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6 days ago

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A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a

nd the deck is μs = 0.73. The coefficient of kinetic friction is μk = 0.59
Write an expression for the force Fv that must be applied to keep the block moving at a constant velocity.
What is the magnitude of the force Fv in newtons?

1 answer:

Sav [1K]6 days ago

4 0

Answer: $F_{v} =\mu_{k} mg$

The force required has a magnitude of 2601.9 N

Explanation:

m = 450 kg

Static friction coefficient μs = 0.73

Kinetic friction coefficient μk = 0.59

The force necessary to initiate movement of the crate is $F_{s} =\mu_{s} mg$ .

Once the crate begins to move, the frictional force decreases to $F_{v} =\mu_{k} mg$ .

To maintain the motion of the crate at a steady velocity, we must lower the pushing force to $F_{v} =\mu_{k} mg$ .

Subsequently, the pushing force aligns with the frictional force stemming from kinetic friction, enabling balanced forces and consistent velocity.

<pMagnitude of the force

$F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N$

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An ideally efficient heat pump delivers 1000 J of heat to room air at 300 K. If it extracted heat from 260 K outdoor air, how mu

Yuliya22 [1153]

Answer:

Wnet, in, = 133.33J

Explanation:

Provided that

Pump heat QH = 1000J

Hot temperature TH= 300K

Cold temperature TL= 260K

Given the heat pump is entirely reversible, the performance coefficient expression is formulated as follows:

According to the first law of thermodynamics,

COP(HP, rev) = 1/(1-TL/TH)

COP(HP, rev) = 1/(1-260/300)

COP(HP, rev) = 1/(1-0.867)

COP(HP, rev) = 1/0.133

COP(HP, rev) = 7.5

The power necessary to operate the heat pump is given by

Wnet, in = QH/COP(HP, rev)

Wnet, in = 1000/7.5

Wnet, in = 133.333J. QED

Thus, the 133.33J represents the initial work input during the heat transfer process.

<padditionally...><pbased on="" the="" first="" law="" rate="" at="" which="" heat="" is="" extracted="" from="" lower="" temperature="" reservoir="" calculated="" as="">

QL=QH-Wnet, in

QL=1000-133.333

QL=866.67J

</pbased></padditionally...>

5 0

1 day ago

In a given city, the permissible limit of CO (carbon monoxide) in the air is 100 parts per million (ppm). The city monitors the

Sav [1095]

The city evaluates the continuous increase of carbon monoxide from different origins each year. According to calculations, in the year "C: 2019"<span> (rounded to the closest whole number), the concentration of CO will surpass the allowed threshold.

If this is not correct, feel free to inform me and I will find out the right answer. However, I am confident this is accurate.:) </span>

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2 days ago

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On a caterpillars map all distances are marked in kilometers . The caterpillars map shows the distance between two milkweed plan

ValentinkaMS [1144]

Answer:

The equivalent distance in kilometers is 4012 × $10^{-6}$ km.

Explanation:

It's known that 1 millimeter converts to $10^{-3}$ meters. Then, 1 meter converts to $10^{-3}$ kilometers. Therefore, the conversion for 1 millimeter to kilometers can be stated as

1 mm = $10^{-3}$ m

1 m = $10^{-3}$ km

Thus, 1 mm = $10^{-3}$ × $10^{-3}$ km = $10^{-6}$ km.

Given the distance of 4012 mm, the corresponding distance in kilometers will be

4012 mm = 4012 × $10^{-6}$ km.

The distance therefore is 4012 × $10^{-6}$ km.

5 0

13 days ago

An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

Maru [1053]

1) For x = 6.6 cm, $E_x=3.47\cdot 10^6 N/C$

2) For x = 6.6 cm, $E_y=0$

3) For x = 1.45 cm, $E_x=-3.76\cdot 10^6N/C$

4) For x = 1.45 cm, $E_y=0$

5) Surface charge density at b = 4 cm: $+62.75 \mu C/m^2$

6) At x = 3.34 cm, the x-component of the electric field equals zero

7) Surface charge density at a = 2.9 cm: $+65.25 \mu C/m^2$

8) None of these regions

Explanation:

1)

The electric field from an infinite charge sheet is perpendicular to it:

$E=\frac{\sigma}{2\epsilon_0}$

where

$\sigma$ is the surface charge density

$\epsilon_0=8.85\cdot 10^{-12}F/m$ represents vacuum permittivity

Outside the slab, the electric field behaves like that of an infinite sheet.

Consequently, the electric field at x = 6.6 cm (situated to the right of both the slab and sheet) results from the combination of the fields from both:

$E=E_1+E_2=\frac{\sigma_1}{2\epsilon_0}+\frac{\sigma_2}{2\epsilon_0}$

where

$\sigma_1=-2.5\mu C/m^2 = -2.5\cdot 10^{-6}C/m^2\\\sigma_2=64 \muC/m^2 = 64\cdot 10^{-6}C/m^2$

The field from the sheet points left (negative, inward), and the slab’s field points right (positive, outward).

