The -is the process by which gases in the atmosphere absorb and reradiate heat?

mixas84 [53]

I believe its the third answer

6 0

2 years ago

A rotating fan completes 1200 revolutions every minute. Consider the tip of a blade, at a radius of 0.15 m. (a) Through what dis

olga55 [171]

Answer:

(a) 0.942 m

(b) 18.84 m/s

(c) 2366.3 m/s²

(d) 0.05 s

Explanation:

(a) In one revolution, it travels through one circumference, 2πr = 2 × 3.14 × 0.15 m = 0.942 m.

(b) Its frequency, f, is 1200 rev/min = $\dfrac{1200}{60}$ rev/s = 20 rev/s.

Its angular frequency, ω = 2πf = 2π × 20 = 40π

The speed is given by

v = ωr = 40π × 0.15 = 6π = 18.84 m/s

(c) Its acceleration is given by, a = ω²r = (40π)² × 0.15 = 2366.3 m/s²

(d) The period is the inverse of the frequency because it is the time taken to complete one revolution.

$T = \dfrac{1}{f}$

T = 1/20 = 0.05 s

6 0

3 years ago

You drop a ball from a window located on an upper floor of a building. It strikes the ground with speed v. You now repeat the dr

bulgar [2K]

Answer:

y = y₀ (1 - ½ g y₀ / v²)

Explanation:

This is a free fall problem. Let's start with the ball that is released from the window, with initial velocity vo = 0 and a height of the window i

y = y₀ + v₀ t - ½ g t²

y = y₀ - ½ g t²

for the ball thrown from the ground with initial velocity v₀₂ = v

y₂ = y₀₂ + v₀₂ t - ½ g t²

in this case y₀ = 0

y₂2 = v t - ½ g t²

at the point where the two balls meet, they have the same height

y = y₂

y₀ - ½ g t² = vt - ½ g t²

y₀i = v t

t = y₀ / v

since we have the time it takes to reach the point, we can substitute in either of the two equations to find the height

y = y₀ - ½ g t²

y = y₀ - ½ g (y₀ / v)²

y = y₀ - ½ g y₀² / v²

y = y₀ (1 - ½ g y₀ / v²)

with this expression we can find the meeting point of the two balls

6 0

4 years ago

Physics question, please answer this ASAP, thanks!

miv72 [106K]

(a) The gravitational force received by each 1 kg mass is 8.66 N.

(b) The magnitude of gravitational acceleration is 8.66 m/s².

(c) The orbital speed of the ISS is 7,663.6 m/s.

(d) The time take for the ISS to orbit round the Earth is 5,558.75 s = 1.54 hours.

<h3>Gravitational force received by each 1 kg mass</h3>

The gravitational force received by each 1 kg mass is calculated as follows;

F = Gm₁m₂/r²

where;

m₁ is mass of Earth
m₂ is mass of ISS
r is the distance between the ISS and center of Earth

F = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ x 1) / (6780,000)²

F = 8.66 N

<h3>Magnitude of gravitational acceleration</h3>

mg = GMm/r²

g = GM/r²

where;

M is mass of Earth

g = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ ) / (6780,000)²

g = 8.66 m/s²

<h3>Orbital Speed of the ISS</h3>

v = √GM/r

v = √(6.67 x 10⁻¹¹ x 5.97 x 10²⁴ / 6780,000)

v = 7,663.6 m/s

<h3>Time of motion of the ISS round the Earth</h3>

T = 2πr/v

T = (2π x 6780,000) / (7663.6)

T = 5,558.75 seconds

1 hour = 3600 seconds

= 5,558.75/3600

= 1.54 hours

Thus, the gravitational force received by each 1 kg mass is 8.66 N.

The magnitude of gravitational acceleration is 8.66 m/s².

The orbital speed of the ISS is 7,663.6 m/s.

The time take for the ISS to orbit round the Earth is 5,558.75 s = 1.54 hours.

Learn more about orbital speed here: brainly.com/question/22247460

#SPJ1

4 0

2 years ago

What is the relationship between distance, time and acceleration

Evgesh-ka [11]

Answer:

total distance travelled

Explanation:

6 0

3 years ago