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Annette [7]
3 years ago
15

How could gravitational potential energy be useful

Physics
1 answer:
Tpy6a [65]3 years ago
6 0
Hydroelectric dams convert the gravitational potential energy of falling water into electricity.
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other than serving as a conduction pathway what is a major function of the pons why is the medulla the most vital part of the br
irga5000 [103]
1. What is the major function of the pons? 
         -Other than serving as a conduction pathway, the pons also contains a nucleus that is essential for breathing.
2.  Why is the medulla the most vital part of the brain?       
         -Medulla is very important because it function for the involuntary movement and processes that happen in the body without giving a thought about it. It performs crucial tasks like regulating the blood pressure.
3 0
3 years ago
A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c
Mashcka [7]

Answer with Explanation:

We are given that

Diameter of pipe,d_1=0.573 m

v_1=13.5 m/s

v_2=5.83 m/s

Volume flow rate of the petroleum along the pipe=Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})

Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s

By equation of continuity

A_1v_1=A_2v_2

\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2

d^2_2=\frac{v_1}{v_2}d^2_1

d_2=\sqrt{\frac{v_1}{v_2}}d_1

d_2=0.573\sqrt{\frac{13.5}{5.83}}

d_2=0.87 m

d_2=0.87\time 100=87 cm

1 m=100 cm

4 0
3 years ago
A pitcher throws a baseball that reaches the catcher in 0.75 s. The ball curves because it is spinning at an average angular vel
7nadin3 [17]

The change in angular displacement as a function of time is the definition given for angular velocity, this is mathematically described as

\omega = \frac{\theta}{t}

Here,

\theta = Angular displacement

t = time

The angular velocity is given as

\omega = 230rev/min

PART A) The angular velocity in SI Units will be,

\omega = 230rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega = \frac{23}{3}\pi rad/s \approx 24.08rad/s

PART B) From our first equation we can rearrange to find the angular displacement then

\theta = \omega t

Replacing,

\theta = (24.08)(0.75)

\theta = 18.06 rad

4 0
3 years ago
A patient comes in and tells you that he has arthritis and needs Cerebyx. What do you do? A. You know that Cerebyx and Celebrex
Flura [38]
B. truthfully it's the only one that makes sense
6 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
Read 2 more answers
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