Answer:
U = 0.413 J
Explanation:
the potential energy between two charges q1 and q2 is given by the following formula:
(1)
k: Coulomb's constant = 8.98*10^9 NM^2/C^2
q1: first charge = 4.6 μC = 4.6*10^-6 C
q2: second charge = 1.0 μC*10^-6 C
r: distance between charges = 10.0 cm = 0.10 m
You replace the values of all variables in the equation (1):
Hence, the energy between charges is 0.413 J
Answer:
Explanation:
Given that,
Weight of jet
W = 2.25 × 10^6 N
It is at rest on the run way.
Two rear wheels are 16m behind the front wheel
Center of gravity of plane 10.6m behind the front wheel
A. Normal force entered on the ground by front wheel.
Taking moment about the the about the real wheel.
Check attachment for better understanding
So,
Clock wise moment = anti-clockwise moment
W × 5.4 = N × 16
2.25 × 10^6 × 5.4 = 16•N
N = 2.25 × 10^6 × 5.4 / 16
N = 7.594 × 10^5 N
B. Normal force on each of the rear two wheels.
Using the second principle of equilibrium body.
Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces
ΣFy = 0
Nr + Nr + N — W = 0
2•Nr = W—N
2•Nr = 2.25 × 10^6 — 7.594 × 10^5
2•Nr = 1.491 × 10^6
Nr = 1.491 × 10^6 / 2
Nr = 7.453 × 10^5 N
Answer:
3.06m/s² to the east
Explanation:
Given parameters:
Mass of car = 2.5 x 10³kg
Force acting on the car = 7.65 x 10³N
Unknown:
Acceleration of the car = ?
Solution:
From Newton's second law of motion:
Force = mass x acceleration
Acceleration = =
= 3.06m/s² to the east
Answer:
He should run at least at 1.5 m/s
The diver will enter the water at an angle of 87° below the horizontal.
Explanation:
Hi there!
The position and velocity of the diver are given by the following vectors:
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
v = (v0x, v0y + g · t)
Where:
r = position vector at time t
x0 = initial horizontal position
v0x = initial horizontal velocity
t = time
y0 = initial vertical position
v0y = initial vertical velocity
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)
v = velocity vector at time t
Please, see the attached figure for a description of the problem. Notice that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0.
We know that, to clear the rocks, the position vector r final (see figure) should be:
r final = ( > 5.0 m, -50 m)
So let´s find first at which time the y-component of the vector r final is - 50 m:
y = y0 + v0y · t + 1/2 · g · t²
-50 m = 2.1 m/s · t - 1/2 · 9.8 m/s² · t²
0 = -4.9 m/s² · t² + 2.1 m/s · t + 50 m
Solving the quadratic equation
t = 3.4 s
Now, we can calculate the initial horizontal velocity using the equation of the x-component of the position vector knowing that at t =3.4 the horizontal component should be greater than 5.0 m:
x = x0 + v0x · t (x0 = 0)
5.0 m < v0x · 3.4 s
v0x > 5.0 m / 3.4 s
v0x > 1.5 m/s
The initial horizontal velocity should be greater than 1.5 m/s
To find the angle at which the diver enters the water, we have to find the magnitude of the final velocity (vector vf in the figure). We already know the magnitude of the x-component of the vector vf, since the horizontal velocity is constant. So:
vfx > 1.5 m/s
Now, let´s calculate vfy:
vfy = v0y + g · t
vfy = 2.1 m/s - 9.8 m/s² · 3.4 s
vfy = -31 m/s
Let´s calculate the minimum magnitude that the final velocity will have if the diver safely clears the rocks. Let´s consider the smallest value allowed for vfx: 1.5 m/s. Then:
Then the final velocity of the diver will be greater or equal than 31 m/s.
To find the angle, we have to use trigonometry. Notice in the figure that the vectors vf, vfx and vy form a right triangle in which vf is the hypotenuse, vfx is the adjacent side and vfy is the opposite side to the angle. Then:
cos θ = adjacent / hypotenuse = vfx / vf = 1.5 m/s / 31 m/s
θ = 87°
The diver will enter the water at an angle of 87° below the horizontal.
