Answer:
The direction of the field is downward, and negatively charged particles will experience an upwards force due to the field.
F = N e E where E is the value of the field and N e the charge Q
M g = N e E and M g is the weight of the drop
N = M g / (e E)
N = 1.1E-4 * 9.8 / (1.6E-19 * 370) = 1.1 * 9.8 / (1.6 * 370) * E15 = 1.82E13
.00011 kg is a very large drop
Q = N e = M g / E = .00011 * 9.8 / 370 = 2.91E-6 Coulombs
Check: N = Q / e = 2.91E-6 / 1.6E-19 = 1.82E13 electrons
Answer:
7.5s
Explanation:
Given parameters:
Velocity = 30m/s
Deceleration = 4m/s²
Unknown:
Time it takes for the car to come to complete rest = ?
Solution:
To solve this problem, we use the kinematics expression below:
v = u + at
Since this is a deceleration
v = u - at
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time taken
v - u = -at
0 - 30 = -4 x t
-30 = -4t
t = 7.5s
Answer:
b. equal in size and opposite in direction
Answer:
h = 618.64 m
Explanation:
First we need to calculate the height gained by rocket while the fuel is burning. We use 2nd equation of motion for that purpose:
h₁ = Vit + (1/2)at²
where,
h₁ = height gained during the burning of fuel
Vi = Initial Velocity = 0 m/s
t = time = 7 s
a = acceleration = 8 m/s²
Therefore,
h₁ = (0 m/s)(7 s) + (1/2)(8 m/s²)(7 s)²
h₁ = 196 m
Now we use 1st equation of motion to find final speed Vf:
Vf = Vi + at
Vf = 0 m/s + (8 m/s²)(7 s)
Vf = 56 m/s
Now, we calculate height covered in free fall motion. Using 3rd equation of motion:
2ah₂ = Vf² - Vi²
where,
a = - 3.71 m/s²
h₂ = height gained during free fall motion = ?
Vf = Final Velocity = 0 m/s (since, rocket will stop at highest point)
Vi = 56 m/s
Therefore,
(2)(-3.71 m/s²)h₂ = (0 m/s)² - (56 m/s)²
h₂ = 422.64 m
So the total height gained will be:
h = h₁ + h₂
h = 196 m + 422.64 m
<u>h = 618.64 m</u>
