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3 years ago

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A woman walks into a carpet store wearing high-heeled shoes with a circular heel of diameter 0.987 cm. To the dismay of the stor

e manager, she balances on one heel on an expensive carpet sample. If she has a mass of 53.0 kg, determine the pressure she exerts on the carpet sample.

1 answer:

jenyasd209 [6]3 years ago

3 0

To solve this problem it is necessary to apply the concepts related to the Pressure which describes the amount of Force made on an area unit.

Its mathematical expression can be defined as

$P = \frac{F}{A}$

Where,

F= Force

A = Area

Our values are given as

$d = 0.987*10^{-2}m$

With the diameter we can find the Area of the circular heel, that is

$A = \pi (\frac{0.987*10^{-2}}{2})^2$

$A = 7.6511*10^{-5}m^2$

To find the force by weight we need to apply the Newton's second Law

$F = mg$

$F = 53*9.8$

$F = 519.4N$

Finally the pressure would be

$P = \frac{F}{A}$

$P = \frac{519.4}{7.6511*10^{-5}}$

$P = 6788566.35Pa$

$P = 6.788MPa$

Therefore the pressure that she exerts on the carpet sample is 6.788Mpa

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Two identical resistors are connected in parallel across a 26-V battery, which supplies them with a total power of 7.1 W. While

Gnom [1K]

Answer:

A) R = 190.42 Ω

B) P = 5.325 W

Explanation:

We are given;

Total power;P_tot = 7.1 W

Voltage;V = 26 V

A)We are told that while the battery is still connected, one of the resistors is heated, so that its resistance doubles.

Thus, the power is doubled.

Now, formula for power is;

P = IV

Thus, since power is doubled, we have;

P = 2(IV)

Now, formula for current is; I = V/R

So, P = 2V²/R

Making R the subject, we have;

R = 2V²/P

In this question, P is p_total = 7.1 W

Thus;

R = (2 × 26²)/7.1

R = 190.42 Ω

B) Now, the resistance of the resistors are R and 2R.

Formula for power in this context is;

P = V²/R

Thus,

Total power delivered to the resistors is;

P = V²/R + V²/2R

P = 3V²/2R

P = (3 × 26²)/(2 × 190.42)

P = 5.325 W

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3 years ago

A jet airliner moving initially at 406 mph (with respect to the ground) to the east moves into a region where the wind is blowin

astraxan [27]

Answer:

966 mph

Explanation:

Using as convention:

- East --> positive x-direction

- North --> Positive y-direction

The x- and y- components of the initial velocity of the jet can be written as

$v_{1x} = 406 mph\\v_{1y} = 0$

While the components of the velocity of the wind are

$v_{2x} = (568)(cos 15^{\circ})=548.6 mph\\v_{2y} = (568)(sin 15^{\circ})=147.0 mph$

So the components of the resultant velocity of the jet are

$v_x = v_{1x}+v_{2x}=406+548.6=954.6 mph\\v_y = v_{1y}+v_{2y}=0+147.0=147.0 mph$

And the new speed is the magnitude of the resultant velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(954.6)^2+(147.0)^2}=965.8 mph \sim 966 mph$

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3 years ago

Unlike acceleration and velocity, speed does not need to specify

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2 years ago

An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode

sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell $v_{0}$ in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

$v_{x}=v_{0} \times \cos \theta$

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of $v_{x}$ and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

$v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}$

$v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}$

For motion with constant acceleration, we know

$s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}$

Along the horizontal, x-axis, we might write this as

$x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}$

Measuring distances relative to the firing point means

$x_{0}=0$

we know that,

$a_{x}=0$

or,

$v_{x}=v_{x 0}=\text { constant }$

By applying the values, we get,

$x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}$

The acceleration of gravity is vertically downward and is $g=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

$y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}$

we know, $y_{0}=0$ and $a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , so,

$y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)$

y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

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3 years ago