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"Suppose an object falling in the atmosphere has mass m=15kg and the drag coefficient is γ=9kg/s. Recall that the differential e

Art [367]

Answer:

a. v(t)= -6.78 $e^{-16.33t}$ + 16.33 b. 16.33 m/s

Step-by-step explanation:

The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)= $e^{\int\limits^ {}k \, dt }$ = $e^{kt}$ . We now multiply both sides of the equation by the integrating factor.

μv' + μkv = μg ⇒ $e^{kt}$ v' + k $e^{kt}$ v = g $e^{kt}$ ⇒ [v $e^{kt}$ ]' = g $e^{kt}$ . Integrating, we have

∫ [v $e^{kt}$ ]' = ∫g $e^{kt}$

v $e^{kt}$ = $\frac{g}{k}$ $e^{kt}$ + c

v(t)= $\frac{g}{k}$ + c $e^{-kt}$ .

From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have

9.55 = 9.8 × 15/9 + c $e^{-16.33 * 0}$ = 16.33 + c

c = 9.55 -16.33 = -6.78.

So, v(t)= 16.33 - 6.78 $e^{-16.33t}$ . m/s = - 6.78 $e^{-16.33t}$ + 16.33 m/s

b. Velocity of object at time t = 0.5

At t = 0.5, v = - 6.78 $e^{-16.33 x 0.5}$ + 16.33 m/s = 16.328 m/s ≅ 16.33 m/s

6 0

2 years ago

6th grade math help me please :))

sergij07 [2.7K]

Answer:

The answer is option D.

3u + 1 + 7y

All the terms here are different and cannot be combined

Hope this helps you

4 0

3 years ago

DanielleElmas [232]

30/3=10
10*2=20
the volume of the smaller figure is 20 cubic cm

3 0

3 years ago

NEED HELP ASAP!!

Solnce55 [7]

Answer:

Only Cory is correct

Step-by-step explanation:

The gravitational pull of the Earth on a person or object is given by Newton's law of gravitation as follows;

$F =G\times \dfrac{M \cdot m}{r^{2}}$

Where;

G = The universal gravitational constant

M = The mass of one object

m = The mass of the other object

r = The distance between the centers of the two objects

For the gravitational pull of the Earth on a person, when the person is standing on the Earth's surface, r = R = The radius of the Earth ≈ 6,371 km

Therefore, for an astronaut in the international Space Station, r = 6,800 km

The ratio of the gravitational pull on the surface of the Earth, F₁, and the gravitational pull on an astronaut at the international space station, F₂, is therefore given as follows;

$\dfrac{F_1}{F_2} = \dfrac{ \dfrac{M \cdot m}{R^{2}}}{\dfrac{M \cdot m}{r^{2}}} = \dfrac{r^2}{R^2} = \dfrac{(6,800 \ km)^2}{(6,371 \ km)^2} \approx 1.14$

∴ F₁ ≈ 1.14 × F₂

F₂ ≈ 0.8778 × F₁

Therefore, the gravitational pull on the astronaut by virtue of the distance from the center of the Earth, F₂ is approximately 88% of the gravitational pull on a person of similar mass on Earth

However, the International Space Station is moving in its orbit around the Earth at an orbiting speed enough to prevent the Space Station from falling to the Earth such that the Space Station falls around the Earth because of the curved shape of the gravitational attraction, such that the astronaut are constantly falling (similar to falling from height) and appear not to experience gravity

Therefore, Cory is correct, the astronauts in the International Space Station, 6,800 km from the Earth's center, are not too far to experience gravity.

6 0

3 years ago

Need help asap!!!!!!!!!!!

salantis [7]

Answer:

the answer should be d for tour answer

3 0

3 years ago