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3 years ago

SPEED AND VELOCITY Plz someone help

Physics

1 answer:

vodomira [7]3 years ago

7 0

1. it changes its velocity though out its circular motion because of its constant change in direction and also in its last 20 second due to a decrease in speed

2. 800kmhr^-1
v = s / t
= 2000/2.5
=800kmhr^-1

3. 1000kmhr^-1
v= s/t
= 2000/2
=1000kmhr^-1

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A charge of −20 µC is distributed uniformly over the surface of a spherical conductor of radius 11.0 cm. Determine the electric

Alex73 [517]

Answer:

(a) $-6.76\times 10^{12}\ N/C$

(b) $-1.352\times 10^{13}\ N/C$

(c) $-7.2\times 10^{11}\ N/C$

Explanation:

(a)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 5 cm = 0.05 m

Coulomb's constant (k) = $9\times 10^{9}\ Nm^2/C^2$

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.05\\\\E_{in}=-1.352\times 10^{14}\times 0.05\\\\E_{in}=-6.76\times 10^{12}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

(b)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 10 cm = 0.10 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.10\\\\E_{in}=-1.352\times 10^{14}\times 0.10\\\\E_{in}=-1.352\times 10^{13}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

(c)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 50 cm = 0.50 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r > R' from the center of sphere is given as:

$E=\dfrac{kQ}{r^2}$

Plug in the given values and solve for 'E'. This gives,

$E_{out}=(\frac{9\times 10^{9}\times -20}{(0.50)^2})\\\\E_{out}=-7.2\times 10^{11}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

8 0

4 years ago

In Case 1, a mass M hangs from a vertical spring having spring constant k and is at rest in its equilibrium position. In Case 2

evablogger [386]

Answer: hello your question is incomplete attached below is the complete question

answer : 1/2 KD^2 ( option A )

Explanation:

P.E ( potential energy ) = mgd

In case 1 P.E = 0 i.e. mgd = 0

Given that in case 2 the Mass M had moved through the Distance D by the compression of the spring

<u>The potential energy of the M in case 2 </u>

= P.E of M at rest + P.E of the spring

= 0 + 1/2 KD^2

4 0

3 years ago

What would increase the force of gravity between two objects?

yulyashka [42]

By Newton's Law of Universal Gravitation.

F = GMm/r²

Where F is Force of Gravitation, M = Mass of first object, m = mass of second object, r = distance of separation

From the formula, you can see that if the masses, M and m, increased, the value of F would definitely increase as well.

And if r increased the value of F would be reduced because you would be dividing by a bigger number, but when the value of r is decreased the value of F would be increased, because you would then be dividing by something smaller. Note the r is at the denominator of the formula.

So F would increase if there was increase in Masses and decrease in distance.

So the answer is C. a and b.

3 0

3 years ago

Please help quick please

zloy xaker [14]

Answer:

I think it is a or b

any of those two sha

cause it is confusing

4 0

3 years ago

Select all the correct answers.

Andre45 [30]

Yo sup??

the 1st, 3rs and 4th statement are true because electromagnetic waves can travel large distances and electromagnetic waves are visible (only 400-700 nm wavelenght)

Hope this helps

8 0

4 years ago