<span> Let’s determine the initial momentum of each car.
#1 = 998 * 20 = 19,960
#2 = 1200 * 17 = 20,400
This is this is total momentum in the x direction before the collision. B is the correct answer. Since momentum is conserved in both directions, this will be total momentum is the x direction after the collision. To prove that this is true, let’s determine the magnitude and direction of the total momentum after the collision.
Since the y axis and the x axis are perpendicular to each other, use the following equation to determine the magnitude of their final momentum.
Final = √(x^2 + y^2) = √(20,400^2 + 19,960^2) = √814,561,600
This is approximately 28,541. To determine the x component, we need to determine the angle of the final momentum. Use the following equation.
Tan θ = y/x = 19,960/20,400 = 499/510
θ = tan^-1 (499/510)
The angle is approximately 43.85˚ counter clockwise from the negative x axis. To determine the x component, multiply the final momentum by the cosine of the angle.
x = √814,561,600 * cos (tan^-1 (499/510) = 20,400</span>
<span>t^2 = 1/4.9 </span>
<span>t = 0.45 sec
answer:</span><span>1 - 4.9t^2 = 0 </span>
LAPA HDIDOSHSUWJWVWIHDHDOSSHSVWIME
Answer:
1.) 274.5v
2.) 206.8v
Explanation:
1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.
The potential difference and charge across EACH capacitor will be
V = Voe
Where Vo = initial voltage
e = natural logarithm = 2.718
For the first capacitor 2.50 µF,
V = Vo × 2.718
746 = Vo × 2.718
Vo = 746/2.718
Vo = 274.5v
To calculate the charge, use the below formula.
Q = CV
Q = 2.5 × 10^-6 × 274.5
Q = 6.86 × 10^-4 C
For the second capacitor 6.80 µF
V = Voe
562 = Vo × 2.718
Vo = 562/2.718
Vo = 206.77v
The charge on it will be
Q = CV
Q = 6.8 × 10^-6 × 206.77
Q = 1.41 × 10^-3 C
B.) Using the formula V = Voe again
165 = Vo × 2.718
Vo = 165 /2.718
Vo = 60.71v
Q = C × 60.71
Q = C
Answer:
0.68 m
Explanation:
We know that the speed of sound in air is a product of frequency and wavelength. Taking speed of sound in air as 340 m/s
V=frequency*wavelength
Then wavelength is given by 350/500=0.68 m
Therefore, to repeat constructive interference at the listener's ear, a distance of 0.68 m is needed
