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sesenic [268]
3 years ago
8

Are the Sun's rays like mechanical or electromagnetic waves

Physics
2 answers:
EleoNora [17]3 years ago
7 0
They are electromagnetic waves.
Dmitry_Shevchenko [17]3 years ago
5 0
If the sun generates any mechanical waves, none of them
have ever reached us, because only electromagnetic waves
can travel through empty space.
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A concert loudspeaker suspended high off the ground emits 28.0 W of sound power. A small microphone with a 0.700 cm2 area is 55.
grandymaker [24]

Explanation:

It is known that wave intensity is the power to area ratio.

Mathematically,    I = \frac{P}{A}

As it is given that power is 28.0 W and area is 7 \times 10^{-5} m^{2}.

Therefore, sound intensity will be calculated as follows.

             I = \frac{P}{A}

               = \frac{28.0 W}{4 \times 3.14 \times 7 \times 10^{-5} m^{2}}

                = 0.318 \times 10^{5} W/m^{2}

or,             = 3.18 \times 10^{4} W/m^{2}

Thus, we can conclude that sound intensity at the position of the microphone is 3.18 \times 10^{4} W/m^{2}.

7 0
3 years ago
Temperature danger zone <br> what is the biggest cause of foodborne illness
Nina [5.8K]
Foodborne illnesses are caused by eating contaminated food and drinking contaminated water. The contaminations are caused mostly by bacteria, virus, and parasites.
5 0
3 years ago
Long ago, it was believed that maggots came from meat. In the late 1600s, Francesco Redi hypothesized that maggots came from fli
Lunna [17]
D) Both A and B.

Francisco Redi must use 1)a covered, unrefrigerated meat and 2) an uncovered, refrigerated meat to experiment and test his hypothesis that maggots came from flies rather than from meat.
5 0
3 years ago
Read 2 more answers
6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres
sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

2ah=v_f^2-v_0^2

Where:

\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}

If we assume the initial velocity to be 0 we get:

2ah=v_f^2

The acceleration is the acceleration due to gravity:

2gh=v_f^2

Now, we take the square root to both sides:

\sqrt{2gh}=v_f

Now, we substitute the values:

\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f

solving the operations:

280\frac{m}{s}=v

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

F_d=\frac{1}{2}C\rho_{air}Av^2

Where:

\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

F_d=mg

Now, we determine the mass of the raindrop using the following formula:

m=\rho_{water}V

Where:

\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}

The volume is the volume of a sphere, therefore:

m=\rho_{water}(\frac{4}{3}\pi r^3)

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)

Solving the operations:

m=1.39\times10^{-5}kg

Now, we substitute the values in the formula for the drag force:

F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})

Solving the operations:

F_d=1.36\times10^{-4}N

Now, we substitute in the formula:

1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2

Now, we solve for the velocity:

\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2

Now, we substitute the values. We will use the area of a circle:

\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2

Substituting the radius:

\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2

Solving the operations:

70.67\frac{m^2}{s^2}=v^2

Now, we take the square root to both sides:

\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\  \\ 8.4\frac{m}{s}=v \\  \end{gathered}

Therefore, the velocity is 8.4 m/s

7 0
1 year ago
A rotating water pump works by taking water in at one side of a rotating wheel, and expelling it from the other side. If a pump
Xelga [282]

Answer:

Explanation:

initial angular velocity, ωo = 0 rad/s

angular acceleration, α = 30.5 rad/s²

time, t = 9 s

radius, r = 0.120 m

let the velocity is v after time 9 s.

Use first equation of motion for rotational motion

ω = ωo + αt

ω = 0 + 30.5 x 9

ω = 274.5 rad/s

v = rω

v = 0.120 x 274.5

v = 32.94 m/s

8 0
3 years ago
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