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2 years ago

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An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its

speed, v, and obeys the equation Fdrag=(12.0N⋅s/m)v+(4.00N⋅s2/m2)v2. What is the terminal speed of this object?

1 answer:

Alja [10]2 years ago

7 0

Answer:

Terminal velocity of object = 12.58 m/s

Explanation:

We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.

Gravitational force = mg = 80 * 9.8 = 784 N

Drag force = $12.0v+4.00v^2$

Equating both, we have

$784=12.0v+4.00v^2\\ \\ v^2+3v-196=0\\ \\ (v-12.58)(v+15.58)=0$

So v = 12.58 m/s or v = -15.58 m/s ( not possible)

So terminal velocity of object = 12.58 m/s

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A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

Lyrx [107]

Answer:

a) 27.3 m/s

b) 3.78 s

c) 10.576 s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

a)

$v=u+at\\\Rightarrow v=0+4.2\times 6.5\\\Rightarrow v=27.3\ m/s$

Top speed of the gazelle is 27.3 m/s

b)

$s=ut+\frac{1}{2}at^2\\\Rightarrow 30=0t+\frac{1}{2}\times 4.2\times t^2\\\Rightarrow t=\sqrt{\frac{30\times 2}{4.2}}\\\Rightarrow t=3.78\ s$

The gazelle would take 3.78 seconds to win a 30 m race.

c)

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 6.5+\frac{1}{2}\times 4.2\times 6.5^2\\\Rightarrow s=88.725\ m$

The gazelle would go 88.725 m with the acceleration after which there will be no acceleration.

Time = Distance / Speed

$\text{Time}=\frac{200-88.725}{27.3}=4.076\ s$

Total time taken by the gazelle to cover 200 m would be 6.5+4.076 = 10.576 seconds

4 0

2 years ago

A wire along the z axis carries a current of 4.9 A in the z direction Find the magnitude and direction of the force exerted on a

AURORKA [14]

Answer:

<h2>0.069 N, in the X direction</h2>

Explanation:

According to Flemming's left hand rule, it sates that if the first three fingers of the left hand are held mutually at right angles to one another, the fore finger will point in the direction of magnetic field, the middle finger will point in direction of current, while the thumb will point to the direction of force.

Mathematically the law is stated as

F= BIL

given data

Magnetic field B= 0.43T

Current I= 4.9 A

length of conductor L= 3.3cm to meter , 3.3/100= 0.033 m

Applying the formula the force is calculated as

F= 0.43*4.9* 0.033= 0.069 N

According to Flemming's rule the direction of all parameters are mutually perpendicular to one another, then the Force is in the X direction

5 0

2 years ago

Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.340 m and carries a current

sladkih [1.3K]

Answer:

The y-value of the line in the xy-plane where the total magnetic field is zero $U = 0.1355 \ m$

Explanation:

From the question we are told that

The distance of wire one from two along the y-axis is y = 0.340 m

The current on the first wire is $I_1 = (27.5i) A$

The force per unit length on each wire is $Z = 295 \mu N/m = 295*10^{-6} N/m$

Generally the force per unit length is mathematically represented as

$Z = \frac{F}{l} = \frac{\mu_o I_1I_2}{2\pi y}$

=> $\frac{\mu_o I_1I_2}{2\pi y} = 295$

Where $\mu_o$ is the permeability of free space with a constant value of $\mu_o = 4\pi *10^{-7} \ N/A2$

substituting values

$\frac{ 4\pi *10^{-7} 27.5 * I_2}{2\pi * 0.340} = 295 *10^{-6}$

=> $I_2 = 18.23 \ A$

Let U denote the line in the xy-plane where the total magnetic field is zero

So

So the force per unit length of wire 2 from line U is equal to the force per unit length of wire 1 from line (y - U)

So

$\frac{\mu_o I_2 }{2 \pi U} = \frac{\mu_o I_1 }{2 \pi(y - U) }$

substituting values

$\frac{ 18.23 }{ U} = \frac{ 27.5 }{(0.34 - U) }$

$6.198 -18.23U = 27.5U$

$6.198=45.73U$

$U = 0.1355 \ m$

5 0

1 year ago

Nitrogen (n2) gas within a piston–cylinder assembly undergoes a compression from p1 = 20 bar, v1 = 0.5 m3 to a state where v2 =

Bingel [31]

Part a)

As we know that

$P_1V_1^{1.35} = P_2V_2^{1.35}$

here we know that

P1 = 20 bar

V1 = 0.5 m^3

V2 = 2.75 m^3

from above equation

$20* 0.5^{1.35} = P * (2.75)^{1.35}$

$P = 2 bar$

so final state pressure will be 2 bar

Part b)

now in order to find the work done

$W = \int PdV$

$W = \int \frac{c}{V^{1.35}}dV$

$W = c\frac{V^{-0.35}}{-0.35}$

$W = \frac{P_1V_1 - P_2V_2}{0.35}$

$W = \frac{20* 0.5 - 2 * 2.75}{0.35}* 10^5 = 12.86 * 10^5 J$

3 0

2 years ago

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At the instant a ball is thrown horizontally with a large force, an identical ball is dropped from the same height. which ball h

sergejj [24]

The one that was dropped

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2 years ago

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