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NNADVOKAT [17]
2 years ago
5

Suppose a 20-foot ladder is leaning against a building, reaching to the bottom of a second-floor window 15 feet above the ground

. What angle, in radians, does the ladder make with the ground Round your answer to two decimal places
Physics
1 answer:
Papessa [141]2 years ago
7 0

Answer:

The answer is β=0,85 rads

Explanation:

As the ladder is leaning against the building, we can imagine there´s a triangle where 20ft is the hypotenuse and 15ft is the maximum vertical distance between the ladder and the ground, it means, the leg opposite to β which is the angle we need

Let β(betha) be the angle between the ladder and the ground

We also know that sin(betha)=(leg opposite)/(hypotenuse)

In this case we will need to find β, this way:

betha=sin^-1((15ft/20ft))

Then β=48,6°  

We also have that 2πrads is equal to 360°, in this way we find how much β is in radians:

betha=(48,6°)*(2pirads/360°)

then we find β=0,85rads

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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}.

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