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Katen [24]
2 years ago
6

A ball is falling at terminal velocity. Terminal velocity occurs when the ball is in equilibrium and the forces are balanced. Wh

ich free body diagram shows the ball falling at terminal velocity? A free body diagram with one force pointing downward labeled F Subscript g Baseline 20 N. A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N. A free body diagram with 2 forces. The first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 5 N. A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 30 N.
Physics
2 answers:
Tema [17]2 years ago
8 0

Answer:

b

Explanation:

Greeley [361]2 years ago
4 0

Answer:

A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N.

Explanation:

This is because at terminal velocity, the ball stops accelerating and the net force on the ball is zero. For the net force to be zero, equal and opposite forces must act on the ball, so that their resultant force is zero. That is F₁ + F₂ = 0 ⇒ F₁ = -F₂

Since F₁ = 20 N, then F₂ = -F₁ = -20 N

So, if F₁ points upwards since it is positive, then F₂ points downwards since it is negative.

So, a free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N best describes the ball falling at terminal velocity.

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A person fishing from a pier observes that 6 wave crests pass by in 8.0 s and estimates the distance between two successive cres
Talja [164]

Answer:

The speed of the wave is 3 m/s.

Explanation:

Given that,

Number of waves = 6

Time = 8.0 s

Successive crests \lambda = 4.8\ m

We need to calculate the frequency

Using formula of frequency

f=\dfrac{Number\ of \ cycles}{t}

Put the value into the formula

f=\dfrac{6-1}{8.0}

f=0.625\ cycle/sec

We need to calculate the speed of the wave

Using formula of speed

v=\lambda\times f

Put the value into the formula

v=4.8\times0.625

v=3\ m/s

Hence, The speed of the wave is 3 m/s.

7 0
2 years ago
A microwave oven operates at 2.60 ghz . what is the wavelength of the radiation produced by this appliance?
cestrela7 [59]
<span>10.3 cm The wavelength will be the distance that light travels in 1 second divided by the frequency of the radiation. Since the over operates at 2.60 ghz, the frequence is 2.6 billion times per second, or 2.60 x 10^9. The speed of light is defined as 299792458 m/s exactly. So 299792458 m/s / 2.60 x 10^9 1/s = 0.10337671 m = 10.337671 cm Since we only have 3 significant digits, the answer rounds to 10.3 cm</span>
8 0
2 years ago
The maximum tension that a 0.80 m string can tolerate is 15 N. A 0.35-kg ball attached to this string is being whirled in a vert
zimovet [89]

Answer:

v=5.86 m/s

Explanation:

Given that,

Length of the string, l = 0.8 m

Maximum tension tolerated by the string, F = 15 N

Mass of the ball, m = 0.35 kg

We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

F=\dfrac{mv^2}{r}

v is the maximum speed

v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{15\times 0.8}{0.35}} \\\\v=5.86\ m/s

Hence, the maximum speed of the ball is 5.86 m/s.

3 0
2 years ago
What units are given to the right of the equals sign
zhuklara [117]
The answer

2y + 14 = 17

The 17 is to the right of the = sign
It is also the answer
7 0
2 years ago
In the 25-ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of int
Ugo [173]

Answer:

a. 8.33 x 10 ⁻⁶ Pa

b. 8.19 x 10 ⁻¹¹ atm

c. 1.65 x 10 ⁻¹⁰ atm

d. 2.778 x 10 ⁻¹⁴ kg / m²

Explanation:

Given:

a.

I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s

P rad = I / us

P rad  = 2500 W / m² / 3.0 x 10 ⁸ m/s

P rad = 8.33 x 10 ⁻⁶ Pa

b.

P rad = 8.33 x 10 ⁻⁶ Pa *[  9.8 x 10 ⁻⁶ atm / 1 Pa ]

P rad = 8.19 x 10 ⁻¹¹ atm

c.

P rad = 2 * I / us = ( 2 * 2500 w / m²) / [ 3.0 x 10 ⁸ m /s ]

P rad = 1.67 x 10 ⁻⁵ Pa

P₁ = 1.013 x 10 ⁵ Pa /atm

P rad = 1.67 x 10 ⁻⁵ Pa / 1.013 x 10 ⁵ Pa /atm = 1.65 x 10 ⁻¹⁰ atm

d.

P rad  = I / us

ΔP / Δt = I / C² = [ 2500 w / m² ] / ( 3.0 x 10 ⁸ m/s)²

ΔP / Δt = 2.778 x 10 ⁻¹⁴ kg / m²

3 0
2 years ago
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