Gayle cooks in her micwave

EM waves consist of changing electric and magnetic fields moving perpendicular with respect to each other. What kind of wave is

Serhud [2]

Answer:

Transverse

Explanation:

Electromagnetic waves don't depend on the medium they travel through like a mechanical wave does, so they aren't mechanical. They don't oscillate (move back in forth) in the direction they travel either, ruling out compressional and longitudinal waves.

That leaves tranverse waves, the ones we're most used to, since they look very "wavelike," with smooth peaks and valleys. Electromagnic waves behave like these, oscillating in a plane perpendicular to the direction they're traveling in.

5 0

3 years ago

A zone reconnaissance involves a directed effort to obtain detailed information on all routes, obstacles, terrain, enemy forces,

son4ous [18]

Answer:

It is True

Explanation:.

A commander assigns a zone reconnaissance mission when he seeks additional information on a zone before committing other forces in the zone. It is appropriate when the enemy situation is vague, existing knowledge of the terrain is limited, or combat operations have altered the terrain. A zone reconnaissance could include several route or area reconnaissance missions assigned to subordinate units.

7 0

3 years ago

Define impulse and momentum.<br>No spam

Doss [256]

Impulse: a certain amount of force you apply for an amount of time.

Impulse: F*t where F= Force & t=time

Momentum: increasing forward motion.

A ball rolling down a slide gains momentum

p=mv where m=mass and v=velocity

Hope it helps!

~Just an emotional teen who listens to music

4 0

2 years ago

Use the impulse-momentum theorem to find how long a stone falling straight down takes to increase its speed from 3.7 m/s to 9.90

11111nata11111 [884]

<h2>Time taken is 0.632 seconds</h2>

Explanation:

Impulse momentum theorem is change in momentum is impulse.

Change in momentum = Impulse

Final momentum - Initial momentum = Impulse

Mass x Final velocity - Mass x Initial Velocity = Force x Time

Mass x Final velocity - Mass x Initial Velocity =Mass x Acceleration x Time

Final velocity - Initial Velocity = Acceleration x Time

Final velocity = 9.9 m/s

Initial Velocity = 3.7 m/s

Acceleration = 9.81 m/s²

Substituting

9.9 - 3.7 = 9.81 x Time

Time = 0.632 seconds

Time taken is 0.632 seconds

8 0

3 years ago

1. A student lifts a box of books that weighs 185 N. The box is

aksik [14]

1) 148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

$W = \Delta U = (mg) \Delta h$

where

mg is the weight of the object

$\Delta h$ is the change in height

For the box in this problem,

mg = 185 N

$\Delta h = 0.800 m$

Substituting into the equation, we find:

$W=(185)(0.800)=148 J$

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

$W=Fd$

where

F is the magnitude of the force

d is the displacement

In this problem,

F = 825 N is the force applied by the two students together

d = 35 m is the displacement of the car

Substituting,

$W=(825)(35)=28875 J$

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

$W=Fd$

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

The work done previously was

W = 28875 J

Now the force is doubled, so the new work done will be

$W' = 2(28875)=57750 J$

3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

$F = mg$

where

m = 0.180 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

Solving the equation,

$F=(0.180)(9.8)=1.76 N$

Now we find the work done by gravity using the same formula applied before:

$W=Fd$

where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,

$W=(1.76)(2.5)=4.4 J$

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

$\Delta h = 1.2 m$

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

$W=mg \Delta h$

Solving for m, we find

$m=\frac{W}{g \Delta h}$

And substituting the numerical values, we find the mass of the box:

$m=\frac{7000}{(9.8)(1.2)}=595.2 kg$

5) They do the same work

In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:

$W=mg \Delta h$

Where $\Delta h$ is the difference in height between the final position and the initial position of the box.

This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of $\Delta h$ is the same for both of them, so the work they have done is exactly the same.

5 0

3 years ago