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Romashka [77]
3 years ago
8

Questionson cell arrangement

Physics
1 answer:
dedylja [7]3 years ago
7 0
Where??? complete your question please
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The Earth's Radius is 6.3710x106 m and mass is 5.9742x1024 kg. What is the acceleration due to gravity at Mount Everest (elevati
Shalnov [3]

Answer is

9.773m/s^2

-----------------------------------------------------------------------------

Given,

h=8848m

The value of sea level is 9.08m/s^2. So, Let g′ be the acceleration due to the gravity on Mount Everest.

g′=g(1 − 2h/h)

=9.8(1 - 6400000/17696)

=9.8(1 − 0.00276)

9.8×0.99724

=9.773m/s^2

Thus, the acceleration due to gravity on the top of Mount Everest is =9.773m/s^2

-----------------------------------------------------------------------

hope this helps :)

3 0
3 years ago
A laser emits light of frequency 4.74 x 1014 hz. what is the wavelength of the light in nm?
Ulleksa [173]
<span>Use this formula:Wavelength=c/v, where c is the speed of light, and v the frequency. 6.33 X 10^-7=3 X 10^8/v v=3 X 10^8/6.33 X 10^-7 v=4.74 X 10^14 Hertz,</span>
5 0
3 years ago
Two lamps, each having reisistance of 3 Ohms, connect in a series. What current will flow if a voltage source of 5 V is connecte
Ket [755]
Total resistance will be R1 + R2.
And the formula will be I equals V over R.
V=5
I=?
R=6

I= 0,83 amps
6 0
3 years ago
a meteor Falls towards Earth's surface given that the acceleration due to gravity is 9.8 m/s^2 what is the meteors potential ene
ser-zykov [4K]

Answer: It’s 98,000

Explanation: Just did it on apex

5 0
3 years ago
A horizontal spring-mass system has low friction, spring stiffness 160 N/m, and mass 0.3 kg. The system is released with an init
anygoal [31]

Answer:

(a) 0.38 m

(b) 2.78 m/s

(c) 0.11 watt

Explanation:

mass, m = 0.3 kg

spring constant, K = 160 N/m

initial compression, d = 12 cm = 01.2 m

initial speed, u = 3 m/s

(a) Let the maximum stretch is y.

Use conservation of energy

Initial potential energy + initial kinetic energy = final potential energy

0.5 x K x d² + 0.5 x m x u² = 0.5 x K x y²

160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 160 x y²

2.304 + 0.00432 = 160 y²

y = 0.38 m

y = 38 cm

(b) Let v is the maximum speed.

The speed is maximum when the stretch in the spring is zero, so by use of conservation of energy

Initial potential energy + initial kinetic energy = final kinetic energy

0.5 x K x d² + 0.5 x m x u² = 0.5 x m x v²

160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 0.3 x v²

2.304 + 0.00432 = 0.3 v²

v = 2.78 m/s

(c) The time period of the spring mass system is given by

T=2\pi\sqrt{\frac{m}{K}}

T=2\pi\sqrt{\frac{0.3}{160}}

T = 0.272 second

Energy dissipated per cycle = 0.03 J

Power, P = 0.03 / 0.272 = 0.11 Watt

5 0
3 years ago
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