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2 years ago

What travels faster than anything anything else in the universe?

Physics

1 answer:

cluponka [151]2 years ago

4 0

Answer:

The announcement he had made promised to overturn our understanding of the Universe. If the data gathered by 160 scientists working on the project were correct, the unthinkable had been observed. Particles – in this case, neutrinos – had travelled faster than light.

Explanation: Plz Mark brainleist

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A echo occurs when sound does what?

klio [65]

Echoes occur when a reflected soundwave reaches the ear more than 0.1 seconds after the original sound wave was heard. ... There will be an echoinstead of a reverberation. Reflection of sound waves off of surfaces is also affected by the shape of the surface.

(mark as brainly please)

8 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 motorcyclist changes the velocity of his bike from 20.0 meters/second to 35.0 meters/second under a constant acceleration of 4

shutvik [7]

You can use the equation V=Vo+at since the acceleration is constant. Plugging in the values you know, you will get an answer of 3.75 seconds

8 0

3 years ago

What is the measurement of six centimeters equal to?

AlladinOne [14]

Six centimeters equal to about two inches

8 0

3 years ago

Make the curvature radius 0.6 m, the refractive index 1.5, and the diameter 0.6 m. place the lamp so that the source of light is

vodka [1.7K]

We are given
r = 0.6 m
n = 1.5
D = 0.6 m, R1 = 30 cm
R2 = 120 cm

We are asked to get the focal length and the distance of the focal plane from the lens

We use the formula
1 / f = ( n - 1) (1/R1 - 1/R2)
Substituting and solving for f
1/ f = (1.5 - 1) (1/30 - 1/120)
f = 80 cm

The focal length is 80 cm and the distance of the focal plane from the lesn is 80 cm - 30 cm = 50 cm.<span />

5 0

3 years ago