Range, Belt.
Just got it right on e,,d,,g,,e
Answer:
Where is the text?
Explanation:
If you refer to the short sentence you wrote as text, I believe the answer is probably the word "crashes" because it shows how the momentum was transferred.
Answer:
Explanation:
There are a couple of ways you could do this.
The easiest is to use E*R1/(R1 + R2)
- E = 10 volts
- R1 = 590 ohms
- R2 = 840 ohms
So the result would be
E_590 = 10 * 590/(590 + 840)
E_590 = 10 * 590/ (1430)
E_590 = 4.13 volts rounded.
You could do this a slightly longer way.
R = 1430 (total ohms in series.
E = 10 volts
I = ???
I = E/R
I = 10 / 1430
I = 0.00699
Now use this current to figure out the voltage drop.
E = I * R
I = 0.00699 amps
R = 590 ohms
E = 0.00699 * 590
E = 4.13 volts
Pick the way of doing it you like best.
m₁ = 2.3 kg <span>
θ₁ = 70° </span><span>
θ₂ = 17° </span><span>
g = 9.8 m/s²
->The component of the gravitational force on m₁ that is parallel down the incline is: </span><span>
F₁ = m₁ × g × sin(θ₁) </span><span>
F₁ = (2.3
kg) × (9.8 m/s²) × sin(70°) = 21.18 N </span><span>
->The component of the gravitational force on m₂ that is parallel down the incline is: </span><span>
F₂ = m₂ × g × sin(θ₂) </span><span>
F₂ = m₂ × (9.8 m/s²) × sin(70°) = m₂ × (2.86 m/s²) </span><span>
Then the total mass of the system is:
m = m₁ + m₂ </span><span>
m = (2.3 kg) + m₂ </span><span>
If it is given that m₂ slides down the incline, then F₂ must be bigger than F₁, </span><span>
and so the net force on the system must be:
F = m₂×(2.86
m/s²) - (21.18 N) </span><span>
Using Newton's second law, we know that
F = m × a
So if we want the acceleration to be 0.64 m/s², then
m₂×(2.86
m/s²) - (21.18 N) = [(2.3 kg) + m₂] ×
(0.64 m/s²) </span><span>
m₂×(2.86
m/s²) - (21.18 N) = (1.47 N) + m₂×(0.64
m/s²) </span><span>
m₂×(2.22
m/s²) = (22.65 N) </span><span>
m₂<span> = 10.2
kg</span></span>
