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3 years ago

7

The famous black planet, haunch, has a radius of 106 m, a gravitational acceleration at the surface of 4 m/s2 , and the tangenti

al speed of any point on its equator is 103 m/s. a famous haunch ant has a mass of 40 kg and is standing on a spring scale at the equator. what is the reading of the spring scale?

1 answer:

levacccp [35]3 years ago

6 0

Gravity on the surface = 4 m/s^2
Now, the acceleration due to centripetal motion, a = v^2/R

Where,
v= 10^3 m/s, R = 10^6 m
Then,
a = (10^3)^2/(10^6) = 1 m^2/s

The net gravitational acceleration = 4-1 = 3 m/s^2

The reading on the spring scale = ma = 40*3 = 120 N

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3 years ago

The next closest star to Earth, Proxima Centuri, is 4.25 lightyears away. How far is that in meters?

lozanna [386]

Answer:

4.0208 x 10^16 meters

Explanation:

speed of light in m/s * # seconds in a year * 4.25 years =

2.99792 x 10^8 * 3600*24*365.25 * 4.25 = 4.0208 x 10^16 m

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2 years ago

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 30 m/s2 for 35 s , then ru

Gre4nikov [31]

Answer:

a) $h_m=74625\ m$

b) $t=265.55\ s$

Explanation:

Given:

mass of rocket, $m_r=200\ kg$

mass of fuel, $m_f=100\ kg$

acceleration of the rocket consuming fuel, $a=30\ m.s^{-2}$

time after which the fuel exhaust, $t_f=35\ s$

<u>During the phase of fuel exhaustion:</u>

<u>velocity attained by the rocket just as the fuel ends:</u>

$v_f=u+a.t_f$

where:

$u=$ initial velocity of the rocket = 0

$v_f=0+30\times 35$

$v_f=1050\ m.s^{-1}$ this will be the initial velocity for the phase of ascending of the rocket's height under the influence of gravity.

<u>height at which the fuel finishes:</u>

$v_f^2=u^2+2a.h_f$

$1050^2=0^2+2\times 30\times h_f$

$h_f=18375\ m$

<u>During the phase of ascend in height of rocket after the fuel is over:</u>

<u>Time taken to reach the top height after the fuel is over:</u>

$v=v_f+g.t'$

at top v = (final velocity during this course of motion )= 0 $m.s^{-1}$

$0=1050-9.8\times t'$

$t'=107.1429\ s$

<u>Height ascended by the rocket after the fuel is over:</u>

$v^2=v_f^2+2g.h'$

at the top height the velocity is zero

$0^2=1050^2-2\times 9.8\times h'$ (-ve sign denotes that the direction of motion is opposite to that of acceleration)

$h'=56250\ m$

<u>Therefore the maximum altitude attained by the rocket:</u>

$h_m=h_f+h'$

$h_m=18375+56250$

$h_m=74625\ m$

b)

time taken by the rocket to fall back to the earth:

$h_m=v.t_m+\frac{1}{2} g.t_m^2$

where:

$v=$ initial velocity of the rocket during the course of free fall from the top height.

$74625=0+4.9\times t_m^2$

$t_m=123.41\ s$

Now the total time for which the rocket is in the air:

$t=t_f+t'+t_m$

$t=35+107.1429+123.41$

$t=265.55\ s$

4 0

3 years ago

A friend rides, in turn, the rims of three fast merry-go-rounds while holding a sound source that emits isotropically at a certa

Aliun [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

b

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

c

The ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

Explanation:

Mathematically Frequency can be represented as

$F = \frac{v}{\lambda}$

Where $\lambda$ is the wavelength and v is the velocity

Now looking at the diagram we see that

For the first frequency we have

Let the wavelength be $\lambda_1 = \lambda$ , and the frequency $F_1 = F$

For the second frequency

Let the wavelength be $\lambda_2 = 2 \lambda$ , and the frequency $F_2 = \frac{F}{2}$

For the third frequency

Let the wavelength be $\lambda_3 = 2\lambda$ , and the frequency $F_3 = \frac{F}{2}$

To obtain v for each of the frequency we make v the subject in the equation above for each frequency

So,

For the first frequency we have

$v_1 = \lambda_1 F_1 = \lambda F$

For the second frequency

$v_2 = \lambda_2 F_2 = 2 \lambda*\frac{F} {2} = \lambda F$

For the third frequency

$v_3 = \lambda_3 F_3 = 2 \lambda*\frac{F} {2} = \lambda F$

Hence

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

Mathematically angular speed can be represented as

$w = 2 \pi f$

For the first frequency we have

$w_1 = 2\pi F_1 = 2 \pi F$

For the second frequency

$w_2 = 2 \pi F_2 = 2 \pi \frac{F}{2} = \pi F$

For the third frequency

$w_3 = 2 \pi F_3 = 2 \pi \frac{F}{2} = \pi F$

Hence

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

Mathematically the relationship between the angular velocity and the linear velocity can be represented as

$v = wr$

=> $r = \frac{v}{w}$

Since the linear velocity is constant we have that

$r \ \alpha \ \frac{1}{w}$

This means that r varies inversely to the angular velocity ,What this means for ranking due to the radius is that the ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

5 0

3 years ago

What’s better csp or pv ?

BARSIC [14]

Answer:

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Explanation:

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8 0

1 year ago