Answer:
option B
Explanation:
given,
height of building = 0.1 km
ball strikes horizontally to ground at = 65 m
speed at which the ball strike = ?
vertical velocity = 0 m/s
time at which the ball strike
![s = \dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=s%20%3D%20%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
![t = \sqrt{\dfrac{2s}{g}}](https://tex.z-dn.net/?f=t%20%3D%20%5Csqrt%7B%5Cdfrac%7B2s%7D%7Bg%7D%7D)
![t = \sqrt{\dfrac{2\times 100}{9.8}}](https://tex.z-dn.net/?f=t%20%3D%20%5Csqrt%7B%5Cdfrac%7B2%5Ctimes%20100%7D%7B9.8%7D%7D)
t = 4.53 s
vertical velocity at the time 4.53 s = g × t = 9.8 × 4.53 = 44.39 m/s
horizontal velocity =
=14.35 m/s
speed of the ball =
= 46.65 m/s
hence, the speed of the ball just before it strike the ground = 47 m/s
The correct answer is option B
Answer:
Explanation:
Given:
Force, f = 5 N
Velocity, v = 5 m/s
Power, p = energy/time
Energy = mass × acceleration × distance
Poer, p = force × velocity
= 5 × 5
= 25 W.
Note 1 watt = 0.00134 horsepower
But 25 watt,
0.00134 hp/1 watt × 25 watt
= 0.0335 hp.
Answer:
A lens placed in a transparent liquid becomes invisible because when refractive index of the material of the lens is equal to the refractive index of the liquid in which lens is placed under this condition no bending of light takes place when it travels from liquid to the lens, so both will start behaving like both are same things.
Explanation:
hope it helps :))
Answer:
(A) It will take 22 sec to come in rest
(b) Work done for coming in rest will be 0.2131 J
Explanation:
We have given the player turntable initially rotating at speed of ![33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec](https://tex.z-dn.net/?f=33%5Cfrac%7B1%7D%7B3%7Drpm%3D33.333rpm%3D%5Cfrac%7B2%5Ctimes%203.14%5Ctimes%2033.333%7D%7B60%7D%3D3.49rad%2Fsec)
Now speed is reduced by 75 %
So final speed ![\frac{3.49\times 75}{100}=2.6175rad/sec](https://tex.z-dn.net/?f=%5Cfrac%7B3.49%5Ctimes%2075%7D%7B100%7D%3D2.6175rad%2Fsec)
Time t = 5.5 sec
From first equation of motion we know that '
![\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%5Cfrac%7B%5Comega%20-%5Comega%20_0%7D%7Bt%7D%3D%5Cfrac%7B2.6175-3.49%7D%7B4%7D%3D-0.158rad%2Fsec%5E2)
(a) Now final velocity ![\omega =0rad/sec](https://tex.z-dn.net/?f=%5Comega%20%3D0rad%2Fsec)
So time t to come in rest ![t=\frac{0-3.49}{-0.158}=22sec](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B0-3.49%7D%7B-0.158%7D%3D22sec)
(b) The work done in coming rest is given by
![\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DI%5Cleft%20%28%20%5Comega%20%5E2-%5Comega%20_0%5E2%20%5Cright%20%29%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%200.035%5Ctimes%20%280%5E2-3.49%5E2%29%3D0.2131J)
Would be A 1012 N/C because The magnitude of the electric field at distance r from a point charge q is E=k
e
q/r
2
, so
E=
(5.11×10
−11
m)
2
(8.99×10
9
N.m
2
/C
2
)(1.60×10
−19
C)
=5.51×10
11
N/C∼10
1
2N/C
making (e) the best choice for this question.