Answer:
It will require a force of 1/5, answer A.
Explanation:
In the attached image we can see an example of an array of pulleys that will lift a 100 kg-f load.
If we analyze pulleys A,B, and C as in the image we can check a force in the cable of 20kg-f.
In the pulley D we have three forces of 20 kg-f each and those forces plus the forces in the pulley B, sum a total of 100 kg-f (60+40). This matches the mechanical advantage (100/5) = 20 kg-f
It would be the lithosphere
I think this happens because during melting or boiling at constant temperature, the entropy dramatically increases because energy is removed from the system during the change. Entropy increases with increase in temperature, entropy increases steadily with increasing temperature up to the melting point of the solid, where the entropy increases further due to addition of freedom of molecules which increases the randomness of the substance. At boiling point of liquid, another abrupt increase in entropy occurs, this results from increased volume available to the atoms or molecules as they enter the gaseous state.
Answer:
1) The angle of deflection will be less than 45° ( C )
2) The angle of deflection will be greater than 45° but less than 90° ( E )
Explanation:
1) Assuming that the force applied has a direction which is perpendicular to the Earth's magnetic field
∴ Fearth > Fapplied hence the angle of deflection will be < 45°
2) when the Fearth < Fapplied
the angle of deflection will be : > 45° but < 90°
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>
