3. A
4.B
3.Since the Earth rotates through two tidal “bulges” every lunar day, we experience two high and two low tides every 24 hours and 50 minutes
4.An eclipse of the Sun happens when the New Moon moves between the Sun and Earth, blocking out the Sun's rays and casting a shadow on parts of Earth
Work = Force x Distance = 500 x 4 = 2000 Nm = 2000 J
Answer:
1.5 m
Explanation:
Let the distance from the box to the pivot be c.
Let the distance from the pivot to the effort be y.
From the question given above, the following data were obtained:
Effort force (Fₑ) = 7 N
Force of resistance (Fᵣ) = 14 N
Distance from the box to the pivot (c) = 0.75 m
Distance from the pivot to the effort (y) =?
Clockwise moment = Fₑ × y
Anticlock wise moment = Fᵣ × c
Clockwise moment = Anticlock wise moment
Fₑ × y = Fᵣ × c
7 × y = 14 × 0.75
7 × y = 10.5
Divide both side by 7
y = 10.5 / 7
y = 1.5 m
Therefore, the distance from the pivot to the effort is 1.5 m
The formula of net Force is:F = mawhere m is the mass of the objecta is the acceleration of the object
thus, if we triple the net force applied to the object:
3F = maa = 3F / m
The acceleration is also tripled since the force is directly proportional to the acceleration.
Answer:
The highest electric field is experienced by a 2 C charge acted on by a 6 N electric force. Its magnitude is 3 N.
Explanation:
The formula for electric field is given as:
E = F/q
where,
E = Electric field
F = Electric Force
q = Charge Experiencing Force
Now, we apply this formula to all the cases given in question.
A) <u>A 2C charge acted on by a 4 N electric force</u>
F = 4 N
q = 2 C
Therefore,
E = 4 N/2 C = 2 N/C
B) <u>A 3 C charge acted on by a 5 N electric force</u>
F = 5 N
q = 3 C
Therefore,
E = 5 N/3 C = 1.67 N/C
C) <u>A 4 C charge acted on by a 6 N electric force</u>
F = 6 N
q = 4 C
Therefore,
E = 6 N/4 C = 1.5 N/C
D) <u>A 2 C charge acted on by a 6 N electric force</u>
F = 6 N
q = 2 C
Therefore,
E = 6 N/2 C = 3 N/C
E) <u>A 3 C charge acted on by a 3 N electric force</u>
F = 3 N
q = 3 C
Therefore,
E = 3 N/3 C = 1 N/C
F) <u>A 4 C charge acted on by a 2 N electric force</u>
F = 2 N
q = 4 C
Therefore,
E = 2 N/4 C = 0.5 N/C
The highest field is 3 N, which is found in part D.
<u>A 2 C charge acted on by a 6 N electric force</u>
