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KatRina [158]
3 years ago
11

Define mechanical energy useing the example of the skateboarder on the half-pipe explain how kinetic and potential energy relate

to the mechanical energy at diffrent positions
Physics
1 answer:
sergejj [24]3 years ago
7 0
When some one balance skate board on pipe the potential energy is responsible to make it slide beacause of m mass and g gravity and height thus the skate boarder concieves mgh potential energy and moving on pipe with balancing himself using kinetic enrgy ie;1/2 mv^2 and thus gauning speed ..for illustrating well there is a another example the two wheeeler gets balanced while moving instead of standing idle or stationary ie; kinetic energy is responsible for balancing
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A 60 year old person has a threshold of hearing of 79.0 dB for a sound with frequency f=10,000 Hz. By what factor must the inten
liubo4ka [24]

They provided the intensity in decibels for the problem, but they are unsure by what factor to increase it (I) to make the sound loud enough for the elderly person to hear.

Neglect f entirely.

The following equation must be used to convert decibels (dB) to I:

I=(10^(dB/10))*10^-12

Divide the elder person's dB by the younger person's dB after doing this for each dB.

After putting values and solving

we get,

1.8372093023.

So, the final answer is 1.8372093023.

learn more about frequency brainly.com/question/254161

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6 0
1 year ago
A temperature of 50F is equal to c
Hitman42 [59]

10.00 °C this is the right answer need more question feel free to post

5 0
3 years ago
A charge +1.9 μC is placed at the center of the hollow spherical conductor with the inner radius 3.8 cm and outer radius 5.6 cm.
Archy [21]

To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.

PART A)

The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:

Q_{int}=-Q1=-1.9*10^{-6} C. This is the total charge on the inner surface of the conducting shell.

PART B)

The positive charge (of the same value) on the external surface of the conducting shell is:

Q_{ext}=+Q_1=1.9*10^{-6} C

The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,

Q_{ext, Total}=Q_2+Q_{ext}

Q_{ext, Total}=1.9+3.8

Q_{ext, Total}=5.7 \mu C

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3 years ago
I will give brainliest if you sub and stay subbed to JD Outdoors & Gaming​
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First Stan txt (tomorrow by together) and stream freeze on YT Hybe labels !!!!!!!!!!!!!

5 0
3 years ago
A point charge q1q1 is held stationary at the origin. A second charge q2q2 is placed at point aa, and the electric potential ene
Yuri [45]

Explanation:

The given data is as follows.

            U_{a} = 5.4 \times 10^{-8} J

        W_{/text{a to b}} = -1.9 \times 10^{-8} J

        Electric potential energy (U_{b}) = ?

Formula to calculate electric potential energy is as follows.

            U_{b} = U_{a} - W_{/text{a to b}}

                        = 5.4 \times 10^{-8} J - (-1.9 \times 10^{-8} J)

                        = 7.3 \times 10^{-8} J

Thus, we can conclude that the electric potential energy of the pair of charges when the second charge is at point b is 7.3 \times 10^{-8} J.

6 0
3 years ago
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