Answer:
Magnification, m = -0.42
Explanation:
It is given that,
Height of diamond ring, h = 1.5 cm
Object distance, u = -20 cm
Radius of curvature of concave mirror, R = 30 cm
Focal length of mirror, f = R/2 = -15 cm (focal length is negative for concave mirror)
Using mirror's formula :
, f = focal length of the mirror


v = -8.57 cm
The magnification of a mirror is given by,


m = -0.42
So, the magnification of the concave mirror is 0.42. Thew negative sign shows that the image is inverted.
Explanation:
The 11Ω, 22Ω, and 33Ω resistors are in parallel. That combination is in series with the 4Ω and 10Ω resistors.
The net resistance is:
R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)
R = 20Ω
Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:
V = IR
120 V = I (20Ω)
I = 6 A
So the voltage drops are:
V = (4Ω) (6A) = 24 V
V = (10Ω) (6A) = 60 V
That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:
V = 120 V − 24 V − 60 V
V = 36 V
So the currents are:
I = 36 V / 11 Ω = 3.27 A
I = 36 V / 22 Ω = 1.64 A
I = 36 V / 33 Ω = 1.09 A
If we wanted to, we could also show this using Kirchhoff's laws.
Your answer would be C. Hope this helps!!
Answer:
Explanation:
Given that
The window height is 2m
And the window is 7.5m from the ground
Then the total height of the window from the ground is 7.5+2=9.5m
It takes the ball 0.32sec travelled pass the window.
When the ball get to the window, it has an initial velocity (u') and when it gets to the top of the window it has a final velocity ( v')
Now using the equation of free fall during this window travels
S=ut-½gt² against motion.
S=2, g=9.81, t=0.32sec
Then,
S=u't-½gt²
2=u'×0.32-½×9.81×0.32²
2=0.32u'-0.5023
2+0.5032=0.32u'
Then, 0.32u'=2.5032
u'=2.5032/0.32
u'=7.82m/s
This is the initial velocity as the ball got the the window
Now, let analyse from the window bottom to the ground which is a distance of 7.5m
Using the equation of free fall again
v²=u²-2gH
In this case the final velocity (v) is the velocity when the ball reach the bottom of the window i.e u'=7.82m/s,
While u is the original initial velocity from the throw of the ball
Then,
u'²=u²-2gH
7.82²=u²-2×9.81×7.5
61.146=u²-147.15
61.146+147.15=u²
Then, u²=208.296
So, u=√208.296
u=14.43m/s
The initial velocity of the ball form the throw is 14.43m/s