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Bezzdna [24]
3 years ago
11

In a thunder and lightning storm there is a rule of thumb that many people follow. After seeing the lightning, count seconds to

yourself. If it takes 5 seconds for the sound of the thunder to reach you, then the lightning bolt was 1 mile away from you. Sound travels at a speed of 331 meters/second. How accurate is the rule of thumb? Express your answer as a percent error.
Physics
1 answer:
lubasha [3.4K]3 years ago
3 0

Answer:

2.837% less than actual value.

Explanation:

Based on given information let's calculate our value.

S = Vxt = 331m/s x 5s = 1655m, that is the total distance that sound would travel in 5 seconds.

1mile = 1609.34meters.

percentage error is.

\frac{actual-calculated}{actual} *100 = \frac{1609.34-1655}{1609.34} *100 = -2.83%

negative indicates less than actual value.

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A dart is thrown from 1.50 m high at 10.0 m/s toward a target 1.73 m from the ground. At what angle was the dart thrown?
Triss [41]

Answer:

The angle of projection is 12.26⁰.

Explanation:

Given;

initial position of the dart, h₀ = 1.50 m

height above the ground reached by the dart, h₁ = 1.73 m

maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m

velocity of the dart, u = 10 m/s

The maximum height reached by the projectile is calculated as;

H_m = \frac{u^2sin^2 \theta}{2g}

where;

θ is angle of projection

g is acceleration due to gravity = 9.8 m/s²

H_m = \frac{u^2sin^2 \theta}{2g}\\\\sin^2 \theta = \frac{H_m \ \times \ 2g}{u^2} \\\\sin^2 \theta = \frac{0.23 \ \times \ 2(9.8)}{10^2} \\\\sin ^2\theta =0.04508\\\\sin \theta = \sqrt{0.04508} \\\\sin \theta = 0.2123\\\\\theta  = sin^{-1}(0.2123)\\\\\theta  = 12.26^0

Therefore, the angle of projection is 12.26⁰.

6 0
2 years ago
Two parallel plates 19 cm on a side are given equal and opposite charges of magnitude 2.0 ✕ 10^−9 C. The plates are 1.8 mm apart
Romashka [77]

Answer:

E   = 5291.00 N/C

Explanation:

Expression for capacitance is

C = \frac{\epsilon  A}{d}

where

A is area of square plate

D = DISTANCE BETWEEN THE PLATE

C = \frac{\epsilon\times(19\times 10^{-2})^2}{1.5\times 10^{-3}}

C = 24.06 \epsilon

C = 24.06\times 8.854\times 10^{-12} F

C =2.1\times 10^{-10} F

We know that capacitrnce and charge is related as

V = \frac{Q}{C}

 = \frac{2\tiimes 10^{-9}}{2.\times 10^{-10}}

v = 9.523 V

Electric field is given as

E = \frac{V}{d}

   = \frac{9.52}{1.8*10^{-3}}

E   = 5291.00 V/m

E   = 5291.00 N/C

5 0
3 years ago
Two coaxial conducting cylindrical shells have equal and opposite charges. The inner shell has charge +q and an outer radius a,
Leviafan [203]

Answer:

\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}

Explanation:

As we know that the charge per unit length of the long cylinder is given as

\lambda = \frac{q}{L}

here we know that the electric field between two cylinders is given by

E = \frac{2k\lambda}{r}

now we know that electric potential and electric field is related to each other as

\Delta V = - \int E.dr

\Delta V = -\int_a^b (\frac{2k\lambda}{r})dr

\Delta V = -2k \lambda ln(\frac{b}{a})

\Delta V = \frac{\lambda ln(\frac{b}{a})}{2\pi \epsilon_0}

\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}

7 0
3 years ago
44. A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down
Vladimir [108]

Answer:

18.4 m

Explanation:

(a)

The known variables in this problem are:

u = 1.40 m/s is the initial vertical velocity (we take downward direction as positive direction)

t = 1.8 s is the duration of the fall

a = g = 9.8 m/s^2 is the acceleration due to gravity

(b)

The vertical distance covered by the life preserver is given by

d=ut + \frac{1}{2}at^2

If we substitute all the values listed in part (a), we find

d=(1.40 m/s)(1.8 s)+\frac{1}{2}(9.8 m/s^2)(1.8 s)^2=18.4m

8 0
3 years ago
Select Light for the type of wave, adjust the wavelength so that the light is red, and increase the amplitude of the light to th
Sergeu [11.5K]

Answer:

here as we increase the distance the intensity will decrease and hence the amplitude of the electric field will decrease and vice-versa

Explanation:

As wee know that the amplitude of the wave will decide the energy of the wave

Here we know that energy density of electromagnetic wave is given as

u = \frac{1}{2}\epsilon_0E_0^2

now we have

\frac{I}{c} = \frac{1}{2}\epsilon_0 E_0^2

so here we can say that intensity of the wave at the given distance from the source is given by formula

I = \frac{P}{4\pi r^2}

so here as we increase the distance the intensity will decrease and hence the amplitude of the electric field will decrease and vice-versa.

5 0
3 years ago
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