The size of the forces between you and the planet you're on is
your weight on that planet.
Don't forget that you pull the planet with a force equal to the force
that the planet pulls on you. Your weight on Earth is the same as
the Earth's weight on you !
Answer:
351 ohm
720 ohm
Explanation:
When c and d are open:
Terminals c and d are open.. If you redraw the circuit as below, you can see that the two resistors in the first column are in parallel as, they are connected together at both pairs of terminals (due to the short).
Hence, we have a pair of parallel resistors:
Req1 = (R1*R2)/ (R1 + R2) = 360*540/(360+540) = 216 ohms
Req2 = (R3*R4)/ (R3 + R4) = 180*540/(180+540) = 135 ohms
Now these two sets are in series with another Hence,
Req = Req1 + Req2 = 216 + 135 = 351 ohms
Answer: 351 ohms
When c and d are shorted:
The current will flow through the least resistant path naturally from resistors R3 and R1 or R4.
Both of these resistor lie in a single path placing the resistors in series to one another, hence
Req = R3 + R1 = 180 + 540 = 720 ohms
Answer:720 ohms
when a hole is made at the bottom of the container then water will flow out of it
The speed of ejected water can be calculated by help of Bernuolli's equation and Equation of continuity.
By Bernoulli's equation we can write

Now by equation of continuity


from above equation we can say that speed at the top layer is almost negligible.

now again by equation of continuity


here we have


now speed is given by


Complete Question
Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
Answer:
The value is
Explanation:
From the question we are told that
The semi - major axis of the rocky debris 
The semi - major axis of Planet D is 
The orbital period of planet D is 
Generally from Kepler third law

Here T is the orbital period while a is the semi major axis
So

=>
=> ![T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }](https://tex.z-dn.net/?f=T_R%20%20%3D%2018.164%20%20%2A%20%20%5B%5Cfrac%7B%2045%7D%7B60%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D)
=>
Answer:
The constant angular acceleration of the centrifuge = -252.84 rad/s²
Explanation:
We will be using the equations of motion for this calculation.
Although, the parameters of this equation of motion will be composed of the angular form of the normal parameters.
First of, we write the given parameters.
w₀ = initial angular velocity = 2πf₀
f₀ = 3650 rev/min = (3650/60) rev/s = 60.83 rev/s
w₀ = 2πf₀ = 2π × 60.83 = 382.38 rad/s
θ = 46 revs = 46 × 2π = 289.14 rad
w = final angular velocity = 0 rad/s (since the centrifuge come rest at the end)
α = ?
Just like v² = u² + 2ay
w² = w₀² + 2αθ
0 = 382.38² + [2α × (289.14)]
578.29α = -146,214.4644
α = (-146,214.4644/578.29)
α = - 252.84 rad/s²
Hope this Helps!!!