Answer:
1. B. Increase
2. A. Decrease
Explanation:
To understand this issue, we need to put some values and using the ohm's law we can corroborate the two situations.
Ohm's law:
![V = I*R\\where:\\I = current[A] ampers\\R = resistance [ohms]\\V = voltage [volts]\\](https://tex.z-dn.net/?f=V%20%3D%20I%2AR%5C%5Cwhere%3A%5C%5CI%20%3D%20current%5BA%5D%20ampers%5C%5CR%20%3D%20resistance%20%5Bohms%5D%5C%5CV%20%3D%20voltage%20%5Bvolts%5D%5C%5C)
Now for the voltage we will use V = 110 [V], for resistance R = 10 [ohms]
Replacing the values we have:
![I = \frac{V}{R} \\\\I = \frac{110}{10}\\I= 11 [amp]](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BV%7D%7BR%7D%20%5C%5C%5C%5CI%20%3D%20%5Cfrac%7B110%7D%7B10%7D%5C%5CI%3D%2011%20%5Bamp%5D)
Now let's double the voltage 220 [V]:
Therefore the current will be increased.
Let's do the same for the resistance if originally we have R = 10 [ohms]
![I = \frac{V}{R}\\I = \frac{110}{10} \\I = 11 [amp]\\](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BV%7D%7BR%7D%5C%5CI%20%3D%20%5Cfrac%7B110%7D%7B10%7D%20%5C%5CI%20%3D%2011%20%5Bamp%5D%5C%5C)
Now let's double the resistance 20 [ohms]:
![I=\frac{110}{20}\\I = 5.5 [amp]](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B110%7D%7B20%7D%5C%5CI%20%3D%205.5%20%5Bamp%5D)
Therefore the current will be decreased.
Answer:
Force of gravity
Explanation:
when the force of gravity pulls large gas clouds and dust together, the concentrated gas clouds and dust collapse under the force of gravity forming stars.
There are many galaxies out there in the universe, each galaxy has its own solar systems, stars, and collection of gas and dust. We (earth) belong to the Milky Way galaxy, our galaxy got this name from the Romans. They called in 'via lactea', which directly translates to 'road of milk' because of the milky patch they saw at night.
Hi there!
In this instance, the centripetal force experienced by the moon is equivalent to the gravitational force.
Thus:

Centripetal acceleration, according to Newton's Second Law:

Therefore:

Answer:
Momentum = 1.534 kgm/s
Explanation:
Using the equations of motion, we can obtain the velocity of the ball as it hits the ground.
g = 9.8 m/s²
y = 12 m
u = initial velocity = 0 m/s, since the ball was released from rest
v = final velocity befor the ball hits the ground.
v² = u² + 2ay
v² = 0 + 2×9.8×12 = 235.2
v = 15.34 m/s
The momentum at any point is given as mass × velocity at that point
Mass = 100 g = 0.1 kg, velocity = 15.34 m/s
Momentum = 0.1 × 15.34 = 1.534 kgm/s