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Lynna [10]
3 years ago
9

A wire with mass 50 g is stretched so that it’s ends are tied down at points 100 cm apart. The wire vibrates in its fundamental

mode with frequency 60 Hz and an amplitude of 0.5 cm at the antinodes. What is the tension in the wire?
2500 N
720 N
120 N
1250 N
Physics
2 answers:
ivolga24 [154]3 years ago
6 0
I think the answer is 2500 N
KIM [24]3 years ago
5 0

Answer:

720 N

Explanation:

Most of these answers  on here are wrong but I just tool the test ! This is the right answer!

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A trapeze artist swings in simple harmonic motion with a period of 3.8 s.
pogonyaev
As we know that time period of simple pendulum is given as

T = 2π √L/g

here we know that

T = 3.8 s

now from above equation we know that

T² = 4π² (L/g)

now on rearranging the above equation we will have

L = gT² / 4π²

now plug in all data into it

L = (9.8) (3.8)² / (4) (3.14)²

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7 0
3 years ago
The y-axis on a distance-time graph would be ____________. time distance speed
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Answer:

Distance

Explanation:

distance is in vertical axis,or y-axis and time is on the horizontal axis,or x-axis.

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2 years ago
A 55.0-g piece of copper wire is heated, and the temperature of the wire changes from 19.0°C to 86.0°C. The amount of heat absor
Alinara [238K]
The specific heat of a metal or any element or compound can be determined using the formula Cp = delta H / delta T / mass. delta pertains to change. That is change in enthalpy and change in temperature. From the given data, Cp is equal to 343 cal per (86-19) c per 55 grams. This is equal to 0.093 cal / g deg. Celsius
4 0
2 years ago
A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne
Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0
3 years ago
(6) A
liraira [26]

Answer:

cyguu and I will be like that and I

Explanation:

GHG and I will be like them and I will be like this in the best of luck in the pulley system and the same your time and consideration and I will be like them and I will be like them and I will be like them and TFT capacitive touchscreen and I will be in Phone

4 0
3 years ago
Read 2 more answers
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