The brightness of the lamp is proportional to the current flowing through the lamp: the larger the current, the brighter the lamp.
The current flowing through the lamp is given by Ohm's law:

where
V is the potential difference across the lamp, which is equal to the emf of the battery, and R is the resistance of the lamp.
The problem says that the battery is replaced with one with lower emf. Looking at the formula, this means that V decreases: if we want to keep the same brightness, we need to keep I constant, therefore we need to decrease R, the resistance of the lamp.
Answer:
P = 33.6 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
F = forces [N]
m = mass = 14 [kg]
a = acceleration = 6 [m/s²]
![F = 14*6\\F = 84 [N]](https://tex.z-dn.net/?f=F%20%3D%2014%2A6%5C%5CF%20%3D%2084%20%5BN%5D)
In the second part of this problem we must find the work done, where the work in physics is known as the product of force by distance, it is important to make it clear that force must be applied in the direction of movement.

where:
W = work [J]
F = force = 84 [N]
d = displaciment = 40 [m]
![W = 84*40\\W = 3360 [J]](https://tex.z-dn.net/?f=W%20%3D%2084%2A40%5C%5CW%20%3D%203360%20%5BJ%5D)
Finally, the power can be calculated by the relationship between the work performed in a given time interval.

where:
P = power [W]
W = work = 3360 [J]
t = time = 100 [s]
Now replacing:
![P=3360/100\\P=33.6[W]](https://tex.z-dn.net/?f=P%3D3360%2F100%5C%5CP%3D33.6%5BW%5D)
The power is given in watts
Answer:
As much wood as a woodchuck could chuck, If a woodchuck could chuck wood.
Explanation:
Answer:
The answer <em><u>is C. Mars</u></em>. Mars and Mercury are both smaller than Earth's core. Hope this helps you :)
Explanation:
An perfect mass less spring, attached at one end and with a free mass attached at the other end, will have a distinct frequency of oscillation depending on its constant spring and mass. On the other hand, a spring with mass along its length will not have a characteristic frequency of oscillation.
Alternatively, based on its spring constant and mass per length, it will now have a wave Speed. It would be possible to use all wavelengths and frequencies, as long as the component fλ= S, where S is the spring wave size. If that sounds like longitudinal waves, like solid sound waves.