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mafiozo [28]
2 years ago
5

Lightning flashes one mile (1609 m) away from you. How much time does it take the light to travel that distance?

Physics
1 answer:
hichkok12 [17]2 years ago
7 0

Answer:

t=5.36\times 10^{-6}\ s

Explanation:

Given that,

Lightning flashes one mile (1609 m) away from you.

We need to find the time it take the light to travel that distance. Let the time be t. We know that,

speed = distance/time

t=\dfrac{d}{v}\\\\t=\dfrac{1609}{3\cdot10^{8}}\\\\=5.36\times 10^{-6}\ s

So, the required time is 5.36\times 10^{-6}\ s

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A baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby has a mass of 4kg. The carriage ha
Ira Lisetskai [31]

Answer:

E=252J

Explanation:

The total mechanical energy of an object or system is given by:

E mech=K+U

Where K is the kinetic energy of the object and U is the potential energy of the object. The carriage, sitting motionless at the top of the hill, has only potential energy in the form of gravitational potential energy.

Gravitational potential energy is given by:

Ug=mgh

Where m is the mass of the object, g is the gravitational acceleration constant, and h is the height of the object above some specific reference point, in this case the ground 21 m below.

The weight of a stationary object at the surface of the earth is equal to the force of gravity acting on the object.

W=→Fg=mg

We are given that the carriage weighs 12 N, therefore mg=12N.

Ug=12N⋅21m

⇒Ug=252Nm=252J

Hope it helped, God bless you!

5 0
3 years ago
An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the
Andrei [34K]

Answer:

a) 2nd case rate of rotation gives the greater speed for the ball

b) 1534.98 m/s^2

c) 1515.04 m/s^2

Explanation:

(a) v = ωR

when R = 0.60, ω = 8.05×2π

v = 0.60×8.05×2π = 30.34 m/s

Now in 2nd case

when R = 0.90, ω = 6.53×2π

v = 0.90×6.53×2π = 36.92 m/s

6.35 rev/s gives greater speed for the ball.

(b) a = ω^2 R = (8.05×2π)^2 )(0.60) = 1534.98 m/s^2

(c) a = ω^2 R = (6.53×2π)^2 )(0.90) = 1515.05 m/s^2

7 0
3 years ago
As part of his work for NASA, Dr. Murdock was asked to find out what percentage of people in the continental united States saw H
mario62 [17]

Answer:

Please, in the Explanation section you will find the explanation of the answer.

Explanation:

The exercise shows the continental United States and 3 cities used in the study carried out by Murdock. It can be said that the sample taken is part of the objective. There are several inconsistencies in Murdock's argument: the first has to do with the fact that the sample that was taken cannot represent the entire American population. A much larger, scientifically calculated sample would be required. The second is that their study did not take into account small cities or people living in the interior of the United States.

5 0
3 years ago
A ship sets sail from Rotterdam, The Netherlands, intending to head due north at 6.5 m/s relative to the water. However, the loc
sattari [20]

Answer:

Explanation:

velocity of ship with respect to water = 6.5 m/s due north

\overrightarrow{v}_{s,w}=6.5 \widehat{j}

velocity of water with respect to earth = 1.5 m/s at 40° north of east

\overrightarrow{v}_{w,e}=1.5\left ( Cos40\widehat{i} +Sin40\widehat{j}\right)

velocity of ship with respect to water = velocity of ship with respect to earth - velocity of water with respect to earth

\overrightarrow{v}_{s,w} = \overrightarrow{v}_{s,e} - \overrightarrow{v}_{w,e}

\overrightarrow{v}_{s,e} = 6.5 \widehat{j}- 1.5\left (Cos40\widehat{i} +Sin40\widehat{j}  \right )

\overrightarrow{v}_{s,e} = - 1.15 \widehat{i}+5.54\widehat{j}

The magnitude of the velocity of ship relative to earth is \sqrt{1.15^{2}+5.54^{2}} = 5.66 m/s

5 0
3 years ago
A square wave has amplitude 0 V for the low voltage and 4 V for the high voltage. Calculate the average voltage by integrating o
Margarita [4]

Answer:

V_{average} = \frac{1}{2}  V_o  ,     V_{average} = 2 V

Explanation:

he average or effective voltage of a wave is the value of the wave in a period

            V_average = ∫ V dt

in this case the given volage is a square wave that can be described by the function

           V (t) = \left \{ {{V=V_o \ \ \  t<  \tau /2} \atop {V=0 \ \  \ \  t> \tau /2 }   } \right.

to substitute in the equation let us separate the into two pairs

             V_average = \int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt

             V_average = V_o \ \int\limits^{1/2}_0 {} \, dt

             V_{average} = \frac{1}{2}  V_o

we evaluate  V₀ = 4 V

             V_{average} = 4 / 2)

             V_{average} = 2 V

6 0
3 years ago
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