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Hatshy [7]
3 years ago
12

What type of wave could measure 5 meters in length?

Physics
1 answer:
Tatiana [17]3 years ago
4 0

-- A sound wave at 68 Hz . . . like from the piano key 2 octaves below middle-C .

-- A radio wave at 60 MHz . . . the upper edge of the old analog TV channel-2 .

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2. A pebble is dropped down a well and hits the water 1.5 s later. Using the equations for motion with constant acceleration, de
kow [346]

Answer:

S = 11.025 m

Explanation:

Given,

The time taken by the pebble to hit the water surface is, t = 1.5 s

Acceleration due to gravity, g = 9.8 m/s²

Using the II equations of motion

                          S = ut + 1/2 gt²

Here u is the initial velocity of the pebble. Since it is free-fall, the initial velocity

                                u = 0

Therefore, the equation becomes

                            S = 1/2 gt²

Substituting the given values in the above equation

                              S = 0.5 x 9.8 x 1.5²

                                 = 11.025 m

Hence, the distance from the edge of the well to the water's surface is, S = 11.025 m

3 0
3 years ago
Define work What is the unit for work? For Physics, thanks.
Minchanka [31]

Answer:

Work, in physics, measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement

Explanation:

5 0
3 years ago
In the diagram, what is happening to the temperature at Point B? Question 6 options: A. The temperature is rising as the molecul
raketka [301]
<span> B. The temperature is not rising because the heat is being used to break the connections between the molecules </span>
6 0
3 years ago
Read 2 more answers
A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a lar
horrorfan [7]

Answer:

W= 4.89 KJ

Explanation:

Lets take

temperature of hot water T₁ = 100⁰C

T₁ = 373 K

Temperature of cold ice T₂= 0⁰C

T₂ = 273 K

The latent heat of ice LH= 334 KJ

The heat rejected by the engine Q= m .LH

Q₂= 0.04 x 334

Q₂= 13.36 KJ

Heat gain by engine = Q₁

For Carnot engine

Q_1=\dfrac{T_1}{T_2}Q_2

Q_1=\dfrac{373}{273}\times 13.36

Q₁  = 18.25 KJ

The work W= Q₁  - Q₂

W= 18.25 - 13.36 KJ

W= 4.89 KJ

7 0
3 years ago
Read 2 more answers
Arrow_forward
garri49 [273]

Explanation:

(a) Hooke's law:

F = kx

7.50 N = k (0.0300 m)

k = 250 N/m

(b) Angular frequency:

ω = √(k/m)

ω = √((250 N/m) / (0.500 kg))

ω = 22.4 rad/s

Frequency:

f = ω / (2π)

f = 3.56 cycles/s

Period:

T = 1/f

T = 0.281 s

(c) EE = ½ kx²

EE = ½ (250 N/m) (0.0500 m)²

EE = 0.313 J

(d) A = 0.0500 m

(e) vmax = Aω

vmax = (0.0500 m) (22.4 rad/s)

vmax = 1.12 m/s

amax = Aω²

amax = (0.0500 m) (22.4 rad/s)²

amax = 25.0 m/s²

(f) x = A cos(ωt)

x = (0.0500 m) cos(22.4 rad/s × 0.500 s)

x = 0.00919 m

(g) v = dx/dt = -Aω sin(ωt)

v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)

v = -1.10 m/s

a = dv/dt = -Aω² cos(ωt)

a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)

a = -4.59 m/s²

3 0
3 years ago
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