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klemol [59]
3 years ago
14

Anyone know what the answer is?..

Physics
1 answer:
Licemer1 [7]3 years ago
8 0

Answer:

sometimes harmful and sometimes beneficial

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Two identical space probes are orbiting Jupiter. Scientists determine that one of the space probes has a larger gravitational fo
galina1969 [7]

If the probes are identical, then the one that feels a larger gravitational
force is orbiting closer to Jupiter than the other one is.

If they're not identical, then the one with greater mass will feel more
gravitational force than the one with less mass, even if they're both
the same distance from Jupiter.  (We know this from the experimental
observation that fatter people weigh more, even on Earth.)
6 0
3 years ago
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Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B caries a charge of-2q. Sphere C
miskamm [114]
<h2>20. How much charge is on sphere B after A and B touch and are separated?</h2><h3>Answer:</h3>

\boxed{q_{B}=+2q}

<h3>Explanation:</h3>

We'll solve this problem by using the concept of electric potential or simply called potential V, which is <em>the energy per unit charge, </em>so the potential V at any point in an electric field with a test charge q_{0} at that point is:

V=\frac{U}{q_{0}}

The potential V due to a single point charge q is:

V=k\frac{q}{r}

Where k is an electric constant, q is value of point charge and r is  the distance from point charge to  where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius r. Before the sphere A and B touches we have:

V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r

When they touches each other the potential is the same, so:

V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}

From the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant. </em>So:

q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}

Therefore:

(1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}

So after A and B touch and are separated the charge on sphere B is:

\boxed{q_{B}=+2q}

<h2>21. How much charge ends up on sphere C?</h2><h3>Answer:</h3>

\boxed{q_{C}=+1.5q}

<h3>Explanation:</h3>

First: A and B touches and are separated, so the charges are:

q_{A}=q_{B}=+2q

Second:  C is then touched to sphere A and separated from it.

Third: C is to sphere B and separated from it

So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:

Here q_{A}=+2q and C carries no net charge or q_{C}=0. Also, r_{A}=r_{C}=r

V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}

Applying the same concept as the previous problem when sphere touches we have:

k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}

For the principle of conservation of charge:

q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q

Finally, from the third step:

Here q_{B}=+2q \ and \ q_{C}=+q. Also, r_{B}=r_{C}=r

V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}

When sphere touches we have:

k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}

For the principle of conservation of charge:

q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q

So the charge that ends up on sphere C is:

q_{C}=+1.5q

<h2>22. What is the total charge on the three spheres before they are allowed to touch each other.</h2><h3>Answer:</h3>

+4q

<h3>Explanation:</h3>

Before they are allowed to touch each other we have that:

q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0

Therefore, for the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant, </em>then this can be expressed as:

q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q

Lastly, the total charge on the three spheres before they are allowed to touch each other is:

+4q

8 0
3 years ago
An object of irregular shape has a characteristic length of L = 0.5 m and is maintained at a uniform surface temperature of Ts =
goblinko [34]

Answer:

The value of the average convection coefficient is 20 W/Km².

Explanation:

Given that,

For first object,

Characteristic length = 0.5 m

Surface temperature = 400 K

Atmospheric temperature = 300 K

Velocity = 25 m/s

Air velocity = 5 m/s

Characteristic length of second object = 2.5 m

We have same shape and density of both objects so the reynold number will be same,

We need to calculate the value of the average convection coefficient

Using formula of  reynold number for both objects

R_{1}=R_{2}

\dfrac{u_{1}L_{1}}{\eta_{1}}=\dfrac{u_{2}L_{2}}{\eta_{2}}

\dfrac{h_{1}L_{1}}{k_{1}}=\dfrac{h_{2}L_{2}}{k_{2}}

Here, k_{1}=k_{2}

h_{2}=h_{1}\times\dfrac{L_{1}}{L_{2}}

h_{2}=\dfrac{q}{T_{2}-T_{1}}\times\dfrac{L_{1}}{L_{2}}

Put the value into the formula

h_{2}=\dfrac{10000}{400-300}\times\dfrac{0.5}{2.5}

h_{2}=20\ W/Km^2

Hence, The value of the average convection coefficient is 20 W/Km².

7 0
3 years ago
Why won’t a magnet help you separate a mixture of salt and water
Gennadij [26K]

Answer: nor are magnetic

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3 years ago
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Lincoln weighs 400 newtons. What’s his mass rounded to the nearest kilogram? Assume that acceleration due to gravity is 9.8 N/kg
Yanka [14]
Weight equals mass times gravitational acceleration=400N, so mass=400/9.8=41kg approx.
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