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4 years ago

5

how does the force of attraction between large masses compare with the force of attraction between small masses at the same dist

ance?

1 answer:

Veseljchak [2.6K]4 years ago

4 0

The force and amount of attracton is great for larger masses

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Which term defines the energy of motion?

Korvikt [17]

D kinetic energy✨
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3 years ago

Why are many adults and youth in the United states overweight?

Anna11 [10]

Answer:

I'd imagine fast food is a lot more popular than most countries over there and easier to get hold of.

Explanation:

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3 years ago

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Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameters

Vesnalui [34]

Answer:

C. $\frac{3F}{8}$

Explanation:

Let initial charges on both spheres be, $q$

$F=\frac{Kq^2}{d^2} \ \ \ \ \ \ \ \ \ \ \_i$

When the sphere C is touched by A, the final charges on both will be, $\frac{q}{2}$

#Now, when C is touched by B, the final charges on both of them will be:

$q_c=q_d=\frac{q/2+q}{2}\\\\=\frac{3q}{4}\\$

Now the force between A and B is calculated as:

$F\prime=\frac{k\times\frac{q}{2}\times \frac{3q}{4}}{d^2}\\F\prime=\frac{3F}{8}$

Hence the electrostatic force becomes 3F/8

4 0

3 years ago

An electron travels with speed 6.0×10^6 m/s between the two parallel charged plates shown in the figure. The plates are separate

svlad2 [7]

Answer:

B= 3.33 m T

Explanation:

Given that

Speed ,C= 6 x 10⁶ m/s

d= 1 cm = 0.01 m

V= 200 V

The electric field E given as

V= E .d

E=Electric field

d=Distance

V=Voltage

200 = 0.01 x E

E=20000 V/m

The relationship between magnetic and electric field given as

E= C x B

20000 = 6 x 10⁶ x B

B =3333.333 x 10⁻⁶ T

B= 3.33 x 10⁻³ T

B= 3.33 m T

Therefore the magnetic filed will be 3.33 m T.

5 0

3 years ago

The force P is applied to the 45-kg block when it is at rest. Determine the magnitude and direction of the friction force exerte

cluponka [151]

Answer:

Check attachment for free body diagram of the question.

I used the free body diagram and the angles given in the diagram missing in the question above, but I used the data given in the above question.

Explanation:

Let frictional force be Fr acting down the plane

Let analyze the structure before inserting values

Using Newton's second law along the y-axis

ΣFy = Fnet = m•ay

Since the body is not moving in the y-direction, then ay = 0

N+PSinβ — WCosθ = 0

N+PSin20—441.45Cos15 = 0

N+PSin20—426.41 = 0

N = 426.41 — PSin20 , equation 1

The maximum Frictional force to be overcome is given as

Fr(max) = μsN

Fr(max) = 0.25(426.41 — PSin20)

Fr(max)= 106.6 —0.25•PSin20

Fr(max) = 106.6 — 0.08551P, equation 2

This is the maximum force that must be overcome before the body starts to move

Using Newton's law of motion in the x direction

Note, we took the upward direction up the plane as the direction of motion since the force want to move the block upward

Fnetx = ΣFx

Fnetx = P•Cosβ —W•Sinθ — Fr

Fnetx = P•Cos20—441.45•Sin15—Fr

Fnetx = 0.9397P — 114.256 — Fr

Equation 3

When Fnetx is positive, then, the body is moving up the plane, if Fnetx is negative, then, the maximum frictional force has not yet being overcome and the object is still i.e. not moving

a. When P = 0

From equation 2

Fr(max) = 106.6 — 0.08551P

Fr(max) = 106.6 — 0.08551(0)

Fr(max)= 106.6 N

So, 106.6N is the maximum force to be overcome

So, here the only force acting on the body is the weight and it acting down the plane, trying to move the body downward.

Wx = WSinθ

Wx = 441.45× Sin15

Wx = 114.256 N.

Since the force trying to move the body downward is greater than the maximum static frictional force, then the body is not in equilibrium, it is moving downward.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — PSin20 , equation 1

N = 426.41 N, since P=0

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 426.41

Fr = 93.81 N.

b. Now, when P = 190N

From equation 2

Fr(max) = 106.6 — 0.08551(190)

Fr(max) = 106.6 —16.2469

Fr(max)= 90.353 N

So, 90.353 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 190Cos20 — 441.45Sin15

Fnetx = 64.29N

So the force moving the body up the incline plane is 64.29N

Fnetx < Fr(max)

Then, the frictional force has not being overcome yet.

Then, the body is in equilibrium.

Then, applying equation 3.

Fnetx = 0.9397P — 114.256 — Fr

Fnetx = 0, since the body is not moving

0 = 0.9397(190) —114.246 — Fr

Fr = 64.297 N

Fr ≈ 64.3N

c. When, P = 268N

From equation 2

Fr(max) = 106.6 — 0.08551(268)

Fr(max) = 106.6 —16.2469

Fr(max)= 83.68 N

So, 83.68 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 268Cos20 — 441.45Sin15

Fnetx = 137.58 N

So the force moving the body up the incline plane is 137.58 N

Fnetx > Fr(max)

Then, the frictional force has being overcome.

Then, the body is not equilibrium.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — 268Sin20 , equation 1

N = 334.75 N, since P=268N

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 334.75

Fr = 73.64 N

d. The required force to initiate motion is the force when the block want to overcome maximum frictional force.

So, Fnetx = Fr(max)

Px — Wx = Fr(max)

From equation 1

Fr(max) = 106.6 — 0.08551P,

P•Cosβ-W•Sinθ = 106.6 — 0.08551P

P•Cos20 — 441.45•Sin15 = 106.6 — 0.08551P

P•Cos20—114.256=106.6 - 0.08551P

PCos20+0.08551P =106.6 + 114.256

1.025P=220.856

P = 220.856/1.025

P = 215.43 N

3 0

3 years ago