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3 years ago

7

The equation of a transverse wave traveling along a string is given by y=0.3sin (0.5x-50t) find the maximum displacement of the

particles for the equilibrium position

1 answer:

uysha [10]3 years ago

5 0

Answer:

0.3cm

Explanation:

Y = 0.3 sin(0.5x - 50t) compare with,

y = A sin(kx - wt)

A = 0.3m

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which planet should punch travel to if his goal is to weigh in at 118 lb? refer to the table of planetary masses and radii given

Harrizon [31]

The planet that Punch should travel to in order to weigh 118 lb is Pentune.

<h3 /><h3 /><h3>The given parameters:</h3>

Weight of Punch on Earth = 236 lb
Desired weight = 118 lb

The mass of Punch will be constant in every planet;

$W = mg\\\\m = \frac{W}{g}\\\\m = \frac{236}{g}$

The acceleration due to gravity of each planet with respect to Earth is calculated by using the following relationship;

$F = mg = \frac{GmM}{R^2} \\\\g = \frac{GM}{R^2}$

where;

M is the mass of Earth = 5.972 x 10²⁴ kg
R is the Radius of Earth = 6,371 km

For Planet Tehar;

$g_T =\frac{G \times 2.1M}{(0.8R)^2} \\\\g_T = 3.28(\frac{GM}{R^2} )\\\\g_T = 3.28 g$

For planet Loput:

$g_L =\frac{G \times 5.6M}{(1.7R)^2} \\\\g_L = 1.94(\frac{GM}{R^2} )\\\\g_L = 1.94g$

For planet Cremury:

$g_C =\frac{G \times 0.36M}{(0.3R)^2} \\\\g_C = 4(\frac{GM}{R^2} )\\\\g_C = 4 g$

For Planet Suven:

$g_s =\frac{G \times 12M}{(2.8R)^2} \\\\g_s = 1.53(\frac{GM}{R^2} )\\\\g_s = 1.53 g$

For Planet Pentune;

$g_P =\frac{G \times 8.3 }{(4.1R)^2} \\\\g_P = 0.5(\frac{GM}{R^2} )\\\\g_P = 0.5 g$

For Planet Rams;

$g_R =\frac{G \times 9.3M}{(4R)^2} \\\\g_R = 0.58(\frac{GM}{R^2} )\\\\g_R = 0.58 g$

The weight Punch on Each Planet at a constant mass is calculated as follows;

$W = mg\\\\W_T = mg_T\\\\W_T = \frac{236}{g} \times 3.28g = 774.08 \ lb\\\\W_L = \frac{236}{g} \times 1.94g =457.84 \ lb\\\\ W_C = \frac{236}{g}\times 4g = 944 \ lb \\\\ W_S = \frac{236}{g} \times 1.53g = 361.08 \ lb\\\\W_P = \frac{236}{g} \times 0.5 g = 118 \ lb\\\\W_R = \frac{236}{g} \times 0.58 g = 136.88 \ lb$

Thus, the planet that Punch should travel to in order to weigh 118 lb is Pentune.

<u>The </u><u>complete question</u><u> is below</u>:

Which planet should Punch travel to if his goal is to weigh in at 118 lb? Refer to the table of planetary masses and radii given to determine your answer.

Punch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and "weighs in" at 236 lb. Unhappy with his recent bouts, Punch decides to go to a different planet where he would weigh in at 118 lb so that he can compete with the bantamweights who are not allowed to exceed 118 lb. His plan is to travel to Xobing, a newly discovered star with a planetary system. Here is a table listing the planets in that system (<em>find the image attached</em>).

<em>In the table, the mass and the radius of each planet are given in terms of the corresponding properties of the earth. For instance, Tehar has a mass equal to 2.1 earth masses and a radius equal to 0.80 earth radii.</em>

Learn more about effect of gravity on weight here: brainly.com/question/3908593

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2 years ago

Fifteen grams of substance X at 95 degrees Celsius is mixed with 45 grams of substance Z at 85 degrees Celsius in a container wh

Viefleur [7K]

According to the Law of Conservation of Energy, energy is neither created nor destroyed. They are just transferred from one system to another. To obey this law, the energy of the substances inside the container must be equal to the substance added to it. The energy is in the form of heat. There can be two types of heat energy: latent heat and sensible heat. Sensible heat is energy added or removed when a substance changes in temperature. Latent heat is the energy added or removed at a constant temperature during a phase change. Since there is no mention of phase change, we assume the heat involved here is sensible heat. The equation for sensible heat is:

H = mCpΔT
where
m is the mass of the substance
Cp is the specific heat of a certain type of material or substance
ΔT is the change in temperature.

So the law of conservation of heat tells that:

Sensible heat of Z + Sensible heat of container = Sensible heat of X

Since we have no idea what these substances are, there is no way of knowing the Cp. We can't proceed with the calculations. So, we can only assume that in the duration of 15 minutes, the whole system achieves equilibrium. Therefore, the equilibrium temperature of the system is equal to 32°C. The answer is C.

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3 years ago

The proper IUPAC naming convention for the compound symbol 'KI' is

Darya [45]

Proablyly a or c idk

6 0

3 years ago

Lila is a track and field athlete. She must complete four laps around a circular track. The track itself is a 400 meter track an

Aleonysh [2.5K]

Answer:

Her speed is 1.1 m/s, and her velocity is 0 m/s

Explanation:

Speed = Distance covered/Time

Given

Distance = 400m

Time = 6minutes = 6*60 = 360 secs

Substitute the given parameter into the formula;

Speed = 400/360

Speed = 1.1m/s

Since the track is a circular track, the displacement will be zero. She is only moving in a circular path (no direction)

Velocity = Displacement/Time

Velocity = 0/3600

Velocity = 0m/s

Hence her speed is 1.1 m/s, and her velocity is 0 m/s

6 0

3 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

3 years ago