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KiRa [710]
3 years ago
13

Which of the following is NOT a type of electromagnetic wave

Physics
2 answers:
Mars2501 [29]3 years ago
6 0

Answer:

where are the options so we can answer

(free point thanks)

Explanation:

shtirl [24]3 years ago
4 0

Answer:

Where's the options?

Explanation:

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Which term is defined as the change in the direction of light when it goes from one medium into a different medium?
love history [14]

Answer:

refraction

Explanation:

3 0
2 years ago
Joe was asked to describe the movements the Earth makes as it rotates on its axis and revolves around the sur
____ [38]

Answer:

A:

Explanation:

The earth rotates on it's axis every 24 hours not revolves.

Imagine rotation as a top spinning on the ground. revolving means if you tied a string to the top and spun it around your body in a circle. (hope this helps:)

7 0
4 years ago
Read 2 more answers
HELPPP ASAP!!!!!!
fomenos

Answer:

The acceleration of the player is - 4.9 m/s²

Explanation:

The given is:

1. The mass of the player is 55 kg

2. His initial speed is 4.6 m/s

3. The coefficient of the kinetic fraction between the player and the

    ground is 0.50

We need to find the player acceleration

According to Newton's Law

→ ∑ forces in direction of motion = mass × acceleration

There is only the friction force opposite to the motion

→ Friction force = μR

where μ is the coefficient of friction and R is the normal reaction

→ The normal reaction R = mg

where m is the mass and g is the acceleration of gravity

→ m = 55 kg , g = 9.8 m/s²

→ R = 55 × 9.8 = 539 N

→ ∑ F = - μR

→ - μR = m × a

→ μ = 0.5 , R = 539 N , m = 55

→ -(0.5)(539) = 55 × a

→ - 269.5 = 55 a

Divide both sides by 55

→ a = - 4.9 m/s²

The acceleration of the player is - 4.9 m/s²

Learn more:

You can learn more about Newton's law in brainly.com/question/11911194

#LearnwithBrainly

3 0
4 years ago
A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0∘above the horizontal by a rope exerting a 72.0-N pull
Elden [556K]

Answer:

(A) 374.4 J

(B) -332.8 J

(C) 0 J

(D) 41.6 J

(E)  351.8 J

Explanation:

weight of carton (w) = 128 N

angle of inclination (θ) = 30 degrees

force (f) = 72 N

distance (s) = 5.2 m

(A) calculate the work done by the rope

  • work done = force x distance x cos θ
  • since the rope is parallel to the ramp the angle between the rope and

        the ramp θ will be 0

       work done = 72 x 5.2 x cos 0

       work done by the rope = 374.4 J

(B) calculate the work done by gravity

  • the work done by gravity = weight of carton x distance x cos θ
  • The weight of the carton = force exerted by the mass of the carton = m x g
  • the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.

      work done by gravity = 128 x 5.2  x cos 120

      work done by gravity = -332.8 J

(C) find the work done by the normal force acting on the ramp

  • work done by the normal force = force x distance x cos θ
  • the angle between the normal force and the ramp is 90 degrees

       

         work done by the normal force = Fn x distance x cos θ

         work done by the normal force = Fn x 5.2 x cos 90

         work done by the normal force = Fn x 5.2 x 0

         work done by the normal force = 0 J

(D)  what is the net work done ?

  • The net work done is the addition of the work done by the rope,       gravitational force and the normal force

     net work done = 374.4 - 332.8 + 0 =  41.6 J  

(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal

  • work done by the rope= force x distance x cos θ
  • the angle of inclination will be 50 - 30 = 20 degrees, this is because the ramp is inclined at 30 degrees to the horizontal and the rope is inclined at 50 degrees to the horizontal and it is the angle of inclination of the rope with respect to the ramp we require to get the work done by the rope in pulling the carton on the ramp

work done = 72 x 5.2 x cos 20

work done = 351.8 J

5 0
4 years ago
The Tevatron acceleator at the Fermi National Accelerator Laboratory (Fermilab) outside Chicago boosts protons to 1 TeV (1000 Ge
Eva8 [605]

Answer:

a) v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s

b) v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s

c) v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s

d) v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s

e) v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s

Explanation:

At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.

KE=mc^{2}(\gamma -1) = mc^{2}(\frac{1}{\sqrt{1-\beta^{2}}} -1)           (1)

Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.

Let's solve (1) for β.

\beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}}

We can write the mass of a proton in MeV/c².

m_{p}=938.28 MeV/c^{2}

Now we can calculate the speed in each stage.

a) Cockcroft-Walton (750 keV)

\beta = \sqrt{1-\frac{1}{\left (\frac{0.75 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.04

v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s

b) Linac (400 MeV)

\beta = \sqrt{1-\frac{1}{\left (\frac{400 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.71

v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s

c) Booster (8 GeV)

\beta = \sqrt{1-\frac{1}{\left (\frac{8000 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.994

v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s

d) Main ring or injector (150 Gev)

\beta = \sqrt{1-\frac{1}{\left (\frac{150000 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.999

v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s

e) Tevatron (1 TeV)

\beta = \sqrt{1-\frac{1}{\left (\frac{1000000 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.9999

v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s

Have a nice day!

4 0
4 years ago
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