Answer:
"2Ω" is the net resistance in the circuit.
Explanation:
The given resistors are:
R1 = 3Ω
R2 = 6Ω
The net resistance will be:
⇒ ![\frac{1}{R_{net}} =\frac{1}{R_1} +\frac{1}{R_2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7Bnet%7D%7D%20%3D%5Cfrac%7B1%7D%7BR_1%7D%20%2B%5Cfrac%7B1%7D%7BR_2%7D)
On substituting the values, we get
⇒ ![\frac{1}{R_{net}} =\frac{1}{3} +\frac{1}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7Bnet%7D%7D%20%3D%5Cfrac%7B1%7D%7B3%7D%20%2B%5Cfrac%7B1%7D%7B6%7D)
On taking L.C.M, we get
⇒ ![\frac{1}{R_{net}} =\frac{2+1}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7Bnet%7D%7D%20%3D%5Cfrac%7B2%2B1%7D%7B6%7D)
⇒ ![\frac{1}{R_{net}} =\frac{3}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7Bnet%7D%7D%20%3D%5Cfrac%7B3%7D%7B6%7D)
⇒ ![\frac{1}{R_{net}} =\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7Bnet%7D%7D%20%3D%5Cfrac%7B1%7D%7B2%7D)
On applying cross-multiplication, we get
⇒ ![R_{net}=2 \Omega](https://tex.z-dn.net/?f=R_%7Bnet%7D%3D2%20%5COmega)
Kinetic energy = (1/2)*mass*velocity^2
KE = (1/2)mv^2
KE = (1/2)(478)(15)^2
KE = 53775J
Answer:
The work done on the wagon is 37 joules.
Explanation:
Given that,
The force applied by Charlie to the right, F = 37.2 N
The force applied by Sara to the left, F' = 22.4 N
We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :
![F_n=F-F'](https://tex.z-dn.net/?f=F_n%3DF-F%27)
![F_n=37.2-22.4](https://tex.z-dn.net/?f=F_n%3D37.2-22.4)
![F_n=14.8\ N](https://tex.z-dn.net/?f=F_n%3D14.8%5C%20N)
Work done on the wagon is given by the product of net force and displacement. It is given by :
![W=F_n\times d](https://tex.z-dn.net/?f=W%3DF_n%5Ctimes%20d)
![W=14.8\ N\times 2.5\ m](https://tex.z-dn.net/?f=W%3D14.8%5C%20N%5Ctimes%202.5%5C%20m)
W = 37 Joules
So, the work done on the wagon is 37 joules. Hence, this is the required solution.
The answer is C.) mass is the matter of an object
The car will take 300 m before it stops due to applying break.
<h3>What's the relation between initial velocity, final velocity, acceleration and distance?</h3>
- As per Newton's equation of motion, V² - U² = 2aS
- V= final velocity velocity of the object, U = initial velocity velocity of the object, a= acceleration, S = distance covered by the object
- Here, U = 60 ft/sec, V = 0 m/s, a= -6 ft/sec²
- So, 0² - 60² = 2×6× S
=> -3600 = -12S
=> S = 3600/12 = 300 m
Thus, we can conclude that the distance covered by the car is 300 m before it stopped.
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 6 ft/sec². How long will it take before the car stops?
Learn more about the Newton's equation of motion here:
brainly.com/question/8898885
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