Someone claps his or her hands is an example of… motion energy to sound energy sound energy to potential energy motion energy to radiant energy

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Answer 1

Answer: motion energy to sound energy i think

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A child attaches a 0.6 kg toy to a 0.9 meter length of string and spins it around in uniform circular motion. Calculate the tension in the string if the period of rotation is 4.1 seconds.

Answers

Answer:

1.26812 N

Explanation:

$m$ = Mass of toy = $0.6\ \text{kg}$

$r$ = Length of string = $0.9\ \text{m}$

$t$ = Period of rotation = $4.1\ \text{s}$

Time period is given by

$t=(2\pi r)/(v)\n\Rightarrow v=(2\pi r)/(t)\n\Rightarrow v=(2\pi 0.9)/(4.1)\n\Rightarrow v=1.3792\ \text{m/s}$

The rotational velocity is 1.3792 m/s

The tension in the rope will be the centripetal force acting on the toy

$T=(mv^2)/(r)\n\Rightarrow T=(0.6* 1.3792^2)/(0.9)\n\Rightarrow T=1.26812\ \text{N}$

The tension in the string is 1.26812 N.

What is the force required to accelerate 1 kilogram of mass at 1 meter per second per second.1 Newton

1 pound

1 kilometer

1 gram

Answers

Answer:

it's answer is 1 newton

A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound produced by the pipe, in SI units, is closest to:

Answers

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

$f_n=(nv_s)/(4L)$

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

$f_(3)=((3)(340m/s))/(4(2.5m))=102\ Hz$

hence, the frequency of the third overtone is 102 Hz

Two narrow slits separated by 0.30 mm are illuminated with light of wavelength 496 nm. (a) How far are the first three bright fringes from the center of the pattern if observed on the screen 130 cm distant? (b) How far are the first three dark fringes from the center of the pattern?

Answers

Answer:

Explanation:

d = separation of the slits = 0.30 mm = 0.30 x 10⁻³ m

λ = wavelength of the light = 496 nm = 496 x 10⁻⁹ m

n = order of the bright fringe

D = screen distance = 130 cm = 1.30 m

$x_(n)$ = Position of nth bright fringe

Position of nth bright fringe is given as

$x_(n) =( n D \lambda )/(d)$

For n = 1

$x_(1) =( (1) (1.30)(496* 10^(-9)))/(0.30* 10^(-3))$

$x_(1) = 2.15* 10^(-3)m$

For n = 2

$x_(2) =( (2) (1.30)(496* 10^(-9)))/(0.30* 10^(-3))$

$x_(2) = 4.30* 10^(-3)m$

For n = 3

$x_(2) =( (2) (1.30)(496* 10^(-9)))/(0.30* 10^(-3))$

$x_(2) = 6.45* 10^(-3)m$

Position of nth dark fringe is given as

$y_(n) =( (2n+1) D \lambda )/(2d)$

For n = 1

$y_(1) =( (2(1)+1) (1.30)(496* 10^(-9)))/(2(0.30* 10^(-3)))$

$y_(1) = 3.22* 10^(-3)m$

For n = 2

$y_(2) =( (2(2)+1) (1.30)(496* 10^(-9)))/(2(0.30* 10^(-3)))$

$y_(2) = 5.4* 10^(-3)m$

For n = 3

$y_(3) =( (2(3)+1) (1.30)(496* 10^(-9)))/(2(0.30* 10^(-3)))$

$x_(3) = 7.5* 10^(-3)m$

NASA scientists suggest using rotating cylindrical spacecraft to replicate gravity while in a weightless environment. Consider such a spacecraft that has a diameter of d = 148 m. What is the speed v, in meters per second, the spacecraft must rotate at its outer edge to replicate the force of gravity on earth?

Answers

The speed of the spacecraft at its outer edge is 26.93 m/s.

