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3 years ago

A good way to de-magnetize something is to___.

Physics

2 answers:

pentagon [3]3 years ago

8 0

Answer:

C.) heat it.

Explanation:

Dominik [7]3 years ago

4 0

HEYA MATE

YOUR ANSWER IS D.PLACE IT IN AN ELECTRIC FIELD

BECAUSE IT makes sense you can use alternating current to remove magnetism

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Projectile SHOW WORK WILL MARK BRANLIEST (Draw Picture and Label)

m_a_m_a [10]

a) The horizontal distance covered by the projectile is 600 m

b) The projectile reaches its maximum height after 3.00 s

c) The altitude of the highest point is 44.1 m

Explanation:

The motion of the projectile consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

In this part A, we just need to analyze the horizontal motion. We know that:

The projectile travels horizontally with a constant velocity ov $v_x = 100 m/s$
The total time of flight of the projectile is $t=6.00 s$

Therefore, the horizontal distance covered by the projectile is given by

$x=v_x t$

And substituting, we find

$x=(100)(6.0) = 600 m$

For this part, we need to analyze the vertical motion of the projectile.

First, we want to find the initial vertical velocity. We can do it by using the equation for the vertical displacement:

$s=u_y t + \frac{1}{2}at^2$

where:

$u_y$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

t is the time

s is the vertical displacement

We know that the total time of flight is t = 6.00 s: this means that when t=6, the projectile returns to its initial vertical position, so s = 0. Substituting and solving for $u_y$ , we get

$u_y = - \frac{1}{2}at=-\frac{1}{2}(-9.8)(6)=29.4 m/s$

The vertical velocity then as a function of t is given by

$v_y = u_y + at$

And at the maximum height, it becomes zero: $v_y = 0$ . Substituting and solving for t, we find the time at which the projectile reaches the maximum height:

$t=-\frac{u_y}{a}=-\frac{29.4}{-9.8}=3.00 s$

To find the altitude of the highest point in the path, we use again the equation:

$s=u_y t + \frac{1}{2}at^2$

where

$u_y = 29.4 m/s$ is the initial vertical velocity

t = 3.00 s is the time at which the projectile reaches the highest point

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

Substituting the values, we find

$s=(29.4)(3.00)+\frac{1}{2}(-9.8)(3.00)^2=44.1 m$

So, the highest point is at 44.1 m above the ground.

Learn more about projectiles:

brainly.com/question/8751410

#LearnwithBrainly

8 0

3 years ago

Physics help, thank you guys so much!

katrin2010 [14]

Answer:

Δt = 5.85 s

Explanation:

For this exercise let's use Faraday's Law

emf = $- \frac{d \phi}{dt}$ - d fi / dt

$\phi$ = B. A

\phi = B A cos θ

The bold are vectors. It indicates that the area of the body is A = 0.046 m², the magnetic field B = 1.4 T, also iindicate that the normal to the area is parallel to the field, therefore the angle θ = 0 and cos 0 =1.

suppose a linear change of the magnetic field

emf = - A $\frac{B_f - B_o}{ \Delta t}$

Dt = - A $\frac{B_f - B_o}{emf}$

the final field before a fault is zero

let's calculate

Δt = - 0.046 (0- 1.4) / 0.011

Δt = 5.85 s

4 0

3 years ago

In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strike

zysi [14]

a) $t=\sqrt{\frac{2h}{g}}$

b) $v=\frac{d}{\sqrt{\frac{2h}{g}}}$

c) $v=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

d) $\theta=tan^{-1}(\frac{2h}{d})$ (radians)

Explanation:

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion (free fall) along the vertical direction

The time of flight of the cup is entirely determined by the vertical motion, therefore we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where here:

$s=h$ (the vertical displacement is the height of the counter)

$u=0$ (the initial vertical velocity of the cup is zero)

$a=g$ (the vertical acceleration is the acceleration of gravity)

Solving for t, we find the time of flight of the cup:

$h=0+\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}$

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

For a uniform motion, the speed is given by

$v=\frac{d}{t}$