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A spring is stretched 5 cm from its equilibrium position. If this stretching requires 30 J of work,

Triss [41]

Answer:

Best answer will get Brainliest!!!

What is the volume scaled down by a factor of 1/10

Measurements:

Top: 7 in, both sides: 12 in, front: 12 in, back: 12 in, bottom: 7 in

Please help!

Explanation:

3 0

3 years ago

Two transverse waves travel along the same taut string inopposite directions. the waves are described by following equations use

Umnica [9.8K]

Answer: y'=2Asin(kx)cos(wt)

Explanation:

Let y1=A sin (kx + wt) be the first wave

y2=A sin (kx - wt) be the second wave in the opposite direction (which we showed by putting a negative sign between the terms kx and wt)

Please do note that both wave have the same attributes (that's Amplitude, wave number and angular frequency) because they are formed on the same medium by the same source just that their directions are opposite.

By super imposing these 2 waves, we have a resulting singular wave representing both wave (law of superimposition) with a resulting value of vertical displacement y'.

Thus y' = y1 + y2.

Let us do the math.

y'=A sin (kx + wt) + A sin (kx - wt)

By factoring A out, we have that

y' = A [ sin (kx + wt) + sin (kx - wt)]

For simplicity let us use the substitution

Let (kx + wt) = a and (kx - wt) =b

Hence we have that

y' = A [sin a + sin b].

From trigonometric ratio

sin a + sin b = 2sin[(a+b)/2] * cos [(a - b)/2]

By recalling that (kx + wt) = a and (kx - wt) =b

sin a + sin b = 2sin [(kx +wt +kx-wt) /2] * cos [(kx +wt - (kx-wt))/2]

Thus we have that

sin a + sin b = 2sin [(kx+wt+kx-wt)/2] * cos[(kx+wt-kx+wt)/2]

By collecting like terms in the bracket we have that

sin a + sin b = 2sin[2kx/2] * cos [2wt/2]

By dividing

sin a + sin b = 2sin(kx) cos(wt)

Now let us get the final resultant vertical displacement (y')

Recall that

y' = A [sin a + sin b]. and we already deduced that

sin a + sin b = 2sin(kx) cos(wt)

Finally,

y' = A [2sin(kx) cos(wt)] which is

y'=2Asin(kx)cos(wt)...... Final answer

4 0

4 years ago

An object of mass m travels along the parabola yequalsx squared with a constant speed of 5 units/sec. What is the force on the

Dmitriy789 [7]

Explanation:

The object is moving along the parabola y = x² and is at the point (√2, 2). Because the object is changing directions, it has a centripetal acceleration towards the center of the circle of curvature.

First, we need to find the radius of curvature. This is given by the equation:

R = [1 + (y')²]^(³/₂) / |y"|

y' = 2x and y" = 2:

R = [1 + (2x)²]^(³/₂) / |2|

R = (1 + 4x²)^(³/₂) / 2

At x = √2:

R = (1 + 4(√2)²)^(³/₂) / 2

R = (9)^(³/₂) / 2

R = 27 / 2

R = 13.5

So the centripetal force is:

F = m v² / r

F = m (5)² / 13.5

F = 1.85 m

7 0

3 years ago

True or false: Everything in the universe works with harmony because of balanced energy

Eduardwww [97]

Answer:

Explanation:

True

7 0

3 years ago

Read 2 more answers

For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate.

Kamila [148]

Answer:

A) E = 4.96 x 10³ eV

B) E = 4.19 x 10⁴ eV

C) E = 3.73 x 10⁹ eV

Explanation:

A)

For photon energy is given as:

$E = hv$

$E = \frac{hc}{\lambda}$

where,

E = energy of photon = ?

h = 6.625 x 10⁻³⁴ J.s

λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Therefore,

$E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}$

$E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})$

E = 4.96 x 10³ eV

B)

The energy of a particle at rest is given as:

$E = m_{0}c^2$

where,

E = Energy of electron = ?

m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (9.1 x 10^{-31} kg)(3 x 10^8 m/s)^2\\$

$E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\$

E = 4.19 x 10⁴ eV

C)

The energy of a particle at rest is given as:

$E = m_{0}c^2$

where,

E = Energy of alpha particle = ?

m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (6.64 x 10^{-27} kg)(3 x 10^8 m/s)^2\\$

$E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\$

E = 3.73 x 10⁹ eV

8 0

3 years ago