(a) 764.4 N
The weight of the astronaut on Earth is given by:
where
m is the astronaut's mass
g is the acceleration due to gravity
Here we have
m = 78.0 kg
g = 9.8 m/s^2 at the Earth's surface
So the weight of the astronaut is
(b) 21.1 N
The spacecraft is located at a distance of
from the center of Earth.
The acceleration due to gravity at a generic distance r from the Earth's center is
where G is the gravitational constant and M is the Earth's mass.
We know that at a distance of r = R (at the Earth's surface) the value of g is 9.8 m/s^2, so we can write:
(1)
the acceleration due to gravity at r=6R instead will be
And substituting (1) into this formula,
So the weight of the astronaut at the spacecratf location is
Answer:
The distance between their eye and the ceiling is 33.06 m.
Explanation:
Given that,
The ceiling of a large symphony hall is covered with acoustic tiles which have small holes that are 4.35 mm center to center, D = 4.35 mm
Diameter of the eye of pupil, d = 5.1 mm
Wavelength, 
We need to find the distance between their eye and the ceiling. Using Rayleigh criteria, we get the distance as follows :
So, the distance between their eye and the ceiling is 33.06 m.
Answer:
3. if you increase your mass you also increase the gravitational pull
6. Radiant energy doesn't depend on a medium and sound energy is dependent on a medium.
Explanation:
i hope this helps-
A few different ways to do this:
Way #1:
The current in the series loop is (12 V) / (total resistance) .
(Turns out to be 2 Amperes, but the question isn't asking for that.)
In a series loop, the current is the same at every point, so it's
the same current through each resistor.
The power dissipated by a resistor is (current)² · (resistance),
and the current is the same everywhere in the circuit, so the
smallest resistance will dissipate the least power. That's R1 .
And by the way, it's not "drawing" the most power. It's dissipating it.
Way #2:
Another expression for the power dissipated by a resistance is
(voltage across the resistance)² / (resistance) .
In a series loop, the voltage across each resistor is
[ (individual resistance) / (total resistance ] x battery voltage.
So the power dissipated by each resistor is
(individual resistance)² x [(battery voltage) / (total resistance)²]
This expression is smallest for the smallest individual resistance.
(The other two quantities are the same for each individual resistor.)
So again, the least power is dissipated by the smallest individual resistance.
That's R1 .
Way #3: (Einstein's way)
If we sat back and relaxed for a minute, stared at the ceiling, let our minds
wander, puffed gently on our pipe, and just daydreamed about this question
for a minute or two, we might have easily guessed at the answer.
===> When you wire up a battery and a light bulb in series, the part
that dissipates power, and gets so hot that it radiates heat and light, is
the light bulb (some resistance), not the wire (very small resistance).
