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Dennis_Churaev [7]
3 years ago
14

A 144-g baseball is dropped from a tree 11.0m above the ground.A.With what speed would it hit the ground if air resistance could

be ignored?Express your answer to three significant figures and include the appropriate units.B.If it actually hits the ground with a speed of 7.50m/s , what is the magnitude of the average force of air resistance exerted on it?Express your answer to three significant figures and include the appropriate units.
Physics
1 answer:
nikitadnepr [17]3 years ago
5 0

Answer:

a. 14.68m/s

b. 1.043N

Explanation:

a.

Given

Mass = m = 144g = 0.144kg

Height = h = 11m

Using energy conservation,

½mv² = mgh where g = 9.8m/s² ----- divide both sides by m

½v² = gh ---- divide both sides by ½

v² = 2gh

v = √2gh

v = √(2 * 9.8 * 11)

v = 14.68332387438212m/s

v = 14.68m/s ----- Approximated

b.

v = 7.5m/s

By definition, “the average force of air resistance,”

When work done against air resistance is solved by multiplying height by the average force of resistance ------ by definition

i.e. W = -Fh

Minus is used because F and h points to different direction (when viewed as vectors).

This is proved by:

W = Fhcos(180) because h points toward the direction of motion (downward) while F points opposite.

W = Fh * -1

W = Fh

Using energy conservation,

mgh - Fh = ½mv² ---- Make Fh, the subject of formula

Fh = mgh - ½mv² ---- Divide through by h

F = mg - mv²/2h

F = 0.144 * 9.8 - 0.144 * 7.5²/(2 * 11)

F = 1.043N

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Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

- The radius of the earth,  R = 6.37 * 10 ^6 m

- The time period for 1 revolution T = 24 hrs

Find:

What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

- The angular speed w of the earth can be related with the Time period T of the earth revolution by:

                                  w = 2π / T

                                  w = 2π / 24*3600

                                  w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

                                 v1 = R*w

                                 v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

                                 v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

                                 π/2  ........... s

                                 x     ............ 1/5 s

                                 x = π/2*5 = 18°    

- The radius of the earth R' at point where θ = 18° from the equator is:

                                R' = R*cos(18)

                                R' = (6.37 * 10 ^6)*cos(18)

                                R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

                              v2 = R'*w

                              v2 = (6058230.0088)*(7.27 * 10^-5)

                              v2 = 440.433 m/s

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Hydraulic engineers often use, as a unit of volume of water, the "acre-foot", defined as the volume of water that will cover 1 a
Alex17521 [72]

Answer:

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Explanation:

Given:

Rain = 1.7 in

Time = 30 min

Area = 29 km²

Find:

Volume in acre-feet

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1 km = 1,000 m

1 m = 3.28 feet

1 km² = 247.105 acre

d = 1.7 in = 1.7 / 12 = 0.14167 ft

Area = 29 × 247.105 = 7,166.045 acre

Volume = 7,166.045 acre × 0.14167 ft

Volume = 1,015 acre-feet (Approx)

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The surface of the Earth changes from processe such as erosion. Which of these changes to Earth's surface is an example of erosi
Ad libitum [116K]

Answer:

D. the wind picking up dust and carrying it

Explanation:

Erosion is a process in which an agent transfer the top soil to another region, thereby exposing the lower soil. These agents have the ability to move the top layer of soil and deposit it at another place. The major agents in this case are; a running or flowing body of water and wind.

Therefore, the change to the Earth's surface that is an example of erosion is the wind picking up dust and carrying it. Thereby exposing the lower layers.

6 0
2 years ago
The maximum wavelength For photoelectric emissions in tungsten is 230 nm. What wavelength of light must be use in order for elec
notka56 [123]

Answer:

λ = 1.8 x 10⁻⁷ m = 180 nm

Explanation:

First we find the work function of tungsten by using the following formula:

∅ = hc/λmax

where,

∅ = work function = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λmax = maximum wavelength for photoelectric emission = 230 nm

λmax = 2.3 x 10⁻⁷ m

Therefore,

∅ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.3 x 10⁻⁷ m)

∅ = 8.64 x 10⁻¹⁹ J

Now we convert Kinetic Energy of electron into Joules:

K.E = (1.5 eV)(1.6 x 10⁻¹⁹ J/1 eV)

K.E = 2.4 x 10⁻¹⁹ J

Now, we use Einstein's Photoelectric Equation:

Energy of Photon = ∅ + K.E

Therefore,

Energy of Photon = 8.64 x 10⁻¹⁹ J + 2.4 x 10⁻¹⁹ J

Energy of Photon = 11.04 x 10⁻¹⁹ J

but,

Energy of Photon = hc/λ

where,

λ = wavelength of light = ?

Therefore,

11.04 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(11.04 x 10⁻¹⁹ J)

<u>λ = 1.8 x 10⁻⁷ m = 180 nm</u>

5 0
3 years ago
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