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timofeeve [1]
4 years ago
8

Radio signals travel at a rate of 3 × 108 meters per second. how many seconds would it take for a radio signal to travel from a

satellite to the surface of earth if the satellite is orbiting at a height of 9.6 × 106 meters? (hint: time is distance divided by speed.)
Physics
1 answer:
nlexa [21]4 years ago
6 0
<span>3.2x10^-2 seconds (0.032 seconds)
   This is a simple matter of division. I also suspect it's an exercise in scientific notation, so here is how you divide in scientific notation:

   9.6 x 10^6 m / 3x10^8 m/s

   First, divide the significands like you would normally.
 9.6 / 3 = 3.2

   And subtract the exponent. So
 6 - 8 = -2

   So the answer is 3.2 x 10^-2
 And since the significand is less than 10 and at least 1, we don't need to normalize it.

   So it takes 3.2x10^-2 seconds for the radio signal to reach the satellite.</span>
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ZanzabumX [31]

Answer: mass x height x gravitational field strength (g)

note: gravitational field strength (g) = 10 N/Kg

55 x 15 x 10 = 8250

gpe = 8250j

Explanation:

4 0
3 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
3 years ago
How do you find the velocity after a collision
Evgen [1.6K]

Answer:

In a collision, the velocity change is always computed by subtracting the initial velocity value from the final velocity value. If an object is moving in one direction before a collision and rebounds or somehow changes direction, then its velocity after the collision has the opposite direction as before.

Explanation:

7 0
4 years ago
What is the momentum of a 1400 kg car traveling at 25 m/s?
sveticcg [70]
<h2>Hello</h2>

The answer is:

35000Kg.\frac{m}{s}=35000N

<h2>Why?</h2>

Momentum is the quantity of movement of an object, and it's calculated using the mass and the velocity of the object. Momentum is expressed by the following formula:

p=m*v

Where:

m=mass\\v=velocity

So, calculating we have:

p=m*v=1400Kg*\frac{25m}{s}=35000Kg.\frac{m}{s}=35000N

Remember,

1N=1Kg.\frac{m}{s}

Have a nice day!

5 0
3 years ago
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Explain how a raccoon is helping the environment when it eats the fruit of a plant.
torisob [31]

Answer: When a raccoon eats a fruit it eats the seeds that are in the fruit. When the raccoon goes to the bathroom it expels the seeds that are in the feces which plants them.

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3 years ago
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