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BartSMP [9]
3 years ago
7

Solve for M₂

Physics
1 answer:
soldi70 [24.7K]3 years ago
5 0

Explanation:

M₂ = Fr²/GM₁

M₂ = [(132N)(.243m)²]/[(6.67*10^-11N*m²/kg)(1.175*10^4kg)]

M₂ = (7.79N*m²)/(7.84*10^-7N*m²)

M₂ = 9.94*10^6 kg

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A 4. 0 nc positive point charge is located at point a in the figure. (figure 1) what is the electric potential at point b?.
algol [13]

The electric potential at point b experienced by the charge cab be determined using the formulas given.

<h3>Electric potential</h3>

The electric potential of a point charge is the work done in moving the charge from infinity to certain point against the electric field.

V = Ed

V = (F/q)d

V = (Fd)/q

where;

  • V is the electric potential
  • F is electric force
  • E is the electric field
  • q is the charge

Thus, the electric potential at point b experienced by the charge cab be determined using the formulas given.

Learn more about electric potential here: brainly.com/question/14306881

7 0
2 years ago
Listed following are the names and mirror diameters for six of the world’s greatest reflecting telescopes used to gather visible
ziro4ka [17]

Answer:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope

Explanation:

How much light a telescope can collect depends on its diameter, since in a bigger area more photons will be collected.    

Remember that in a circle the area is defined as:

A = \pi r^{2}  (1)

Where A is the area and r is its radius.

However, the radius can be determined by means of its diameter.

     

d = 2r

r = \frac{d}{2} (1)

Where d is its diameter.

An example of this is when a person is collecting raindrops with a bucket and with a cup. Since the bucket has a bigger area than the cup, it will collect more raindrops by unit of time. In this scenario the raindrops represent the photons.  

   

To determine the light collecting area of each telescope, equation 2 will be replaced in equation 1.

A = \pi (\frac{d}{2})^{2}  (3)

Case for Large binocular telescope:

A_{mirror1} = \pi (\frac{8.4m}{2})^{2}    

A_{mirror1} = 55.41m        

For the second mirror will be the same value

A = A_{mirror1}+A_{mirror2}  

A = 55.41m+55.41m

A= 110.82m

Case for Keck 1 telescope:

A = \pi (\frac{10m}{2})^{2}    

A = 78.53m  

Case for Hobby-Ebberly telescope:

A = \pi (\frac{9.2m}{2})^{2}    

A = 66.47m  

Case for Subaru telescope:

A = \pi (\frac{8.3m}{2})^{2}    

A = 54.10m  

Case for Gemini North telescope:

A = \pi (\frac{8m}{2})^{2}    

A = 50.26m  

Case for Magellan 2 telescope:

A = \pi (\frac{6.5m}{2})^{2}    

A = 33.18m  

Hence, they may be rank in the following way:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope.

<em>Key term:</em>

<em>Photons: particles that constitute light. </em>

3 0
3 years ago
Does coefficient of linear expansion depends on length​
nlexa [21]

Answer:

yes

Explanation:

As the formula is α= ΔL/L*ΔT  where alpha (α) is the sign of coefficient of linear expansion

5 0
3 years ago
Two astronauts, each having a mass of 74 kg, are connected by a 8.53 m rope of negligible mass. They are isolated in space, orbi
AveGali [126]

Answer:

a) 2575 kgm²/s

b) 1.23 kJ

c) 0.478 rad/s

Explanation:

Given

Mass of astronauts, m = 74 kg

Length of rope, l = 8.53 m

Speed of orbit, v = 4.08 m/s

L = m1v1.x1(i) + m2v2.x2(i) = 2mv(d/2)

Thus, L = 2.m.v.(d/2)

L = 2 * 74 * 4.08 * (8.53/2)

L = 2 * 74 * 4.08 * 4.265

L = 2575.38 kgm²/s

Rotational Energy of the system

K(i) = 1/2m1v1(i)² + 1/2m2v2(i)²

K(i) = 2(1/2) * 74 * 4.08²

K(i) = 74 * 16.6464

K(i) = 1231.83 J = 1.23 kJ

Angular momentum is conserved, thus, angular velocity, w = v/r

w = 4.08 / 8.53

w = 0.478 rad/s

7 0
3 years ago
Based on the law of conservation of energy, how can we reasonably improve a machine’s ability to do work?
Sholpan [36]
You will have to redefine the machine's boundaries.
6 0
3 years ago
Read 2 more answers
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