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3 years ago

Doubling an object’s height will have what effect on its potential energy due to gravity?

Physics

1 answer:

Gekata [30.6K]3 years ago

3 0

Potential energy due to gravity = Ep = mgh [symbols have their usual meaning

Evidently, HALVING the mass will make Ep , HALF its previous value. So, It will be halved.

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s344n2d4d5 [400]

Answer:

Explanation:

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3 years ago

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Which scientist's work led to our understanding of how planets move around

Nataly [62]

Answer:

<h2><em><u>Johannes Kepler</u></em></h2>

Explanation:

He made rules about planetary motion.
The scientist Johannes Kepler was a German astronomer.
He found out that the planets evolved around the Sun.
He also made the laws of planetary motion.

Hope this helped,

Kavitha

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3 years ago

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Aluminium has a work function of 4. 08 ev. (a) find the cutoff wavelength and cutoff frequency for the photoelectric effect?

Oksanka [162]

Aluminium has a work function of 4. 08 eV. The cutoff wavelength and cutoff frequency for the photoelectric effect is 303.9nm and 911.7× 10¹⁷ s⁻¹ respectively.

Work Function is the minimum energy required to eject an electron from a photoelectric material.

Cutoff Wavelength is the maximum wavelength below which electron will be ejected

Cutoff Frequency is the minimum frequency required to eject electron.

Let the work function of Aluminium be Φ

Given, work function Φ = 4.08eV

We know that hc/λ = Φ

where, h is Planck constant

c is speed of light

λ is wavelength of light used

Hence, on subsitution

1240 / λ = 4.08 eV (hc = 1240)

⇒ λ = 303.9 nm

Hence, cutoff wavelength used is 303.9 nm

We know that ν = c/λ

ν = 3 × 10⁸ / 303.9

⇒ ν = 911.7 × 10¹⁷ s⁻¹

Hence, cutoff frequency used is 911.7 × 10¹⁷ s⁻¹

Learn more about Work function here, brainly.com/question/24180170

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2 years ago

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching this

allsm [11]

Answer:

$\theta_2 \approx 40.5^{\circ}$

Explanation:

<u>Given:</u>

velocity of the ball first and ball second, $v_1=v_2=v\,m.s^{-1}$

angle of projection of the first ball, $\theta_1= 72^{\circ}$

∵The balls should land at the same point,

∴their range of projectile,

$R_1=R_2=R\,m$

As we know for the range of projectile:

$R=\frac{v^2}{g}.sin\,2\theta$

∵ we have equal range in both the cases

$\therefore \frac{v_1^2}{g}.sin\,2\theta_1=\frac{v_2^2}{g}.sin\,2\theta_2$

$\Rightarrow \frac{v^2}{g}.sin\,(2\times 72) =\frac{v^2}{g}.sin\,2\theta_2$

$\Rightarrow sin 144^{\circ}=sin\,2\theta_2$

$2\theta_2=sin^{-1}[sin 144^{\circ}]$

$\theta_2=\frac{1}{2}\times sin^{-1}[sin 144^{\circ}]$

$\theta_2 \approx 40.5^{\circ}$

8 0

3 years ago

Using a series circuit, remove one battery from the circuit by re-wiring. How did it change the brightness of the bulbs? How doe

morpeh [17]

The higher the voltage in series, the brighter the bulb gets. so removing one battery from the circuit causes a reduction in the brightness of the bulb.
hope this helps

8 0

3 years ago