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3 years ago

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A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Complete the following statemen

t: The centripetal acceleration of the ball can be increased by a factor of 4 by Entry field with incorrect answer keeping the radius fixed and increasing the period by a factor of 4. keeping the radius fixed and increasing the speed by a factor of 4. keeping the radius fixed and decreasing the period by a factor of 4. keeping the speed fixed and decreasing the radius by a factor of 4. keeping the speed fixed and increasing the radius by a factor of 4.

1 answer:

Eva8 [605]3 years ago

7 0

Answer:

Keeping the speed fixed and decreasing the radius by a factor of 4

Explanation:

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :

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

We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"

It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,

R' = R/4

New centripetal acceleration will be,

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

$a'=\dfrac{v^2}{R/4}$

$a'=4\times \dfrac{v^2}{R}$

$a'=4\times a$

So, the centripetal acceleration of the ball can be increased by a factor of 4.

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Scientists want to place a 3400.0 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2480.0

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Answer:

Part a)

$r = 6.96 \times 10^6 m$

Part b)

$F_g = 3.004 \times 10^3 N$

Part c)

$a = 0.88 m/s^2$

Part d)

$v = \frac{2480}{2} = 1240 m/s$

Explanation:

Part a)

As we know that the orbital speed of the satellite is given as

$v = 2480 m/s$

now we will have

$v = \sqrt{\frac{GM_{mars}}{r}}$

now we have

$M_{mars} = 6.4191 \times 10^{23} kg$

$R_{mars} = 3.397 \times 10^6 m$

now we have

$2480 = \sqrt{\frac{(6.67 \times 10^{-11})(6.4191 \times 10^{23})}{r}}$

$r = 6.96 \times 10^6 m$

Part b)

Here force between mars and satellite is the gravitational attraction force which is given as

$F_g = \frac{GM_{mars} m}{r^2}$

$F_g = \frac{(6.67\times 10^{-11})(6.4191 \times 10^{23})(3400)}{(6.96\times 10^6)^2}$

$F_g = 3.004 \times 10^3 N$

Part c)

Acceleration of satellite is the ratio of gravitational force and mass of the satellite

$a = \frac{F_g}{m}$

$a = \frac{3004}{3400}$

$a = 0.88 m/s^2$

Part d)

As we know by III law of kepler

$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$

here we know that T2 = 8 T1

$(\frac{1}{8})^2= \frac{r_1^3}{r_2^3}$

$\frac{r_1}{r_2} = (\frac{1}{2})^2$

so we have

$r_2 = 4r_1$

as we know that speed is given as

$v = \sqrt{\frac{GM}{r}}$

so we can say since radius is orbit becomes 4 times so the orbital speed must be half

$v = \frac{2480}{2} = 1240 m/s$

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