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nika2105 [10]
3 years ago
7

A wire, 1.0 m long, with a mass of 90 g, is under tension. A transverse wave is propagated on the wire, for which the frequency

is 890 Hz, the wavelength is .10m, and the amplitude is 6.5 mm. The tension in the line, in SI units, is closest to
Physics
1 answer:
Mice21 [21]3 years ago
7 0

Answer:

T = 712.9 N

Explanation:

First, we will find the speed of the wave:

v = fλ

where,

v = speed of the wave = ?

f = frequency = 890 Hz

λ = wavelength = 0.1 m

Therefore,

v = (890 Hz)(0.1 m)

v = 89 m/s

Now, we will find the linear mass density of the wire:

\mu = \frac{m}{L}

where,

μ = linear mass density of wie = ?

m = mass of wire = 90 g = 0.09 kg

L = length of wire = 1 m

Therefore,

\mu = \frac{0.09\ kg}{1\ m}

μ = 0.09 kg/m

Now, the tension in wire (T) will be:

T = μv² = (0.09 kg/m)(89 m/s)²

<u>T = 712.9 N</u>

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Answer:

m = 28.7[kg]

Explanation:

To solve this problem we must use the definition of kinetic energy, which can be calculated by means of the following equation.

E_{k}=\frac{1}{2}*m*v^{2}\\

where:

Ek = kinetic energy = 1800 [J]

m = mass [kg]

v = 11.2 [m/s]

1800=\frac{1}{2}*m*(11.2)^{2}\\m = 28.7[kg]

7 0
3 years ago
The largest watermelon ever grown had a mass of 118 kg. Suppose this watermelon were exhibited on a platform 5.00 m above the gr
WINSTONCH [101]

Answer: height = 3.98m

Explanation: by placing the watermelon at a height above the ground, it has a potential energy of the formulae

p = mgh

p = potential energy = 4.61kJ = 4610J

m = mass of watermelon = 118 kg

g = acceleration due gravity = 9.8 m/s²

4610 = 118 * 9.8 * h

h = 4610/ 118 * 9.8

h = 4610/ 1156.4

h = 3.98m

6 0
3 years ago
Which statement correctly differentiates between transmitters and receivers?
jeka94

Answer:

Transmitters send radio waves, and receivers capture radio waves.

Explanation:

Let us look at each of the choices one by one:

(1).Transmitters have antennas, and receivers do not have antennas.

Nope. To send signals transmitters need antennas, and to receive signals   the receivers need antennas as well.

(2). Transmitters send radio waves, and receivers capture radio waves.

This is true. Transmitters are for transmitting and receivers are for     receiving EM signals.

(3). Transmitters have demodulators, and receivers have modulators.

No, it is the other way around. Transmitters have modulators, and          receivers have demodulators.

(4). Transmitters do not have amplifiers, and receivers have amplifiers.

Nope. Both the transmitters and the receivers need amplifiers.    Transmitters need them to increase the power of the broadcast, and   receivers need them to amplify the signal for processing.

Therefore, only the 2nd statement "Transmitters send radio waves, and receivers capture radio waves." is correct.

7 0
3 years ago
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Which component of a loudspeaker produces sound waves? a solenoid coil a thin membrane a permanent magnet
zimovet [89]

Answer: A thin membrane.

Explanation:

When a sound signal is allowed to pass through the voice coil suspended between permanent magnet, magnetic field will be induced which will cause vibration in the diaphragm - a thin membrane causing disturbance of air in the surrounding of membranes which eventually produce sound waves.

Therefore, we can conclude that a thin membrane of the loud speaker produces sound waves

4 0
3 years ago
You wad up a piece of paper and throw it into the wastebasket. How far will
vitfil [10]

The range of the piece of paper is C) 1.4 m

Explanation:

The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:

- A uniform motion along the horizontal direction, with constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, g=9.8 m/s^2)

From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

d=\frac{u^2 sin 2\theta}{g}

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u is the initial velocity

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g is the acceleration of gravity

For the piece of paper in this problem,

u = 4.3 m/s

\theta=65^{\circ}

Substituting,

d=\frac{(4.3)^2 sin(2\cdot 65^{\circ})}{9.8}=1.45 m \sim 1.4 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0
3 years ago
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