5.8x10^3
3.02x10^8
4.5x10^5
8.6x10^10
Answer:
The average force has a magnitude 6524 N due north.
Explanation:
The average net force F = ma where m = mass of car = 1400 kg and a = acceleration.
a = (v - u)/t where u = initial velocity of car = 0 m/s (since it starts from rest)
v = final velocity of car = 27 m/s due north and t = time of motion = 5.8 s
a = (27 m/s - 0 m/s)/5.8 s = 27 m/s ÷ 5.8 s = 4.66 m/s
Since the direction of the velocity change is the direction of the acceleration, the acceleration is 4.66 m/s due north.
The average force, F = ma = 1400 kg × 4.66 m/s = 6524 N
Since the acceleration is due north, the average force takes the direction of the acceleration.
So the direction of the average force is due north
The average force has a magnitude 6524 N due north.
What a delightful little problem !
Here's how I see it:
When 'C' is touched to 'A', charge flows to 'C' until the two of them are equally charged. So now, 'A' has half of its original charge, and 'C' has the other half.
Then, when 'C' is touched to 'B', charge flows to it until the two of <u>them</u> are equally charged. How much is that ? Well, just before they touch, 'C' has half of an original charge, and 'B' has a full one, so 1/4 of an original charge flows from 'B' to 'C', and then each of them has 3/4 of an original charge.
To review what we have now: 'A' has 1/2 of its original charge, and 'B' has 3/4 of it.
The force between any two charges is:
F = (a constant) x (one charge) x (the other one) / (the distance between them)².
For 'A' and 'B', the distance doesn't change, so we can leave that out of our formula.
The original force between them was 3 = (some constant) x (1 charge) x (1 charge).
The new force between them is F = (the same constant) x (1/2) x (3/4) .
Divide the first equation by the second one, and you have a proportion:
3 / F = 1 / ( 1/2 x 3/4 )
Cross-multiply this proportion:
3 (1/2 x 3/4) = F
F = 3/2 x 3/4 = 9/8 = <em>1.125 newton</em>.
That's my story, and I'm sticking to it.
Save money on fuel and go for longer
<span>k = 1.7 x 10^5 kg/s^2
Player mass = 69 kg
Hooke's law states
F = kX
where
F = Force
k = spring constant
X = deflection
So let's solve for k, the substitute the known values and calculate. Don't forget the local gravitational acceleration.
F = kX
F/X = k
115 kg* 9.8 m/s^2 / 0.65 cm
= 115 kg* 9.8 m/s^2 / 0.0065 m
= 1127 kg*m/s^2 / 0.0065 m
= 173384.6154 kg/s^2
Rounding to 2 significant figures gives 1.7 x 10^5 kg/s^2
Since Hooke's law is a linear relationship, we could either use the calculated value of the spring constant along with the local gravitational acceleration, or we can simply take advantage of the ratio. The ratio will be both easier and more accurate. So
X/0.39 cm = 115 kg/0.65 cm
X = 44.85 kg/0.65
X = 69 kg
The player masses 69 kg.</span>
