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3 years ago

6

Recall that force is a change in momentum over a change in time, the force due to radiation pressure reflected off of a solar sa

il can be calculated as 2 times the radiative momentum striking the sail per second. What is the approximate magnitude of the pressure on the sail in the vicinity of Earth’s Orbit?

1 answer:

Zarrin [17]3 years ago

4 0

Answer:

magnitude of the pressure on the sail in the vicinity of Earth’s Orbit= $\frac{2I}{c}$

Explanation:

The momentum of a photon is:

p = E/c

E = the photon energy

c = the speed of light.

take the time derivative (gives the force)

F = dp/dt = (dE/dt)/c

F = 2(dE/dt)/c (is doubled for complete reflection of the light)

Intensity has the units of energy per unit time per unit area

= I

then,

Force/unit area = 2I/c

magnitude of the pressure on the sail in the vicinity of Earth’s Orbit= $\frac{2I}{c}$

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The pilot of an aircraft wishes to fly due west in a wind blowing at 45 km/h toward the south. If the speed of the aircraft in t

DiKsa [7]

Answer:

164.2°

Explanation:

speed of wind (w) = 45 km/h due south

speed of aircraft (a) = 165 km/h

in what direction (in degrees) should the aircraft head in so as to fly due west?

To get the direction the pilot should fly, we can form a triangle with the data available where

the direction of the wind (due south) serves as the opposite side
the direction the pilot would have to fly so he can end up at he west serves as the hypothenuse
θ is the angle between the direction the pilot would have to fly and the direction the pilot wishes to fly.
the direction the pilot wishes to fly ( west) will serve as the adjacent side
all this can be seen from the attached diagram.

now sin θ = $\frac{speed of the wind}{speed of the aircraft}$

sin θ = $\frac{45}{165}$

θ = $sin^{-1}$ 0.2728

θ = 15.8°

since we are to use the counter-clockwise from east convention our measurement would have to be taken anticlockwise from the east direction, therefore the direction of the aircraft (Ф) = 180-15.8 = 164.2°

3 0

3 years ago

A spring (k=15.19kN/m)is is compresses 25cm and held in place on a 36.87° incline. A block (M=10kg) is placed on the spring. Whe

Savatey [412]

Answer:

The maximum vertical displacement is 2.07 meters.

Explanation:

We can solve this problem using energy. Since there is a frictional force acting on the block, we need to consider the work done by this force. So, the initial potential energy stored in the spring is transferred to the block and it starts to move upwards. Let's name the point at which the block leaves the ramp "1" and the highest point of its trajectory in the air "2". Then, we can say that:

$E_0=E_1\\\\U_e_0=K_1+U_g_1+W_f_1$

Where $U_e_0$ is the elastic potential energy stored in the spring, $K_1$ is the kinetic energy of the block at point 1, $U_g_1$ is the gravitational potential energy of the block at point 1, and $W_f_1$ is the work done by friction at point 1.

Now, rearranging the equation we obtain:

$\frac{1}{2}kx^{2}=\frac{1}{2}mv_1^{2}+mgh_1+\mu Ns_1$

Where $k$ is the spring constant, $x$ is the compression of the spring, $m$ is the mass of the block, $v_1$ is the speed at point 1, $g$ is the acceleration due to gravity, $h_1$ is the vertical height of the block at point 1, $\mu$ is the coefficient of kinetic friction, $N$ is the magnitude of the normal force and $s_1$ is the displacement of the block along the ramp to point 1.

Since the force is in an inclined plane, the normal force is equal to:

$N=mg\cos\theta$

Where $\theta$ is the angle of the ramp.

We can find the height $h_1$ using trigonometry:

$h_1=s_1\sin\theta$

Then, our equation becomes:

$\frac{1}{2}kx^{2}=\frac{1}{2}mv_1^{2}+mgs_1\sin\theta+\mu mgs_1\cos\theta\\\\\implies v_1=\sqrt{\frac{2(\frac{1}{2}kx^{2}-mgs_1\sin\theta-\mu mgs_1\cos\theta)}{m}}=\sqrt{\frac{kx^{2}}{m}-2gs_1(\sin\theta+\mu \cos\theta)}$

Plugging in the known values, we get:

$v_1=\sqrt{\frac{(15190N/m)(0.25m)^{2}}{10kg}-2(9.8m/s^{2})(1.12m)(\sin36.87\°+(0.300) \cos36.87\°)}\\\\v_1=8.75m/s$

Now, we can obtain the height from point 1 to point 2 using the kinematics equations. We care about the vertical axis, so first we calculate the vertical component of the velocity at point 1:

$v_1_y=v_1\sin\theta=(8.75m/s)\sin36.87\°=5.25m/s$

Now, we have:

$y=\frac{v_1_y^{2}}{2g}\\\\y=\frac{(5.25m/s)^{2}}{2(9.8m/s^{2})}\\\\y=1.40m$

Finally, the maximum vertical displacement $h_2$ is equal to the height $h_1$ plus the vertical displacement $y$ :

$h_2=h_1+y=s_1\sin\theta +y\\\\h_2=(1.12m)\sin36.87\°+1.40m\\\\h_2=2.07m$

It means that the maximum vertical displacement of the block after it becomes airborne is 2.07 meters.

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3 years ago