Answer:
mass of the planet: 
Explanation:
When a moon keeps a circular orbit around a planet, it is the force of gravity the one that provides the centripetal force to keep it in its circular trajectory of radius R. So if we can write that in such cases (being the mass of the planet "M" and the mass of the moon "m"), we can form an equation by making the centripetal force on the moon equal the force of gravity (using the Newton's Universal Law of Gravity):
where we used here the tangential velocity (v) of the moon around the planet. This equation can be further simplified by dividing both sides by "m" and multiplying both sides by the orbital radius R:
Notice that the mass of the moon has actually disappeared from the equation, which tells us that the orbiting velocity and period do not depend on the mass of the moon, but on the mass of the actual planet.
We know the orbital radius R (
, the value of the Universal Gravitational constant G, and we can estimate the value of the tangential velocity of the moon since we know it period: 36.3 hrs = 388800 seconds.
We know that the moon makes a full circumference (
) in 388800 seconds, therefore its tangential velocity is:
where we rounded the velocity to one decimal.
Notice that we have converted all units to the SI system, so when using the formula to solve for the mass of the planet, the answer comes directly in kg.
Now we use this value for the tangential velocity to estimate the mass of the planet in the first equation we made and simplified:
Answer:
v = 2.21 m/s
Explanation:
The foreman had released the box from rest at a height of 0.25 m above the ground.
We need to find the speed of the crate when it reaches the bottom of the ramp. Let v is the velocity at the bottom of the ramp. It can be calculated using conservation of energy as follows :
So, its velocity at the bottom of the ramp is 2.21 m/s.
Answer:
5.5 × 10^14 Hz or s^-1
no orange light has less frequency so no photoelectric effect
Explanation:
hf = hf0 + K.E
HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s
f is frequency of incident photon and f0 is threshold frequency
hf0 = hf- k.E
6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20
6.63 × 10 ^-34 × f0 = 3.64158×10^-19
f0 = 3.64158×10^-19/ 6.63 × 10 ^-34
f0 = 5.4925 × 10^14
f0 =5.5 × 10^14 Hz or s^-1
frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Answer:
The pressure at this point is 0.875 mPa
Explanation:
Given that,
Flow energy = 124 L/min
Boundary to system P= 108.5 kJ/min
We need to calculate the pressure at this point
Using formula of pressure
Here, 
Where, v = velocity
Put the value into the formula
Hence, The pressure at this point is 0.875 mPa
