A helicopter is hovering above the ground. Jim reaches out of the copter (with a safety harness on) at 180 m above the ground. A package is launched upward, from a point on a roof 10 m above the ground. The initial velocity of the package is 50.5 m/s. Consider all quantities as positive in the upward direction. Does Jim Bond have a chance to catch the package? (calculate how high will it go)

Answers

Answer 1
Answer:

Answer:

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

Explanation:

Hi there!

The equation of height and velocity of the package are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the package at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² because we consider the upward direction as positive).

v = velocity of the package at a time t.

First, let´s find the time it takes the package to reach the maximum height. For this, we will use the equation of velocity because we know that at the maximum height, the velocity of the package is zero. So, we have to find the time at which v = 0:

v = v0 + g · t

0 = 50.5 m/s - 9.8 m/s² · t

Solving for t:

-50.5 m/s / -9.81 m/s² = t

t = 5.15 s

Now, let´s find the height that the package reaches in that time using the equation of height. Let´s place the origin of the frame of reference on the ground so that the initial position of the package is 10 m above the ground:

h = h0 + v0 · t + 1/2 · g · t²

h = 10 m + 50.5 m/s · 5.15 s - 1/2 · 9.81 m/s² · (5.15 s)²

h = 140 m

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.


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Answers

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You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.09 mm and place your screen 8.61 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.53 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength ???? expressed in nanometers?

Answers

Answer:

λ = 5.734 x 10⁻⁷ m = 573.4 nm

Explanation:

The formula of the Young's Double Slit experiment is given as follows:

\Delta x = (\lambda L)/(d)\n\n\lambda = (\Delta x d)/(L)

where,

λ = wavelength = ?

L = distance between screen and slits = 8.61 m

d = slit spacing = 1.09 mm = 0.00109 m

Δx = distance between consecutive bright fringes = (4.53\ cm)/(10) = 0.00453 m

Therefore,

\lambda = ((0.00453\ m)(0.00109\ m))/(8.61\ m)

λ = 5.734 x 10⁻⁷ m = 573.4 nm

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Answers

Answer:

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OFFERING 60 POINTS IF YOU CAN SHOW THE WORK!!!!A 1000 kg roller coaster begins on a 10 m tall hill with an initial velocity of 6m/s and travels down before traveling up a second hill. As the coaster moves from its initial height to its lowest position, 1700J of energy is transformed to thermal energy by friction.

Answers

Answer; 10.6 i think

Explanation:

(a) At the top of the hill, the coaster has total energy (potential and kinetic)

E = (1000 kg) g (10 m) + 1/2 (1000 kg) (6 m/s)² = 116,000 J

As it reaches its lowest position, its potential energy is converted to kinetic energy, and some is lost to friction, making its speed v such that

1/2 (1000 kg) v ² = 116,000 J - 1700 J = 114,300 J

===>   v ≈ 15.2 m/s

If no energy is lost to friction as the coaster makes its way up the second hill, all of its kinetic energy would be converted to potential energy at the maximum possible height H.

1/2 (1000 kg) (15.2 m/s)² = (1000 kg) gH

===>   H11.7 m

(b) At the top of the second hill with minimum height h, and with maximum speed 4.6 m/s, the coaster has energy

E = P + K = (1000 kg) gh + 1/2 (1000 kg) (4.6 m/s)²

Assuming friction isn't a factor again, the energy here should match the energy at the lowest point in part (a), 114,300 J.

(1000 kg) g h + 1/2 (1000 kg) (4.6 m/s)² = 114,300 J

===>   h10.6 m

What If? Fluoride ions (which have the same charge as an electron) are initially moving with the same speed as the electrons from part (a) through a different uniform electric field. The ions come to a stop in the same distance d. Let the mass of an ion be M and the mass of an electron be m. Find the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a). (Use the following as necessary: d, K, m, M, and e for the charge of the electron.)

Answers

Answer:

E₁ / E₂ = M / m

Explanation:

Let the electric field be E₁ and E₂ for ions and electrons respectively .

Force on ions = E₁ e where e is charge on ions .

Acceleration on ions a = E₁ e / M . Let initial velocity of both be u . Final velocity v = 0

v² = u² - 2as

0 = u² - 2 x E₁ e d  / M  

u² = 2 x E₁ e d  / M

Similarly for electrons

u² = 2 x E₂ e d  / m

Hence

2 x E₁ e d  / M =  2 x E₂ e d  / m

E₁ / E₂ = M / m

Final answer:

The ratio of the magnitude of the electric field the ions travel through to the magnitude of the electric field found in part (a) is M/m.

Explanation:

The ratio of the magnitude of the electric field the ions travel through to the magnitude of the electric field found in part (a) can be determined using the concept of mechanical energy conservation. Since the ions come to a stop, their initial kinetic energy must be equal to the work done by the electric field on them. The work done is given by the equation:

Work = Change in kinetic energy

The change in kinetic energy can be calculated using the formula:

Change in kinetic energy = (1/2)Mv2 - (1/2)mv2

where M and m are the masses of the ions and electrons respectively, and v is their initial speed. Solving for the ratio, we get:

Ratio = (1/2)M/(1/2)m = M/m

So, the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a) is M/m.

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A beam of light, which is traveling in air, is reflected by a glass surface. Does the reflected beam experience a phase change, and if so, by how much is the phase of the beam changed?

Answers

The reflected beam experienced a phase change of about 180°.

What is reflection in the glass surface?

According to Snell's law, the light that incident on the glass surface will be reflected and transmitted at an angle equals to the angle of incidence.  

By the observation of refractive index of the glass for the normal incidence only 4% of the light is transmitted or reflected.

The light passing through glass is not only reflected on the front surface, but also on the back. For several times the light will gets reflected back and forth. So, the total reflectance through a glass window can be calculated as

                                    2·R / (1+R).

Thus, A light wave travelling in air is reflected by a glass barrier will undergo a  phase change of 180°, while light travelling in glass will not undergo a phase change if it is reflected by a boundary with air.

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Answer:

180 degree phase change

Explanation: