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worty [1.4K]
4 years ago
5

Rewrite this formula to solve for GGi: DX=fL/GGi

Physics
1 answer:
sashaice [31]4 years ago
3 0
Comment if you need clarification!

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A neon sign that requires a voltage of 11,000 v is plugged into a 120-v wall outlet. what turns ratio (secondary/primary) must a
Arisa [49]
The equation that relates the voltages and the number of turns in a transformer is
\frac{V_s}{V_p}= \frac{N_s}{N_p}
where V_s = 120 V is the voltage in the secondary coil, V_p=11000 V is the voltage in the primary coil, and N_s and N_p are the number of turns on the secondary and primary coils.

Using the numbers, we find the ratio between the number of turns:
\frac{N_s}{N_p}= \frac{V_s}{V_p}= \frac{120 V}{11000 V}=0.0109
8 0
3 years ago
Something is thrown in the air and 5.35 seconds later it lands on something 2.1 meters from the ground. How fast was it thrown u
UNO [17]

Answer:

26.6 m/s

Explanation:

Given:

Δy = 2.1 m

t = 5.35 s

a = -9.8 m/s²

Find: v₀

Δy = v₀ t + ½ at²

(2.1 m) = v₀ (5.35 s) + ½ (-9.8 m/s²) (5.35 s)²

v₀ = 26.6 m/s

5 0
3 years ago
What color does the star alpha centauri appear to be?
myrzilka [38]
B.yellow-white is the answer.
3 0
3 years ago
Read 2 more answers
A satellite is in circular orbit around the earth. The orbit the satellite is at a height of 420 km above the earth's surface. F
astraxan [27]

Answer:

7661.06 m/s

Explanation:

R = Radius of Earth = 6.38\times 10^6\ m

h = Distance from the Earth = 420000 m

G = Gravitational constant = 6.674\times 10^{-11} N m^2/kg^2

M = Mass of Earth = 5.98\times 10^{24}\ kg

V=\sqrt{g{\frac{R^2}{R + h}}}\\\Rightarrow V=\sqrt{\frac{GM}{R^2}{\frac{R^2}{R + h}}}\\\Rightarrow V=\sqrt{\frac{6.674\times 10^{-11}\times 5.98\times 10^{24}}{(6.38\times 10^6)^2}{\frac{(6.38\times 10^6)^2}{6.38\times 10^6 + 420000}}}\\\Rightarrow V=7661.06\ m/s

The orbital speed of the satellite is 7661.06 m/s

6 0
4 years ago
A 0.676 m long section of cable carrying current to a car starter motor makes an angle of 57.7º with the Earth’s 5.29 x 10^−5T f
Studentka2010 [4]

Answer:

The current in the wire under the influence of the force is 216.033 A

Solution:

According to the question:

Length of the wire, l = 0.676 m

\theta = 57.7^{\circ}

Magnetic field of the Earth, B_{E} = 5.29\times 10^{- 5} T

Forces experienced by the wire, F_{m} = 6.53\times 10^{-3} N

Also, we know that the force in a magnetic field is given by:

F_{m} = IB_{E}lsin\theta

I = \frac{F_{m}}{B_{E}lsin\theta}

I = \frac{6.53\times 10^{-3}}{5.29\times 10^{- 5}\times 0.676sin57.7^{\circ}

I = 216.033 A

5 0
3 years ago
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