Answer:
He will complete the race in total time of T = 10 s
Explanation:
Total distance moved by the sprinter in 2.14 s is given as



now the distance remaining to move

now he will move with uniform maximum speed for the remaining distance
so we will have


so the total time to complete the race is given as

Answer:
Data:-m=0.88kg ,g=9.8m/sec² ,P.E=96J ,h=?
Explanation:
solution ,P.E=mgh here we have to find h so h=P.E/mg ,h=96/0.88×9.8 ,h=96/8.624=11.131m and if you want to verify so just put the value of h in same formula, likewise :-P.E=mgh ,P.E=0.88×9.8×11.131=96J so we got the same value of P.E as it is given the question (verified).
Explanation:
Let
is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.
The magnetic force is balanced by the centripetal force acting on the proton as :

r is the radius of path,

Time period is given by :


Frequency of proton is given by :

The wavelength of radiation is given by :


So, the wavelength of radiation produced by a proton is
. Hence, this is the required solution.
Answer:
a_total = 2 √ (α² + w⁴)
, a_total = 2,236 m
Explanation:
The total acceleration of a body, if we use the Pythagorean theorem is
a_total² = a_T²2 +
²
where
the centripetal acceleration is
a_{c} = v² / r = w r²
tangential acceleration
a_T = dv / dt
angular and linear acceleration are related
a_T = α r
we substitute in the first equation
a_total = √ [(α r)² + (w r² )²]
a_total = 2 √ (α² + w⁴)
Let's find the angular velocity for t = 2 s if we start from rest wo = 0
w = w₀ + α t
w = 0 + 1.0 2
w = 2.0rad / s
we substitute
a_total = r √(1² + 2²) = r √5
a_total = r 2,236
In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m
a_total = 2,236 m
Answer:
Explanation:
Yes , their displacement may be equal .
Suppose the displacement is AB where A is starting point and B is end point .
The car is covering the distance AB by going from A to B on straight line . On the other hand plane goes from A to C , then from C to D and then from D to B . In this way plane reaches B from A on a different path which is longer than path of the car . In the second case also displacement of plane is AB . In the second case distance covered is longer but displacement is same that is AB .