Look at bitesizes physics section, they have all the information you need to complete this question.
<h2>Question:</h2>
In this circuit the resistance R1 is 3Ω, R2 is 7Ω, and R3 is 7Ω. If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be?
Answer:
9.1Ω
Explanation:
The circuit diagram has been attached to this response.
(i) From the diagram, resistors R1 and R2 are connected in parallel to each other. The reciprocal of their equivalent resistance, say Rₓ, is the sum of the reciprocals of the resistances of each of them. i.e
![\frac{1}{R_X} = \frac{1}{R_1} + \frac{1}{R_2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_X%7D%20%3D%20%5Cfrac%7B1%7D%7BR_1%7D%20%2B%20%5Cfrac%7B1%7D%7BR_2%7D)
=>
------------(i)
From the question;
R1 = 3Ω,
R2 = 7Ω
Substitute these values into equation (i) as follows;
![R_{X} = \frac{3 * 7}{3 + 7}](https://tex.z-dn.net/?f=R_%7BX%7D%20%3D%20%5Cfrac%7B3%20%2A%207%7D%7B3%20%2B%207%7D)
![R_{X} = \frac{21}{10}](https://tex.z-dn.net/?f=R_%7BX%7D%20%3D%20%5Cfrac%7B21%7D%7B10%7D)
Ω
(ii) Now, since we have found the equivalent resistance (Rₓ) of R1 and R2, this resistance (Rₓ) is in series with the third resistor. i.e Rₓ and R3 are connected in series. This is shown in the second image attached to this response.
Because these resistors are connected in series, they can be replaced by a single resistor with an equivalent resistance R. Where R is the sum of the resistances of the two resistors: Rₓ and R3. i.e
R = Rₓ + R3
Rₓ = 2.1Ω
R3 = 7Ω
=> R = 2.1Ω + 7Ω = 9.1Ω
Therefore, the combination of the resistors R1, R2 and R3 can be replaced with a single resistor with an equivalent resistance of 9.1Ω
Answer with Explanation:
We are given that
Initial velocity,u=4.5 m/s
Time=t =0.5 s
Final velocity=v=0m/s
We have to find the deceleration and estimate the force exerted by wall on you.
We know that
Acceleration=![\frac{v-u}{t}](https://tex.z-dn.net/?f=%5Cfrac%7Bv-u%7D%7Bt%7D)
Using the formula
Acceleration=![a=\frac{0-4.5}{0.5}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B0-4.5%7D%7B0.5%7D)
deceleration=a=![-9m/s^2](https://tex.z-dn.net/?f=-9m%2Fs%5E2)
We know that
Force =ma
Using the formula and suppose mass of my body=m=40 kg
The force exerted by wall on you
Force=![40\times (9)=360N](https://tex.z-dn.net/?f=40%5Ctimes%20%289%29%3D360N)
Explanation:
There are three forces on the bicycle:
Reaction force Rp pushing up at P,
Reaction force Rq pushing up at Q,
Weight force mg pulling down at O.
There are four equations you can write: sum of the forces in the y direction, sum of the moments at P, sum of the moments at Q, and sum of the moments at O.
Sum of the forces in the y direction:
Rp + Rq − (15)(9.8) = 0
Rp + Rq − 147 = 0
Sum of the moments at P:
(15)(9.8)(0.30) − Rq(1) = 0
44.1 − Rq = 0
Sum of the moments at Q:
Rp(1) − (15)(9.8)(0.70) = 0
Rp − 102.9 = 0
Sum of the moments at O:
Rp(0.30) − Rq(0.70) = 0
0.3 Rp − 0.7 Rq = 0
Any combination of these equations will work.