PHYSICS COLLEGE

A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine provides anet thrust of 0.795 N perpendicular tothe tethering wire.(a) Find the torque the net thrust producesabout the center of the circle.
N·m

(b) Find the angular acceleration of the airplane when it is inlevel flight.
rad/s2

(c) Find the linear acceleration of the airplane tangent to itsflight path.
m/s2

Answers

Answer 1

Answer:

(a) 24.6 Nm

The torque produced by the net thrust about the center of the circle is given by:

$\tau = F r$

where

F is the magnitude of the thrust

r is the radius of the wire

Here we have

F = 0.795 N

r = 30.9 m

Therefore, the torque produced is

$\tau = (0.795 N)(30.9 m)=24.6 N m$

(b) $0.035 rad/s^2$

The equivalent of Newton's second law for a rotational motion is

$\tau = I \alpha$

where

$\tau$ is the torque

I is the moment of inertia

$\alpha$ is the angular acceleration

If we consider the airplane as a point mass with mass m = 0.741 kg, then its moment of inertia is

$I=mr^2 = (0.741 kg)(30.9 m)^2=707.5 kg m^2$

And so we can solve the previous equation to find the angular acceleration:

$\alpha = (\tau)/(I)=(24.6 Nm)/(707.5 kg m^2)=0.035 rad/s^2$

(c) $1.08 m/s^2$

The linear acceleration (tangential acceleration) in a rotational motion is given by

$a=\alpha r$

where in this problem we have

$\alpha = 0.035 rad/s^2$ is the angular acceleration

r = 30.9 m is the radius

Substituting the values, we find

$a=(0.035 rad/s^2)(30.9 m)=1.08 m/s^2$

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An electromagnetic wave with frequency 65.0hz travels in an insulating magnetic material that has dielectric constant 3.64 and relative permeability 5.18 at this frequency. the electric field has amplitude 7.20×10−3v/m.what is the intensity of the wave in a medium\?

Answers

The intensity of the electromagnetic wave which travels in an insulating magnetic material in a medium is 5.766×10⁻⁸ W/m².

What is the intensity of the wave?

The intensity of a wave is the total power delivered per unit area. It can be given as,

$I=(P)/(A)$

It can also be given as,

$I=(E^2)/(2)\sqrt{(k\varepsilon_o)/(\mu_r\mu_o)}$

Here, ( $\mu_r$ ) is relative permeability, ( $\mu_0$ ) is physical constant, (k) is dielectric constant, (E) is the amplitude of electric field, and $\varepsilon_o$ is the permittivity of free space.

Here, the electromagnetic wave with frequency 65.0hz travels in an insulating magnetic material that has dielectric constant 3.64 and relative permeability 5.18 at this frequency.

As the electric field has amplitude 7.20×10−3v/m. Thus, put the values in the above formula to find the intensity as,

$I=((7.20*10^(-3))^2)/(2)\sqrt{(3.64*8.85*10^(-12))/(5.18*(4\pi*10^(-7)))}\nI=5.766*10^(-8)\rm W/m^2$

Hence, the intensity of the electromagnetic wave which travels in an insulating magnetic material in a medium is 5.766×10⁻⁸ W/m².

Learn more about the intensity of the wave here;

brainly.com/question/18109453

Ans: Intensity = I = $5.8 * 10^(-8) W/m^2$

Explanation:
First you need to find out the speed of Electromagnetic wave:

Since
v = $\sqrt{ (1)/(\mu\epsilon) }$

v = $\sqrt{ (1)/(\mu_r\mu_ok\epsilon_o) }$
$\mu_r = 5.18 \n\mu_o = 4 \pi * 10^(-7) \nk = 3.64 \n\epsilon_o = 8.85 * 10^(-12)$

Plug in the values:
v = $\sqrt{ (1)/((5.18)(4\pi*10^(-7))(3.64)(8.85*10^(-12))) }$
v = $6.9 * 10^7$ m/s

Now that we have "v", we can use the following formula to find the intensity of wave:

$I = (k\epsilon_o*v*E_(max)^2)/(2)$

$I = ((3.64)(8.85*10^(-12))*(6.9*10^7)*(7.20*10^(-3))^2)/(2)$

Intensity = I = $5.8 * 10^(-8) W/m^2$

Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 4.90 m away from the slits.a. What is the distance Δ ymax-max between the first maxima (on the same side of the central maximum) of the two patterns?
b. What is the distance Δymax-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?

Answers

Answer:

a)Δy = 81.7mm

b)Δy = 32.7cm

Explanation:

To calculate the distance between any point of the interference pattern, simply use the trigonometric ratio of the tangent:

$Tan \theta = (y)/(D)$

where D is the separation between the slits and the screen where the interference pattern is observed.

a) In this case:

Δy = |y1max (λ1) − y1max (λ2)|

Δy = $|(D\lambda _1)/(d) - (D\lambda _2)/(d) |$

Δy = $D |(d/20)/(d) - (d/15)/(d) |$

Δy = $D |(1)/(20) - (1)/(15) |$

Δy = $4.90 |(1)/(20)- (1)/(15) |$

Δy = 81.7mm

The separation between these maxima is 81.7 mm

Δy = |y₂max (λ1) − y₂max (λ2)|

Δy = $D|(2(d/20))/(d) - (5(d/15))/(2d) |$

Δy = $4.90|(1)/(10) - (1)/(6) |$

Δy = 32.7cm

The separation between the maximum interference of the 2nd order (2nd maximum) of the pattern produced by the laser 1 and the minimum of the 2nd order (3rd minimum) of the pattern produced by the laser 2 is 32.7 cm.

