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3 years ago

How do magnetic forces repel or attract? Plz make it as simple as possible

1 answer:

8 0

In magnets the molecules are uniquely arranged so that their electrons spin in the same direction. This arrangement of atoms creates two poles in a magnet, a North-seeking pole and a South-seeking pole. The magnetic force in a magnet flows from the North pole to the South pole.so if there both south or both north the atoms can’t flow through causing them to stop flowing

Good luck

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The unit for measuring electric power is the

bazaltina [42]

C.) The measuring unit of "Electrical Power" is "Watt"

Hope this helps!

4 0

3 years ago

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$

Which has 2 solutions:

x = 0 (origin)

and

$tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft$

So, the distance d down the hill at which the shell strikes the hill is

$d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m$

In order to find how long the mortar shell remain in the air, we can use the equation:

$t=\frac{x}{u cos \theta}$

where:

x = 1322 ft is the final position of the shell when it strikes the hill

$u=156 ft/s$ is the initial velocity of the shell

$\theta=49.0^{\circ}$ is the angle of projection of the shell

Substituting these values into the equation, we find the time of flight of the shell:

$t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s$

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

$v_x = u cos \theta =(156)(cos 49^{\circ})=102.3 ft/s$

Instead, the vertical component of the velocity is given by

$v_y=usin \theta -gt$

And substituting at t = 12.9 s (time at which the shell strikes the hill),

$v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s$

Therefore, the final speed of the shell is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0

3 years ago

To a good approximate, the only external force that does work on a cyclist moving on level ground is the force of air resistance

mart [117]

Answer:

a. 120 W

b. 28.8 N

Explanation:

To a good approximate, the only external force that does work on a cyclist moving on level ground is the force of air resistance. Suppose a cyclist is traveling at 15 km/h on level ground. Assume he is using 480 W of metabolic power.

a. Estimate the amount of power he uses for forward motion.

b. How much force must he exert to overcome the force of air resistance?

(a) He is 25% efficient, therefore the cyclist will be expending 25% of his power to drive the bicycle forward

Power = efficiency X metabolic power

= 0.25 X 480

= 120 W

(b)

power if force times the velocity

P = Fv

convert 15 km/h to m/s

v = 15 kmph = 4.166 m/s

F = P/v

= 120/4.166

= 28.8 N

definition of terms

power is the rate at which work is done

force is that which changes a body's state of rest or uniform motion in a straight line

velocity is the change in displacement per unit time.

3 0

3 years ago

A capacitor consists of two square plates, 8.7 cm on a side, separated by a 2.0 mm air gap. How much energy would be stored in t

slavikrds [6]

Answer:

122.84 J

Explanation:

Since plate is square, area, A is given by $(8.7/100)^{2}=0.007569m^{2}$

The distance between plates, d, is given in the question as 2mm=0.002m

Charge on plate, Q, as given in the question is 240 $\mu c$

Assuming mica dielectric constant, k of 7

Capacitance, C is given by

C= $\frac {k\epsilon_{o}A}{d}=\frac {(7)(8.85*10^{-12})(0.007569)}{0.002}=2.34*10*^{-10}F$

Stored energy, E is given by

E= $\frac {Q^{2}}{2C}=\frac {(240*10^{-6})^{2}}{2*(2.34*10^{-10})}=122.84J$

Therefore, the stored energy is 122.84 J

5 0

3 years ago

Which of the following are results of the force of gravity?

docker41 [41]

Hello,

Here is your answer:

The proper answer for this question is option B "When released,a book falls to the ground". That's because of gravity the book will hit the ground!

Your answer is B.

If you need anymore help feel free to ask me!

Hope this helps.

7 0

4 years ago

Read 2 more answers