348.34 m/s. When Superman reaches the train, his final velocity will be 348.34 m/s.
To solve this problem, we are going to use the kinematics equations for constant aceleration. The key for this problem are the equations
and
where
is distance,
is the initial velocity,
is the final velocity,
is time, and
is aceleration.
Superman's initial velocity is
, and he will have to cover a distance d = 850m in a time t = 4.22s. Since we know
,
and
, we have to find the aceleration
in order to find
.
From the equation
we have to clear
, getting the equation as follows:
.
Substituting the values:

To find
we use the equation
.
Substituting the values:

Seven days after a waxing gibbous moon, the moon will be waning gibbous, and at some point during that seven days, it will have been Full.
OPTION C The car is accelerating because the direction of velocity is changing explains why a race car going around a curve is accelerating, even if the speed is constant
- When a body is in uniform circular motion ( constant speed ), it will continuously cheanges its direction and so the body is accelerating
- The rate at which an item changes its velocity is known as acceleration, a vector variable. If an object's velocity is changing, it is accelerating.
- As a vector quantity, acceleration has a direction attached to it. The acceleration vector's direction is determined by two factors: if the thing is slowing down or speeding up the direction the thing is travelling in (+ or -)
- The following general rule is used to calculate acceleration:
An object's acceleration will be in the opposite direction of its velocity if it is slowing down.
You may use this basic concept to determine if an object's acceleration is positive or negative, to the right or left, up or down, etc.
To know more about acceleration visit : brainly.com/question/12550364
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Electric field between the plates of parallel plate capacitor is given as

here area of plates of capacitor is given as


also the maximum field strength is given as

now we will plug in all data to find the maximum possible charge on capacitor plates


so the maximum charge that plate will hold will be given by above