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jasenka [17]
3 years ago
12

A 1.1-kg uniform bar of metal is 0.40 m long and has a diameter of 2.0 cm. When someone bangs one end of this bar, a 1.5 MHz sho

ck wave is travels along the length of the bar and reaches the other end in 0.12 ms. What is the wavelength of the shock wave in the metal?
Physics
1 answer:
lyudmila [28]3 years ago
6 0

Answer:

\lambda = 2.22\times 10^{-3}\ m

Explanation:

Given,

mass of the bar = 1.1 Kg

length of rod, l = 0.40 m

diameter of the rod, d = 2 cm

frequency, f = 1.5 MHz

time, t = 0.12 ms

wavelength of the shock wave = ?

Speed of the wave =\dfrac{L}{t}=\dfrac{0.40}{0.12\times 10^{-3}}

   v = 3333.33 m/s

wavelength of the wave

\lambda = \dfrac{v}{f}= \dfrac{3333.33}{1.5\times 10^6}

\lambda = 2.22\times 10^{-3}\ m

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Fiesta28 [93]

Answer:

a) 1.75s b) 17.2 m/s (down)

Explanation:

d1= 15m d2= 0m (because it hits ground)

a= -9.81 m/s^2 t=???

Equation

the triangle means change in so d2-d1

Δd= v1 * t + 1/2 * a * t^2

0m-15m= v1*t + 1/2 a t^2

-15 m= 0m/s*t (goes away) + 1/2* a *t^2

-15mx2= t^2

-15mx2/a= t^2

Square root (-30/-9.81m/s^2)

t=1.75 s

b) now v2!!

Im going to use v2= v1 + a*t

v2= 0m/s + -9.81 x 1.75s

v2 = -17.2 m/s or you can say 17.2 m/s down!!!

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3 years ago
Is condensation adding thermal energy
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Svetradugi [14.3K]

the more pressure put on the string, the more frequency and higher pitch.

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4 years ago
A sled slides down a hill, reaches the level surface, and eventually comes to a stop. Which fact ultimately explains why the sle
Kitty [74]

Answer:

B. The presence of an unbalanced force(e.g friction) causes a moving object to stop.

Explanation:

As the friction is that force that can stop the sled upon reaching the levelled surface so the option b is correct.

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3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
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