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weqwewe [10]
3 years ago
9

What is diffrence between ammeter and voltmeter

Physics
1 answer:
Daniel [21]3 years ago
3 0
VOLTMETER : voltage
AMMETER: current
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In the Bohr model of hydrogen, the electron moves in a circular orbit around the nucleus. (a) Determine the orbital frequency of
Airida [17]

Answer:

(a) 6.567 * 10^15 rev/s or hertz

(b) 8.21 * 10^14 rev/s or hertz

Explanation:

Fn= 4π^2k^2e^4m * z^2/(h^3*n^3)

Where Fn is frequency at all levels of n.

Z = 1 (nucleus)

e = 1.6 * 10^-19c

m = 9.1 * 10^-31 kg

h = 6.62 * 10-34

K = 9 * 10^9 Nm2/c2

(a) for groundstate n = 1

Fn = 4 * π^2 * (9*10^9)^2*(1.6*10^-19)^4* (9.1 * 10^-31) * 1 / (6.62 * 10^-31)^3 = 6.567 * 10^15 rev/s

(b) first excited state

n = 1

We multiple the groundstate answer by 1/n^3

6.567 * 10^15 rev/s/ 2^3

F2 = 8.2 * 10^ 14 rev/s

3 0
3 years ago
Which graph shows an object that is dropped?
Olegator [25]

Answer:

I'm pretty sure it's the third one where velocity goes from positive to negative

Explanation:

the positive velocity is before the object hits the ground and the negative is after

8 0
3 years ago
1. A pendulum has a period of 3 seconds. What's its frequency? 2. A pendulum has a frequency of 0.25 Hz. What's its period? 3. A
RideAnS [48]

Answer:

(1) 0.333 Hz

(2) 4 sec

(3) 2 sec, 0.5 Hz

Explanation:

(1) We have given time period of pendulum is 3 sec

So T = 3 sec

Frequency will be equal to f=\frac{1}{T}=\frac{1}{3}=0.333Hz

(2) Frequency of the pendulum is given f = 0.25 Hz

Time period is equal to T=\frac{1}{f}=\frac{1}{0.25}=4sec

(3) It is given that a pendulum makes 10 back and forth swings in 20 seconds

So time taken to complete 1 back and forth swings = =\frac{20}{10}=2sec

So time period T = 2 sec

Frequency will be equal to f=\frac{1}{T}=\frac{1}{2}=0.5Hz

6 0
3 years ago
Read 2 more answers
How much charge is on each plate of a 3.00-μF capacitor when it is connected toa 15.0-V battery? b) If this same capacitor is c
Sauron [17]

Answer:

(a) 45 micro coulomb

(b) 6 micro Coulomb

Explanation:

C = 3 micro Farad = 3 x 10^-6 Farad

V = 15 V

(a) q = C x V

where, q be the charge.

q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb

(b)

V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad

q = C x V

where, q be the charge.

q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb

6 0
3 years ago
A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.
MakcuM [25]

Answer:

b. 9.5°C

Explanation:

m_i = Mass of ice = 50 g

T_i = Initial temperature of water and Aluminum = 30°C

L_f = Latent heat of fusion = 3.33\times 10^5\ J/kg^{\circ}C

m_w = Mass of water = 200 g

c_w = Specific heat of water = 4186 J/kg⋅°C

m_{Al} = Mass of Aluminum = 80 g

c_{Al} = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C

The final equilibrium temperature is 9.50022°C

4 0
3 years ago
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