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2 years ago

If an electromagnetic wave has components Ey = E0 sin(kx - ωt) and Bz = B0 sin(kx - ωt), in what direction is it traveling?

Physics

1 answer:

fomenos2 years ago

8 0

Answer:

Its traveling in the +x direction

Explanation:

The E-field is in the +y-direction, and the B-field is in the +z-direction, so it must be moving along the +x-direction, since the E-field, B-field and the direction of moving are all at right angles to each other.

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What is the instantaneous velocity of a freely falling object 9.0 s after it is released from a position of rest? Express your a

Umnica [9.8K]

Answer:

So instantaneous velocity after 9 sec will be 88.2 m/sec

Explanation:

We have given time t = 9 sec

As the object is released from rest so its initial velocity u = 0 m/sec

We have to find its final velocity v

Acceleration due to gravity $g=9.8m/sec^2$

From first equation of motion we know that $v=u+gt$

$v=0+9.8\times 9=88.2m/sec$

So instantaneous velocity after 9 sec will be 88.2 m/sec

5 0

2 years ago

Consider a simple ideal Rankine cycle with fixed turbine inlet conditions. What is the effect of lowering the condenser pressure

mr Goodwill [35]

Answer:

The effect of lowering the condenser pressure on different parameters is explained below.

Explanation:

The simple ideal Rankine cycle is shown in figure.

Effect of lowering the condenser pressure on

(a). Pump work input :- By lowering the condenser pressure the pump work increased.

(b) Turbine work output :- By lowering the condenser pressure the turbine work increased.

(c). Heat supplied :- Heat supplied increases.

(d). Heat rejected :- The heat rejected may increased or decreased.

(e). Efficiency :- Cycle efficiency is increased.

(f). Moisture content at turbine exit :- Moisture content increases.

8 0

1 year ago

Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0

2 years ago

The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f

zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

584 = Ha + Lg

an also this mean that torques acting on the body is balanced

So, 0.410 Ha = 0.840 Lg

Making Lg the subject of formula in the equation above we

Lg = 0.4881 Ha

Considering the first equation and replacing Lg with this recent equation we have

584 = Ha + 0.4881 Ha

Therefore Ha = 392.44 N

This value obtained is for both hands for each hand we divide by 2

Therefore we have for each hand $= 392.44/2$ =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

Lg = 0.4881 × 392.44

= 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

191.22/2 = 95.8 N.

8 0

2 years ago

A 900 kg steel beam is supported by the two ropes shown in (Figure 1) . Calculate the tension in the rope.

Rzqust [24]

Let T1 and T2 be tension in ropes1 and 2 respectively.
since system is stationary (equilibrium), considering both ropes + beam as a system 

for horizontal equilibrium (no movement in that direction, so resultant force must be zero horizontally) 
T1sin(20) = T2sin(30) 
=> T1 = T2sin(30) / sin(20) 

for vertical equilibrium, (no movement in this direction, so resultant force must be zero vertically) 
T1cos(20) + T2cos(30) = mg 

m = 900kg, substituting for T1 
T2sin(30)*cos(20)/sin(20) + T2cos(30) = 900g 
2.328*T2 = 900*9.8 
T2 = 3788.65N 
so T1 from (1) 
T1 = 5535.21N

8 0

2 years ago