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Crank
3 years ago
5

Two charged objects, A and B, are exerting an electric force on each other. What will happen if the charge on A is increased?

Physics
1 answer:
ozzi3 years ago
6 0

Answer: The forces acting on both of them will increase in magnitude.

Explanation:

According to Coulomb's law, the electrostatic force between two bodies is proportional to the product of their two charges. If the charge on A is increased this product increases in size (it must have been non-zero to begin with, since there was a force between them at first). Thus, the force between them rises.

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A sample of helium has a volume of 12.7 m3. The temperature is raised to 323 K at which time the gas occupies 32.5 m3? Assume pr
jasenka [17]

Answer: The original temperature was

T_{1}=126.51K

Explanation:

Let's put the information in mathematical form:

V_{1}=12.7m^{3}

T_{1}=?

V_{2}=32.5m^{3}

T_{2}=323K

P_{1}=P_{2}=3atm

If we consider the helium as an ideal gas, we can use the Ideal Gas Law:

PV=nRT

were <em>R</em> is the gas constant. And <em>n</em> is the number of moles (which we don't know yet)

From this, taking R=0.08205746\frac{atm.l}{mol.K},  we have:

n=\frac{P_{2}V_{2}}{RT_{2}}

⇒n=3.67mol

Now:

T_{1}=\frac{P_{1}V_{1}}{nR}

⇒T_{1}=126.51K

7 0
3 years ago
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Which student is doing work?
Vilka [71]

Answer:

d

hope this helps

Explanation:

8 0
3 years ago
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Which best explains why people see carrots as orange
Andru [333]

Beta Carotene is the reason why a carrot is orange. Carotene is a color pigment that is found in large quantities in carrots and it has health benefits as well. The main health benefit of carotene is that our body converts it into vitamin A which is essential.

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3 years ago
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The spring in the muzzle of a child's spring gun has a spring constant of 730 N/m. To shoot a ball from the gun, first the sprin
Korolek [52]

Answer:

a. V=11.84 m/s

b.x=0.052m

Explanation:

a).

Given

K=730 N/m,m=0.053kg, h=1.90m.

v_f^2=v_i^2+2*g*h

v_i^2=2*g*h=2*9.8m/s^2*1.9m

v_i=\sqrt{2*9.8m/s^2*1.9m}=\sqrt{37.24 m^2/s^2}

v_i=6.1 m/s

v_i=V*sin(31)

V=\frac{v_i}{sin(31)}=\frac{6.1m/s}{sin(31)}

V=11.84 m/s

b).

K_k=\frac{1}{2}*K*x^2

No friction on the ball so:

x^2=\frac{2*K_k}{K}

x=\sqrt{\frac{2*0.053kg*9.8m/s^2*1.9m}{730N/m}}

x=\sqrt{2.7x10^{-3}m^2}=0.052m

5 0
3 years ago
What is the initial velocity of a go-kart traveling at a uniform acceleration of 0.5 m/s^2 for 5s as it slows down to a stop?
Arlecino [84]

The go-kart's velocity v after time t is given by

v=v_0+at

where v_0 is its initial velocity and a is its acceleration. After t=5\,\mathrm s, the go-kart stops completely, so

0\,\dfrac{\mathrm m}{\mathrm s}=v_0+\left(-0.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(5\,\mathrm s)\implies v_0=2.5\,\dfrac{\mathrm m}{\mathrm s}

where a because we know the go-kart is slowing down.

7 0
3 years ago
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