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LenKa [72]
1 year ago
14

A wave traveling in a string has a wavelength of 35 cm, an amplitude of 8. 4 cm, and a period of 1. 2 s. What is the speed of th

is wave?.
Physics
1 answer:
AlekseyPX1 year ago
3 0

0.29 m/s (wave velocity = wavelength (lamda)/period (T) in metres)

35 / 1.2 = 29.16

29.16 ÷ 100 = 0.29

Wave velocity in string:

The properties of the medium affect the wave's velocity in a string. For instance, if a thin guitar string is vibrated while a thick rope is not, the guitar string's waves will move more quickly. As a result, the linear densities of the two strings affect the string's velocity. Linear density is defined as the mass per unit length.

Instead of the sinusoidal wave, a single symmetrical pulse is taken into consideration in order to comprehend how the linear mass density and tension will affect the wave's speed on the string.

Learn more about density here:

brainly.com/question/15164682

#SPJ4

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When a potential difference of 13 V is placed across a resistor, the current in the resistor is 1.4
vitfil [10]
Just divide the two numbers with each other.
I mean 13/1.4=9.2857...
8 0
3 years ago
Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e
Nataliya [291]

Answer:

q=1.95*10^{-7}C

Explanation:

According to the free-body diagram of the system, we have:

\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)

So, we can solve for T from (1):

T=\frac{mg}{cos(15^\circ)}(3)

Replacing (3) in (2):

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e

The electric force (F_e) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)

Finally, we replace (5) in (4) and solving for q:

mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C

5 0
3 years ago
Which of the following types of particles can be categorized into different kind of elements?
KatRina [158]

The answer would be atoms

4 0
2 years ago
Read 2 more answers
What is the gravitational potential energy of a 0.550-kg projectile flying with 335 m/s, 72 meters above the ground?
Fofino [41]

Answer:

GPE = 388.08 Joules.

Explanation:

Given the following data;

Mass = 0.550kg

Speed = 335 m/s

Height = 72 meters

We know that acceleration due to gravity, g is equal to 9.8 m/s²

To find the gravitational potential energy;

Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.

Mathematically, gravitational potential energy is given by the formula;

G.P.E = mgh

Where;

G.P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Substituting into the formula, we have;

G.P.E = 0.550 * 9.8 * 72

GPE = 388.08 Joules.

4 0
3 years ago
Which of the following layers in the earth has the highest density?
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If we are being specific, the inner core has the highest density, but if not then the core in general
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