Answer:
Part a)
a = - 8.45 m/s/s
Part b)
![d = 66 m](https://tex.z-dn.net/?f=d%20%3D%2066%20m)
Part c)
![d = 14.16 m](https://tex.z-dn.net/?f=d%20%3D%2014.16%20m)
Explanation:
Part a)
when car apply brakes then the friction force on the car in front of us is given as
![F_f = \mu mg](https://tex.z-dn.net/?f=F_f%20%3D%20%5Cmu%20mg)
here we need to find deceleration due to friction
![a = \frac{F_f}{m}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BF_f%7D%7Bm%7D)
![a = -\mu g](https://tex.z-dn.net/?f=a%20%3D%20-%5Cmu%20g)
![a = -8.45 m/s^2](https://tex.z-dn.net/?f=a%20%3D%20-8.45%20m%2Fs%5E2)
Part b)
Braking distance of the car is the distance that it move till it stops
so we will have
![v_f^2 - v_i^2 = 2ad](https://tex.z-dn.net/?f=v_f%5E2%20-%20v_i%5E2%20%3D%202ad)
![0 - 33.4^2 = 2(-8.45)d](https://tex.z-dn.net/?f=0%20-%2033.4%5E2%20%3D%202%28-8.45%29d)
![d = 66 m](https://tex.z-dn.net/?f=d%20%3D%2066%20m)
Part c)
Since we know that average reaction time for human is 0.424 s
now we know that during reaction time our car will travel at uniform speed
so we will have
![d = vt](https://tex.z-dn.net/?f=d%20%3D%20vt)
![d = (33.4) (0.424)](https://tex.z-dn.net/?f=d%20%3D%20%2833.4%29%20%280.424%29)
![d = 14.16 m](https://tex.z-dn.net/?f=d%20%3D%2014.16%20m)
Answer:
F = 274.68[N]
Explanation:
The gravitational force is equal to the weight of a body, or this case that of a person. Weight can be calculated by means of the product of mass by gravitational acceleration. In this way we have the following equation:
![F=m*g](https://tex.z-dn.net/?f=F%3Dm%2Ag)
where:
F = force or weight [N]
m = mass = 28 [kg]
g = gravity acceleration = 9.81 [m/s²]
Now replacing:
![F=28*9.81\\F=274.68[N]](https://tex.z-dn.net/?f=F%3D28%2A9.81%5C%5CF%3D274.68%5BN%5D)
Answer:
Force A=-−2,697.75 N
Force B=13, 488.75 N
Explanation:
Taking moments at point A, the sum of clockwise and anticlockwise moments equal to zero.
25 mg-20Fb=0
25*1100g=20Fb
Fb=25*1100g/20=1375g
Taking g as 9.81 then Fb=1375*9.81=13,488.75 N
The sum of upward and downward forces are same hence Fa=1100g-1375g=-275g
-275*9.81=−2,697.75. Therefore, force A pulls downwards
Note that the centre of gravity is taken to be half the whole length hence half of 50 is 25 m because center of gravity is always at the middle
Answer:
Tides on our planet are caused by the gravitational pull of the Moon and Sun. Earth's oceans "bulge out" because the Moon's gravity pulls a little harder on one side of our planet (the side closer to the Moon) than it does on the other. The Sun's gravity raises tides, too, but lunar tides are twice as big.
1) Current in each bulb: 0.1 A
The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:
![R_{eq}=R_1 + R_2 = 400 \Omega + 800 \Omega=1200 \Omega](https://tex.z-dn.net/?f=R_%7Beq%7D%3DR_1%20%2B%20R_2%20%3D%20400%20%5COmega%20%2B%20800%20%5COmega%3D1200%20%5COmega)
And so, the current through the circuit is (using Ohm's law):
![I=\frac{V}{R_{eq}}=\frac{120 V}{1200 \Omega}=0.1 A](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BV%7D%7BR_%7Beq%7D%7D%3D%5Cfrac%7B120%20V%7D%7B1200%20%5COmega%7D%3D0.1%20A)
And since the two bulbs are connected in series, the current through each bulb is the same.
2) 4 W and 8 W
The power dissipated by each bulb is given by the formula:
![P=I^2 R](https://tex.z-dn.net/?f=P%3DI%5E2%20R)
where I is the current and R is the resistance.
For the first bulb:
![P_1 = (0.1 A)^2 (400 \Omega)=4 W](https://tex.z-dn.net/?f=P_1%20%3D%20%280.1%20A%29%5E2%20%28400%20%5COmega%29%3D4%20W)
For the second bulb:
![P_1 = (0.1 A)^2 (800 \Omega)=8 W](https://tex.z-dn.net/?f=P_1%20%3D%20%280.1%20A%29%5E2%20%28800%20%5COmega%29%3D8%20W)
3) 12 W
The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so:
![P_{tot} = P_1 + P_2 = 4 W + 8 W=12 W](https://tex.z-dn.net/?f=P_%7Btot%7D%20%3D%20P_1%20%2B%20P_2%20%3D%204%20W%20%2B%208%20W%3D12%20W)