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alisha [4.7K]
3 years ago
12

A Cathode of initial mass 10.00g weigh to 10.05g

Physics
1 answer:
Leto [7]3 years ago
6 0

Answer:

i'm sorry i'm not a physics student

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If a transformer has 50 turns in the primary winding and 10 turns on the secondary winding, what is the reflected resistance in
Elza [17]

Answer:

The reflected resistance in the primary winding is 6250 Ω

Explanation:

Given;

number of turns in the primary winding, N_P = 50 turns

number of turns in the secondary winding, N_S = 10 turns

the secondary load resistance, R_S = 250 Ω

Determine the turns ratio;

K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5

Now, determine the reflected resistance in the primary winding;

\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms

Therefore, the reflected resistance in the primary winding is 6250 Ω

6 0
3 years ago
Two parallel plates of area 0.155 m2<br> are separated by 0.00100 m. What<br> is their capacitance?
KonstantinChe [14]

Answer:

1.37 x 10^-9

Explanation:

Trust me bro.

4 0
2 years ago
A construction worker puts 20J of energy in to one strike of his hammer on the head of a nail. The energy transferred to driving
Slav-nsk [51]

Answer:

efficiancy=40 percent

Explanation:

efficiency=energy output/energy input×100

efficiancy=8J/20J×100

efficiancy=0.4×100

efficiancy=40 percent

Mark brianliest if my answer suit your question..

3 0
2 years ago
Read 2 more answers
the illuminance on a surface is 6 lux and the surface is 4 meters from the light source. What is the intensity of the saurce?
Lelechka [254]
L = illuminance
A = surface
i = intensity

L = i / A ==: i = L * A

i = 6 lux * 4 m^2 = 24 lumen
8 0
2 years ago
A mass spectrometer was used in the discovery of the electron. In the velocity selector, the electric and magnetic fields are se
Mama L [17]

Answer:

Explanation:

Radius of dee, r = 8 mm = 0.008 m

Electric field, e = 400 V/m

Magnetic field, B = 4.7 x 10^-4 T

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

(a) Let v is the speed of electrons.

v = \frac{Bqr}{m}

v = \frac{4.7\times 10^{-4}\times 1.6\times 10^{-19}\times 0.008}{9.1 \times 10^{-31}}

v = 661098.9 = 661099 m/s

(b)

\frac{e}{m}=\frac{1.6 \times 10^{-19}}{9.1\times 10^{-31}}

e / m = 1.76 x 10^14 C / kg

(c) Let K be the kinetic energy

K = 0.5 x mv²

K = 0.5 x 9.1 x 10^-31 x 661099 x 661099

K = 1.99 x 10^-19 J

K = 1.24 eV

So, the potential difference is

V = 1.24 V

(d) if the acceleration voltage is doubled

V = 2 x 1.24 = 2.48 V

So, Kinetic energy

K = 2.48 eV

K = 2.48 x 1.6 x 10^-19 = 3.968 x 10^-19 J

Let v is the speed

K = 0.5 x mv²

3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²

v = 933856.5 m/s

Let the new radius is r.

r=\frac{mv}{Bq}

r=\frac{9.1\times 10^{-31}\times 933856.5}{4.7\times 10^{-4}\times 1.6\times 10^{-19}}

r = 0.0113 m = 1.13 cm

7 0
3 years ago
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