What is the minimum value of force acting between two charges placed at 1 m apart from each other?

You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond

luda_lava [24]

Answer:

a) L = 0.75m f₁ = 113.33 Hz , f₃ = 340 Hz, b) L=1.50m f₁ = 56.67 Hz , f₃ = 170 Hz

Explanation:

This resonant system can be simulated by a system with a closed end, the tile wall and an open end where it is being sung

In this configuration we have a node at the closed end and a belly at the open end whereby the wavelength

With 1 node λ₁ = 4 L

With 2 nodes λ₂ = 4L / 3

With 3 nodes λ₃ = 4L / 5

The general term would be λ_n= 4L / n n = 1, 3, 5, ((2n + 1)

The speed of sound is

v = λ f

f = v / λ

f = v n / 4L

Let's consider each length independently

L = 0.75 m

f₁ = 340 1/4 0.75 = 113.33 n

f₁ = 113.33 Hz

f₃ = 113.33 3

f₃ = 340 Hz

L = 1.5 m

f₁ = 340 n / 4 1.5 = 56.67 n

f₁ = 56.67 Hz

f₃ = 56.67 3

f₃ = 170 Hz

8 0

2 years ago

A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m

madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂ = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

4 0

2 years ago

A car is traveling with speed v0 when it begins to speed up at a rate of Δv every second. After t1 seconds, the car travels with

Rainbow [258]

Answer:

d = Δv(t2-t1)

Explanation:

Speed is defined as the change of displacement with respect to time. It is expressed as shown;

Speed = change in displacement/change in time

Δv = d/Δt

d = Δv*Δt

d = ΔvΔt

Δt = t2-t1

d = Δv(t2-t1)

Δv is the change in rate of speed

Δt = change in time

The correct expression for the displacement of the car during this motion is d = Δv(t2-t1)

8 0

2 years ago

How much heat Q2Q2 is transferred to the skin by 25.0 gg of steam onto the skin? The heat of vaporization for steam is L=2.256×1

astraxan [27]

Answer:

56400Joules

Explanation:

The quantity of heat required is expressed as;

Q = mL

m is the mass = 25g = 0.025kg

L is the latent heat of vaporization for steam = 2.256×10^6J/kg

Substitute into the formula as shown;

Q = 0.025×2.256×10^6

Q = 56400Joules

Hence the quantity of hear required is 56400Joules

3 0

2 years ago

A 2.0-kg block sliding on a rough horizontal surface is attached to one end of a horizontal spring (k = 250 N/m) which has its o

Burka [1]

Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.

There are two forces performing work on the block: the restoring force of the spring and kinetic friction.

By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:

$W_{\rm total}=\Delta K$

or

$W_{\rm friction}+W_{\rm spring}=0-K=-K$

where K is the block's kinetic energy at the equilibrium point,

$K=\dfrac12\left(2.0\,\mathrm{kg}\right)\left(2.6\dfrac{\rm m}{\rm s}\right)^2=6.76\,\mathrm J$

Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is

$W_{\rm spring}=-\dfrac12\left(250\dfrac{\rm N}{\rm m}\right)(0.20\,\mathrm m)^2=-5.00\,\mathrm J$

Compute the work performed by friction:

$W_{\rm friction}-5.00\,\mathrm J=-6.76\,\mathrm J \implies W_{\rm friction}=-1.76\,\mathrm J$

By Newton's second law, the net vertical force on the block is

∑ F = n - mg = 0 ==> n = mg

where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.

So we have

$W_{\rm friction}=-f(0.20\,\mathrm m)$

$\implies -1.76\,\mathrm J=-\mu\left(2.0\,\mathrm{kg}\right)\left(9.8\dfrac{\rm m}{\mathrm s^2}\right)(0.20\,\mathrm m)$

$\implies \boxed{\mu\approx0.45}$

4 0

2 years ago