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A rider on a bike with the combined mass of 100 kg attains a terminal speed of 15 m/s on a 12% slope. Assuming that the only forces affecting the speed are the weight and the drag, calculate the drag coefficient. The frontal area is 0.9 m2.

Answers

Answer 1

Answer:

Answer:

7.84

Explanation:

Draw free body diagram and put all forces on it. Forces are

Gravity in direction of slope = $(12)/(100) *100*12$ =117.6newton
Viscous force(opposite to slope) after attaining terminal speed =k*V=k*15

As the bike+man system has attained a terminal velocity thus acceleration is zero .

Both forces are opposite then equate them

117.6=k.15

k=7.84

Here k is drag coefficient.

Answer 2

Answer:

Answer:

0·95

Explanation:

Given the combined mass of the rider and the bike = 100 kg

Percent slope = 12%

∴ Slope = 0·12

Terminal speed = 15 m/s

Frontal area = 0·9 m²

Let the slope angle be β

tanβ = 0·12

As it attains the terminal speed, the forces acting on the combined rider and the bike must be balanced and therefore the rider must be moving download as the directions of one of the component of weight and drag force will be in opposite directions

The other component of weight will get balance by the normal reaction and you can see the figure which is in the file attached

From the diagram m × g × sinβ = drag force

Drag force = 0·5 × d × $C_(D)$ × v² × A

where d is the density of the fluid through which it flows

$C_(D)$ is the drag coefficient

v is the speed of the object relative to the fluid

A is the cross sectional area

As tanβ = 0·12

∴ sinβ = 0·119

Let the fluid in this case be air and density of air d = 1·21 kg/m³

m × g × sinβ = 0·5 × d × $C_(D)$ × v² × A

100 × 9·8 ×0·119 = 0·5 × 1·21 × $C_(D)$ × 15² × 0·9

∴ $C_(D)$ ≈ 0·95

∴ Drag coefficient is approximately 0·95

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Một vật chuyển động tròn đều có chu kì T = 0,25 s. Tính tần số chuyển động f của vật?

Answers

Answer:8pi

Explanatio

omega=2pi/T

Answer:

0000

Explanation:

The wind-chill index is modeled by the function W = 13.12 + 0.6215T − 11.37v0.16 + 0.3965Tv0.16 where T is the temperature (°C) and v is the wind speed (km/h). When T = 12°C and v = 18 km/h, by how much would you expect the apparent temperature W to drop if the actual temperature decreases by 1°C? (Round your answers to two decimal places.)

Answers

This question involves the concepts of derivative, apparent temperature, actual temperature,and wind speed.

The drop in apparent temperature will be "1.25°C".

The apparent temperature (W) is given in terms of actual temperature (T) and wind speed (v) is given by the following function:

$W = 13.12 + 0.6215\ T-11.37\ v^(0.16)+0.3965\ Tv^(0.16)$

Taking the derivative with respect to actual temperature, we get:

$(dW)/(dT)=0.6215+0.3965\ v^(0.16)\n\n$

where,

dW = drop in apparent temperatures = ?

dT = drop in actual temperature = - 1°C

v = wind speed = 18 km/h

Therefore,

$dW=(-1)(0.6215-0.3965(18)^(0.16))$

dW = - 1.25°C

Learn more about derivatives here:

brainly.com/question/9964510?referrer=searchResults

Answer:

Δw=1.25°C

Explanation:

Given that

$w=13.12 +0.6215 T-11.37 v^(0.16)+0.3965 T v^(0.16)$

Given that T= 12°C and v=19 km/h

Now to find the drop in the apparent temperature w

$(dw)/(dT)=0.6215 +0.3965v^(0.16)$

$(\Delta w)/(\Delta T)= 0.6215 +0.3965 v^(0.16)$

Now by putting the values v=19 km/hr and ΔT=1

$(\Delta w)/(1)=0.6215 +0.3965* 18^(0.16)$

Δw=1.25°C

So we can say that when temperature is decrease by 1°C then apparent temperature will decrease by 1.25°C at given velocity.

Suppose that a particle accelerator is used to move two beams of particles in opposite directions. In a particular region, electrons move to the right at 6020 m/s and protons move to the left at 1681 m/s. The particles are evenly spaced with 0.0476 m between electrons and 0.0662 m between protons. Assuming that there are no collisions and that the interactions between the particles are negligible, what is the magnitude of the average current in this region

Answers

Answer:

$2.429783984* 10^(-14)\ A$

Explanation:

Velocity of electron = 6020 m/s

Velocity of proton = 1681 m/s

Electron space = 0.0476 m

Proton space = 0.0662 m

e = Charge of particle = $1.6* 10^(-19)\ C$

Number of electrons passing per second

$n_e=(6020)/(0.0476)\n\Rightarrow n_e=126470.588$

Number of protons passing per second

$n_p=(1681)/(0.0662)\n\Rightarrow n_p=25392.749$

Current due to electrons

$I_e=n_ee\n\Rightarrow I_e=126470.588* 1.6* 10^(-19)\n\Rightarrow I_e=2.0235* 10^(-14)\ A$

Current due to protons

$I_p=n_pe\n\Rightarrow I_p=25392.749* 1.6* 10^(-19)\n\Rightarrow I_p=4.06283984* 10^(-15)\ A$

Total current

$I=2.0235* 10^(-14)+4.06283984* 10^(-15)\n\Rightarrow I=2.429783984* 10^(-14)\ A$

The average current is $2.429783984* 10^(-14)\ A$

Why We can’t Cure Aging? Support your answer?

Answers

We can’t cure aging because it is inevitable. Aging is a part of life and it leads to death which is also inevitable and inescapable. We can’t stop our aging because it is natural and normal to age

A grandfather clock keeps time using a pendulum consisting of a light rod connected to a small heavy mass. With a rod of length L, the period of oscillation is 2.00 s. What should the length of the rod be for the period of the oscillations to be 1.00 s?

Answers

Answer:

The length of the rod should be

$(L)/(4) \n$

Explanation:

Period of simple pendulum is given by

$T=2\pi\sqrt{(l)/(g)} \n$

We have

$(T_1^2)/(T_2^2)=(l_1)/(l_2)\n\n(2^2)/(1^2)=(L)/(l_2)\n\nl_2=(L)/(4) \n$

The length of the rod should be

$(L)/(4) \n$

A power P is required to do work W in a time interval T. What power is required to do work 3W in a time interval 5T? (a) 3P (b) 5P (c) 3P/5 (a) P (e) 5P/3