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4 years ago

Trade winds are found ________.

1 answer:

5 0

Selection (b) is the best answer. These are the winds found between the 25-degree north and south latitudes, converging at the doldrums just north of the equator. They have been used for centuries to navigate much more quickly around the oceans.

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X rays of wavelength 0.0100 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

Fofino [41]

Answer:

a) $=4.84*10^{-12}$

b) $= -2.76*10^{-14} J$

c) $i.e -2.76*10^{-14} J$

d)= 0 and the direction of motion is equal to zero

Explanation:

a) compton shift

$\Delta\lambda = \frac{h}{mc} (1-cos\theta)$

$\Delta\lambda = \frac{6.626*10^{-34}}{9.11*10^{-11}3*10^8} (1-cos180)$

$=4.84*10^{-12}$

b) the new wavelength

$\lambda' = 10.0*10^{-12} +4.84^10^{-12}$

$=14.84*10^{-12} m$

$\Delta E = E' - E$

$=hc[\frac{1}{\lambda'}-\frac{1}{\lambda}]$

$\Delta E = 6.626*10^{-34}*(3*10^8)[\frac{1}{14.84*10^{-12}}-\frac{1}{4.8*10^{-12}}]$

$= -2.76*10^{-14} J$

C)By conservation of energy, the kinetic energy of recoiling electron is equal to the magnitude of energy between the photon energy

$i.e -2.76*10^{-14} J$

d) the angle between the positive direction of motion

$sin\phy = \frac{\lambda_t sin\theta}{\lambda'}$

$=\frac{2.43*10^{-12}sin180}{14.84*10^{-12}}$

= 0

the direction of motion is equal to zero.

4 0

4 years ago

A sock stuck to the inside of the clothes dryer spins around the drum once every 2.0 s at a distance of 0.50 m from the center o

Rashid [163]

a) 1.57 m/s

The sock spins once every 2.0 seconds, so its period is

T = 2.0 s

Therefore, the angular velocity of the sock is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{2.0}=3.14 rad/s$

The linear speed of the sock is given by

$v=\omega r$

where

$\omega$ is the angular velocity

r = 0.50 m is the radius of the circular path of the sock

Substituting, we find:

$v=(3.14)(0.50)=1.57 m/s$

B) Faster

In this case, the drum is twice as wide, so the new radius of the circular path of the sock is twice the previous one:

$r' = 2r = 1.00 m$

At the same time, the drum spins at the same frequency as before, therefore the angular frequency as not changed:

$\omega' = \omega = 3.14 rad/s$

Therefore, the new linear speed would be:

$v'=\omega' r' = \omega (2r)$

And substituting,

$v'=(3.14)(1.00)=3.14 rad/s = 2v$

So, we see that the linear speed has doubled.

8 0

3 years ago

A gyroscope flywheel of radius 3.21 cm is accelerated from rest at 13.2 rad/s2 until its angular speed is 2450 rev/min.

SIZIF [17.4K]

Answer:

Part a)

$a_t = 0.423 m/s^2$

Part b)

$a_c = 2113 m/s^2$

Part c)

$d = 80 m$

Explanation:

Part a)

as we know that angular acceleration of the wheel is given as

$\alpha = 13.2 rad/s^2$

now the radius of the wheel is given as

R = 3.21 cm

so the tangential acceleration is given as

$a_t = R\alpha$

$a_t = (0.0321)(13.2)$

$a_t = 0.423 m/s^2$

Part b)

frequency of the wheel at maximum speed is given as

$f = 2450 rev/min$

$f = \frac{2450}{60} = 40.8 rev/s$

now we know that

$\omega = 2\pi f = 2\pi(40.8) = 256.56 rad/s$

now radial acceleration is given as

$a_c = \omega^2 r$

$a_c = (256.56)^2(0.0321) = 2113 m/s^2$

Part c)

total angular displacement of the point on rim is given as

$\Delta \theta = \omega_0 t + \frac{1}{2}\alpha t^2$

here we know that

$\omega = \omega_0 + \alpha t$

$256.56 = 0 + 13.2 t$

$t = 19.4 s$

now angular displacement will be

$\Delta \theta = 0 + \frac{1}{2}(13.2)(19.4)^2$

$\Delta \theta = 2493.3 rad$

now the distance moved by the point on the rim is given as

$d = R\theta$

$d = (0.0321)(2493.3)$

$d = 80 m$

7 0

3 years ago

The rate at which a candle burns in millimeters per minute is:

anzhelika [568]

since both components, length and time, are measurable
since Rate = length ÷ time 
∴ rate is also measurable and ∴ quantitative.

3 0

3 years ago

An air track car with a mass of 0.75 kg and a velocity of 8.5 m/s to the right collides elastically with a 0.65kg car moving to

Sunny_sXe [5.5K]

We can do this with the conservation of momentum. The fact it is elastic means no KE is lost so we don't have to worry about the loss due to sound energy etc.

Firstly, let's calculate the momentum of both objects using p=mv:

Object 1:
p = 0.75 x 8.5 = 6.375 kgm/s

Object 2 (we will make this one negative as it is travelling in the opposite direction):
p = 0.65 x -(7.2) = -4.68 kgm/s

Based on this we know that the momentum is going to be in the direction of object one, and will be 6.375-4.68=1.695 kgm/s

Substituting this into p=mv again:

1.695 = (0.75+0.65) x v
Note I assume here the objects stick together, it doesn't specify - it should!

1.695 = 1.4v
v=1.695/1.4 = 1.2 m/s to the right (to 2sf)

8 0

3 years ago