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4 years ago

15

Susan is making an electromagnet in her science class today. First, she takes a nail and winds coils of copper wire around it th

en, she
connects the ends of the wire to a battery. She uses the electromagnet to pick up paper clips, but the paper clips drop when she
disconnects the wires from the battery, Which kind of energy does Susan change into magnetic energy in this project
chemical
electrical
heat
light

2 answers:

matrenka [14]4 years ago

5 0

Answer:

Electrical

Explanation:

She uses a battery, which is electrical.

It doesn't operate using chemicals, heat, or light

Pie4 years ago

4 0

Answer:

it's chemical and electrical because she uses her circuit in which the both forms take place...........so according to me the most probably the answer is electrical.....

Explanation:

because she uses her circuit in which the both forms take place...........so according to me the most probably the answer is electrical.....

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3 years ago

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Answer:

a) v= 1732.05m/s

b) d=250000m

c) v= 1414.214m/s

Explanation:

Notation

M= mass of the asteroid

m= mass of the particle moving upward

R= radius

v= escape speed

G= Universal constant

h= distance above the the surface

Part a

For this part we can use the principle of conservation of energy. for the begin the initial potential energy for the asteroid would be $U_i =-\frac{GMm}{R}$ .

The initial kinetic energy would be $\frac{1}{2}mv^2$ . The assumption here is that the particle escapes only if is infinetely far from the asteroid. And other assumption required is that the final potential and kinetic energy are both zero. Applying these we have:

$-\frac{GMm}{R}+\frac{1}{2}mv^2=0$ (1)

Dividing both sides by m and replacing $\frac{GM}{R}$ by $a_g R$

And the equation (1) becomes:

$-a_g R+\frac{1}{2} v^2=0$ (2)

If we solve for v we got this:

$v=\sqrt{2 a_g R}=\sqrt{2x3\frac{m}{s^2}x500000m}=1732.05m/s$

Part b

When we consider a particule at this surface at the starting point we have that:

$U_i=-\frac{GMm}{R}$

$K_i=\frac{1}{2}mv^2$

Considering that the particle is at a distance h above the surface and then stops we have that:

$U_f=-\frac{GMm}{R+h}$

$K_f=0$

And the balance of energy would be:

$-\frac{GMm}{R}+\frac{1}{2}mv^2 =-\frac{GMm}{R+h}$

Dividing again both sides by m and replacing $\frac{GM}{R}$ by $a_g R^2$ we got:

$-a_g R+\frac{1}{2}v^2 =-\frac{a_g R^2}{R+h}$

If we solve for h we can follow the following steps:

$R+h=-\frac{a_g R^2}{-a_g R+\frac{1}{2}v^2}$

And subtracting R on both sides and multiplying by 2 in the fraction part and reordering terms:

$h=\frac{2a_g R^2}{2a_g R-v^2}-R$

Replacing:

$h=\frac{2x3\frac{m}{s^2}(500000m)^2}{2(3\frac{m}{s^2})(500000m)-(1000m/s)^2}- 500000m=250000m$

Part c

For this part we assume that the particle is a distance h above the surface at the begin and start with 0 velocity so then:

$U_i=-\frac{GMm}{R+h}$

$K_i=0$

And after the particle reach the asteroid we have this:

$U_f=-\frac{GMm}{R}$

$K_f=\frac{1}{2}mv^2$

So the balance of energy would be:

$-\frac{GMm}{R+h}=-\frac{GMm}{R}+\frac{1}{2}mv^2$

Replacing again $a_g R^2$ instead of GM and dividing both sides by m we have:

$-\frac{a_g R^2}{R+h}=-a_g R+\frac{1}{2}v^2$

And solving for v:

$a_g R-\frac{a_g R^2}{R+h}=\frac{1}{2}v^2$

Multiplying both sides by two and taking square root:

$v=\sqrt{2a_g R-\frac{2a_g R^2}{R+h}}$

Replacing

$v=\sqrt{2(3\frac{m}{s^2})(500000m)-\frac{2(3\frac{m}{s^2}(500000m)^2}{500000+1000000m}}=1414.214m/s$

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Https://courses.physics.illinois.edu/phys102/sp2013/lectures/lecture2.pdf

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