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svetlana [45]
3 years ago
7

Consult Multiple-Concept Example 5 in preparation for this problem. The velocity of a diver just before hitting the water is -10

.5 m/s, where the minus sign indicates that her motion is directly downward. What is her displacement during the last 1.17 s of the dive?
Physics
1 answer:
LenaWriter [7]3 years ago
7 0
Time (- 10.5 m/s = displacement / 1.17 s)

so, 1.17 s moves to the other side, so the equation would look like

(- 10.5 m/s) (1.17 s) = - 12.285

hope this helps
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fiasKO [112]

Answer:

Yes

Explanation:

Any transparent surface in practical is neither a perfect absorber of electromagnetic waves neither a perfect reflector. Generally all the transparent surfaces reflect some amount of irradiation and the other parts are absorbed and transmitted.

<u>That is given by as relation:</u>

\alpha+\rho+\tau=1

where:

\alpha= absorptivity which is defined as the ratio of the absorbed radiation to the total irradiation

\rho= reflectivity is defined as the ratio of reflected radiation to the total irradiation

\tau= transmittivity is defined as the ratio of total transmitted radiation to the total irradiation

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3 years ago
During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau
Luden [163]

10.8 seconds is the correct answ

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3 years ago
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
3 years ago
Eric has a mass of 19.0 kg on the earth. What is Eric's weight on earth? What is Eric's weight on Mars? where the acceleration o
Tom [10]

Answer:

Weight on Earth = We = 186.2 N

Weight on Mars = Wm = 70.94 N

Explanation:

The weight of an object is defined as the force applied on the object by the gravitational field. The magnitude of weight is given by the following formula:

W = mg

were,

W= Weight of Eric

m = mass of Eric

g = acceleration due to gravity

ON EARTH:

W = We = Eric's Weight on Earth = ?

m = Eric's Mass on Earth = 19 kg

ge = acceleration due to gravity on Earth = 9.8 m/s²

Therefore,

We = (19 kg)(9.8 m/s²)

<u>We = 186.2 N</u>

<u></u>

ON MARS:

W = Wm = Eric's Weight on Mars = ?

m = Eric's Mass on Mars = 19 kg

gm = acceleration due to gravity on Mars = 0.381(ge) = (0.381)9.8 m/s² = 3.733 m/s²

Therefore,

Wm = (19 kg)(3.733 m/s²)

<u>Wm = 70.94 N</u>

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Select the correct answer.
natali 33 [55]

The wave speed completely depends on the characteristics and properties of the medium  . . . physical properties for mechanical waves, electrical properties for electromagnedtic waves.

So if you want to change the speed of a wave, you have to change the medium . . . shoot it through some different kind of stuff. <em>(B) </em>

4 0
3 years ago
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