All categories

3 years ago

A guitar string transmits waves at

Physics

1 answer:

vichka [17]3 years ago

5 0

Answer:

The wavelength of the waves on the string is found using v = λf:

λ = v/f = 315/370 = 0.85135m

The first harmonic (fundamental) standing wave formed on the string has nodes only at the ends (see diagram in link) so:

String length = λ/2 = 0.85135/2 = 0.426m (to 3 sig. figs.)

You might be interested in

These types of electromagnetic waves are listed in order of decreasing

viktelen [127]

The answer you’re looking for is
D. Gamma rays
Enjoy the rest of your day/night :)

4 0

2 years ago

Read 2 more answers

What potential difference is required to cause 4.00 a to flow through a resistance of 330 ω?

Alisiya [41]

We can solve the problem by using Ohm's law, which states that an Ohmic conductor the following relationship holds:
$\Delta V = I R$
where
$\Delta V$ is the potential difference applied to the resistor
I is the current flowing through it
R is the resistance

In our problem, I=4.00 A and $R=330 \omega$ , so the potential difference is
$\Delta V = IR=(4.00 A)(330 \omega)=1320 V$

7 0

3 years ago

How many moles of MgCl2 are there in 315 g of the compound? mol

jeka94

315g/95gmol-1
3.315 moles of MgCl2

4 0

3 years ago

Read 2 more answers

1-A car moves toward east 12km is represented as A and it turns towards south 16km is represented as B. What is the resultant ve

hjlf

1. A-20 km south east

The car's displacement consists of two components into two different directions. Using a system of coordinates in which x represents the east direction and y represents the south direction, the two displacements are:

$d_x = 12 km$ east

$d_y = 16 km$ south

Since the two components are orthogonal to each other, we can find the resultant displacement by using Pythagorean's theorem:

$d=\sqrt{d_x^2+d_y^2}=\sqrt{(12 km)^2+(16 km)^2}=\sqrt{400}=20 km$

and the direction is between the two original directions, so south-east.

2. D. 10 m/s

First of all, we need to calculate the total time the stone took to hit the ground. Since the vertical distance covered is S = 78.4 m, and since the motion is an accelerated motion with constant acceleration g=9.8 m/s^2, we have

$S=\frac{1}{2}gt^2$

From which we find the total time of the fall, t:

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(78.4 m)}{9.8 m/s^2}}=4 s$

Now we can consider the horizontal motion of the stone: we know that the stone travels for d = 40 m in a time of t = 4 s, therefore the horizontal velocity of the stone is

$v=\frac{d}{t}=\frac{40 m}{4 s}=10 m/s$

3. B=32.32 m

As in the previous problem, we have to calculate the total time it takes for the stone to reach the river first. Since the vertical distance covered is S = 20 m, we have

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(20 m)}{9.8 m/s^2}}=2.0 s$