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3 years ago

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A piece of tape is pulled from a spool and lowered toward a 120-mg scrap of paper. Only when the tape comes within 8.0 mm is the

electric force magnitude great enough to overcome the gravitational force exerted by Earth on the scrap and lift it.

1 answer:

Anton [14]3 years ago

6 0

Answer:

$1.176\times 10^{-3} N$

Upward

Explanation:

We are given that

Mass of scarp paper, $m=120 mg=120\times 10^{-6}kg$

1mg= $10^{-6} mkg$

Distance,d =8 mm= $8\times 10^{-3} m$

Magnitude of electric force = $F_E= w=mg$

Where $g=9.8 m/s^2$

Substitute the values

$F_E=120\times 10^{-6}\times 9.8$

$F_E=1.176\times 10^{-3} N$

Gravitational force act in downward direction.

The electric force acts in opposite direction and magnitude of electric force is equal to gravitational force.

Hence, the direction of electric force is upward.

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A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1

aivan3 [116]

Answer:

Part a)

$F_n = 306 N$

Part b)

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

Explanation:

Part a)

By force equation on the rider at the position of the hump we can say

$mg - F_n = ma_c$

now we will have

$mg - F_n = \frac{mv^2}{R}$

$F_n = mg - \frac{mv^2}{R}$

now we have

$F_n = 100(9.81) - \frac{100(9^2)}{12}$

$F_n = 981 - 675$

$F_n = 306 N$

Part b)

At the top of the loop if the minimum speed is required so that it remains in contact so we will have

$F_n + mg = ma_c$

$F_n = 0$ at minimum speed

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = \sqrt{15 \times 9.81}$

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

5 0

3 years ago

Liquid droplets that fall from the sky are?<br> A) rain<br> B)snow<br> C) sleet <br> D) hail

alukav5142 [94]

The answer is a)rain

6 0

3 years ago

Read 2 more answers

Increasing which two things would increase gravitational potential energy?

sukhopar [10]

Answer:

Mass and height

Explanation:

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about $9.8 m/s2.$

Which is represented as;

$PE_g=mgh$

$PE_g$ stands for gravitational potantial energy,

m stands for mass of object,

g is the gravitational constant and

h is the height.

Here we see that mass of object and height is directly proportional to the gravitational potential energy.

That means increasing in mass and height will result in increasing gravitational potential energy.

7 0

3 years ago

A 2.74 g coin, which has zero potential energy at the surface, is dropped into a 12.2 m well. After the coin comes to a stop in

VikaD [51]

Answer:

B. - 0.328

Explanation

Potential Energy:<em> This is the energy of a body due to position.</em>

<em>The S.I unit of potential energy is Joules (J).</em>

<em>It can be expressed mathematically as</em>

<em>Ep = mgh........................... Equation 1</em>

<em>Where Ep = potential energy, m = mass of the coin, h = height, g = acceleration due to gravity,</em>

<em>Given: m = 2.74 g = 0.00274 kg, h = 12.2 m, g = 9.8 m/s²</em>

Substituting these values into equation 1

Ep = 0.00274×12.2×9.8

Ep = 0.328 J.

Note: Since the potential energy at the surface is zero, the potential Energy with respect to the surface = -0.328 J

The right option is B. - 0.328

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7 0

3 years ago

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 8.0 s. At thi

tatiyna

Answer:

50 revolutions

Explanation:

Data provided:

case I: From rest to top spin

The initial angular speed of the washer, ωi = 0 rev /s

Final angular speed of the washer ωf = 5 rev /s

Time taken, t₁ = 8 s

now,

The angular displacement or the number of revolutions taken (θ₁) is calculated as:

θ₁ = ωi t₁ + (1/2)α₁t₁²

where,

α is the angular acceleration

The angular acceleration can be calculated as:

ωf - ωi = α₁t₁

on substituting the values, we get

8α₁ = 5 - 0

or

α₁ = 0.625 rev/s²

substituting the values in the equation for the number of revolutions, we get

θ₁ = 0 + (1/2) (0.625)(8)²

or

θ₁ = 20 revolutions

also,

For the case II: From top spin to rest

we have

The initial angular speed, ωi = 5 rev /s

and the final angular speed, ωf = 0 rev /s

Total time taken, t₂ = 12 s

Now, angular acceleration for this case

ωf - ωi = α₂t₂

on substituting the values, we have

12α₂ = 0 - 5

α₂ = - 0.4166 rev/s²

Therefore, the number of revolutions ( i.e angular displacement )

θ₂ = ωit₂ + (1/2)α₂t₂²

on substituting the values, we have

θ₂ = 5 × 12 + (1/2)(-0.4166)(12)²

or

θ₂ = 30 rev

Hence,

the total number of revolutions made by the washer during the 20s is

θ = θ₁ + θ₂

or

θ = 20 rev + 30 rev

or

θ = 50 revolutions

7 0

3 years ago