Final answer:

We can solve the problem using the concepts of waveinterference and the formulas for maxima and minima positions (i.e., y = L*m*λ/d and y = L*(m+1/2)*λ/d respectively). The difference between the first maxima of the two patterns is 4.9/60 m and the difference between the second maximum of laser 1 and the third minimum of laser 2 is also 4.9/60 m.

Explanation:

The problem described deals with wave interference and can be addressed using the formulas for path difference and phasedifference.

To answer part a, we need to find the difference between the positions of the first maxima for the two lasers. The position of any maxima in an interference pattern can be found using the formula: y = L * m * λ / d, where L is the distance from the slits to the screen, m is the order of the maxima, λ is the wavelength, and d is the slit separation.

So for the first laser (λ1=d/20) the position of the first maxima would be y1 = 4.9m * 1 * (d/20) / d =4.9/20 m.

And for the second laser (λ2 = d/15) the position of the first maxima would be y2= 4.9m * 1 * (d/15) / d =4.9/15 m.

Then, the distance Δ ymax-max between the first maxima of the two patterns is y2-y1= 4.9/15 m - 4.9/20 m = 4.9/60 m.

Answering part b involves finding the positions of the second maximum of laser 1 and the third minimum of laser 2. The position of any minimum in an interference pattern can be calculated using the formula: y = L * (m+1/2) * λ / d. For the second maximum of laser 1, we have y1max2 = 4.9 m * 2 * (d/20) / d = 4.9/10 m. For the third minimum of laser 2, we have y2min3 = 4.9m * (3.5) * (d/15)/d = 4.9*7/30 m. The difference Δymax-min is y2min3-y1max2= 4.9*7/30 m - 4.9/10 m = 4.9/60 m.

Learn more about Wave Interference here:

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With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 41 m

Answers

Answer:

The speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.

Explanation:

Given;

maximum vertical height of the throw, H = 41 m

Apply the following kinematic equation;

V² = U² + 2gH

where;

V is the final speed with which the ball will rise to a maximum height

U is the initial speed of the ball = 0

g is acceleration due to gravity = 0

V² = U² + 2gH

V² = 0² + 2gH

V² = 2gH

V = √2gH

V = √(2 x 9.8 x 41)

V = 28.35 m/s

Therefore, the speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.

As in problem 80, an 76-kg man plans to tow a 128000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that he instead attempts the feat by pulling the cable at an angle of 6.7° above the horizontal. The coefficient of static friction between his shoes and the runway is 0.87. What is the greatest acceleration the man can give the airplane? Assume that the airplane is on wheels that turn without any frictional resistance.

Answers

In order to solve this problem it is necessary to apply the concepts related to Newton's second law and the respective representation of the Forces in their vector components.

The horizontal component of this force is given as

F_x = Fcos(6.7)

While the vertical component of this force would be

F_y = Fsin(6.7)

In the vertical component, the sum of Force indicates that:

$\sum F_y= 0$

The Normal Force would therefore be equivalent to the weight and vertical component of the applied force, therefore:

$N = mg+Fsin(6.7)$

In the horizontal component we have that the Force of tension in its horizontal component is equivalent to the Force of friction:

$\sum F_x = 0$

$F_x = F_(friction)$

$Fcos (6.7) = N\mu$

Using the previously found expression of the Normal Force and replacing it we have to,

$Fcos(6.7)= \mu (mg+Fsin(6.7))$

Replacing,

$Fcos(6.7)= (0.87) (mg+Fsin(6.7))$

$Fcos(6.7) = (0.87)(mg) + (0.87)(Fsin(6.7))$

$Fcos(6.7) -(0.87)(Fsin(6.7)) = 0.87 (mg)$

$F(cos(6.7)-0.87sin(6.7)) = 0.87 (mg)$

$F = (0.87 (mg))/((cos(6.7)-0.87sin(6.7)))$

$F = (0.87(128000*9.8))/((cos(6.7)-0.87sin(6.7)))$

$F = 1.95*10^6N$

Finally the acceleration would be by Newton's second law:

$F = ma$

$a = (F)/(m)$

$a = ( 1.95*10^6)/(128000)$

$a = 15.234m/s^2$

Therefore the greatest acceleration the man can give the airplane is $15.234m/s^2$

The time taken by a mass projected verticallyupwards to reach the maximum height (with air
resistance not neglected) is 10 sec. The time of
descent of the mass from the same height will be

Answers

Answer:

10s

Explanation:

The time to get to the maximum would be the same as the time to get down to the maximum unless somehow gravity’s changes during the duration it goes up to and from maximum height.

A pitcher is in 85° of abduction, holding a 1.4 N baseball at point C, 65 cm from the joint axis at point O • The center of gravity of his arm is 25 cm from the joint axis of shoulder abduction at point O • The weight of the arm W is 0.06 of the pitcher’s weight of 100 N • Deltoids muscles are at an angle θ of 15° with respect to the humerus and insert 15 cm from the joint axis at point A • Determine the force applied by the Deltoid muscles and the joint reaction force at the shoulder joint and its orientation β