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vagabundo [1.1K]

3 years ago

12

At what point on the hill will the car have zero gravitational potential energy?

1 answer:

natali 33 [55]3 years ago

8 0

C) at the bottom of the hill

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Where does the majority of the mass of an atom come from

asambeis [7]

Answer:

The Nucleus

Explanation:

The nucleus contains the majority of an atom's mass because protons and neutrons are much heavier than electrons, whereas electrons occupy almost all of an atom's volume. I hope this helps you :D

5 0

3 years ago

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When jogging outside you accidentaly bumb into a curb. Your feet stop but your body continues to movr foward and you end up on t

sattari [20]

B.) Newtons first law of motion

8 0

3 years ago

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In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

6 0

3 years ago

What is a small body that follows a highly elliptical orbit around the Sun

Ipatiy [6.2K]

That could be a comet, or any one of the billions of meteoroids
moving in a cloud that's actually the remains of a shattered comet.

6 0

3 years ago

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A crane lifts a 545 kg piano up into a high-rise apartment 75.0 meters above the ground. In order to do so, 5350 Newtons were ap

vodka [1.7K]

Answer:

Work done to pull the piano upwards is 401250 J

Explanation:

Work is done against the gravity to pull the piano upwards

So here we can say that work done is

$W = mgH$

here we know that

$mg = 5350 N$

also we know that

H = 75 m

now we have

$W = 5350 \times 75$

$W = 401250 J$

7 0

3 years ago