What is the velocity in meters per second of a runner who runs exactly 110 m toward the beach in 72 seconds.
2 answers:
1.53 m/s toward the beach
Explanation:
The magnitude of the velocity of the runner is given by:
where
d is the displacement of the runner
t is the time taken
In this case, d=110 m and t=72 s, so the velocity of the runner is
Velocity is a vector, so it consists of both magnitude and direction: we already calculate the magnitude, while the direction is given by the problem, toward the beach.
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Answer:
time will elapse before it return to its staring point is 23.6 ns
Explanation:
given data
speed u = 2.45 × m/s
uniform electric field E = 1.18 × N/C
to find out
How much time will elapse before it returns to its starting point
solution
we find acceleration first by electrostatic force that is
F = Eq
here
F = ma by newton law
so
ma = Eq
here m is mass , a is acceleration and E is uniform electric field and q is charge of electron
so
put here all value
9.11 × kg ×a = 1.18 ×
× 1.602 ×
a = 20.75 × m/s²
so acceleration is 20.75 × m/s²
and
time required by electron before come rest is
use equation of motion
v = u + at
here v is zero and u is speed given and t is time so put all value
2.45 × = 0 + 20.75 ×
(t)
t = 11.80 × s
so time will elapse before it return to its staring point is
time = 2t
time = 2 ×11.80 ×
time is 23.6 × s
time will elapse before it return to its staring point is 23.6 ns
Sarah's acceleration is
Explanation:
The force of kinetic friction acting on Sarah has a magnitude which is given by:
where
is the coefficient of kinetic friction
m is Sarah's mass
g is the acceleration of gravity
Moreover, according to Newton's second law of motion, we know that the net force on Sarah is equal to its mass times its acceleration:
where a is the acceleration
Since the force of friction is the only force acting on Sarah, we can say that the net force is equal to the force of friction, therefore:
where the negative sign is due to the fact that the force of friction has a direction opposite to the motion of Sarah. Solving for a, we find
And substituting the following values:
(coefficient of friction)
(acceleration of gravity)
we find:
Learn more about acceleration and forces:
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