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3 years ago

Which chart correctly identifies the properties of electric and magnetic fields?

1 answer:

6 0

The correct answer for the question that is being presented above is this one: "D.
Property Electric Field Magnetic Field
<span>Can be produced by moving electric charge X </span>
Have two sides with opposite characteristics X X
Can make metals, such as nickel, iron, and cobalt, into magnets X X
<span>Can be turned on or off with a switch X"</span>

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A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra

Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

$sin(\alpha) = \frac{Vyi}{Vi}$

$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

$Vy(t) = Vyi -g.t$

where g is the gravity acceleration

$Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t$

This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

$Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}$

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

$Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s$

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

$Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} } .(0.675s)^{2} \\Y(0.675s) =2.236 m$

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

$Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m$

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0

4 years ago

If you want to calculate the acceleration of a car, you should _____ the change in velocity of the car by the time interval.

Neporo4naja [7]

Answer:

Your awnser is B. Increase

Explanation:

5 0

3 years ago

Read 2 more answers

How will unbalanced forces affect the speed and direction of an object

azamat

Answer:

Explanation:

Unbalanced forces will result in the presence of acceleration. The formula

F net = ma

says that if there is a net force present and the object in question has a mass, then an acceleration is present. Now acceleration is constant in this situation because nowhere does it say the acceleration is changing. If acceleration is constant then the velocity is increasing at a steady pace (think linear function!).

The direction of the object depends on the direction that the net force is in. If the net force is to the left, then that object will accelerate to the left.

Hope this helps :)

3 0

4 years ago

Physcs Questions from 8th grade....

scoray [572]

1). Yes. An object can be moving if the net force acting on the object is zero.
A non-zero net force is required to CHANGE the object's motion ... speed it up,
slow it down, or change its direction. But no force is required to keep it moving
at a constant speed in a straight line.

2).a). The motion of the box changes, from not moving to moving.
So the forces acting on it must be unbalanced.

2).b). The motion of the box doesn't change. It goes from not moving
to still not moving. So the forces acting on it must be balanced.

2).c). The motion of the box changes. Its speed decreases.
So the forces acting on it must be unbalanced.

4 0

3 years ago

A ball is thrown eastward into the air from the origin (in the direction of the positive x-axis). The initial velocity is 50 i +

nikklg [1K]

Answer:

The ball lands 154.3 ft from the origin at an angle of 13.6° from the eastern direction toward the south.

Explanation:

Hi there!

The position vector of the ball is described by the following equation:

r = (x0 + v0x · t + 1/2 · ax · t², y0 + v0y · t + 1/2 · ay · t², z0 + v0z · t + 1/2 · g · t²)

Where:

r = poisition vector of the ball at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity (eastward).

t = time.

ax = horizontal acceleration (eastward).

y0 = initial horizontal position.

v0y = initial horizontal velocity (southward).

ay = horizontal acceleration (southward)

z0 = initial vertical position.

v0z = initial vertical velocity.

g = acceleration due to gravity.

We have to find at which time the vertical component of the position vector is zero (the ball is on the ground) and then we can calculate the horizontal distance traveled by the ball at that time, using the equations of the horizontal components of the position vector.

Let´s place the origin of the system of reference at the throwing point so that x0 and y0 and z0 = 0.

y = z0 + v0z · t + 1/2 · g · t² (z0 = 0)

0 = 48 ft/s · t - 1/2 · 32 ft/s² · t²

0 = t (48 ft/s - 16 ft / s² · t) (t= 0, the origin point)

0 = 48 ft/s - 16 ft / s² · t

- 48 ft/s / -16 f/s² = t

t = 3.0 s

Now, we can calculate how much distance the ball traveled in that time.

First, let´s calculate the distance traveled in the eastward direction:

x = x0 + v0x · t + 1/2 · ax · t² (x0 = 0, ax = 0 there is no eastward acceleration)

x = 50 ft/s · 3 s

x = 150 ft

And now let´s calculate the distance traveled in southward direction:

y = y0 + v0y · t + 1/2 · ay · t² (y0 = 0 and v0y = 0, initially, the ball does not have a southward velocity).

y = 1/2 · ay · t²

y = 1/2 · (-8 ft/s²) · (3 s)²

y = -36 ft

Then, the final position vector will be:

r = (150 ft, -36 ft, 0)

The traveled distance is the magnitude of the position vector:

$|r| = \sqrt{(150ft)^{2} + (-36ft)^{2}} = 154.3 ft$

To calculate the angle, we have to use trigonometry (see attached figure):

cos angle = adjacent side / hypotenuse

cos α = x/r

cos α = 150 ft / 154.3 ft

α = 13.6°

The ball lands 154.3 ft from the origin at an angle of 13.5° from the eastern direction toward the south.

8 0

3 years ago