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2 years ago

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 frictionless toy car is placed on a ramp, which is inclined at an unknown angle with respect to the horizontal. Starting from

NikAS [45]

The final speed of the toy car at the end of the given time period is 3.58 m/s.

The given parameters;

distance traveled by the car, s = 1.2 m
time of motion of the car, t = 0.67 s
initial velocity of the car, u = 0

The acceleration of the car is calculated as;

$s = ut + \frac{1}{2} at^2\\\\1.2 = 0 + 0.5\times a\times (0.67)^2\\\\1.2 = 0.225a\\\\a = \frac{1.2}{0.225} \\\\a = 5.33 \ m/s^2$

The final velocity of the toy car is calculated as;

$v_f^2 = u^2 + 2as\\\\v_f^2 = 0 + 2\times 5.33 \times 1.2\\\\v_f^2 = 12.792\\\\v_f = \sqrt{12.792} \\\\v_f = 3.58 \ m/s$

Thus, the final speed of the toy car at the end of the given time period is 3.58 m/s.

Learn more here: brainly.com/question/20352766

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2 years ago

How do exons and introns differ?

Norma-Jean [14]

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2 years ago

In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103V/m is established between the p

Bond [772]

Answer:

$B_0 = 1.69 \times 10^{-4}\ T$

Explanation:

given,

total deflection = 4.12 cm

Electric field = 1.1 ×10³ V/m

plate length = 6 cm

distance between them = 12 cm

using formula

$v_0 = \sqrt{\dfrac{q\epsilon_0d}{ym}(\dfrac{d}{2}+L)}$

q = 1.6 × 10⁻¹⁹ C

m = 9.11 x 10⁻³¹ kg

d = 0.06 m

L = 0.12 m

$v_0 = \sqrt{\dfrac{1.6 \times 10^{-19}\times 1.1 \times 10^{3}\times 0.06}{0.0412\times 9.11 \times 10^{-31} }(\dfrac{0.06}{2}+0.12)}$

v_0 = 6496355.63 m/s

$v_0 = \dfrac{E}{B_0}$

$B_0 = \dfrac{E}{v_0}$

$B_0 = \dfrac{1.1\times 10^{3}}{6496355.63}$

$B_0 = 1.69 \times 10^{-4}\ T$

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Answer:

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