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balandron [24]
3 years ago
15

Two neutral metal spheres on wood stands are touching. A negatively charged rod is held directly above the top of the left spher

e, not quite touching it. While the rod is there, the right sphere is moved so that the spheres no longer touch. Then the rod is withdrawn. which answer is correct?
A) Both the spheres are neutral.
b) Left sphere is negatively charged, another is charged positively.
C) Right sphere is negatively charged, another is charged positively.
D) Both the spheres are charged positively.
E) Both the spheres are charged negatively
Physics
2 answers:
FrozenT [24]3 years ago
7 0

Answer: Option (C) is the correct answer.

Explanation:

As we know that metals are able to conduct electricity so, when a negatively charges rod is kept closer to the left sphere then electrons will enter the sphere.

Since, like charges repel each other. Hence, some of the negative changes from the rod will repel the negative charges of left sphere.

As both left and right spheres are touching each other so, the electrons will move towards the right sphere. As a result, there will be too many electrons (negative charge) present on the right sphere and very less electrons present in the left sphere.

Thus, we can conclude that the statement right sphere is negatively charged, another is charged positively, is true.

Murrr4er [49]3 years ago
4 0

Answer: ITS CCCC

Explanation:

ITS CORRECT

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-<br> ZOOLS<br> 6) The mass of a motorcycle is 250 kg. What is?<br> A) Its weight on Earth in Newtons?<br> B) Its weight on the
marishachu [46]

Answer:

Explanation:

Weight is actually a force. A force can change depending on its location. A mass remains constant no matter where it is.

A)

F = m * a

m = 250 kg

a = 9.81 m/s^2

F = 250 * 9.81 = 2452.5 N

B)

The acceleration due to gravity on the moon is roughly 1/6 what it is on earth. You can check its value in your notes.

a = 9.81 + (1/6) = 1.635

m = 250

F = 250 * 1.635

F = 408.75

C)

The mass is the same anywhere in the universe.

250 kg

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3 years ago
Inertia keeps your vehicle moving until a force slows it down, like __________ ?
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Both answers a and c
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Answer:

The Electric force and nuclear forces are in opposition to each other

Explanation:

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4 years ago
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If Johnny won a 300 meter race in 40 seconds, his speed would be ?
irina1246 [14]

Formula: s = d/t

s = speed

d = distance

t = time

Solve using the values we are given.

s = 300/40

s = 7.5m/s

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Problem 4 A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true ano
Nikitich [7]

Answer:

Part a: The eccentricity is 1.086.

Part b: The altitude at closest approach is 5088 km

Part c: The velocity at perigee is 8.516 km/s

Part d: The turn angle is 134.08 while the aiming radius is 5641.28 km

Explanation:

<h2>Part a </h2>

Specific energy is given by

\epsilon=\frac{v^2}{2}-\frac{\mu}{r}

Here

  • ε is the specific energy
  • v is the velocity which is given as 2.23 km/s
  • μ is the gravitational constant whose value is 398600
  • r is the distance between earth and the meteorite which is 402,000 km

                         \epsilon=\frac{v^2}{2}-\frac{\mu}{r}\\\epsilon=\frac{2.2^2}{2}-\frac{398600}{402,000}\\\epsilon=1.495 km^2/s^2

Value of specific energy is also given as

\epsilon=\frac{\mu}{2a}\\a=\frac{\mu}{2\epsilon}\\a=\frac{398600}{2\times 1.495}\\a=13319 km

Orbit formula is given as

r=a(\frac{e^2-1}{1+ecos \theta})\\ae^2-recos\theta-(a+r)=0

Putting values in this equation and solving for e via the quadratic formula gives

ae^2-recos\theta-(a+r)=0\\(133319)e^2-(402000)(cos 150) e-(133319+402000)=0\\133319e^2+348142.21 e-535319=0\\\\e=\frac{-348142.21 \pm \sqrt{348142.21^2-4(133319)(535319)}}{2 (133319)}\\\\e=1.086 \, or \, -3.69

As the value of eccentricity cannot be negative so the eccentricity is 1.086.

<h2>Part b</h2>

The radius of trajectory at perigee is given as

r_p=a(e-1)\\

Substituting values gives

r_p=133319 (1.086-1)\\r_p=11465.4 km

Now for estimation of altitude z above earth is given as

z=r_p-R_E\\z=11465.4-6378\\z=5087.434\\z\approx 5088 km

So the altitude at closest approach is 5088 km

<h2>Part c</h2>

radius of perigee is also given as

r_p=\frac{h^2}{\mu}\frac{1}{1+e}

Rearranging this equation gives

h=\sqrt{r_p\mu(1+e)}\\h=\sqrt{11465.4 \times 3986000 \times (1+1.086)}\\h=97638.489 km^2/s

Now the velocity at perigee is given as

v_p=\frac{h}{r_p}\\v_p=\frac{97638.489}{11465.4}\\v_p=8.516 km/s\\

So the velocity at perigee is 8.516 km/s

<h2>Part d</h2>

Turn angle is given as

\delta =2 sin^{-1} (\frac{1}{e})

Substituting value in the equation gives

\delta =2 sin^{-1} (\frac{1}{e})\\\delta =2 sin^{-1} (\frac{1}{1.086})\\\delta =134.08

Aiming radius is given as

\Delta =a \sqrt{e^2-1}

Substituting value in the equation gives

\Delta =a \sqrt{e^2-1}\\\Delta =13319 \sqrt{1.086^2-1}\\\Delta=5641.28 km

So the turn angle is 134.08 while the aiming radius is 5641.28 km

3 0
4 years ago
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