Answer:
14 seconds
Explanation:
Distance covered by sports car is measured with 21*t where t is time.
Distance covered by police is measured with at^2/2(t-time, a-acceleration) as it doesnt have starting velocity. If the distances are equal(police catches sport car)
then we have 21t=3/2*t^2 Solving this equation we get t=14 & t=0;
excluding the starting point we have 14 seconds
<u>Answer:</u>
<em>The initial distance between the trains is 1450 m.
</em>
<u>Explanation:</u>
In the question two trains are of equal length 400 m and moves at a uniform speed of 72 km/h. train A is moving ahead of train B. If the train B has to overtake train A it should accelerate.
Train B’s acceleration is 
and it accelerated for 50 seconds.
<em>
</em>
<em>t=50 s
</em>
<em>initial speed u=72km/h
</em>
<em>we have to convert this speed into m/s </em>
<em>
</em>
<em>Distance covered in accelerating phase 
</em>
<em>
</em>
<em>
</em>
If a train is just behind another, the distance covered by the train located behind during overtaking phase will be equal to the sum of the lengths of the trains.
<em>Here length of train A+length of train 
</em>
<em>Hence the initial distance between the trains = 
</em>
Answer:
Answer:
5.2307 %
Explanation:
(acutal mass- estimated mass) / ( estimated mass)
Answer:
Explanation:
Average speed is distance traveled over time taken to do so
home to station
d = 60 km
t = 60 km/60 km/hr = 1 hr
station to shop
d = 60 km
t = 60 km / 120 km/hr = ½ hr
speed
s = (60 + 60) / (1 + ½) = 120/1.5 = 80 km/h
Answer:
Explanation:
λ=c x²
c = λ / x²
λ is mass / length
so its dimensional formula is ML⁻¹
x is length so its dimensional formula is L
c = λ / x²
= ML⁻¹ / L²
= ML⁻³
B )
We shall find out the mass of the rod with the help of given expression of mass per unit length and equate it with given mass that is M
The mass in the rod is symmetrically distributed on both side of middle point.
we consider a small strip of rod of length dx at x distance away from middle point
its mass dm = λdx = cx² dx
By integrating it from -L to +L we can calculate mass of whole rod , that is
M = ∫cx² dx
= [c x³ / 3] from -L/2 to +L/2
= c/3 [ L³/8 + L³/8]
M = c L³/12
c = 12 M L⁻³
C ) Moment of inertia of rod
∫dmx²
= ∫λdxx²
= ∫cx²dxx²
= ∫cx⁴dx
= c x⁵ / 5 from - L/2 to L/2
= c / 5 ( L⁵/ 32 +L⁵/ 32)
= (2c / 160)L⁵
= (c / 80) L⁵
= (12 M L⁻³/80)L⁵
= 3/20 ML²
=
=
