Answer:
an astronomer studies planets, stars, moons, etc, or objects that are outside the field of Earth.
Answer:
Solution
Verified by Toppr
Correct option is
C
3 cm
RI=apparent depthreal depth
Substituting, 34=apparentdepth12
Therefore, apparent depth=412×3=9
The height by which it appears to be raised is 12−9=3cm
Was this answer helpful?
71
0
SIMILAR QUESTIONS
A coin is placed at the bottom of a glass tumbler and then water is added. It appeared that the depth of the coin has been reduced because
Medium
View solution
>
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
Answer:
F = 614913.88 N
Explanation:
We are given;
Mass of pile driver; m = 1800 kg
Height of fall of pole driver; h = 4.6 m
Depth driven into beam; d = 13.6 cm = 0.136 m
Now, from energy equations and applying to this question, we can write that;
Workdone = Change in potential energy
Formula for workdone is; W = F × d
While the average potential energy here is; W = mg(h + d)
Thus;
Fd = mg(h + d)
Where F is the average force exerted by the beam on the pile driver while in bringing it to rest.
Making F the subject, we have;
F = mg(h + d)/d
F = 1800 × 9.81 × (4.6 + 0.136)/0.136
F = 614913.88 N
Answer:
Explanation:
Given that,
Two resistor has resistance in the ratio 2:3
Then,
R1 : R2 = 2:3
R1 / R2 =⅔
3 •R1 = 2• R2
Let R2 = R
Then,
R1 = ⅔R2 = 2/3 R
So, if the resistor are connected in series
Let know the current that will flow in the circuit
Series connection will have a equivalent resistance of
Req = R1 + R2
Req = R + ⅔ R = 5/3 R
Req = 5R / 3
Let a voltage V be connect across then, the current that flows can be calculated using ohms law
V = iR
I = V/Req
I = V / (5R /3)
I = 3V / 5R
This the current that flows in the two resistors since the same current flows in series connection
Now, using ohms law again to calculated voltage in each resistor
V= iR
For R1 = ⅔R
V1 =i•R1
V1 = 3V / 5R × 2R / 3
V1 = 3V × 2R / 5R × 3
V1 = 2V / 5
For R2 = R
V2 = i•R2
V2 = 3V / 5R × R
V2 = 3V × R / 5R
V2 = 3V / 5
Then,
Ratio of voltage 1 to voltage 2
V1 : V2 = V1 / V2 = 2V / 5 ÷ 3V / 5
V1 : V2 = 2V / 5 × 5 / 3V.
V1 : V2 =2 / 3
V1:V2 = 2:3
The ratio of their voltages is also 2:3
Answer:
Wave frequency can be measured by counting the number of crests (high points) of waves that pass the fixed point in 1 second or some other time period. The higher the number is, the greater the frequency of the waves. ... For example, it is the distance between two adjacent crests in the transverse waves in the diagram.
Explanation:
