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vichka [17]
3 years ago
6

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?
Physics
1 answer:
statuscvo [17]3 years ago
3 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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A. because I had this question yesterday.
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We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What mu
liubo4ka [24]

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}

\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}

f = 30 cm

using lens formula

\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})

R_1 = R\ and\ R_2 = -R

\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})

R = (n -1)\ f

R = 2(1.5 -1)\ 30

R = 30 cm

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3 0
3 years ago
I push a box with mass 10 kg across a 15 m room. If I apply a force of 20N to do this how much work is being done on the box?
devlian [24]
You could use the formula

W=Fd

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8 0
3 years ago
A light beam approaches a plain mirror at an incident angle of 39
Paul [167]
Angle is 31 adding all sides
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3 years ago
When an old lp turntable was revolving at 33.3 rpm, it was shut off and uniformly slowed down and stopped in 5.5 seconds. throug
velikii [3]
First, let's convert the initial angular speed from rpm (rev/min) into rad/s, keeping in mind that 1 rev = 2 \pi rad and 1 min=60 s:
\omega_i = 33  \frac{rev}{min}=33\frac{rev}{min} \cdot  \frac{2 \pi rad}{60 s}=3.45 rad/s

Now we can find the angular acceleration of the lp, keeping in mind that the final speed is zero:
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Now we can find the angle covered by the LP from the beginning to the end of its motion:
\theta (t)= \omega_i t + \frac{1}{2}\alpha t^2 = (3.45 rad/s)(5.5 s)+ \frac{1}{2}(-0.63 rad/s^2)(5.5 s)^2=
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And finally, we can convert it into number of revolutions:
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3 years ago
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