The image will form in the vicinity of F. Its nature will be small and inverted
Answer:
Explanation:
We shall solve this problem on the basis of pinciple that water is incompressible so volume of flow will be equal at every point .
rate of volume flow of one stream
= cross sectional area x velocity
= 8.4 x 3.5 x 2.2 = 64.68 m³ /s
rate of volume flow of other stream
= 6.6 x 3.6 x 2.7
= 64.15 m³ /s
rate of volume flow of rive , if d be its depth
= 11.2 x d x 2.8
= 31.36 d
volume flow of river = Total of volume flow rate of two streams
31.36 d = 64.15 + 64.68
31.36 d = 128.83
d = 4.10 m /s .
Answer: (a) The magnitude of its temperature change in degrees Celsius is
.
(b) The magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is
.
Explanation:
(a) Expression for change in temperature is as follows.

= 15.1 K
= 
= 
= 
Therefore, the magnitude of its temperature change in degrees Celsius is
.
(b) Change in temperature from Celsius to Fahrenheit is as follows.
F = 1.8C + 32
C = 
Since, K = C + 273
or, 

= 1.8 (15.1)
= 
or, = 
Thus, we can conclude that the magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is
.
Answer:
1.98s
Explanation:
The time taken to hit the ground is given by
h=ut+ 1/2 at^2
but u =0
so we have
h=1/2at^2
making t the subject
t=√2h/g
√2×19.6/10
1.98s
Answer:
52 mm/s (approximately)
Explanation:
Given:
Initial speed of the projectile is, 
Angle of projection is, 
Time taken to land on the hill is, 
In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.
So, the velocity in the horizontal direction always remains the same.
The horizontal component of initial velocity is given as:

Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.
Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.