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Volgvan
3 years ago
5

1. Maples, dogwoods, and oaks are all examples oftrees, which shed their leaves every year.​

Physics
1 answer:
yulyashka [42]3 years ago
4 0

Answer:

deciduous

Explanation:

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What is hydroelectric power ?<br><br> Answer quickly..!
GenaCL600 [577]

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<u>Hydroelectric power,</u> also called hydropower is the electricity produced from generators driven by turbines that convert the potential energy of falling or fast-flowing water into mechanical energy.

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3 years ago
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One type of potential energy is gravitational potential energy. What is necessary for an object to have gravitational potential
galben [10]

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Gravitational potential energy is energy an object possesses because of its position in a gravitational field. Since the force required to lift it is equal to its weight, it follows that the gravitational potential energy is equal to its weight times the height to which it is lifted.

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3 years ago
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Hannah walks 0.30 km to class in 5.0 min. what is her average speed in m/s?
OverLord2011 [107]

Answer:

1.0 m/s

Explanation:

First, convert to SI units.

0.30 km × (1000 m / km) = 300 m

5.0 min × (60 s / min) = 300 s

Speed is distance divided by time:

300 m / 300 s = 1.0 m/s

3 0
3 years ago
Help me with this???
Vikentia [17]

Yo sup??

Average velocity=total distance covered/total time taken

total distance covered=4 + 8=12 miles

total time taken=6 hours

Therefore

average velocity=12/6

=2 miles/hour

Hope this helps

8 0
3 years ago
Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

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

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0
3 years ago
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