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3 years ago

13

When a guitar string is plucked, what part of the standing wave is found at the fixed ends of the string?(1 point)

1 answer:

LenKa [72]3 years ago

5 0

The nodes are found at the fixed ends of the strings

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After school, several students skateboard at the nearby park. Your best friend, Erwin, is a good skateboarder and happens to fal

Alchen [17]

Answer:

A person who never gives up.

Explanation:

due to his passion for skateboarding he try's his never gives up until he will finally learns the trick.

3 0

3 years ago

What is the strength of the electric field ep 0.90 mm from a proton?

dem82 [27]

The electric field strength is inversely related to the square of the distance.so the strength of the electric field is given by

$E=\frac{1}{4\pi \epsilon _{0} } \frac{q}{r^{2} } = \frac{k q}{r^{2} }$

Here, $\frac{1}{4\pi \epsilon _{0} } = k$ is constant depend upon medium and its value is $9.0 \times10^{9} \ N m^2/C^2$ and q is charge and r is the distance.

Given $r = 0.90 mm = 9.0 \times 10^{-4} m$ and we know the charge of proton, $q = 1.6\times 10^{-19} \ C$ .

Therefore,

$E=\frac{9.0 \times10^{9} \ N m^2/C^2 \times1.6\times 10^{-19} \ C }{(9.0 \times 10^{-4} m)^2} = 0.177 \times 10^{-2} \ N/C \\\\ E= 1.77 \times 10^{-3} N/C$

4 0

3 years ago

Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p

ira [324]

Answer:

The force exerted on the $q_1$ is $F = 2.25*10^{3} \ N$

Explanation:

From the question we are told that

The area is $A = 2.34*10^{-3} \ m^2$

The magnitude of charge placed on them is $q = 7.07 * 10^{-7} C$

The charge placed between the plate is $q_1 = 6.62 *10^{-5} C$

The electric field generated around the plate is mathematically represented as

$E = \frac{q}{A \epsilon_o}$

Substituting values

$E = \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}$

$E = 34*10^{6} \ V/m$

The force exerted the charge $q_1$ is mathematically represented as

$F = q_1 * E$

Substituting values

$F = 6.62 *10^{-5} * 34*10^{6}$

$F = 2.25*10^{3} \ N$

8 0

3 years ago

A body in simple harmonic motion has a displacement x that varies in time t according to the equation x = 5cos(π t + π/3) , wher

zhuklara [117]

Answer:

1/2 Hz

Explanation:

A simple harmonic motion has an equation in the form of

$x(t) = Acos(\omega t - \phi)$

where A is the amplitude, $\omega = 2\pi f$ is the angular frequency and $\phi$ is the initial phase.

Since our body has an equation of x = 5cos(π t + π/3) we can equate $\omega = \pi$ and solve for frequency f

$2\pi f = \pi$

f = 1/2 Hz

7 0

3 years ago

Read 2 more answers

If the mass of the spacecraft were doubled, how much would the force of gravity would change?

vodka [1.7K]

F=mg
if m doubled, F would double as well

5 0

3 years ago