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VARVARA [1.3K]
3 years ago
11

What happens to white light when it passes through a prism

Physics
1 answer:
love history [14]3 years ago
6 0

Answer:

Refraction splits the light into its component wavelengths, and you see a rainbow.

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A satellite system that has unique properties is the ____ satellite, which is used by governments for spying and by scientific a
agasfer [191]

Answer: Highly-elliptical-earth-orbit (heo)

Explanation: Highly-elliptical-earth-orbit (heo) satellite system has unique properties

which is used by governments for spying and by scientific agencies for observing celestial bodies. It is an extremely elongated orbit that is useful for communication satellites which creates signals between a source transmitter and a receiver at different locations on earth. They are used by government for spying, scientific agencies for observing celestial bodies, for the internet and telephone communications.

6 0
3 years ago
Read 2 more answers
A train locomotive is pulling two cars of the same mass behind it. Determine the ratio of the tension in the coupling (think of
Anna007 [38]

Answer:

The ratio is  \frac{F_{T1}}{F_{T2}}  =  2

Explanation:

The diagram for this question is shown on the first uploaded image

Here we are assume the acceleration of the train is a

which makes the acceleration of each car a

From the question we are told that

      Considering the second car

 The force causing it s movement  is mathematically represented as

       F_{T2} =  ma

 Considering the first car

 The force causing it s movement  is mathematically represented as

      F  = F_{T1} -F_{T2} = ma

=>   F_{T1} -ma  = ma

=>   F_{T1} =  2 ma

=> \frac{F_{T1}}{ma}  =  2

=> \frac{F_{T1}}{F_{T2}}  =  2

7 0
3 years ago
A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of th
Yuri [45]

Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

E=-\frac{d}{dt}(BACOS\alpha )\\

where B is the magnitude of the field and A is the area of the circular coil.

First, let determine the area using \pi r^{2} \\ where r is the radius of 5cm or 0.05m

A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\

since we no that the angle is at 0^{0}

we determine the magnitude of the magnetic filed

B=0.5e^{-t} \\t=2s

E=-(0.5e^{-2} * 0.00785)

E=-0.000532v\\

the Magnitude of the voltage is 0.000532V

Next we determine the current using ohm's law

V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A

I=2.6mA

6 0
3 years ago
1)Determine, in terms of unit vectors, the resultant of the five forces illustrated in the figure, Consider F1=20 N, F2= 12 N, F
LiRa [457]

Explanation:

1) F₁ lies in a plane perpendicular to the xy plane, 60° from the x axis.  The angle between F₁ and the +z axis is 30°.  Therefore, the vector is:

<F₁> = 20 (sin 30° cos 60° i + sin 30° sin 60° j + cos 30° k)

<F₁> = 20 (¼ i + ¼√3 j + ½√3 k)

<F₁> = 5 i + 5√3 j + 10√3 k

F₂ is in the xy plane.  Its slope is -24/7.  The vector is:

<F₂> = 12 (-⁷/₂₅ i + ²⁴/₂₅ j + 0 k)

<F₂> = -3.36 i + 11.52 j

F₃ is parallel to the +x axis.  The vector is:

<F₃> = 17 (i + 0 j + 0 k)

<F₃> = 17 i

F₄ is parallel to the -z axis.  The vector is:

<F₄> = 15 (0 i + 0 j − k)

<F₄> = -15 k

F₅ is in the xy plane.  It forms a 15° angle with the -y axis.  The vector is:

<F₅> = 9 (-sin 15° i − cos 15° j + 0 k)

<F₅> = -9 sin 15° i − 9 cos 15° j

The resultant vector is therefore:

<F> = (5 − 3.36 + 17 − 9 sin 15°) i + (5√3 + 11.52 − 9 cos 15°) j + (10√3 − 15) k

<F> = 16.31 i + 11.49 j + 2.32 k

2) Sum of forces at point B in the x direction:

∑F = ma

Tbc cos 40° − ¹⁵/₁₇ Tab = 0

Tbc cos 40° = ¹⁵/₁₇ Tab

Tbc = 1.15 Tab

Sum of forces at point B in the y direction:

∑F = ma

Tbc sin 40° + ⁸/₁₇ Tab − mAg = 0

Tbc sin 40° + ⁸/₁₇ Tab = (2 kg) (10 m/s²)

(1.15 Tab) sin 40° + ⁸/₁₇ Tab = 20 N

1.21 Tab = 20 N

Tab = 16.52 N

Tbc = 19.02 N

Sum of forces at point C in the x direction:

∑F = ma

Tcd sin 25° − Tbc cos 40° = 0

Tcd sin 25° = Tbc cos 40°

Tcd = 1.81 Tbc

Tcd = 34.48 N

3(a) When the crane is on the verge of tipping, the center of gravity is directly over point F.  Relative to point A:

3.7 m = [ (390 kg) (0.9 m) + (90 kg) (9 m cos θ + 1.7 m) + (80 kg) (9 m cos θ + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + 810 kgm cos θ + 153 kgm + 720 kgm cos θ + 232 kgm

1336 kgm = 1530 kgm cos θ

θ = 29.17°

3(b) 3.7 m = [ (390 kg) (0.9 m) + (90 kg) (x + 1.7 m) + (80 kg) (x + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + (90 kg) x + 153 kgm + (80 kg) x + 232 kgm

1336 kgm = (170 kg) x

x = 7.86 m

4) Find the lengths of the cables.

Lab = √((2 m)² + (3 m)² + (5 m)²)

Lab = √38 m

Lac = √((2 m)² + (3 m)² + (5 m)²)

Lac = √38 m

Lde = √((2 m)² + (3 m)²)

Lde = √13 m

Sum of forces in the x direction:

∑F = ma

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Sum of forces in the y direction:

∑F = ma

2/√38 Fab − 2/√38 Fac = 0

Fab = Fac

Sum of forces in the z direction:

∑F = ma

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

Sum of moments about the y-axis:

∑τ = Iα

(3/√38 Fab) (5 m) + (3/√38 Fac) (5 m) + (3/√13 Fde) (2 m) − (mg) (2 m) = 0

Substitute Fab = Fac and simplify:

6/√38 Fab + 3/√13 Fde − mg = 0

30/√38 Fab + 6/√13 Fde − 2mg = 0

Double first equation:

12/√38 Fab + 6/√13 Fde − 2mg = 0

Subtract from the second equation:

28/√38 Fab = 0

Fab = 0

Fac = 0

Solve for Fde:

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

3/√13 Fde = mg

3/√13 Fde = (1.7 kg) (10 m/s²)

Fde = 20.43 N

Solve for Rx:

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Rx = 2/√13 Fde

Rx = 11.33 N

8 0
3 years ago
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D. distance

A light-year is the distance light would travel in 1 year.
6 0
3 years ago
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