Answer:
Explanation:
Here image distance is fixed .
In the first case if v be image distance
1 / v - 1 / -25 = 1 / .05
1 / v = 1 / .05 - 1 / 25
= 20 - .04 = 19.96
v = .0501 m = 5.01 cm
In the second case
u = 4 ,
1 / v - 1 / - 4 = 1 / .05
1 / v = 20 - 1 / 4 = 19.75
v = .0506 = 5.06 cm
So lens must be moved forward by 5.06 - 5.01 = .05 cm ( away from film )
Answer:
The minimum speed = ![\sqrt{\frac{2GM}{R} }](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B2GM%7D%7BR%7D%20%7D)
Explanation:
The minimum speed that the rocket must have for it to escape into space is called its escape velocity. If the speed is not attained, the gravitational pull of the planet would pull down the rocket back to its surface. Thus the launch would not be successful.
The minimum speed can be determined by;
Escape velocity = ![\sqrt{\frac{2GM}{R} }](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B2GM%7D%7BR%7D%20%7D)
where: G is the universal gravitational constant, M is the mass of the planet X, and R is its radius.
If the appropriate values of the variables are substituted into the expression, the value of the minimum speed required can be determined.
The amount of fluid displaced by a submerged object depends on its volume.
Answer:
2 seconds
Explanation:
The function of height is given in form of time. For maximum height, we need to use the concept of maxima and minima of differentiation.
![h(t)=-16t^{2}+64t+112](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E%7B2%7D%2B64t%2B112)
Differentiate with respect to t on both the sides, we get
![\frac{dh}{dt}=-32t+64](https://tex.z-dn.net/?f=%5Cfrac%7Bdh%7D%7Bdt%7D%3D-32t%2B64)
For maxima and minima, put the value of dh / dt is equal to zero. we get
- 32 t + 64 = 0
t = 2 second
Thus, the arrow reaches at maximum height after 2 seconds.
Answer:
Q = 282,000 J
Explanation:
Given that,
The mass of liquid water, m = 125 g
Temperature, T = 100°C
The latent heat of vaporization, Hv = 2258 J/g.
We need to find the amount of heat needed to vaporize 125 g of liquid water. We can find it as follows :
![Q=mH_v\\\\Q=125\ g\times 2285\ J/g\\\\Q=282250\ J](https://tex.z-dn.net/?f=Q%3DmH_v%5C%5C%5C%5CQ%3D125%5C%20g%5Ctimes%202285%5C%20J%2Fg%5C%5C%5C%5CQ%3D282250%5C%20J)
or
Q = 282,000 J
So, the required heat is 282,000 J
.