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3 years ago

12

Help me?!?!?!?!?!?!?!?!?!?!?!?!?!

2 answers:

fredd [130]3 years ago

8 0

Before you can pick the correct one, you have to review in your mind what
you know about the electric field. (Or what you SHOULD know about it.)

-- The field lines show where a teeny tiny POSITIVE charge
wants to go when you put it down anywhere.

-- A teeny tiny POSITIVE charge always wants to get AWAY from other
positive charges, and it wants to go TOWARD negative charges.

That's all you need. Now let's look at the pictures.
I want to look at them from the bottom up:

(D). ... No.
The arrows show that the tiny positive test charge wants to go AWAY from
the negative (blue) charge and TOWARD the positive (red) charge.
That's wrong.

(C). ... No.
The arrows show that the tiny positive test charge wants to go AWAY from
any negative charges. That's wrong.

(B). ... No.
The arrows show that the tiny positive test charge wants to go AWAY from
BOTH the negative charge AND the positive charge. That's wrong.

(A). ... YES !
The arrows show that the tiny test charge wants to go away from the positive
charge, and toward the negative charge. That's exactly right.

(That's why I wanted to do them backwards ... because
the first picture on the page is the correct one.)

marishachu [46]3 years ago

3 0

The correct field line would be A.

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It’s 4 because a coiled springs is closely spaced then widen

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How to represent milligram in kilogram by standard formula?

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Answer:

0.000001 kg

Explanation:

because 1 kg equal 1,000,000 milligrams

we take $\frac{1}{1,000,000}$ which equals 0.000001 kg

4 0

3 years ago

Help please, I need it

love history [14]

Answer:

1.97×10⁻²¹ J

Explanation:

Use ideal gas law to find temperature.

PV = nRT

(9 atm) (9 L) = (83.3 mol) (0.0821 L·atm/mol/K) T

T = 11.9 K

The average kinetic energy per atom is:

KE = 3/2 kT

KE = 3/2 (1.38×10⁻²³ J/K) (11.9 K)

KE = 2.46×10⁻²² J

For a mass of 5.34×10⁻²⁶ kg, the kinetic energy is:

KE = (5.34×10⁻²⁶ kg) (1 mol / 0.004 kg) (6.02×10²³ atom/mol) (2.46×10⁻²² J)

KE = 1.97×10⁻²¹ J

5 0

3 years ago

mario62 [17]

It's D. If it absorbed it would be turning to steam. I am taking honors chem in high school we are learning this.

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3 years ago

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A 3500 kg spaceship is in a circular orbit 220 km above the surface of Earth. It needs to be moved into a higher orbit of 360 km

Vitek1552 [10]

Answer:

The work done is calculated as $0.044\times 10^{11}\ J$

Solution:

As per the question:

Mass of the spaceship, m = 3500 kg

Height of the orbit, h = 220 km

Mass of the earth, $M_{e} = 5.97\times 10^{24}\ kg$

Height of the higher orbit, h' = 360 km

Radius of the orbit, $R = 6.37\times 10^{6}\ km$

Now,

The work done is given by the change in the gravitational potential energy:

Gravitational Potential Energy, U = $- \frac{GM_{e}m}{R + h}$

Now, for the lower orbit:

$U_{l} = - \frac{GM_{e}m}{R + h}$

$U_{l} = - \frac{6.67\times 10^{- 11}\times 5.97\times 10^{24}\times 3500}{6.37\times 10^{6} + 220\times 10^{3}}$

$U_{l} = - 2.115\times 10^{11}\ J$

For upper orbit:

$U_{U} = - \frac{GM_{e}m}{R + h}$

$U_{u} = - \frac{6.67\times 10^{- 11}\times 5.97\times 10^{24}\times 3500}{6.37\times 10^{6} + 360\times 10^{3}}$

$U_{U} = - 2.071\times 10^{11}\ J$

Change in the gravitational Potential energy:

$\Delta U = U_{U} - U_{l} = - 2.071\times 10^{11} - (- 2.115\times 10^{11}) = 0.044\times 10^{11}\ J$

Therefore, the work done:

$W = 0.044\times 10^{11}\ J$

7 0

3 years ago