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Ksju [112]
3 years ago
6

A teacher stands well back from an outside doorway 0.88 m wide, and blows a whistle of frequency 780 Hz. Ignoring reflections, e

stimate at what angle(s) it is not possible to hear the whistle clearly on the playground outside the doorway.
Physics
1 answer:
AlladinOne [14]3 years ago
5 0

the wavelength is to be 56.67 cm and I know that i should be applying the double slit equation for destructive interference.

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Social assistance is a synonym for social service.<br> False<br> True
nadezda [96]
The answer is that it is true
6 0
3 years ago
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Explain why a school bus is a compound machine
Zolol [24]
It combines multiplesimple machines working together
3 0
4 years ago
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A 1.05 kg block slides with a speed of 0.865 m/s on a frictionless horizontal surface until it encounters a spring with a force
djyliett [7]

Answer:

a) U = 0 J    

k = 0.393 J

E = 0.393 J

b) U = 0.0229J

k = 0.370 J

E = 0.393 J

c) U = 0.0914 J

k = 0.302 J

E = 0.393 J

d) U = 0.206 J

k = 0.187 J

E = 0.393 J

e) U = 0.366 J

k = 0.027 J

E = 0.393 J

Explanation:

Hi there!

The equations of kinetic energy and elastic potential energy are as follows:

k = 1/2 · m · v²

U = 1/2 · ks · x²

Where:

m = mass of the block.

v = velocity.

ks = spring constant.

x = displacement of the string.

a) When the spring is not compressed, the spring potential energy will be zero:

U = 1/2 · ks · x²

U = 1/2 · 457 N/m · (0 cm)²

U = 0 J

The kinetic energy of the block will be:

k = 1/2 · m · v²

k = 1/2 · 1.05 kg · (0.865 m/s)²

k = 0.393 J

The mechanical energy will be:

E = k + U = 0.393 J + 0 J = 0.393 J

This energy will be conserved, i.e., it will remain constant because there is no work done by friction nor by any other dissipative force (like air resistance). This means that the kinetic energy will be converted only into spring potential energy (there is no thermal energy due to friction, for example).

b) The spring potential energy will be:

U = 1/2 · 457 N/m · (0.01 m)²

U = 0.0229 J

Since the mechanical energy has to remain constant, we can use the equation of mechanical energy to obtain the kinetic energy:

E = k + U

0.393 J = k + 0.0229 J

0.393 J - 0.0229 J = k

k = 0.370 J

c) The procedure is now the same. Let´s calculate the spring potential energy with x = 0.02 m.

U = 1/2 · 457 N/m · (0.02 m)²

U = 0.0914 J

Using the equation of mechanical energy:

E = k + U

0.393 J = k + 0.0914 J

k = 0.393 J - 0.0914 J = 0.302 J

d) U = 1/2 · 457 N/m · (0.03 m)²

U = 0.206 J

E = 0.393 J

k = E - U = 0.393 J - 0.206 J

k = 0.187 J

e) U = 1/2 · 457 N/m · (0.04 m)²

U = 0.366 J

E = 0.393 J

k = E - U = 0.393 J - 0.366 J = 0.027 J.

4 0
3 years ago
A boy throws rocks with an initial velocity of 12m/s [down] from a 20 m bridge into a river. Consider the river to be at a heigh
stira [4]

Answer:

Please find attached the Velocity-Time graph, the Displacement-Time graph and the combined Velocity/Displacement-Time graph, created with Microsoft Excel

Explanation:

The given parameters are;

The initial velocity with which the boy throws the rock, u = 12 m/s

The direction in which he throws the rock = Down

The height from which the rock was thrown, h = 20 m

The height at which the river is located = 0 m

The kinematic equation of motion, of the rock can be given as follows;

h = u·t + 1/2·g·t²

Where;

t = The time of motion of the rock

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the rock is thrown = 20 m

Substituting the known values into given equation, we get;

20 = 12·t + 1/2·9.8·t² = 12·t + 4.9·t²

4.9·t² + 12·t - 20 = 0

t = (-12 ± √(12² - 4×4.9×(-20)))/(2 × 4.9) = (-12 ± √(242))/(9.8)

t ≈ -3.587 seconds or t ≈ 1.138 seconds

Attached please find the Velocity-Time graph, the Displacement-Time graph and the combined Velocity/Displacement-Time graph, created with Microsoft Excel.

4 0
3 years ago
A computer is purchased for $2816 and depreciates at a constant rate to $0 in 8 years. Find a formula for the value, V , of the
marusya05 [52]

Answer:

  • The formula its f(t) \ = \ - \ 352 \ \frac{\$ }{years} \ t \ + \ \$ \ 2816
  • After 5 years, the computer value its $ 1056

Explanation:

<h3>Obtaining the formula</h3>

We wish to find a formula that

  • Starts at 2816. f(0 \ years) \ = \ \$ \ 2816
  • Reach 0 at 8 years. f( 8 \ years) \ = \ \$ \ 0
  • Depreciates at a constant rate. m

We can cover all this requisites with a straight-line equation. (an straigh-line its the only curve that has a constant rate of change) :

f(t) \ = \ m\ t \ + \ b,

where m its the slope of the line and b give the place where the line intercepts the <em>y</em> axis.

So, we can use this formula with the data from our problem. For the first condition:

f ( 0 \ years ) = m \ (0 \ years) + b = \$ \ 2816

b = \$ \ 2816

So, b = $ 2816.

Now, for the second condition:

f ( 8 \ years ) = m \ (8 \ years) + \$ \ 2816 = \$ \ 0

m \ (8 \ years) = \ - \$ \ 2816

m = \frac{\ - \$ \ 2816}{8 \ years}

m = \frac{\ - \$ \ 2816}{8 \ years}

m = \ - \ 352 \frac{\$ }{years}

So, our formula, finally, its:

f(t) \ = \ - \ 352 \ \frac{\$ }{years} \ t \ + \ \$ \ 2816

<h3>After 5 years</h3>

Now, we just use <em>t = 5 years</em> in our formula

f(5 \ years) \ = \ - \ 352 \ \frac{\$ }{years} \ 5 \ years \ + \ \$ \ 2816

f(5 \ years) \ = \ - \$ \ 1760 + \ \$ \ 2816

f(5 \ years) \ = $ \ 1056

4 0
4 years ago
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