During a total solar eclipse, the moon passes between Earth and the sun. This completely blocks out the sun’s light. However, the moon is about 400 times smaller than the sun. How can it block all of that light?
Answer:
Weight is what you get when a certain amount of gravity is acting on that mass, and something, like the surface of a planet, is resisting that action. In space, when falling freely, there's nothing resisting the pull of gravity so weight disappears. Mass however stays.
hope this helps u
Explanation:
Answer:
C: equal to mg
Explanation:
in free-fall, gravity is always the net force on an object
Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:
![E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)](https://tex.z-dn.net/?f=E%2AA%20%3D%20%5Cfrac%7BQ_%7Benc%7D%20%7D%7B%5Cepsilon_%7B0%7D%7D%20%281%29)
- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:
![Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C](https://tex.z-dn.net/?f=Q_%7Benc%7D%20%3D%20-%204e-6%20C%20-%20%28-8e-6%20C%29%20%3D%20%2B%204%20e-6%20C)
- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:
![E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B1%7D%7B4%2A%5Cpi%2A%5Cepsilon_%7B0%7D%20%7D%20%2A%5Cfrac%7BQ_%7Benc%7D%20%7D%7Br%5E%7B2%7D%20%7D%20%3D%20%5Cfrac%7B9e9%20N%2Am2%2FC2%2A4e-6C%7D%7B%280.07m%29%5E%7B2%7D%20%7D%20%3D%207.35e6%20N%2FC)
- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
Part (a):
1- Since the resistors are in series, therefore, the total resistance is the summation of the two resistors.
Therefore:
Rtotal = R1 + R2 = 3.11 + 6.15 = 9.26 ohm
2- Since the two resistors are in series, therefore, the current flowing in both is the same. We will use ohm's law to get the current as follows:
V = I*R
V is the voltage of the battery = 24 v
I is the current we want to get
R is the total resistance = 9.26 ohm
Therefore:
24 = 9.26*I
I = 24 / 9.26
I = 2.59 ampere
Part (b):
To get the voltage across the second resistor, we will again use Ohm's law as follows:
V = I*R
V is the voltage we want to get
I is the current in the second resistor = 2.59 ampere
R is the value of the second resistor = 6.15 ohm
Therefore:
V = I*R
V = 2.59 * 6.15
V = 15.9285 volts
Hope this helps :)