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Ede4ka [16]
3 years ago
14

A mountain climber increases their height from 200 meters to 400 meters. What affect will this have on their potential energy?

Physics
1 answer:
Yanka [14]3 years ago
4 0

Answer:

At 400 m the potential energy of the mountain climber doubled the initial value.

Explanation:

Given;

initial height of the mountain climber = 200 m

final height of the mountain climber, = 400 m

The potential energy of the mountain climber is calculated as;

Potential energy, P.E = mgh

At 200 m, P.E₁ = mg x 200 = 200mg

At 400 m, P.E₂ = mg x 400 = 400mg

Then, at 400 m, P.E₂ = 2 x 200mg = 2 x P.E₁

Therefore, at 400 m the potential energy of the mountain climber doubled the initial value.

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Answer:

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(2.437×10⁴)(6.5411 x 10^9)/(5.37x10^6). write in scientific notation​
Musya8 [376]

Answer:

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2.968456 ×10^7

4 0
3 years ago
In which of these cases do we have enough information to say that the atom is electrically neutral?
Luba_88 [7]
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5 0
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Read 2 more answers
In an experiment you measure a first-order red line for Hydrogen at an angle difference of ΔΘ = 22.78o. The diffraction grating
azamat

Answer:

a) wavelength = 656.3 nm

b)  the value of Rydberg's constant for this measurement is 1.097 × 10⁷ m⁻¹

Explanation:

Given that;

angle of diffraction Θₓ = 22.78°

incident angle Θ₁ = 0

slit separation d  = 5900 lines per cm = 1/5900 cm = 10⁻²/5900 m = 0.01/5900 m

order of diffraction n = 1

wavelength λ = ?

to find the wavelength, we use the expression

λ = d (sinΘ₁ + sinΘₓ) / n

To find the wavelength λ;

λ = 0.01/5900 × (sin0 + sin22.78° )

λ = 6.5626 × 10⁻⁷ m

λ = 656.3 x 10⁻⁹ m

∴ λ = 656.3 nm

b)

According Balnur's  series spectral lines; n₁ = 3, n₂ = 2 and

λ = R [ 1/n₂² - 1/n₁²]

where  R is Rydberg's constant

from λ = R [ 1/n₂² - 1/n₁²]

R = 1/λ [n₂²n₁² / n₁² - n₂²]

R = 10⁹/ 656.3 [ 9 × 4 / 9 - 4 ]

R = 1.097 × 10⁷ m⁻¹

Therefore the value of Rydberg's constant for this measurement is 1.097 × 10⁷ m⁻¹

4 0
3 years ago
Would this officer calculate a higher value of Edef, a lower value of Edef, or the same value of Edef if he used the velocities
NISA [10]

Answer:

<em>The officer would calculate the similar value of Edef.</em>

Explanation:

As Edef is the energy required to deform the body therefore  the reference frame does not affect the calculation of energy. In this context  the value of Edef will remain same irrespective of fact, whichever frame of reference is used.

4 0
3 years ago
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