Answer:
The beam of light is moving at the peed of:
km/min
Given:
Distance from the isalnd, d = 3 km
No. of revolutions per minute, n = 4
Solution:
Angular velocity,
(1)
Now, in the right angle in the given fig.:

Now, differentiating both the sides w.r.t t:

Applying chain rule:


Now, using
and y = 1 in the above eqn, we get:

Also, using eqn (1),


Electric pumps are not useful they clog
Answer:
1. 12 V
2a. R₁ = 4 Ω
2b. V₁ = 4 V
3a. A = 1.5 A
3b. R₂ = 4 Ω
4. Diagram is not complete
Explanation:
1. Determination of V
Current (I) = 2 A
Resistor (R) = 6 Ω
Voltage (V) =?
V = IR
V = 2 × 6
V = 12 V
2. We'll begin by calculating the equivalent resistance. This can be obtained as follow:
Voltage (V) = 12 V
Current (I) = 1 A
Equivalent resistance (R) =?
V = IR
12 = 1 × R
R = 12 Ω
a. Determination of R₁
Equivalent resistance (R) = 12 Ω
Resistor 2 (R₂) = 8 Ω
Resistor 1 (R₁) =?
R = R₁ + R₂ (series arrangement)
12 = R₁ + 8
Collect like terms
12 – 8 =
4 = R₁
R₁ = 4 Ω
b. Determination of V₁
Current (I) = 1 A
Resistor 1 (R₁) = 4 Ω
Voltage 1 (V₁) =?
V₁ = IR₁
V₁ = 1 × 4
V₁ = 4 V
3a. Determination of the current.
Since the connections are in series arrangement, the same current will flow through each resistor. Thus, the ammeter reading can be obtained as follow:
Resistor 1 (R₁) = 4 Ω
Voltage 1 (V₁) = 6 V
Current (I) =?
V₁ = IR₁
6 = 4 × I
Divide both side by 4
I = 6 / 4
I = 1.5 A
Thus, the ammeter (A) reading is 1.5 A
b. Determination of R₂
We'll begin by calculating the voltage cross R₂. This can be obtained as follow:
Total voltage (V) = 12 V
Voltage 1 (V₁) = 6 V
Voltage 2 (V₂) =?
V = V₁ + V₂ (series arrangement)
12 = 6 + V₂
Collect like terms
12 – 6 = V₂
6 = V₂
V₂ = 6 V
Finally, we shall determine R₂. This can be obtained as follow:
Voltage 2 (V₂) = 6 V
Current (I) = 1.5 A
Resistor 2 (R₂) =?
V₂ = IR₂
6 = 1.5 × R₂
Divide both side by 1.5
R₂ = 6 / 1.5
R₂ = 4 Ω
4. The diagram is not complete
Answer: It is both B and D
Select all that apply.
At night, thermal energy moves _____.
from space to the atmosphere
from the land to the atmosphere
from the atmosphere to the land
from the atmosphere to space
Answer:
n the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.
In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.
Explanation:
Velocity is a vector therefore it has magnitude and direction, a change in either of the two is the consequence of an acceleration on the system.
In the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.
= (v₂-v₁)/Δt
In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.
= v2/R
In the general case, both the module and the address change
a = Ra ( a_{t}^2 + a_{c}^2)