The answers to the problem are as follows:
<span>a) use f=ma. Mass of student is 836/9.8=85.3kg, force is 935-836=99N. Put it into the formula you will get a=99/85.3
b)Use same formula. Force is 782-836=-54 so put it into formula a=-54/85.3
c)Stopping would take longer as the acceleration is smaller
I hope my answer has come to your help. God bless and have a nice day ahead!</span>
If Resistors are in series= The equivalent is the sum.
E.g R1 and R2 in series, R = R1 + R2.
If in Parallel, equivalent is Product/sum.
E.g If R1 and R2 in parallel, R = (R1*R2)/(R1+R2)
1.) 60 is parallel with 40 and both are then in series with 20.
60//40 = (60*40)/(60+40) = 2400/100 = 24
Now the 24 is in series with the 20
R = 24 + 20 = 44 ohms.
2.) 80 is in series with 40 and both are then in parallel with 40.
Solving the series, R = 80 + 40 =120.
Parallel: 120//40 = (120*40)/(120+40) = 4800/160 = 30
Equivalent Resistance = 30 ohms.
3.) 100 is in parallel with 100 and both are then in series with the parallel of 50 and 50.
The 1st parallel = (100*100)/(100+100) = 10000/200 = 50
The 2nd parallel = (50*50)/(50+50) = 2500/100 = 25.
Solving the series = 50 + 25 =75 ohms.
Cheers.
Answer:
Explanation:
Let the initial speed be u . final speed = 0
v² = u² - 2gh
v is final speed , u is initial speed , h is height .
0 = u² - 2g x 134
u² = 2 x 9.8 x 134
u = 51.25 m /s .
Answer:
-9.4 kg m/s
Explanation:
Change in momentum = final momentum − initial momentum
Δp = p − p₀
Δp = -4.3 kg m/s − 5.1 kg m/s
Δp = -9.4 kg m/s
Impulse = change in momentum
J = Δp
J = -9.4 kg m/s
Answer:
The work done is -209.42 J.
Explanation:
F(x) = (- 20 - 3 x ) N
x = 0 to x = 6.9 m
Here, the force is variable in nature, so the work done by the variable force is given by
![W =\int F dx\\\\W =\int_{0}^{6.9}(-20- 3x dx )\\\\W= \left [ - 20 x - 1.5 x^2 \right ]_{0}^{6.9}\\\\W = - 20 (6.9 - 0) - 1.5(6.9\times 6.9 - 0)\\\\W =- 138 - 71.42\\\\W = - 209.42 J](https://tex.z-dn.net/?f=W%20%3D%5Cint%20F%20dx%5C%5C%5C%5CW%20%3D%5Cint_%7B0%7D%5E%7B6.9%7D%28-20-%203x%20dx%20%29%5C%5C%5C%5CW%3D%20%5Cleft%20%5B%20-%2020%20x%20-%201.5%20x%5E2%20%5Cright%20%5D_%7B0%7D%5E%7B6.9%7D%5C%5C%5C%5CW%20%3D%20-%2020%20%286.9%20-%200%29%20-%201.5%286.9%5Ctimes%206.9%20-%200%29%5C%5C%5C%5CW%20%3D-%20138%20-%2071.42%5C%5C%5C%5CW%20%3D%20-%20209.42%20%20J)
