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2 years ago

The sound from a bolt of lightning travelled 4.08 km in 12.0 s. What was the speed

Physics

1 answer:

LenaWriter [7]2 years ago

6 0

Answer:

83.67 m/s

Explanation:

Set up a calculation to convert units of measure to what you need.

You have km/s and you need m/s.

4.08km 1000 m 83.67m

----------- X ---------- = --------------- the km will cancel out and you are left

12.0 s 1 km s with m/s

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On James’s MP3 player, he has 12 sad songs and 40 upbeat songs that he wants to put into playlists. He wants to have the same nu

Anna [14]

Answer:

i would say a) two playlists

hope this helps!

Explanation:

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2 years ago

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A stunt driver drives a car horizontally off the edge of a cliff at 3.8m/s and reaches the water below 2.5s later.

andreyandreev [35.5K]

A. The cliff was 30.7 m high
B. I also got 9.5 as the horizontal distance

Here is my work, I find making charts like this one to find knowns and unknowns can be helpful

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2 years ago

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

8 0

2 years ago

Change in speed over a given period of time is

Dvinal [7]

Explanation:

Acceleration is the change in speed over a given time period

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2 years ago

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Compare the maximum rate of heat transfer to the basal metabolic rate by converting a bmr of 88 kcal/hr into watts. what is the

elena-14-01-66 [18.8K]

Explanation :

It is given that,

BMR i.e basal metabolic rate is 88 kcal/hr. So, BMR in watts is converted by the following :

We know that, 1 kilocalorie = 4184 joules

So, $1\ kcal/h=\dfrac{1\times 4184\ J}{3600\ sec}$

$1\ kcal/h=1.16\ J/sec$

J/sec is nothing but watts.

So, $1\ kcal/h=1.16\ watts$

and $88\ kcal/h=88\times 1.16\ watts = 102.08\ watts$

So, it can be seen that the body can accommodate a modes amount of activity in hot weather but strenuous activity would increase the metabolic rate above the body's ability to remove heat.

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2 years ago