁻⁸ )

I = [ 1.17374 × 10⁻⁸ ] × [ 1108.116677 × 0.99999998 ]

I = [ 1.17374 × 10⁻⁸ ] × [ 1108.11665 ]

I = 1.3 × 10⁻⁵ A

Therefore, the current flows through the outer loop is 1.3 × 10⁻⁵ A

7 0
2 years ago
An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electric
ELEN [110]

Answer:

a. 0 W b. ε²/R c. at R = r maximum power = ε²/4r d. For R = 2.00 Ω, P = 227.56 W. For R = 4.00 Ω, P = 256 W. For R = 6.00 Ω, P = 245.76 W

Explanation:

Here is the complete question

An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electrical power output of the source. By conservation of energy, P is equal to the power consumed by R. What is the value of P in the limit that R is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when R = r . What is this maximum P in terms of ε and r? (d) A battery has ε= 64.0 V and r=4.00Ω. What is the power output of this battery when it is connected to a resistor R, for R=2.00Ω, R=4.00Ω, and R=6.00Ω? Are your results consistent with the general result that you derived in part (b)?

Solution

The power P consumed by external resistor R is P = I²R since current, I = ε/(R + r), and ε = e.m.f and r = internal resistance

P = ε²R/(R + r)²

a. when R is very small , R = 0 and P = ε²R/(R + r)² = ε² × 0/(0 + r)² = 0/r² = 0

b. When R is large, R >> r and R + r ⇒ R.

So, P = ε²R/(R + r)² = ε²R/R² = ε²/R

c. For maximum output, we differentiate P with respect to R

So dP/dR = d[ε²R/(R + r)²]/dr = -2ε²R/(R + r)³ + ε²/(R + r)². We then equate the expression to zero

dP/dR = 0

-2ε²R/(R + r)³ + ε²/(R + r)² = 0

-2ε²R/(R + r)³ =  -ε²/(R + r)²

cancelling out the common variables

2R =  R + r

2R - R = R = r

So for maximum power, R = r

So when R = r, P = ε²R/(R + r)² = ε²r/(r + r)² = ε²r/(2r)² = ε²/4r

d. ε = 64.0 V, r = 4.00 Ω

when R = 2.00 Ω, P = ε²R/(R + r)² = 64² × 2/(2 + 4)² = 227.56 W

when R = 4.00 Ω, P = ε²R/(R + r)² = 64² × 4/(4 + 4)² = 256 W

when R = 6.00 Ω, P = ε²R/(R + r)² = 64² × 6/(6 + 4)² = 245.76 W

The results are consistent with the results in part b

8 0
3 years ago
How could you determine the instantaneous speed during a<br> trip?
Fiesta28 [93]
Look at your speedometer for say, a couple of seconds. Depends on whether or not you are moving on average at a constant speed (speedo won't change much) or whether you're in a polluting traffic jam/queue in which case the speedo will go up and down like a yo yo. to determine the speed, you'd probably need to plot the speed on the speedo against the times at which the speedo speeds were read from the speedo.
6 0
3 years ago
Read 2 more answers
You throw a basketball and a tennis ball across the classroom so that each ball has the exact same momentum. Explain how this is
Helga [31]

Answer:

It is possible by increasing the speed of the tennis ball by a factor of (Mass of the tennis ball)/(Mass of the basketball)

Explanation:

The momentum of a body = The bod's mass × The body's velocity

Therefore, the momentum of a given mass of an object, such as a tennis ball can be increased by increasing the velocity or speed of the object. Whereby the speed of the ball, v₁, is increased such that the momentum of the basketball and the tennis ball will be the same, is given by the following equation

Mass of the basketball × v₂ = Mass of the tennis ball × v₁

Therefore, v₁/v₂ = (Mass of the tennis ball)/(Mass of the basketball)

7 0
3 years ago
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