You use the laces to throw the football accurately, but you turn the laces to kick a

What is the most important factor for the formation of our planets

ohaa [14]

The gravitational pull of the Sun the interstellar dust attracting heat away from the protosun the process of nuclear fusion the nebular cloud condensing.

7 0

3 years ago

Points a and b lie in a region where the y-component of the electric field is Ey=α+β/y2. The constants in this expression have t

Drupady [299]

Answer:

V_{a} - V_{b} = 89.3

Explanation:

The electric potential is defined by

$V_{b} - V_{a}$ = - ∫ E .ds

In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.

V_{b} - V_{a} = - ∫ E ds

We substitute

V_{b} - V_{a} = - ∫ (α + β/ y²) dy

We integrate

V_{b} - V_{a} = - α y + β / y

We evaluate between the lower limit A 2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m

V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)

V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33

V_{b} - V_{a} = - 89.3 V

As they ask us the reverse case

V_{b} - V_{a} = - V_{b} - V_{a}

V_{a} - V_{b} = 89.3

3 0

3 years ago

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

Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in

Serggg [28]

Answer:

(a) The force between them quadruples

Explanation:

According to coulomb's law, initial force between the two charged objects is given as;

$F_1=\frac{Kq_1q_2}{r^2}$

where;

k is coulomb's constant

q₁ is the charge on the first object

q₂ is the charge on the second object

r is the distance between the two objects

When the charges on both objects are doubled, then;

q₁ = 2q₁

q₂ = 2q₂

Force between the two charged objects will become

$F_2 = \frac{K2q_12q_2}{r^2} = \frac{4Kq_1q_2}{r^2} = 4(\frac{Kq_1q_2}{r^2}) = 4F_1$

Therefore, the force between them quadruples

4 0

3 years ago

How to do this question

Anna71 [15]

Answer:

(a) 10 m/s

(b) 22.4 m/s

Explanation:

(a) Draw a free body diagram of the car when it is at the top of the loop. There are two forces: weight force mg pulling down, and normal force N pushing down.

Sum of forces in the centripetal direction (towards the center):

∑F = ma

mg + N = mv²/r

At minimum speed, the normal force is 0.

mg = mv²/r

g = v²/r

v = √(gr)

v = √(10 m/s² × 10.0 m)

v = 10 m/s

(b) Energy is conserved.

Initial kinetic energy + initial potential energy = final kinetic energy

½ mv₀² + mgh = ½ mv²

v₀² + 2gh = v²

(10 m/s)² + 2 (10 m/s²) (20.0 m) = v²

v = 22.4 m/s

4 0

3 years ago