To solve this problem it is necessary to apply the concepts related to momentum, momentum and Force. Mathematically the Impulse can be described as
Where,
F= Force
t= time
At the same time the moment can be described as a function of mass and velocity, that is
Where,
m = mass
v = Velocity
From equilibrium the impulse is equal to the momentum, therefore
PART A) Since the body ends at rest, we have the final speed is zero, so the momentum would be
Therefore the magnitude of the person's impulse is 1125Kg.m/s
PART B) From the equation obtained previously we have that the Force would be:
Therefore the magnitude of the average force the airbag exerts on the person is 45000N
The solution would be like
this for this specific problem:
<span>
The force on m is:</span>
<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] ->
1
The force on 2m is:</span>
<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2]
-> 2
From (1), you’ll get M = 2mx^2 / L^2 and from
(2) you get M = m(L - x)^2 / L^2
Since the Ms are the same, then
2mx^2 / L^2 = m(L - x)^2 / L^2
2x^2 = (L - x)^2
xsqrt2 = L - x
x(1 + sqrt2) = L
x = L / (sqrt2 + 1) From here, we rationalize.
x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)
x = L(sqrt2 - 1) / (2 - 1)
x = L(sqrt2 - 1) </span>
= 0.414L
<span>Therefore, the third particle should be located the 0.414L x
axis so that the magnitude of the gravitational force on both particle 1 and
particle 2 doubles.</span>
Answer:
The true course: north of east
The ground speed of the plane: 96.68 m/s
Explanation:
Given:
- = velocity of wind =
- = velocity of plane in still air =
Assume:
- = resultant velocity of the plane
- = direction of the plane with the east
Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.
Let us find the direction of this resultant velocity with respect to east direction:
This means the the true course of the plane is in the direction of north of east.
The ground speed will be the magnitude of the resultant velocity of the plane.
Hence, the ground speed of the plane is 96.68 km/h.
Answer:
b is the answer
Explanation:
tq friend b is the answer
Answer:
3.98V
Explanation:
Given
Pontential difference V as 3v
Energy delivered is 30%,
Recall that Enery E=1/2cv^2 from this E=V^2(since Current C is not provided we can assume a value 2)
So E=V^2
E=3^2=9
At full charge E=9,30%of 9,0.3*9=2.7 energy in capacitor is 9-2.7=6.3
But E=V^2
✓E=V
✓6.3=3.98V