Thus,

$E=\frac{1}{2\epsilon_0}(\sigma_1+\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}+64\cdot 10^{-6})=3.47\cdot 10^6 N/C$

and the negative sign indicates a rightward direction.

2)

Both the sheet’s and slab’s fields are perpendicular to their surfaces, directing along the x-axis, hence there's no y-component for the total field.

<pThus, the y-component totals zero.

This happens because both the sheet and slab stretch infinitely along the y-axis. Choosing any x-axis point reveals that the y-component of the field, generated by a surface element dS of either the sheet or slab, $dE_y$ , will be equal and opposite to the corresponding component from the opposite side, $-dE_y$ . Thus, the combined y-direction field is always zero.

3)

This scenario resembles part 1), but the point here is

x = 1.45 cm

which lies between the sheet and the slab. The fields from both contribute leftward as the slab has a negative charge (resulting in an outward field). Thus, the total field computes to

$E=E_1-E_2$

Replacing with expressions from part 1), we get

$E=\frac{1}{2\epsilon_0}(\sigma_1-\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}-64\cdot 10^{-6})=-3.76\cdot 10^6N/C$

where the negative illustrates a leftward direction.

4)

This portion parallels part 2). Since both fields remain perpendicular to the slab and sheet, no component exists along the y-axis, thus the electric field's y-component is zero.

5)

Notably, the slab behaves as a conductor, signifying charge mobility within it.

The net charge on the slab is positive, indicating a surplus of positive charge. With the negatively charged sheet on the left of the slab, positive charges shift towards the left slab edge (at a = 2.9 cm), while negative charges move to the right edge (at b = 4 cm).

The surface charge density per unit area of the slab is

$\sigma=+64\mu C/m^2$

This average denotes the surface charge density on both slab sides at points a and b:

$\sigma=\frac{\sigma_a+\sigma_b}{2}$ (1)

Additionally, the infinite sheet at x = 0 negatively charged $\sigma_1=-2.5\mu C/m^2$ , induces an opposite net charge on the slab's left surface, thus

$\sigma_a-\sigma_b = +2.5 \mu C/m^2$ (2)

Having equations (1) and (2) allows for solving the surface charge densities at a and b, yielding:

$\sigma_a = +65.25 \mu C/m^2\\\sigma_b = +62.75 \mu C/m^2$

6)

We aim to compute the x-component of the electric field at

x = 3.34 cm

This point lies inside the slab, bounded at

a = 2.9 cm

b = 4.0 cm

In a conducting slab, the electric field remains at zero owing to charge equilibrium; thus, the x-component thereof in the slab is zero

7)

From part 5), we determined the surface charge density at x = a = 2.9 cm is $\sigma_a = +65.25 \mu C/m^2$

8)

As mentioned in part 6), conductors have zero electric fields internally. Since the slab is conductive, the electric field inside remains zero; therefore, the regions where the electric field is null are

2.9 cm < x < 4 cm

Thus, the suitable answer is

"none of these regions"

Learn more about electric fields:

8 0

5 days ago

Compressed air is used to fire a 60 g ball vertically upward from a 0.70-m-tall tube. The air exerts an upward force of 3.0 N on

Yuliya22 [1153]

Answer:

2.87 m

Explanation:

Given parameters:

Mass of the ball (m) = 60 g = 0.06 kg

Height of the tube (h) = 0.70 m

Force applied on the ball by compressed air (F) = 3.0 N

Initial velocity of the ball (u) = 0 m/s (Assumed)

Final velocity of the ball at the tube's exit (v) =?

Acceleration of the ball (a) =?

The ball's weight is derived from multiplying mass and gravity. Therefore,

Weight (W) = $mg=0.06\times 9.8=0.588\ N$

Thus, the total force acting on the ball equals the net of upward force minus the weight.

Net force = Air force - Weight

$F_{net}=F-mg\\F_{net}=3.0-0.588 = 2.412\ N$

According to Newton's second law, net force equals the mass multiplied by acceleration.

$F_{net}=ma\\\\a=\frac{F_{net}}{m}=\frac{2.412\ N}{0.06\ kg}=40.2\ m/s^2$

Acceleration (a) is calculated as 40.2 m/s².

Using the motion equation, we find:

$v^2=u^2+2ah\\\\v^2=0+2\times 40.2\times 0.7\\\\v=\sqrt{56.28}=7.5\ m/s$

Let’s denote the maximum height achieved as 'H'.

Next, we apply the principle of energy conservation from the pipe's peak to the maximum height.

A decrease in kinetic energy equals an increase in potential energy.

$\frac{1}{2}mv^2=mgH\\\\H=\frac{v^2}{2g}$

Substituting the values, we solve for 'H', yielding:

$H=\frac{56.28}{2\times 9.8}\\\\H=\frac{56.28}{19.6}=2.87\ m$

Hence, the ball ascends to a height of 2.87 m above the top of the tube.

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4 days ago