Answer:
The option is B is not true for Hubble telescope.
Answer:
R1 = 5.13 Ω
Explanation:
From Ohm's law,
V = IR............... Equation 1
Where V = Voltage, I = current, R = resistance.
From the question,
I = 2 A, R = R1
Substitute into equation 1
V = 2R1................ Equation 2
When a resistance of 2.2Ω is added in series with R1,
assuming the voltage source remain constant
R = 2.2+R1, and I = 1.4 A
V = 1.4(2.2+R1)................. Equation 3
Substitute the value of V into equation 3
2R1 = 1.4(2.2+R1)
2R1 = 3.08+1.4R1
2R1-1.4R1 = 3.08
0.6R1 = 3.08
R1 = 3.08/0.6
R1 = 5.13 Ω
Answer: 
Explanation:
According to the <u>Third Kepler’s Law</u> of Planetary motion:
(1)
Where;:
is the period of the satellite
is the Gravitational Constant and its value is 
is the mass of the Earth
is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).
On the other hand, the orbital velocity 
is given by:
(2)
Now, from (1) we can find
, in order to substitute this value in (2):
![a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BT%5E%7B2%7DGM%7D%7B4%5Cpi%7D%5E%7B2%7D%7D)
(3)
![a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.26%2810%29%5E%7B4%7Ds%29%5E%7B2%7D%286.674%2810%29%5E%7B-11%7D%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkgs%5E%7B2%7D%7D%29%285.98%2810%29%5E%7B24%7Dkg%29%7D%7B4%5Cpi%7D%5E%7B2%7D%7D)
(4)
(5)
Substituting (5) in (2):
(6)
(7) This is the speed at which the satellite travels
- In order to achieve the desired resistance under the given circumstances, we would connect two 50 Ohms resistors in parallel and then connect it in series with the 20 Ohms resistors.
- In order to get a 35 Ohms resistance under the given circumstances, we would connect two 50 Ohms resistors in parallel and then connect it in series with two 20 Ohms resistors that are connected in parallel.
<h3>How to achieve the desired resistance under these circumstances?</h3>
In order to achieve the desired resistance under the given circumstances, we would connect two 50 Ohms resistors in parallel and then connect it in series with the 20 Ohms resistors.
Mathematically, the total equivalence resistance of two resistors that are connected in parallel is given by:
1/Rt = 1/R₁ + 1/R₂
1/Rt = 1/50 + 1/50
1/Rt = 2/50
1/Rt = 1/25
Rt = 25 Ohms.
Next, we would connect this 25 Ohms resistor in series with the 20 Ohms resistor:
R₃ = 20 + Rt
R₃ = 20 + 25
R₃ = 45 Ohms.
<h3>Part B.</h3>
In order to get a 35 Ohms resistance under the given circumstances, we would connect two 50 Ohms resistors in parallel and then connect it in series with two 20 Ohms resistors that are connected in parallel.
1/Rt = 1/R₁ + 1/R₂
1/Rt = 1/50 + 1/50
1/Rt = 2/50
1/Rt = 1/25
Rt = 25 Ohms.
1/R't = 1/R₁ + 1/R₂
1/R't = 1/20 + 1/20
1/R't = 2/20
1/R't = 1/10
R't = 10 Ohms.
Next, we would connect the 25 Ohms resistor in series with the 10 Ohms resistor:
R₃ = 10 + Rt
R₃ = 10 + 25
R₃ = 35 Ohms.
Read more on resistors in parallel here: brainly.com/question/15121871
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Complete Question:
You need a 45-ω resistor, but the stockroom has only 20-ω and 50-ω resistors.
(a) How can the desired resistance be achieved under these circumstances?
(b) What can you do if you need a 35-ω resistor?
The answer your looking for is, A. alpha particle.
