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AnnyKZ [126]
3 years ago
9

PLEASE HELP REALLY WOULD APPRECIATE THE HELP;) A positively charged glass rod is bought close to a suspended metal needle. What

Physics
1 answer:
Kisachek [45]3 years ago
8 0

Answer:

Explanation:

attracted

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A wye-connected load has a voltage of 480v applied to it. What is the voltage drop across each phase
rodikova [14]

Answer:

Y_A=277.128 \angle 30v

Y_B=277.128 \angle (-150)v

Y_C=277.128 \angle (90)v

Explanation:

From the question we are told that

Voltage V_L_L =480v

Generally in a case of Y_connection V_p_ h is mathematical represented as

V_p_h=\frac{V_l_l}{\sqrt{3}} \angle (\phi-30)v

Generally voltage drop across phase A

Y_A=\frac{408}{\sqrt{3}} \angle -(0-30)

Y_A=277.128 \angle 30v

Generally voltage drop across phase B

Y_B=277.128 \angle (-30-120)

Y_B=277.128 \angle (-150)v

Generally voltage drop across phase C

Y_C=277.128 \angle (-30+120)

Y_C=277.128 \angle (90)v

3 0
3 years ago
A photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 2.5 m away. He then wants to take a p
Temka [501]

Answer:

0.004 m away from the film

Explanation:

u = Object distance

v = Image distance

f = Focal length = 50 mm

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{2.5}\\\Rightarrow \frac{1}{v}=\frac{98}{5} \\\Rightarrow v=\frac{5}{98}=0.051\ m

The image distance is 0.051 m

When u = 50 cm

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{0.5}\\\Rightarrow \frac{1}{v}=18\\\Rightarrow v=\frac{1}{18}=0.055\ m

The image distance is 0.055 m

The lens has moved 0.055-0.051 = 0.004 m away from the film

3 0
3 years ago
A shopper walks eastward 3.2 meters and then westward
Musya8 [376]
11.347 yards or 34.1 ft
4 0
3 years ago
What are tissues made up of?
ELEN [110]

Answer:

tissue are made of cells .

5 0
3 years ago
Read 2 more answers
I drop a quarter from a 2 story house. It takes 1.85 to hit the ground. what was the velocity
Valentin [98]

At the instant you dropped it from your hand, its speed was zero.
The coin gained 9.8 m/s every second it fell.  At the end of 1.85 sec,
just before it went 'tink' onto the ground, its speed was

           (9.8 m/s) x (1.85 sec) = 18.13 m/s .

Its velocity at that final instant was 18.13 m/s downward.

7 0
3 years ago
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