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inna [77]
3 years ago
12

A pole-vaulter converts the kinetic energy of running to elastic potential energy in the pole, which is then converted to gravit

ational potential energy. If a pole-vaulter's center of gravity is 1.3 m above the ground while he sprints at 10.3 m/s, what is the maximum height of his center of gravity during the vault? For an extended object, the gravitational potential energy is U = mgh, where h is the height of the center of gravity
Physics
1 answer:
ZanzabumX [31]3 years ago
6 0

Explanation:

bf means of cost of the day of the day of the day Happy Birthday to you and your family and friends and family members of a na please replay video call rate of your life is not a good night and good night and friends and you are right sir and friends for the day of the day of the day of my friends with you and your family and friends and family is fine for the day of the day of the year and and your family is a great friend of the day and you can see it and I have a good time to talk with you and your friends for the delay of your family is my friend of my friend of my friend and daughter of your mother boad and daughter and I have told you I

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Constants A capacitor is connected across an ac source that has voltage amplitude 59 0 V and frequency 77 0 Hz Part C What is th
Vladimir79 [104]

Answer:

C = 1.77 \times 10^{-4} F

Explanation:

As we know that in AC circuit we have

V = i x_c

here we have

V = 59 V

i = 5.05 A

so we will have

x_c = \frac{59}{5.05}

x_c = 11.68 ohm

also we know that

x_c = \frac{1}{\omega C}

here we will have

11.68 = \frac{1}{(2\pi 77)C}

C = 1.77 \times 10^{-4} F

7 0
3 years ago
A negative charge is moved from point A to point B along an equipotential surface. Which of the following statements must be tru
elena-s [515]

Answer:

C) No work is required to move the negative charge from point A to point B.

Explanation:

An equipotential surface is defined as a surface connecting all the points at the same potential.

Therefore, when a charge moves along an equipotential surface, it moves between points at same potential.

The work done when moving a charge is given by

W=q\Delta V

where

q is the charge

\Delta V is the potential difference between the initial and final point of motion of the charge

However, the charge in this problem moves along an equipotential surface: this means that the potential does not change, so

\Delta V=0

And so, the work done is also zero.

7 0
4 years ago
What is meant by heat energys​
ad-work [718]

Answer:

heat energy is the form of energy produced by heat

when we burn heat a type of enery is came

4 0
3 years ago
A projectile is launched at an angle above the
gtnhenbr [62]
The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.

The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.

v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s

Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.

The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.

Combining this together we get:
(1) vx=40m/s and vy=10m/s

7 0
3 years ago
The universe was 5 percent its current size when light left objects observed now at redshift of ______________
Nuetrik [128]

The redshift of distant galaxy are larger than those of closer galaxies, which indicates that the galaxy is receding at a faster rate.

  • The Universe was 5 percent its current size when light left objects now at redshift of <u>19</u>.

Reasons:

The size of the universe represented as a scale factor with relation to the redshift can be presented as follows;

\displaystyle \frac{a}{a_0}  =\mathbf{ \frac{1}{1 + z}}

Where;

a₀ = The current size of the Universe

a = The size of the early Universe = 5% of a

Therefore;

\displaystyle \frac{a}{a_0}  =5\% = 0.05=  \frac{1}{1 + z}

\displaystyle 0.05  = \frac{1}{1 + z}

0.05 + 0.05·z = 1

\displaystyle z = \mathbf{ \frac{1 - 0.05}{0.05} } = 19

  • The redshift is of the observed light is, z = <u>19</u>

Learn more here:

brainly.com/question/14459434

brainly.com/question/3654558

4 0
2 years ago
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