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masya89 [10]
3 years ago
11

A circuit contains four resistors connected in series. What happens to the equivalent resistance when one of the resistors is re

placed with an ideal wire?
Physics
1 answer:
Gennadij [26K]3 years ago
7 0

Answer:

R = (R1 + R2 + R3 + R4) / 4     equivalent resistance

If one of the resistors is replaced by an ideal wire

R = (R1 + R2 + R3 / 3       equivalent resistance

The problem does not state that the resistors have the same value.

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Which set of terms represent mechanical waves?
Helga [31]
Energy that transfers through the medium
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3 years ago
oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student
12345 [234]

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

3 0
3 years ago
What 2 key vocabulary words are used when you send an object in motion?
ad-work [718]

Answer:

push or pull

Explanation:

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3 years ago
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I am not sure about this one
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Assume air resistance is negligible unless otherwise stated. Calculate the displacement in m and velocity in m/s at the followin
tankabanditka [31]

Answer: 1) t=0.5 s; S=6.225 m and v=14.9 m/s

2) t=1 s; S=14.9 m and v=19.8 m/s

3) t=1.5 s; S=26.05 m and v=24.7 m/s

Explanation:

The displacement S is given by

S=ut+\frac{1}{2} at^{2}

and  final velocity v is given by

v=u+at

where u is the initial velocity

a is acceleration

t is time taken

Case 1: when time is 0.5 s

The displacement is

S=ut+\frac{1}{2} at^{2}\\S=10\times  0.5 +\frac{1}{2}\times 9.8\times 0.5^{2}\\\\S=6.225 m

the velocity is

v=u+at\\v=10+9.8\times 0.5\\v=14.9 m/s

Case 2: when t=1 sec

The displacement is

S=ut+\frac{1}{2} at^{2}\\S=10\times  1 +\frac{1}{2}\times 9.8\times 1^{2}\\\\S=14.9 m

the velocity is

v=u+at\\v=10+9.8\times 1\\v=19.8 m/s

Case 3: t=1.5 s

The displacement is

S=ut+\frac{1}{2} at^{2}\\S=10\times  1.5 +\frac{1}{2}\times 9.8\times 1.5^{2}\\\\S=26.05 m

the velocity is

v=u+at\\v=10+9.8\times 1.5\\v=24.7 m/s

4 0
3 years ago
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