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3 years ago

Which statement is true about the part of the electromagnetic spectrum that human eyes can detect?

Physics

2 answers:

lesantik [10]3 years ago

5 0

I believe its B. it contains only the colors of the rainbow and TV waves
Hope it helped

storchak [24]3 years ago

5 0

Answer:

The correct answer is D) It is divided into seven ranges of wavelengths.

Explanation:

I took the test

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The Earth moving round the Sun in a circular orbit is acted upon by a

zalisa [80]

No I do not agree. It is because work is done when force acting o a body displace or covers certain displacement in the direction of force applied .

3 0

3 years ago

A torque of 50 N*m acts on a wheel of moment of inertia 25 kg*m^2 for 4 s and then is removed.

motikmotik

You have to find the calculate the circumference first then you can just multiply the diameter by π, which is about 3.142. That gives you the distance for each revolution. Then you can multiply by the number of revolutions per minute.

3 0

3 years ago

A uniform cubical crate is 0.740 m on each side and weighs 600 N. It rests on the floor with one edge against a very small, fixe

Sindrei [870]

Answer:

H = 0.673

Explanation:

given,

side of cubical crate = 0.74

weight of the crate = 600 N

magnitude of force = 330 N

the Horizontal distance of its Center of mass

= 0.74/2

= 0.37

Let the required Height be H

By Balancing the Torques, we get

H x 330 N = 0.37 x 600

330 H = 222

H = 0.673

hence, the height above the floor where force is acting is equal to 0.673 m

6 0

3 years ago

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

2 years ago

A 13-kg sled is moving at a speed of 3.0 m/s. At which of the following speeds will the sled have twice as much kinetic energy?

AysviL [449]

K.E. = 1/2 mv²
K.E. is directly proportional to v^2
So, when K.E. increase by 2, K.E. increase by root. 2
v' = 1.41v
original v value was 3 so, final would be:
v' = 1.41*3 = 4.23
After round-off to it's tenth value, it will be:
v' = 4.2

So, option B is your answer!

Hope this helps!

7 0

3 years ago