The given parameters;

diameter of the spacecraft, d = 148 m
radius of the spacecraft, r = 74 m

The speed of the spacecraft at its outer edge is calculated as follows;

$F_g = F_c\n\nmg = (mv^2)/(r) \n\nv^2 = rg\n\nv = √(rg) \n\nv = √((74)(9.8)) \n\nv = 26.93 \ m/s$

Thus, the speed of the spacecraft at its outer edge is 26.93 m/s.

Learn more here:brainly.com/question/20905151

Answer:

Explanation:

Given

diameter of spacecraft $d=148\ m$

radius $r=74\ m$

Force of gravity $F_g$ =mg

where m =mass of object

g=acceleration due to gravity on earth

Suppose v is the speed at which spacecraft is rotating so a net centripetal acceleration is acting on spacecraft which is given by

$F_c=(mv^2)/(r)$

$F_c=F_g$

$(mv^2)/(r)=mg$

$(v^2)/(r)=g$

$v=√(gr)$

$v=√(1450.4)$

$v=38.08\ m/s$

Solenoid A has total number of turns N length L and diameter D. Solenoid B has total number of turns 2N, length 2L and diameter 2D. Inductance of solenoid A is 8 times inductance of solenoid B
1/4 of inductance of solenoid B
same as inductance of solenoid B
1/8 of inductance of solenoid B
four times of inductance of solenoid B

Answers

Answer:

∴Inductance of solenoid A is $\frac18$ of inductance of solenoid B.

Explanation:

Inductance of a solenoid is

$L=N\frac\phi I$

$=N(B.A)/(I)$

$=N(\mu_0NI)/(l.I)A$

$=(\mu_0N^2A)/(l)$

$=(\mu_0N^2)/(l).\pi(\frac d2)^2$

$=\mu_0\pi(N^2d^2)/(4l)$

N= number of turns

$l$ = length of the solenoid

d= diameter of the solenoid

A=cross section area

B=magnetic induction

$\phi$ = magnetic flux

$I$ = Current

Given that, Solenoid A has total number of turns N, length L and diameter D

The inductance of solenoid A is

$=\mu_0\pi(N^2D^2)/(4L)$

Solenoid B has total number of turns 2N, length 2L and diameter 2D

The inductance of solenoid B is

$=\mu_0\pi((2N)^2(2D)^2)/(4.2L)$

$=\mu_0\pi(16 N^2D^2)/(4.2L)$

Therefore,

$\frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=(\mu_0\pi(N^2D^2)/(4L))/(\mu_0\pi(16 N^2D^2)/(4.2L))$

$\Rightarrow \frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=\frac18$

$\Rightarrow {\textrm{Inductance of A}}=\frac18* {\textrm{Inductance of B}}$

∴Inductance of solenoid A is $\frac18$ of inductance of solenoid B.

Hi there!

We can begin by calculating the inductance of a solenoid.

Recall:
$L = (\Phi _B)/(i)$

L = Inductance (H)
φ = Magnetic Flux (Wb)

i = Current (A)

We can solve for the inductance of a solenoid. We know that its magnetic field is equivalent to:
$B = \mu _0 (N)/(L)i$

And that the magnetic flux is equivalent to:
$\Phi _B = \int B \cdot dA = B \cdot A$

Thus, the magnetic flux is equivalent to:
$\Phi _B = \mu _0 (N)/(L)iA$

The area for the solenoid is the # of loops multiplied by the cross-section area, so:
$A_(total)= N * A$

$\Phi _B = \mu _0 (N^2)/(L)iA$

Using this equation, we can find how it would change if the given parameters are altered:
$\Phi_B ' = \mu_0 ((2N)^2)/(2L) i * 4A$

**The area will quadruple since a circle's area is 2-D, and you are doubling its diameter.

$\Phi'_B = (4)/(2) * 4(\mu_0 (N)/(L)iA) = 8\mu_0 (N)/(L)iA$

Thus, Solenoid B is 8 times as large as Solenoid A.

Solenoid A is 1/8 of the inductance of solenoid B.

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