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13 days ago

6

In a later chapter we will be able to show, under certain assumptions, that the velocity v(t) of a falling raindrop at time t is

v(t) = vT(1 − e−gt/vT) where g is the acceleration due to gravity and vT is the terminal velocity of the raindrop. (a) Find lim t→[infinity] v(t).

1 answer:

Maru [1K]13 days ago

8 0

Answer:

$\lim_{t \to \infty} v(t) =vT$

Explanation:

Applying the distributive property:

$v(t)=vT(1-\frac{e^{-gt} }{vT} )=vT-e^{-gt}$

Thus:

$\lim_{t\to \infty} vT-e^{-gt}$

The sum of the limits of two functions equals the limit of their sum, therefore:

$\lim_{t\to \infty} vT-e^{-gt} = \lim_{t\to \infty} vT - \lim_{t\to \infty} e^{-gt}$

For a constant function, the limit remains the same constant, hence:

$\lim_{t\to \infty} vT=vT$

Next, we will evaluate the other limit:

$\lim_{t\to \infty} e^{-gt}=e^{ \lim_{t \to \infty} -gt}$

The limit for a constant multiplied by a function equals the product of the constant and the limit of the function, thus:

$\lim_{t \to \infty} -gt}=-g\lim_{t \to \infty} t}=-g(\infty)$

$-g(\infty)=-\infty$

Consequently:

$e^{(-\infty)} =0$

In conclusion:

$\lim_{t\to \infty} vT-e^{-gt}=vT-0=vT$

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A 1100kg car pulls a boat on a trailer. (a) what total force resists the motion of the car, boat,and trailer, if the car exerts

inna [987]

Refer to the diagram shown below.

m₁ = 1100 kg represents the mass of the car.
m₂ = 700 kg indicates the combined mass of the trailer and boat.
F = 1900 N is the driving force acting on the vehicle.
N₁ denotes m₁g, the normal force on the car.
N₂ corresponds to m₂g, the normal force on the trailer and boat.
Frictional forces are represented by μN₁ and μN₂, where μ is the coefficient of kinetic friction.
T signifies the force in the connection between the car and the trailer.

Part (a)
Let R₁ signify the total resistance acting against the motion of the car, boat, and trailer.
With the acceleration at 0.550 m/s², it follows that
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100 + 700 kg)*(0.55 m/s²) = (1900 - R₁) N.
This leads to the equation 990 = 1900 - R.
Therefore, R₁ = 910 N.

Answer: The total resistive force amounted to 910 N.

Part (b)
The trailer and boat experience 80% of the resisting forces.
Let R₂ denote this resistive force.
Thus,
R₂ = 0.8*R₁ = 728 N.
Assuming T is the tension in the hitch connecting the car and trailer, it follows:
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²).
This leads to T - 728 = 385.
Thus, T equals 1113 N.

Answer: The tension in the hitch is 1113 N.

3 0

6 days ago

Un tren parte de la ciudad A, a las 8 h. con una velocidad de 50 km/h, para llegar a la ciudad B a las 10 h. Allí permanece dura

Keith_Richards [1021]

Response:

AB = 100 km; BC = 80 km; AC = 180 km

Time of arrival = 11:30

Reasoning:

1. Distance from A to B

(a) Duration of travel

Duration = 10:00 - 8:00 = 2.00 hours

(b) Distance

Distance = speed × time = 50 km/h × 2.00 h = 100 km

2. Distance from B to C

Distance = 80 km/h × 1 h = 80 km

3. Summary of Distances

AB = 100 km

BC = 80 km

AC = 180 km

4. Time of Arrival

Departure from A = 08:00

Travel duration to B = 2:00

Arrival at B = 10:00

Waiting time at B = 0:30

Departure from B = 10:30

Travel duration to C = 1:00

Arrival at C = 11:30

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13 days ago

Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet

ValentinkaMS [1144]

Answer:

The rate at which the root beer level is decreasing is 0.08603 cm/s.

Explanation:

The formula for the volume of the cone is:

$V=\frac {1}{3}\times \pi\times r^2\times h$

Where V denotes the cone's volume

r indicates the radius

h signifies the height

The ratio of radius to height remains consistent throughout the cone.

Thus, we have r = d / 2 = 10 / 2 cm = 5 cm

h is 13 cm

Consequently, r / h = 5 / 13

r = {5 / 13} h

$V=\frac {1}{3}\times \frac {22}{7}\times ({{{\frac {5}{13}\times h}}})^2\times h$

$V=\frac {550}{3549}\times h^3$

Additionally, we differentiate the volume expression in relation to time:

$\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}$

Given that $\frac {dV}{dt}$ = -4 cm³/sec (the negative sign indicates outflow)

h equals 10 cm

Hence,

$-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}$

$\frac{55000}{1183}\times \frac {dh}{dt}=-4$

$\frac {dh}{dt}=-0.08603\ cm/s$

The rate at which the root beer level is decreasing is 0.08603 cm/s.

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11 days ago

Two planets having equal masses are in circular orbit around a star. Planet A has a smaller orbital radius than planet B. Which

serg [1189]

Answer:

Explanation:

To approach this problem, we need to understand two key concepts.

First, the gravitational force on an object in orbit equals its mass multiplied by centripetal acceleration.

Secondly, Newton's law of universal gravitation defines the force between two masses: Fg = mMG/r², where Fg denotes gravitational force, m and M signify the masses, G represents the gravitational constant, and r indicates the distance separating the two masses.

Thus:

Fg = m v²/r

mMG/r² = m v²/r

v² = MG/r

Potential energy for each planet is expressed as:

PE = mgr = m (MG/r²) r = mMG/r

Kinetic energy for each planet is computed as:

KE = 1/2 mv² = 1/2 m (MG/r) = 1/2 mMG/r

Total mechanical energy is calculated as:

ME = PE + KE = 3/2 mMG/r

Since both planets share the same mass, the only variable is their orbital radius. Consequently, Planet A, with a smaller radius, possesses greater potential, kinetic, and mechanical energy.

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13 days ago

Simone is walking her dog on a leash. The dog is pulling with a force of 32 N to the right and Simone is pulling backward with a

ValentinkaMS [1144]

Conclusion:

The total net force acting on the objects is 16 N, directed towards the right.

Clarification:

It is stated that,

The force exerted by the dog, $F_1 = 32\ N$ (to the right)

The force exerted by Simone, $F_2 = -16\ N$ (backward)

Here, assume the backward direction is negative and the right direction is positive.

The net force will move in the direction where the larger force is present. The net force can be calculated as:

$F=F_1+F_2$

$F=32+(-16)$

F = 16 N

Thus, the net force amounts to 16 N, acting towards the right.

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3 days ago

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