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aev [14]
3 years ago
6

What electromagnetic waves are used in these applications?

Physics
2 answers:
evablogger [386]3 years ago
4 0

Answer:

Radio Waves

Microwaves

Infrared Waves

Explanation:

Just checked on edge.

andreyandreev [35.5K]3 years ago
3 0

Answer:

Diagnosing illnesses:  Radio Waves

Warming and cooking food:  microwaves

Transmitting data from remote controls to televisions: infrared waves

Explanation:

Radio waves are the only thing that can pass through skin harmlessly. microwaves (Duh. the machine called A MICROWAVE!!!!) and remotes have a red light on the front of them. Hope this helps. also how did you get your question to appear on google search?

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El. A horizontally directed force of 40 N is used to pull a box a distance of 2.5 m across a tabletop. How much work is done by
PIT_PIT [208]

The same formula of work can be applied to all the questions. The answers are:

E1. 100J

E2. 40N

E3. 5m

E4. a.) 360 J           b.) 240 J               c.) 120 J

El. If a horizontally directed force of 40 N is used to pull a box a distance of 2.5 m across a tabletop. The formula to get the much work that will be done by the 40 - N force will be

Work done = force x distance

Work done = 40 x 2.5

Work done = 100 J

E2. If a woman does 160 J of work to move a table 4 m across the floor. We will use the same formula to calculate the magnitude of the force that the woman applied to the table assuming the force is applied in the horizontal direction.

Work done = force x distance

160 = 4F

F = 160/4

Force F = 40 N

E3. Given that a force of 60 N used to push a chair across a room does 300 J of work. Same formula to get how far the chair move in this process.

Work done = force x distance

300 = 60 x distance

distance = 300/60

Distance = 5 m

Therefore, the chair moved 5m away.

E4. Given that a rope applies a horizontal force of 180 N to pull a crate a distance of 2 m across the floor. And a frictional force of 120 N opposes this motion.

a. The work done by the force applied by the rope can  be found by

W = F x S

W = 180 x 2

W = 360 J

b. What is the work done by the frictional force?

W = F_{r} x s

W = 120 x 2

W = 240 J

c. What is the total work done on the crate?

W = (F - F_{r}) x distance

Where  F_{r}  = frictional force

Substitute all the parameters

W = (180 - 120) x 2

W = 60 x 2

W = 120 J

Learn more about work here: brainly.com/question/8119756

4 0
2 years ago
Movies and TV shows sometimes portray a person being thrown backwards a sizable distance as a result of being struck by a bullet
MAXImum [283]

Answer:

 R = 4.24 x 10⁻⁴ m

Explanation:

given,

mass of the person = 60.3-kg

mass of the bullet = 10 gram = 0.01 Kg

velocity of bullet = 389 m/s

angle made with the horizontal = 45°

using conservation of momentum.

M v  + m u  = ( M + m ) V

60.3 x 0 + 0.01 x 389 = (60.3 + 0.01) V

V = \dfrac{3.89}{60.31}

V = \dfrac{3.89}{60.31}

V = 0.0645 m/s

for calculation of range

R = \dfrac{V^2sin 2 \theta}{g}

R = \dfrac{0.0645^2sin 2 (45^0)}{9.8}

     R = 4.24 x 10⁻⁴ m

the distance actor fall is  R = 4.24 x 10⁻⁴ m

6 0
3 years ago
A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c
saw5 [17]

Answer:

The final velocity of the thrower is \bf{3.88~m/s} and the final velocity of the catcher is \bf{0.029~m/s}.

Explanation:

Given:

The mass of the thrower, m_{t} = 70~Kg.

The mass of the catcher, m_{c} = 55~Kg.

The mass of the ball, m_{b} = 0.0480~Kg.

Initial velocity of the thrower, v_{it} = 3.90~m/s

Final velocity of the ball, v_{fb} = 33.5~m/s

Initial velocity of the catcher, v_{ic} = 0~m/s

Consider that the final velocity of the thrower is v_{ft}. From the conservation of momentum,

&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s

Consider that the final velocity of the catcher is v_{fc}. From the conservation of momentum,

&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s

Thus, the final velocity of thrower is 3.88~m/s and that for the catcher is 0.029~m/s.

8 0
2 years ago
Consider the circuit below, which is powered by a 8-v battery. switch s is opened at t = 0 after having been closed for a long t
GrogVix [38]
The battery will be full still a 8v bc of no time comparison
4 0
3 years ago
Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light
motikmotik

Answer:

n_cladding = 1.4764

Explanation:

We are told that θ_max = 5 °

Thus;

θ_max + θ_c = 90°

θ_c = 90° - θ_max

θ_c = 90° - 5°

θ_c = 85°

Now, critical angle is given by;

θ_c = sin^(-1) (n_cladding/n_core)

sin θ_c = (n_cladding/n_core)

n_cladding = (n_core) × sin θ_c

Plugging in the relevant values, we have;

n_cladding = 1.482 × sin 85

n_cladding = 1.4764

7 0
3 years ago
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