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miv72 [106K]
3 years ago
9

A 6 kg cart starting from rest rolls down a hill and at the bottom has a speed of 10 m/s. What is the height of the hill?

Physics
1 answer:
Arisa [49]3 years ago
4 0

Answer:

h = 5.09 m

Explanation:

Applying the Law of conservation of energy to this situation, we can write:

Kinetic\ Energy\ Gained\ by\ the\ Cart = Potential\ Energy\ Lost\ by\ the\ Cart\\\frac{1}{2}mv^2 = mgh\\\\h = \frac{v^2}{2g}

where,

h = height of the hill = ?

v = speed of cart at the end = 10 m/s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

h = \frac{(10\ m/s)^2}{(2)(9.81\ m/s^2)}\\\\

<u>h = 5.09 m</u>

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Answer:

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What kind of frequency do radio waves have?<br><br>A. High frequency <br><br>B. Low frequency ​
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B low frequency it is the lowest frequency
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Thalia is drafting a plan to move a large, perfect sphere concrete sculpture that is in front of her office building. Describe t
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Read 2 more answers
A uniform stick 1.5 m long with a total mass of 250 g is pivoted at its center. A 3.3-g bullet is shot through the stick midway
Andru [333]

Answer:

<em>63.44 rad/s</em>

<em></em>

Explanation:

mass of bullet = 3.3 g = 0.0033 kg

initial velocity of bullet v_{1} = 250 m/s

final velocity of bullet v_{2} = 140 m/s

loss of kinetic energy of the bullet = \frac{1}{2}m(v^{2} _{1} - v^{2} _{2})

==> \frac{1}{2}*0.0033*(250^{2}  - 140^{2} ) = 70.785 J

this energy is given to the stick

The stick has mass = 250 g =0.25 kg

its kinetic energy = 70.785 J

from

KE = \frac{1}{2} mv^{2}

70.785 = \frac{1}{2}*0.25*v^{2}

566.28 = v^{2}

v= \sqrt{566.28} = 23.79 m/s

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5 0
3 years ago
A thin film of MgF₂ (n = 1.38) coats a piece of glass. Constructive interference is observed for the reflection of light with wa
TEA [102]

Answer:

t = 905.8 nm

Explanation:

Given:

- Wavelength λ_1 = 500 nm

- Wavelength λ_2 = 625 nm

- MgF₂ refractive index n = 1.38

Find:

What is the thinnest film for which this can occur?

Solution:

- We have two different wavelength of light that constructively interfere at the surface of the film. We need the minimum thickness of film that would satisfy the condition of constructive interference for both!

-Since the refractive index of glass is greater than that of MgF_2, the expression of constructive interference would be as follows:

                                2*t = m*λ / n

- Since, the orders m are unknown for each wavelength, also different. We will try to determine the first for each wavelength of light.

- Construct two equation:

                                t = m_1*(500 nm ) / (2*1.38 )

                                t = 181.1594203*m_1  nm

                                t = m_2*(625 nm ) / (2*1.38 )

                                t = 226.4492754*m_2  nm

- Now equate the two thicknesses which should be equal:

                               226.4492754*m_2 = 181.1594203*m_1

                               m_2 = 0.8*m_1

- Now we know that m can only take integer values, and m is proportional to thickness t. So for thinnest thickness m's must take the least integer values. Hence, we have:                    

                               m_2 = (4 / 5) * m_1

So,                           m_1 = 5 , m_2 = 4   ..... Least integer values.

- Now that we have m's we can compute the thickness t as follows:

                                t = 181.1594203*m_1  nm

- Substitute m_1 = 5, we have:

                                t = 181.1594203*5  nm

                               t = 905.8 nm

- Substitute m_2 = 4 in:

                                 t = 226.4492754*m_2

                                 t = 226.4492754*4

                                t = 905.8 nm

- Our values of t = 905.8 nm matches for both wavelengths.

                                   

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