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3 years ago

Which electromagnetic wave enables us to see objects?

2 answers:

4 0

I believe the correct answer from the choices listed above is option B. It is the visible light that enables us to see objects. A typical human eye will respond to wavelengths from about 390 to 700 nm which is the visible light. Hope this answers the question.

Valentin [98]3 years ago

3 0

Choose visible light

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A 0.26 kg rod of length 80 cm is suspended by a frictionless pivot at one end. It is held horizontal and released.

Daniel [21]

Answer:

a) $a_{center} = 7.38 ~m/s^2$

b) $a_{end} = 14.77 ~m/s^2$

c) $v_{center} = 2.43~m/s$

Explanation:

a) Immediately after the rod is released, the rod is still horizontal but now subject to gravity. Since one end of the rod is fixed, then the weight of the rod applies a torque. Then by Newton's Second Law, the acceleration can be found.

$\tau =I\alpha$

where I is the moment of inertia of the rod with respect to its fixed end, and α is the angular acceleration.

The net torque of the rod is

$\vec{\tau} = \vec{r} \times \vec{F}\\\tau = rF\sin(90) = rF$

where r is the distance from center of the mass to the fixed end, so r = 0.4 m.

The weight of the rod is w = mg = 0.26 x 9.8 = 2.54 N.

So the net torque is τ = 1.01 Nm.

The moment of inertia of the rod is

$I = \frac{1}{3}mL^2 = \frac{1}{3}(0.26)(0.8)^2 = 0.055~kg m^2$

So, the Newton's Second Law yields

$\tau = I\alpha\\\alpha = \frac{\tau}{I} = \frac{1.01}{0.055} = 18.47$

The relation between angular acceleration and linear acceleration is a = αr

So, the linear acceleration of the rod is

$a = \alpha r = 7.38~m/s^2$

b) Using the same relationship between angular acceleration and linear acceleration, the linear acceleration of the end of the rod can be found.

$a = \alpha L = 14.77~m/s^2$

c) The conservation of energy can be used to find the velocity when the rod is vertical.

$K_1 + U_1 = K_2 + U_2\\0 + mg(L/2) = \frac{1}{2}I\omega^2 + 0\\(0.26)(9.8)(0.4) = \frac{1}{2}(0.055)\omega^2\\\omega = 6.08~rad/s$

The linear velocity is v = ωr, so

v = 2.43 m/s.

4 0

3 years ago

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P

Natasha_Volkova [10]

Answer

given,

distance of first satellite = 48,000 Km

distance of second satellite = 64,000 Km

orbital period = 6.39 day

Using equation of time period

$T = \dfrac{2\pi r^{3/2}}{\sqrt{Gm_{pluto}}}$

now, from the above equation we can say that only variable is Time period and r is the radii of orbit.

from the first satellite

$\dfrac{T_{charon}}{r^{3/2}_{charon}}=\dfrac{T_{sat1}}{r^{3/2}_{sat1}}$

$T_{sat1}=\dfrac{T_{charon}\ r^{3/2}}{r^{3/2}_{charon}}$

$T_{sat1}=\dfrac{6.39\times (48000)^{3/2}}{19600^{3/2}}$

$T_{sat1}=24.5\ days$

for second satellite

$T_{sat2}=\dfrac{T_{charon}\ r^{3/2}_{sat2}}{r^{3/2}_{charon}}$

$T_{sat1}=\dfrac{6.39\times (64000)^{3/2}}{19600^{3/2}}$

$T_{sat1}=37.7\ days$

7 0

3 years ago

Through the process of transpiration, plants do what?

Evgesh-ka [11]

Well, the process of TRANSPIRATION deals with the release of water to the environment by plants, it is mostly in the form of vapour and is released into the atmosphere. Hope i helped, Have a nice day

8 0

3 years ago

Which of these is not a proper hand washing step

aleksandrvk [35]

Which of what? Apologies please list the following so I can help.

8 0

3 years ago

The images output from your new color laser printer seem to be a little too blue. What can you do to fix this?

Svetradugi [14.3K]

Answer:

The images output from your new color laser printer seem to be a little too blue. to fix this problem we need to calibrate the printer.

Explanation:

This can be done by opening the toolbox, clicking in the device setting folder their you get print quality page click on it. Under the print quality option click on the calibrate next to calibrate now. Then click OK unless when the 'your request has been sent to the device' appears on the screen. When the calibration ends again try printing. calibrating is useful for managing the proper alignment of the inkjet cartridge nozzle to the paper and each other, without proper calibration the print quality deteriorates.

3 0

3 years ago