Answer:

Explanation:
First of all, we need to find the pressure exerted on the sphere, which is given by:

where
is the atmospheric pressure
is the water density
is the gravitational acceleration
is the depth
Substituting,

The radius of the sphere is r = d/2= 1.1 m/2= 0.55 m
So the total area of the sphere is

And so, the inward force exerted on it is

Answer:
Your answer would be
A person 40 cm- blows into the left end of the pipe to eject the marshmallow from the right end. ... A strain of sound waves is propagated along an organ pipe and gets reflected from an. play · like-icon ... The velocity of sound in air is 340ms^(-1). ... The two pipes are submerged in sea water, arranged as shown in figure. Pipe.Explanation:
I belive this is the answer sorry if im wrong!
Newton's third law says:
"<span>For every action, there is an equal and opposite reaction. ".
So, the force that Tom does on the sister is equal to force the sister applies on Tom:
</span>

<span>where the label "t" means "on Tom", while the label "s" means "on the sister".
From Newton's second law, we also know
</span>

where m is the mass and a the acceleration. <span>so we can rewrite the first equation as
</span>

<span>And find Tom's acceleration:
</span>

<span>
</span>
Answer with Explanation:
a.Intensity of radiation is directly proportional to the frequency of radiation
When the intensity of radiation increases then the frequency of radiation increases and therefore, the number of photo-electrons emitted by the metal increases.
b.When all of the wavelength in the radiation are increased by the same amount
We know that

Frequency is inversely proportional to the wavelength.
Therefore, the frequency decrease .
When the frequency decreases then the number of photo-electrons emitted by the metal decrease.
c.When the work function of the metal is increased then the gain of kinetic energy decreases .
When energy decreases then the number of photo-electrons emitted by the metal decreases.
Answer:

Explanation:
Let 'F₁' and 'F₂' be the forces applied by left and right wires on the bar as shown in the diagram below.
Now, the horizontal and vertical components of these forces are:

As the system is in equilibrium, the net force in x and y directions is 0 and net torque about any point is also 0. Therefore,

Now, let us find the net torque about a point 'P' that is just above the center of mass at the upper edge of the bar.
At point 'P', there are no torques exerted by the F₁x and F₂x nor the weight of the bar as they all lie along the axis of rotation.
Therefore, the net torque by the forces
will be zero. This gives,

But, 
Therefore,


We know,

∴