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Nostrana [21]
3 years ago
13

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Which way(s) can increase the c

entripetal acceleration of the ball by a factor of 9? Group of answer choices Keeping the speed fixed and decreasing the radius by a factor of 9 Keeping the radius fixed and increasing the speed by a factor of 3 decreasing both the radius and the speed by a factor of 9 Keeping the radius fixed and increasing the speed by a factor of 9 increasing both the radius and the speed by a factor of 9 Keeping the speed fixed and increasing the radius by a factor of 9
Physics
1 answer:
Charra [1.4K]3 years ago
4 0

Explanation:

When an object moves in a circular path, due to the change in its velocity, the object possess centripetal acceleration. The formula for the centripetal acceleration is given by :

a=\dfrac{v^2}{R}

Where

v is the speed of the object

R is the radius of the circle path

Option (1) : Keeping the speed fixed and decreasing the radius by a factor of 9

a=\dfrac{v^2}{R/9}

a=\dfrac{9v^2}{R}

The centripetal acceleration of the ball by a factor of 9.

Option (2) : Keeping the radius fixed and increasing the speed by a factor of 3

a=\dfrac{(3v)^2}{R}

a=\dfrac{9v^2}{R}

The centripetal acceleration increases by a factor of 9.

Option (3) : Decreasing both the radius and the speed by a factor of 9.

a=\dfrac{(v/9)^2}{R/9}

a=\dfrac{(v)^2}{9R}

The centripetal acceleration decreases by a factor of 9.

Option (4) : Keeping the radius fixed and increasing the speed by a factor of 9

a=\dfrac{(3v)^2}{R}

a=\dfrac{9v^2}{R}

The centripetal acceleration increases by a factor of 9.

Option (5) : Increasing both the radius and the speed by a factor of 9

a=\dfrac{(9v)^2}{9R}

a=\dfrac{9v^2}{R}

The centripetal acceleration increases by a factor of 9.

Option (6) : Keeping the speed fixed and increasing the radius by a factor of 9

a=\dfrac{(v)^2}{9R}

a=\dfrac{9v^2}{R}

The centripetal acceleration increases by a factor of 9.

So, as the radius of the circle decreases, its centripetal acceleration increase. Also, if the speed of the object increases, its centripetal acceleration increase. Hence, this is the required solution.

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un avión se mueve en línea recta a una velocidad constante de 400 km/h durante 1,5h de su recorrido ¿que distancia recorrió en e
Amanda [17]

The distance covered by the plane is 600 km

Explanation:

The motion of the plane is a uniform motion, so at constant velocity, therefore we can use the following equation

d=vt

where

d is the distance covered

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t is the time interval considered

For the plane in this problem,

v = 400 km/h

t = 1.5 h

Substituting, we find the distance covered:

d=(400)(1.5)=600 km

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6 0
3 years ago
A loop of wire in the shape of a rectangle rotates with a frequency of 284 rotation per minute in an applied magnetic field of m
yaroslaw [1]

Answer:

a) Magnitude of maximum emf induced = 0.0714 V = 71.4 mv

b) Maximum current through the bulb = 0.00793 A = 7.93 mA

Explanation:

a) The induced emf from Faraday's law of electromagnetic induction is related to angular velocity through

E = NABw sin wt

The maximum emf occurs when (sin wt) = 1

Maximum Emf = NABw

N = 1

A = 4 cm² = 0.0004 m²

B = 6 T

w = (284/60) × 2π = 29.75 rad/s

E(max) = 1×0.0004×6×29.75 = 0.0714 V = 71.4 mV

Note that: since we're after only the magnitude of the induced emf, the minus sign that indicates that the induced emf is 8n the direction opposite to the change in magnetic flux, is ignored for this question.

b) Maximum current through the bulb

E(max) = I(max) × R

R = 9 ohms

E(max) = 0.0714 V

I(max) = ?

0.0714 = I(max) × 9

I(max) = (0.0714/9) = 0.00793 A = 7.93 mA

Hope this Helps!!

8 0
3 years ago
10 points
Nana76 [90]

Answer:

Explanation:

You didn't fill in the proper masses which is why you never got an answer to this. But that's ok...I got you. I happen to know what they are! We will use the universal law of gravitation and the gravitational constant to solve this.

F_g=\frac{Gm_1m_2}{r^2} and filling in:

F_g=\frac{(6.67*10^{-11})(5.98*10^{24})(7.36*10^{22})}{(3.84*10^8)^2} The denominator is the radius of the earth plus the radius of the moon plus the distance between their surfaces, just FYI.

That gives us that

F_g=1.99*10^{20}N Not sure what your choices entail, but I'd have to say, taking into consideration that maybe your problem didn't figure in the distance between the surfaces, you'd be at choice B.

5 0
2 years ago
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