Ask question

Ask question

All categories

3 years ago

11

A rock with mass m = 4.00 kg falls from rest in a viscous medium. The rock is acted on by a net constant downward force of F = 1

7.4 N (a combination of gravity and the buoyant force exerted by the medium) and by a fluid resistance force f=kv, where v is the speed in m/s and k = 2.16 N×s/m .
a. Find the initial acceleration a0.
b. Find the acceleration when the speed is 3.00 m/s.
c. Find the speed when the acceleration equals 0.1 a0.
d. Find the terminal speed vt.
e. Find the coordinate, speed, and acceleration 2.00 s after the start of the motion.
f. Find the time required to reach a speed of 0.9 Vt.

1 answer:

pogonyaev3 years ago

6 0

Answer:

a) 4.35 m/s²

b) 2.73 m/s²

c) 7.25 m/s

d) 8.06 m/s

e) At t = 2 s

x = 16.5 m

v = 7.88 m/s

a = 0.099 m/s²

f) t = 0.743 s

Explanation:

Force balance on the rock

ma = 17.4 - F

4a = 17.4 - kv

4a = 17.4 - 2.16v

a) At the initial instant, F = kv = 0

4a = 17.4

a = 4.35 m/s²

b) When v = 3 m/s

4a = 17.4 - (2.16)(3) = 10.92

a = 2.73 m/s²

c) a₀ = 4.35 m/s²

0.1 a₀ = 0.435 m/s²

4a = 17.4 - 2.16v

4(0.435) = 17.4 - 2.16v

1.74 = 17.4 - 2.16v

2.16v = 15.66

v = 7.25 m/s

d) Terminal speed is when the body stops accelerating in the fluid

When a = 0

0 = 17.4 - 2.16v

2.16 v = 17.4

v = 8.06 m/s

e) 4a = 17.4 - 2.16v

a = 4.35 - 0.54 v

But a = dv/dt

(dv/dt) = 4.35 - 0.54v

∫ dv/(4.35 - 0.54v) = ∫ dt

Integrating the left hand side from 0 to v and the right hand side from 0 to t

- 1.852 In (4.35 - 0.54v) = t

In (4.35 - 0.54v) = - 0.54 t

4.35 - 0.54v = e⁻⁰•⁵⁴ᵗ

0.54v = 4.35 - e⁻⁰•⁵⁴ᵗ

v = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ

Then, v = dx/dt

(dx/dt) = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ

dx = (8.06 - 0.54 e⁻⁰•⁵⁴ᵗ) dt

∫ dx = ∫ (8.06 - 0.54 e⁻⁰•⁵⁴ᵗ) dt

Integrating the left hand side from 0 to x and the right hand side from 0 to t

x = 8.06t + e⁻⁰•⁵⁴ᵗ

Acceleration too can be obtained as a function of time

since v = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ and a = dv/dt

a = 0.54² e⁻⁰•⁵⁴ᵗ = 0.2916 e⁻⁰•⁵⁴ᵗ

At t = 2 s

Coordinate

x = 8.06t + e⁻⁰•⁵⁴ᵗ

x = (8.06)(2) + e^(-1.08) = 16.5 m down into the fluid.

Velocity

v = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ

v = 8.06 - 0.54 e^(-1.08) = 7.88 m/s

Acceleration

a = 0.2916 e⁻⁰•⁵⁴ᵗ

a = 0.2916 e^(-1.08) = 0.099 m/s²

f) t = ? When v = 0.9 × 8.06 = 7.254 m/s

v = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ

7.254 = 8.06 - 0.54e⁻⁰•⁵⁴ᵗ

- 0.806 = - 0.54 e⁻⁰•⁵⁴ᵗ

e⁻⁰•⁵⁴ᵗ = 1.493

0.54t = In 1.493 = 0.401

t = 0.743 s.

You might be interested in

Violet light has a frequency of 7.26 × 1014 Hz and travels at a speed of 3.00 × 108 m/s.

Anuta_ua [19.1K]

Answer:

what's the question

........

8 0

3 years ago

A measurement that has both magnitude and direction

Korvikt [17]

<span>A measurement that both magnitude and direction is a vector quantity. An example of this is a moving car. The car exerts force due to its thrust and weight that runs in it. This will give us the magnitude of the car. The resulting motion of the car in terms of displacement, velocity and acceleration that determines its direction makes it a vector quantity. On the other hand, a measurement that has only magnitude is a scalar quantity. The energy exerted by the engine of the car is a scalar quantity.</span>

7 0

3 years ago

What unit is used for speed?<br> 1. m/s<br> 2. S<br> 3. m/s2<br> 4. m

bixtya [17]

Answer:

The unit of speed is m/s.

7 0

3 years ago

A tennis ball was thrown straight up with initial velocity of 22.5 m/s. How high does the ball rise

kiruha [24]

Answer:

$h_m=25.8\ m$

Explanation:

<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up without taking into consideration friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

$\displaystyle h_m=\frac{v_o^2}{2g}$

The tennis ball was thrown straight up with a speed of v0=22.5 m/s. The acceleration of gravity is g=9.81\ m/s^2, thus:

$\displaystyle h_m=\frac{22.5^2}{2\cdot 9.81}$

$\mathbf{h_m=25.8\ m}$

6 0

3 years ago

(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50

Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, $p=m_Av_A=m_Bv_B=m_Cv_C$ , were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as $m_h$ , and the mass of the first and second coals as $m_1$ and $m_2$ respectively

We start with the transition between parts A and B, so we have:

$m_Av_A=m_Bv_B$

Which means

$m_hv_A=(m_h+m_1)v_B$

And since we want the mass of the first coal thrown ( $m_1$ ) we do:

$m_hv_A=m_hv_B+m_1v_B$

$m_hv_A-m_hv_B=m_1v_B$

$m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}$

Substituting values we obtain

$m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg$

For the transition between parts B and C, we can write:

$m_Bv_B=m_Cv_C$

Which means

$(m_h+m_1)v_B=(m_h+m_1+m_2)v_C$

Since we want the new final speed of the car ( $v_C$ ) we do:

$v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}$

Substituting values we obtain

$v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s$

5 0

3 years ago