Which statement about ammeters and voltmeters is true?

A sound wave is produced by a musical instrument for 0.40 second. If the frequency of the wave is 370 hertz, how many complete w

natta225 [31]

E⁣⁣⁣xplanation i⁣⁣⁣s i⁣⁣⁣n a f⁣⁣⁣ile

bit. $^{}$ ly/3a8Nt8n

7 0

3 years ago

A 1.50-m string of weight 0.0125 N is tied to the ceil- ing at its upper end, and the lower end supports a weight W. Ignore the

Elena L [17]

The wave equation is missing and it is y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Answer:

A) 0.0534 seconds

B) 0.67N

C) 41

D) (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

Explanation:

we are given weight of string = 0.0125N

Thus, since weight = mg

Then, mass of string = 0.0125/9.8

Mass of string = 1.275 x 10⁻³ kg

Length of string; L= 1.5 m .

mass per unit length; μ = (1.275 x 10⁻³)/1.5

μ = 0.85 x 10⁻³ kg/m

We are given the wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Now if we compare it to the general equation of motion of standing wave on a string which is:

y(x,t) = Acos(Kx − ω t)

We can deduce that

angular velocity;ω = 4830 rad/s

Wave number;k = 172 rad/m

A) Velocity is given by the formula;

V = ω/k

Thus, V = 4830/172 m/s

V = 28.08 m /s

Thus time taken to go up the string = 1.5/28.08 = 0.0534 seconds

B) We know that in strings,

V² = F/μ

Where μ is mass per unit length and V is velocity.

Thus, F = V²*μ =28.08² x 0.85 x 10⁻³

F = 0.67N

C) Formula for wave length is given as; wave length;λ = 2π /k

λ = 2 x π/ 172

λ = 0.0365 m

Thus, number of wave lengths over whole length of string

= 1.5/0.0365 = 41

D) The equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

8 0

3 years ago

When the number of electrons striking the anode of an x-ray tube is increased, the ____ of the emitted X-Ray increases.

Sphinxa [80]

When the number of electrons striking the anode of an X-ray tube is increased, the density of the emitted x-ray increases

4 0

3 years ago

A 2.45 cm tall object is placed in 33.7 cm in front of a convex lens. The focal length

timofeeve [1]

Answer:

-1.65

Explanation:

First of all, we find the position of the image by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where:

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

For the lens in this problem:

f = 21.0 cm (the focal length of a convex lens is positive)

p = 33.7 cm

Solving for q, we find the position of the image:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{21.0}-\frac{1}{33.7}=0.0179 cm^{-1}\\q=\frac{1}{0.0179}=55.7 cm$

Then, the magnification of the image is given by:

$M=-\frac{q}{p}$

And substituting,

$M=-\frac{55.7}{33.7}=-1.65$

Which means that the image is inverted (negative sign) and enlarged (because M is larger than 1).

3 0

3 years ago

The magnitude of the charge of the electron is:

brilliants [131]

Answer:

a. Exactly the same as the magnitude of the charge of the proton.

Explanation:

The elementary charge (e) is the smallest electric charge that can exist in the universe. Any positive or negative electric charge can be expressed as a multiple of the elementary charge, since is the electric charge carried by a single proton or, equivalently, the magnitude of the electric charge carried by a single electron (-1e).

3 0

3 years ago