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3 years ago

What is the speed of a wave frequency of 300hz and a wavelength of 25M

Physics

1 answer:

tia_tia [17]3 years ago

8 0

Answer:

7500 m/s

Explanation:

We can use the equation velocity of a wave equals wavelength times frequency. Therefore, v = wavelength*f = (25 m)(300 Hz) = m/s7,500

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A total of 2.0x 1013 electrons pass a given point in a wire in 15s. what is the current in the wire

timurjin [86]

The current in a wire is defined as the amount of charge that passes a given point of the wire in a given time:
$I= \frac{Q}{\Delta t}$
where Q is the charge and $\Delta t$ is the time interval.

Since one electron has a charge of $q=1.6 \cdot 10^{-19}C$ , the total charge of $2.0 \cdot 10^{13}$ electrons is
$Q=qN=(1.6 \cdot 10^{-19} C)(2.0 \cdot 10^{13} )=3.2 \cdot 10^{-6} C$

And since the time interval is $\Delta t=15 s$ , the current in the wire is
$I= \frac{Q}{\Delta t}= \frac{3.2 \cdot 10^{-6} C}{15 s}=2.1 \cdot 10^{-7}A$

4 0

3 years ago

A 400 g ball swings in a vertical cirde at the end of

velikii [3]

Answer:

15.10m/s

Explanation:

The mass of the ball(m)=400g = 0.4kg

The radius of the string is(r)=15m

The tension in the string is(T)=10N

The acceleration due to gravity = $9.8m/s^{2}$

The tension in the string when the body is at the bottom is given by

$T=\frac{mv^{2} }{r}+mg$

To find the speed of the ball, we make v the subject of the formula

Therefore, $v=\sqrt\frac{r(T-mg}{m}$

$v= \sqrt\frac{15(10-0.4*9.8)}{0.4}$

$v=\sqrt\frac{91.2}{0.4} \\$

$v=\sqrt228 = 15.10 m/s$

The speed of the ball = 15.10m/s

8 0

2 years ago

A rod bent into the arc of a circle subtends an angle 2θ at the center P of the circle (see below). If the rod is charged unifor

Zigmanuir [339]

Answer:

Qsinθ/4πε₀R²θ

Explanation:

Let us have a small charge element dq which produces an electric field E. There is also a symmetric field at P due to a symmetric charge dq at P. Their vertical electric field components cancel out leaving the horizontal component dE' = dEcosθ = dqcosθ/4πε₀R² where r is the radius of the arc.

Now, let λ be the charge per unit length on the arc. then, the small charge element dq = λds where ds is the small arc length. Also ds = Rθ.

So dq = λRdθ.

Substituting dq into dE', we have

dE' = dqcosθ/4πε₀R²

= λRdθcosθ/4πε₀R²

= λdθcosθ/4πε₀R

E' = ∫dE' = ∫λRdθcosθ/4πε₀R² = (λ/4πε₀R)∫cosθdθ from -θ to θ

E' = (λ/4πε₀R)[sinθ] from -θ to θ

E' = (λ/4πε₀R)[sinθ]

= (λ/4πε₀R)[sinθ - sin(-θ)]

= (λ/4πε₀R)[sinθ + sinθ]

= 2(λ/4πε₀R)sinθ

= (λ/2πε₀R)sinθ

Now, the total charge Q = ∫dq = ∫λRdθ from -θ to +θ

Q = λR∫dθ = λR[θ - (-θ)] = λR[θ + θ] = 2λRθ

Q = 2λRθ

λ = Q/2Rθ

Substituting λ into E', we have

E' = (Q/2Rθ/2πε₀R)sinθ

E' = (Q/θ4πε₀R²)sinθ

E' = Qsinθ/4πε₀R²θ where θ is in radians

5 0

3 years ago

Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates

IRINA_888 [86]

Answer:

Fc = [ - 4.45 * 10^-8 j ] N

Explanation:

Given:-

- The masses and the position coordinates from ( 0 , 0 ) are:

Sphere A : ma = 80 kg , ( 0 , 0 )

Sphere B : ma = 60 kg , ( 0.25 , 0 )

Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2

Find:-

what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

Determine the angle (α) between vectors rac and rab using cosine rule:

$cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}$

Determine the angle (β) between vectors rbc and rab using cosine rule:

$cos ( \beta ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta ) = 0.6\\\\\beta = 53.13^{\circ \:}$

- Now determine the scalar gravitational forces due to sphere A and B on C:

Between sphere A and C:

Fac = G*ma*mc / rac^2

Fac = (6.674×10−11)*80*0.2 / 0.2^2

Fac = 2.67*10^-8 N

vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N

Between sphere B and C:

Fbc = G*mb*mc / rbc^2

Fbc = (6.674×10−11)*60*0.2 / 0.15^2

Fbc = 3.56*10^-8 N

vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

- The Net gravitational force can now be determined from vector additon of Fac and Fbc:

Fc = vector Fac + vector Fbc

Fc = [ - 2.136 i - 1.602 j ]*10^-8 + [ 2.136 i - 2.848 j ]*10^-8

Fc = [ - 4.45 * 10^-8 j ] N

3 0

3 years ago

What factors might affect how fast a balloon falls to the ground?

wolverine [178]

Answer:

air resistance, gravitational force

5 0

3 years ago