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jeyben [28]
3 years ago
7

A comet is cruising through the solar system at a speed of 50,000 kilometers per hour foe 4 hours time. What is the total distan

ce traveled by the comet during that time
Physics
1 answer:
Triss [41]3 years ago
5 0

Answer:

<em>The distance covered by comet is </em>200,000 km

Explanation:

Speed is defined as the rate of change of distance with time. It is given by the equation speed= \bold{\frac{distance}{time}}

Thus distance= speed*time

In this problem it is given that speed of comet= \frac{50,000km}{hr}

time travelled by the comet= 4 hours

Thus distance= speed*time

                            = 500000*4

                            = \bold{200,000km}

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Q|C (a) You need a 45-ω resistor, but the stockroom has only 20-ω and 50-ω resistors. How can the desired resistance be achieved
NNADVOKAT [17]
  1. 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.
  2. 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

#SPJ4

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?

3 0
2 years ago
Pre-laboratory Assignment: Experiment 20 Reflection and Refraction of Light 1. When light is incident on a reflective surface, w
shusha [124]

Answer:

1) ngle of incidence and reflection are equal,  light carries does not change

2) the angle of this line with respect to the surface is 90º

3) protractor

4)    n₂  sin θ₂ = n_1 sin θ₁,  light ray must have a greater angle than the incident ray ,

Explanation:

1) When light falls on a reflective surface, the angle of incidence and reflection are equal and as it travels in the same medium, the speed that the light carries does not change

2) The normal is a line perpendicular to the point of incidence of light, so the angle of this line with respect to the surface is 90º

3) Angles are measured with a protractor

4) When light passes from one medium to another, the speed of the ray changes due to the difference in the refractive index in each medium, due to this change in speed the transmitted light ray must have a greater angle than the incident ray , since the speed increases as the density of the medium decreases

           \frac{sin \theta _2}{ sin \theta_1} = \frac{v_2}{v_1}

          \frac{c}{v_2} \  sin \theta_2 = \frac{c}{v_1}  \ sin \theta_1

          n₂  sin θ₂ = n_1 sin θ₁

6 0
3 years ago
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