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3 years ago

A Porter carries a 25 kg suitcase a distance of 1 km. Explain why the Porter does no work

Physics

1 answer:

il63 [147K]3 years ago

8 0

The Porter does no work because the force applied is perpendicular to the displacement

Explanation:

The work done by a force applied to an object is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and the displacement of the object

From the formula, we notice that:

- The work done is maximum when the force is parallel to the displacement (because $\theta = 0^{\circ}, cos 0^{\circ}=1$ )

- The work done is zero when the force is perpendicular to the displacement (because $\theta = 90^{\circ}, cos 90^{\circ}=0$ )

In this problem, Porter is carrying the suitcase: this means that the force he is applying is upward (to balance the weight of the suitcase, pulling down), while the displacement is in the horizontal direction: therefore, force and displacement are perpendicular, so the work done is zero.

Learn more about work:

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What is air pressure?

dolphi86 [110]

Answer:

A. Earth's gravity pulling down on air molecules

Explanation:

Air pressure refers to the weight of the air per unit surface area. It is the amount of gravitational force which is pulling down the molecules of air.

The common unit of air pressure is: Pascal, atm

1 atm = 101325 Pa

As the column of the air above increases, the air pressure increase. This is because with the increase in amount of air, the weight increase of the air increases. This is the reason a diver feels immense pressure in the sea and cooking takes a lot of time on hilly areas because of low air pressure.

5 0

3 years ago

Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27

densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

$\rho=n\times\dfrac{m}{V_{c}\times N_{A}}$

$n=\dfrac{\rho\timesV_{c}\times N}{m}$ ....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

$\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100$

$\rho=5.98$

We calculate the mass of the alloy

$m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100$

$m=111.15$

Put the value into the equation (I)

$n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}$

$n=1.99\approx 2\ atoms/cell$

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

5 0

4 years ago

How do you feel about the way land is used around you?

cupoosta [38]

Answer:

I feel that people hurt the land and are destroying. I think they aren't using the land for the right reasons.

5 0

3 years ago

A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If

Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

$\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s$

The induced emf in the shorter coil is calculated as;

$E = NA\frac{\delta B}{\delta t}$

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

$E = NA\frac{\delta B}{\delta t}$

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

8 0

3 years ago

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Answer:

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Explanation:

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6 0

3 years ago