In the Olympic games,an athlete covers a dist | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The average velocity is given by $\bar{v} = \frac{d}{t}$.Given $d = 100 \ m$ and $t = 10 \ s$,we have $\bar{v} = \frac{100}{10} = 10 \ m/s$.Assumi

Vedclass · Accepted Answer

$2,000 \ J - 5,000 \ J$

Vedclass · Answer

(D) The average velocity is given by $\bar{v} = \frac{d}{t}$.Given $d = 100 \ m$ and $t = 10 \ s$,we have $\bar{v} = \frac{100}{10} = 10 \ m/s$.Assuming the mass of an athlete $(m)$ ranges from $50 \ kg$ to $100 \ kg$,we calculate the kinetic energy $(K = \frac{1}{2}mv^2)$:For $m_1 = 50 \ kg$: $K_1 = \frac{1}{2} 	imes 50 	imes (10)^2 = 25 	imes 100 = 2,500 \ J$.For $m_2 = 100 \ kg$: $K_2 = \frac{1}{2} 	imes 100 	imes (10)^2 = 50 	imes 100 = 5,000 \ J$.Thus,the estimated range of kinetic energy is $2,000 \ J - 5,000 \ J$.

In the Olympic games,an athlete covers a distance of $100 \ m$ in $10 \ s$. The estimated range of their kinetic energy is:

Similar Questions

Two bodies of masses $m$ and $4m$ are moving with equal kinetic energy $(K.E.)$. The ratio of their linear momentums is

If the momentum of a body increases by $0.01\%$,its kinetic energy will increase by ........... $\%$

The kinetic energy of a car is doubled when its velocity is increased by $1 \,m/s$. Then the initial velocity of the car is

If the momentum of a body is increased $n$ times,its kinetic energy increases by a factor of:

The kinetic energy of a body of mass $4 \,kg$ and momentum $6 \,Ns$ will be (in $\,J$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module