Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice power work and power work, energy and power physics energy and its forms

If the average power radiated by the star is $10 ^ { 16 } \mathrm { W }$ , the deuteron supply of the star is exhausted in a time of the order of 

  1. $10 ^ { 6 }$ seccond
  2. $10 ^ { 14 }$ second
  3. $10 ^ { 12 }$ second
  4. $10 ^ { 16 }$ second
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is an order-of-magnitude estimation problem. Total energy available in a star's fuel (deuteron) is typically around 10^28 J. Time = Energy / Power = 10^28 / 10^16 = 10^12 seconds.

Multiple choice power work and power work, energy and power physics energy and its forms

A boat moving with constant speed v in still waters experiences a total frictional force F. The power developed by the boat is

  1. $\frac{1}{2}Fv$
  2. $Fv$
  3. $\frac{1}{2}Fv^2$
  4. $Fv^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For an object moving at constant velocity, the driving force must equal the frictional force. Power = Force * velocity = F * v.

Multiple choice power work and power work, energy and power physics energy and its forms

A body of mass m is projected at an angle $\displaystyle \theta $ with the horizontal with an initial velocity $\displaystyle v _{0}.$ The average power of gravitational force over the whole time of flight is

  1. $\displaystyle mg\cos \theta $
  2. $\displaystyle \frac{1}{2}mg\sqrt{u\cos \theta }$
  3. $\displaystyle \frac{1}{2}mgu\sin \theta $

  4. zero
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The projected body again comes back down. Hence the net displacement in the vertical direction will be 0.
$\therefore W=mgh=mg\times 0=0$

Multiple choice power work and power work, energy and power physics energy and its forms

A time varying power $P=2t$ is applied on a particle of mass $m$. Find average power over a time interval from t=0 to t=t :

  1. $\displaystyle P _{av}= t$
  2. $\displaystyle P _{av}= 2t$
  3. $\displaystyle P _{av}= 4t$
  4. $\displaystyle P _{av}= 8t$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Average power over a time interval $t=0$ to $t=t$ is 
$\dfrac{\int _0^tPdt}{\int _0^tdt}$
$=\dfrac{\int _0^t 2tdt}{t}$
$=\dfrac{t^2}{t}=t$
Multiple choice power work and power work, energy and power physics energy and its forms

A car of mass 1000 kg accelerates from rest to $100 : km : h^{-1}$ in 5 seconds. What is the average power of the car ?

  1. $7.71\times 10^5 W$
  2. $7.71\times 10^4 W$
  3. $15.42\times 10^4 W$
  4. $15.42\times 10^5 W$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation


Mass of the car m$=1000\ kg$
Initial velocity of car ${v} _{1}=0$
Final velocity of car ${v} _{2}$$=100km/h=100\times\dfrac{5}{18}m/s=27.78m/s$
Average power of car$=$change in kinetic energy per unit time$=\dfrac{\dfrac{1}{2}m{{v} _{2}}^{2}-\dfrac{1}{2}m{{v} _{1}}^{2}}{t}$,since ${v} _{1}=0$
So,average power of car$=\dfrac{\dfrac{1}{2}m{{v} _{2}}^{2}}{t}$
                                 $=\dfrac{\dfrac{1}{2}\times1000\times{27.78}^{2}}{5}$
                                 $=7.71\times{10}^{4}W$
                                

Multiple choice power work and power work, energy and power physics energy and its forms

A particle of mass $m$ is lying on smooth horizontal table. A constant force $F$ tangential to the surface is applied on it. Find average power over a time interval from $t=0$ to $t=t$.

  1. $\displaystyle \frac{F^{2}t}{3m}$
  2. $\displaystyle \frac{F^{2}t}{2m}$
  3. $\displaystyle \frac{2F^{2}t}{3m}$
  4. $\displaystyle \frac{3F^{2}t}{2m}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Average power of a time interval $t=0$ to $t=t$


is $\dfrac{\int^t _0 Fvdt}{ _0^t\int dt}$


$=\dfrac{F\int _0^t vdt}{t}$

$=\dfrac{F\int _0^t \dfrac{F}{m}tdt}{t}$

$=\dfrac{F^2t}{2m}$

Multiple choice power work and power work, energy and power physics energy and its forms

A block of mass $1$ $kg$ starts moving with constant acceleration $\displaystyle a= 4m/s^{2}.$. Find the average power of the net force in a time interval from $t=0$ to $t=2s$.

  1. $16 W$
  2. $1.6 W$
  3. $15 W$
  4. $1.5 W$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Velocity of an object after time $t$ is given by $v=at$
Average power from time $t _1$ to $t _2$

$=\dfrac{\int _{t _1}^{t _2}Fvdt}{\int _{t _1}^{t _2}dt}$

$=\dfrac{\int _0^t(ma)(at)dt}{\int _0^tdt}$

$=ma^2\dfrac{t}{2}$

$=1\times 4^2\times\dfrac{2}{2}W$

$=16W$
Multiple choice power work and power work, energy and power physics energy and its forms

A given $1700 kg$ car goes from $18 {m}/{s}$ to $0 {m}/{s}$. If this transition took $9 sec$, what was the average power supplied by the force causing this deceleration?
Take the car system to be otherwise isolated (i.e. the decelerating force was the only force acting on the car).

  1. $-1700 W$
  2. $-15300 W$
  3. $-30600 W$
  4. $-61200 W$

  5. Cannot be determined from the information given
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given :   $u =18$ m/s             $u = 0$ m/s                $m = 1700$ kg             $t = 9$ s

Using   $v = u+at$
$\therefore$  $0 = 18 + a \times 9$                 $\implies a = -2$  $m/s^2$
Using       $v^2  - u^2 = 2aS$
$\therefore$  $0 - (18)^2  =2(-2) S$                      $\implies S = 81$  m

Thus work done by decelerating force        $W  = -ma S $
$\therefore$  Power supplied     $P =\dfrac{-maS}{t}  = \dfrac{- 1700 \times 2 \times 81}{9}  =-30600$  W

Multiple choice power work and power work, energy and power physics energy and its forms

Your uncle pushes a $60.0 kg$ crate along a floor with average speed $v=0.65 {m}/{s}$ for $5.0$ seconds as he moves furniture to clean up the garage.
If the coefficient of friction between the floor and the crate is $\mu=0.340$, what is the average power output of your uncle during this time?

  1. $26.0 W$
  2. $130 W$
  3. $383 W$
  4. $650 W$

  5. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given :  $\mu = 0.340$             $v = 0.65$  m/s                $m =60.0$  kg

As the crate moves with constant speed, thus the force applied by uncles must be equal to the frictional force.
$\therefore$   $F  = \mu mg = 0.340 \times 60.0 \times 9.8  = 199.92$  N
Average power output        $P = F v = 199.92 \times 0.65  \approx 130$  W