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Harrizon [31]
1 year ago
11

A source charge of 3 µC generates an electric field of 2.86 × 105 N/C at the location of a test charge. Using k = 8.99 × 109N.m^

2/c^2, what is the distance, to the
Physics
2 answers:
viktelen [127]1 year ago
9 0

0.31 for E2020 users

Nataliya [291]1 year ago
7 0
Variables:

Source charge, Q = 3 micro C = 3 * 10^ - 6 C

E = electric field = 2.86 * 10 ^5 N/C

K = 8.99 * 10^9 N * m^2 / C

d = distance = ?

Formula:

E = K * Q / (d^2) => d^2 = K * Q / E

=> d^2 = 8.99 * 10^9 N * m^2 / C * 3 * 10^ -6 C / (2.86 * 10^ 5 N/C)

d^2 = 9.43 * 10 ^ -2  m^2

=> d = 3.07 * 10^-1 m

Answer: 0.307 m

Note: it is a long distance due to the Electric field is very low
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frez [133]

Answer:

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Explanation:

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3 0
2 years ago
Read 2 more answers
(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind
Vilka [71]
There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
F_f = \mu N
where \mu is the coefficient of friction, while N is the normal force. So we have:
F=\mu N
since we know that F=300 N and \mu=0.3, we can find N, the magnitude of the normal force:
N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is \mu=0.03 due to the presence of the oil. Therefore, we have:
N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N
8 0
1 year ago
In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that
mars1129 [50]

Answer:

The maximum transverse speed of the bead is 0.4 m/s

Explanation:

As we know that the Amplitude of the travelling wave is

A = 3.65 mm

Now the speed of the travelling wave is

v_x = 13.5 m/s

now we know that distance of first antinode from one end is 27.5 cm

so length of the loop of the standing wave is given as

\frac{\lambda}{4} = 27.5 cm

\lambda = 110 cm

now we have

N = \frac{2L}{\lambda}

N = \frac{2(1.65)}{1.10}

N = 3

now we have

R = 2A sin(kx)

R = 2(3.65) sin(\frac{2\pi}{1.10}x)

R = 7.3 sin(1.82 \pi x)

now at x = 13.8 cm

R = 7.3 sin(1.82 \pi (0.138))

R = 5.18 mm

now we have

f = \frac{v}{\lambda}

f = \frac{13.5}{1.1}

f = 12.27 Hz

now maximum speed is given as

v_y = R\omega

v_y = (5.18 \times 10^{-3})(2\pi(12.27))

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4 0
1 year ago
In which direction does the electric field point at a position directly east of a positive charge
lora16 [44]
Field lines always point away from the positive side of a magnet. So i would say east but im not to sure 
7 0
1 year ago
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A block with mass m1 = 4.50 kg and a ball with mass m2 = 7.70 kg are connected by a light string that passes over a frictionless
Allisa [31]
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4 0
2 years ago
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