Answer:
You will feel more weight if it is accelerating out of the planet.
You will feel less weight if it is accelerating towards the planet.
Explanation:
The weight that you are observing or feeling is basically due to the change in acceleration of your fall or rising up in the spaceship. When the acceleration is stationary on the surface, you experience your normal weight due to the gravitational acceleration of that planet.
When the spaceship accelerates above or out of the planet you experience acceleration more than the acceleration of gravity hence more weight.
When the spaceship accelerates towards the planet you experience acceleration less than the acceleration of gravity hence less weight.
If the spaceship is free falling at the gravitational acceleration you experience a zero weight
Answer:![59.78\times 10^3\ Pa](https://tex.z-dn.net/?f=59.78%5Ctimes%2010%5E3%5C%20Pa)
Explanation:
Given
Diver is at a depth of
inside the water![(\rho=10^3\ kg/m^3)](https://tex.z-dn.net/?f=%28%5Crho%3D10%5E3%5C%20kg%2Fm%5E3%29)
As we go down pressure increases due to the weight of liquid column above us
Pressure difference is given by ![P=\rho \times g \times h](https://tex.z-dn.net/?f=P%3D%5Crho%20%5Ctimes%20g%20%5Ctimes%20h)
Where
=density of liquid
h=depth
g=acceleration due to gravity
So external-internal pressure difference is
![\Delta P=10^3\times 9.8\times 6.1](https://tex.z-dn.net/?f=%5CDelta%20P%3D10%5E3%5Ctimes%209.8%5Ctimes%206.1)
![\Delta P=59.78\times 10^3\ Pa](https://tex.z-dn.net/?f=%5CDelta%20P%3D59.78%5Ctimes%2010%5E3%5C%20Pa)
Answer:
Rage :)
Raging lets ur anger out :)
These forces work together to break down and carry away the surficial materials ... Wind, water, ice, or gravity transport these products from their site of origin
Answer:
27.5Ω
Explanation:
Given parameters:
Potential difference in the circuit = 11V
Current in the circuit = 0.4A
Unknown:
Resistance in the circuit = ?
Solution:
According to ohm's law;
V = IR
V is the voltage
I is the current
R is the resistance
Since the unknown is R, we find it;
11 = 0.4 x R
R =
= 27.5Ω