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3 years ago

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A common acoustic phenomenon is thunder, the rapid heating and supersonic expansion of air created by a lightning stroke. Becaus

e the speed of light reaches almost any point on Earth instantly while sound travels more slowly, the time lag between lightning and thunder can be used to estimate distance to the lightning. This delay time is ______ seconds per kilometer (5 sec/mi).

1 answer:

Oksana_A [137]3 years ago

4 0

Answer:

Δt ≈ 2.9137 ≈ 3 seconds per kilometer

Explanation:

The speed of light is approximately 300000 km /s, while the speed fo the sound in the air is 0.3432 km/s.

The light takes therefore this time to travel one kilometer

$x = vt$ $t = \frac{x}{v}$

$t = \frac{1}{300000}$

$t = 3.33 x 10^{-6} s$

On the other hand the sound takes this time to travel one kilometer

$t = \frac{1}{0.3432}$

t = 2.9137 s

Then the delay time is 2.9137 - $3.33 x 10^{-6} s$

Δt ≈ 2.9137 ≈ 3 s

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A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

<u>Now the horizontal component of velocity:</u>

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

<u>The vertical component of the velocity:</u>

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

<u>Time taken by the projectile to travel the distance of 30 m:</u>

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

<u>Vertical position of the projectile at this time:</u>

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

<u>We observe that </u> $t>t'$ <u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

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4 years ago

in a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallw

madam [21]

Answer:

μ=0.151

Explanation:

Given that

m= 3.5 Kg

d= 0.96 m

F= 22 N

v= 1.36 m/s

Lets take coefficient of kinetic friction = μ

Friction force Fr=μ m g

Lets take acceleration of block is a m/s²

F- Fr = m a

22 - μ x 3.5 x 10 = 3.5 a ( take g =10 m/s²)

a= 6.28 - 35μ m/s²

The final speed of the block is v

v= 1.36 m/s

We know that

v²= u²+ 2 a d

u= 0 m/s given that

1.36² = 2 x a x 0.96

a= 0.963 m/s²

a= 6.28 - 35μ m/s²

6.28 - 35μ = 0.963

μ=0.151

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3 years ago

A block with a mass of 8.7 kg is dropped from rest from a height of 8.7 m, and remains at rest after hitting the ground. 1)If we

Harlamova29_29 [7]

To solve this problem we will apply the concepts related to gravitational potential energy.

This can be defined as the product between mass, gravity and body height.

Mathematically it can be expressed as

$\Delta P = mgh$

$\Delta P = (8.7)(9.8)(3)$

$\Delta P = 255.78J$

Therefore the change in the internal energy of the system is 255.78

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3 years ago

It is very efficient to drive a large truck to and from school. Please select the best answer from the choices provided T F I th

cestrela7 [59]

It's 100% false, it wastes a ton (sorry for the bad pun!) of gas.

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3 years ago

Read 2 more answers

A ball is thrown downward at 12 m/s from a windowsill 35 m above the ground. At the same time, another ball is thrown upward at

wariber [46]

Answer:

The second ball lands 1.5 s after the first ball.

Explanation:

Given;

initial velocity of the ball, u = 12 m/s

height of fall, h = 35 m

initial velocity of the second, v = 12 m/s

Time taken for the first ball to land;

$t = \sqrt{\frac{2h}{g} }\\\\t =\sqrt{ \frac{2*35}{9.8}}\\\\t = 2.67 \ s$

determine the maximum height reached by the second ball;

v² = u² -2gh

at maximum height, the final velocity, v = 0

0 = 12² - (2 x 9.8)h

19.6h = 144

h = 144 / 19.6

h = 7.35 m

time to reach this height;

$t_1 = \sqrt{\frac{2h}{g} }\\\\t_1 = \sqrt{\frac{2*7.35}{9.8}}\\\\t_1 = 1.23 \ s$

Total height above the ground to be traveled by the second ball is given as;

= 7.35 m + 35m

= 42.35 m

Time taken for the second ball to fall from this height;

$t_2 = \sqrt{\frac{2h}{g} }\\\\t_2 = \sqrt{\frac{2*42.35}{9.8} }\\\\t_2 = 2.94 \ s$

total time spent in air by the second ball;

T = t₁ + t₂

T = 1.23 s + 2.94 s

T = 4.17 s

Time taken for the second ball to land after the first ball is given by;

t = 4.17 s - 2.67 s

T = 1.5 s

Therefore, the second ball lands 1.5 s after the first ball.

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3 years ago