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Morgarella [4.7K]
3 years ago
10

The number of moons a planet has orbiting around it is determined by...

Physics
1 answer:
Pavlova-9 [17]3 years ago
7 0

We have no idea.  There are probably many things involved ... the planet's mass,
the availability of small bodies in the neighborhood that can be captured, etc.


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Which option gives an
9966 [12]

Answer:

option b is correct

Explanation:

the SI  unit of temperature is kelvin

so option gives an objects temperature in SI unit

6 0
3 years ago
The maximum permissible workday dose for occupational exposure to radiation is 18 mrem. A 54-kg laboratory technician absorbs 2.
professor190 [17]
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6 0
3 years ago
A 200g particle oscillating in SHM travels 66cm between the two extreme points in its motion with an average speed of 110cm/s. F
IgorC [24]
For a:v = d / Δt 
110 = 0.66 / Δt 
Δt = 0.66 / 110 
Δt = 0.006 s 
the period is: 
T = 2Δt 
T = 2*0.006 
T = 0.012 s 
the frequency is the inverse of the period. so: f = 1 / T 
f = 83.3333333 Hz (about; Hz = 1/s) 
b. T = 2π√(m/k) 
being the mass m = 200g = 0.2 kg = 2*10^-1 kg, π = 3.14 (about) and T = 0.012, k is equal to: 
0.012 = 6.28√(2*10^-1 / k) 
0.012 / 6.28 = √(2*10^-1 / k) 
0.00191082803 = √(2*10^-1 / k) 
2*10^-1/ k = 0.000003
2*10^-1 / k = 3*10^-6 
k = 2*10^-1 / 3*10^-6
k = 6.67*10^-5

now using hooke's law:
F = -kx 
F = - 6.67*10^-5* 3.3*10^-1
F = -2.20x10^-5m
F = -0.22 *10^4 N 
4 0
3 years ago
A disk with mass 1.64 kg and radius 0.61 meters is spinning counter-clockwise with an angular velocity of 17.6 rad/s. A rod of m
Masja [62]

Answer:

The loss in rotational kinetic energy due to the collision is 36.585 J.

Explanation:

Given;

mass of the disk, m₁ = 1.64 kg

radius of the disk, r = 0. 61 m

angular velocity of the disk, ω₁ = 17.6 rad/s

mass of the rod, m₂ = 1.51 kg

length of the rod, L = 1.79 m

angular velocity of the rod, ω₂ =  5.12 rad/s (clock-wise)

let the counter-clockwise be the positive direction

let the clock-wise be the negative direction

The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;

m₁ω₁  + m₂ω₂ = ωf(m₁ + m₂)

where;

ωf is the common final angular velocity

1.64 x 17.6    + 1.51(-5.12) = ωf(1.64 + 1.51)

21.1328 = ωf(3.15)

ωf = 21.1328 / 3.15

ωf = 6.709 rad/s

The moment of inertia of the disk is calculated as follows;

I_{disk} = \frac{1}{2} mr^2\\\\I_{disk}  = \frac{1}{2} (1.64)(0.61)^2\\\\I_{disk}  = 0.305 \ kgm^2

The moment of inertia of the rod about its center is calculated as follows;

I_{rod} = \frac{1}{12} mL^2\\\\I_{rod} = \frac{1}{12} \times 1.51 \times 1.79^2\\\\I _{rod }= 0.4032\ kgm^2

The initial rotational kinetic energy of the disk and rod;

K.E_i = \frac{1}{2} I_{disk}\omega _1 ^2 \ \ + \ \  \frac{1}{2} I_{rod}\omega _2 ^2 \\\\K.E_i=  \frac{1}{2} (0.305)(17.6) ^2 \ \ + \ \  \frac{1}{2} (0.4032)(-5.12) ^2\\\\K.E_i = 52.523 \ J

The final rotational kinetic energy of the disk-rod system is calculated as follows;

K.E_f = \frac{1}{2} I_{disk}\omega _f ^2 \ \ + \ \  \frac{1}{2} I_{rod}\omega _f ^2\\\\K.E_f = \frac{1}{2} \omega _f ^2(I_{disk} + I_{rod})\\\\K.E_f = \frac{1}{2} (6.709) ^2(0.305+ 0.4032)\\\\K.E_f = 15.938 \ J

The loss in rotational kinetic energy due to the collision is calculated as follows;

\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 15.938 J  \ - \ 52.523 J\\\\\Delta K.E = - 36.585 \ J

Therefore, the loss in rotational kinetic energy due to the collision is 36.585 J.

8 0
2 years ago
You stare at a bright red screen for so long that your red cones become saturated and no longer function. The red screen is then
adell [148]

Answer: you'll see cyan color on the screen

Explanation:

Saturating the red cone causes them to stop functioning, hence you can't perceive the red part of white light. White light is made up of three main colors which are blue, red and green. When one can no longer perceive the red part of light, one is left with the grean and blue part. The green and blue part of light will superimpose to give a cyan color.

5 0
3 years ago
Read 2 more answers
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