(a) The pitcher must throw the ball at 27.7 m/s
The momentum of an object is given by:
where
m is the mass of the object
v is the object's velocity
Let's calculate the momentum of the bullet, which has a mass of
m = 2.70 g = 0.0027 kg
and a velocity of
Its momentum is:
The pitcher must throw the baseball with this same momentum. The mass of the ball is
m = 0.146 kg
So the velocity of the ball must be
So, the pitcher must throw the ball at 27.7 m/s.
(b) a. The bullet has greater kinetic energy
The kinetic energy of an object is given by
where m is the mass of the object and v is its speed.
For the bullet, we have:
For the ball:
So, the bullet has greater kinetic energy.
"Anti-Lock" brake systems release the brakes momentarily when wheel speed sensors indicate a locked wheel during braking and traction.
<u>Explanation:</u>
The safety anti-skid braking system is known as "anti-lock braking system" having huge application on land vehicles like one, two and multiple wheeler vehicles and aircraft. During braking, it avoids wheels to get locked by building tractive contacts to the road's surface.
This seems to be an automated system work on the principles of techniques - threshold and cadence braking. The wheel velocity sensors are utilized by ABS to find whether one or more than one wheels chose to get lock while braking.
1) 333.6 C
In order to have breakdown, the electric field at the surface of the cloud must be equal to the breakdown electric field:
The electric field strength at the surface of a charged sphere is given by
where
is the vacuum permittivity
Q is the charge on the sphere
R is the radius of the sphere
Here we have a cloud of radius
So we can re-arrange the previous equation in order to find the charge on the cloud:
2) 
excess electrons
The total charge of the cloud must be (in magnitude)
Q = 333.3 C
We know that one electron carries a charge of
The total charge is just given by the charge of each electron multiplied by the number of excess electrons in the cloud:
where
N is the number of excess electrons
Solving for N, we find:
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