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jonny [76]
3 years ago
12

462.3 cm - 25.74 cm

Physics
1 answer:
Murrr4er [49]3 years ago
8 0

Answer:

436.56 centimeters

Explanation:

heart this! My answer is correct, I search it.

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what happens to the current across a circuit when the voltage is doubled while the resistance is held back ​
Lynna [10]

Since the current is inversely propotional to its resistance, when the voltage is doubled the current will be one-half

7 0
3 years ago
Read 2 more answers
Question 1
sineoko [7]
Velocity = displacement / time
Displacement = 2.3 km
Time = 5.78 mins
Velocity = 2.3/5.78
Velocity = 0.398 to the nearest 3 significant figures
4 0
3 years ago
A wire with a current of 5.2 A is at an angle of 45° relative to a magnetic field of 0.87 T. What is the force exerted on a 1.2
julsineya [31]

Answer:

F = I L X B = I L B sin theta

F = 5.2 * 1.2 * .87 sin 45 = 3.84 N   in the direction of (L X B)

Note: L and B must be at right angles for   F = I L B

7 0
3 years ago
A uniform electric field exists in the region between two oppositely charged plane-parallel plates. An electron is released from
Zigmanuir [339]

Answer:

Explanation:

  • given S = distance from the first = 3.20cm = 0.032m, t = 1.30×10−8 s
  • q = 1.6 x 10_19C
  • using S = at^2/2
  • acceleration = 0.032 X 2 /(1.30×10−8)^2

a = 3.79 x 10^14m/s^2

  • From F = ma
  • F = qE
  • ma = qE

E = ma /q = 9.11 x 10^-31 x 3.79 x 10^14 / 1.6 x 10^-19

E = magnitude of this electric field. = 2156.3N/C

b) Find the speed of the electron when it strikes the second plate ; V^2 = 2as

= 2 X 3.79 x 10^14 X 0.032

= 4.92 X 10^6m/s

5 0
3 years ago
Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of
Pachacha [2.7K]

Answer:

<em>The frequency changes by a factor of  0.27.</em>

<em></em>

Explanation:

The frequency of an object with mass m attached to a spring is given as

f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }

where f is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

f_{n} = \frac{1}{2\pi } \sqrt{\frac{k}{14m} }

simplifying, we have

f_{n} = \frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }

f_{n} = \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }

if we divide this final frequency by the original frequency, we'll have

==> \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }  ÷  \frac{1}{2\pi } \sqrt{\frac{k}{m} }

==> \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }  x  2\pi \sqrt{\frac{m}{k} }

==> 1/3.742 = <em>0.27</em>

7 0
3 years ago
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