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Varvara68 [4.7K]

3 years ago

8

A Shaolin monk of mass 60 kg is able to do a ‘finger stand’: he supports his whole weight on his two index fingers, giving him a

total contact area of 4 cm 2 with the ground. Calculate the pressure he exerts on the ground (include units), and write your answer to two significant figures.

1 answer:

ss7ja [257]3 years ago

8 0

Answer:

P = 1471500 [Pa]

Explanation:

We must remember that pressure is defined as the relationship between Force over the area.

$P=F/A$

where:

P = pressure [Pa] (units of pascals)

F = force [N] (units of Newtons)

A = area of contact = 4 [cm²]

But first we must convert from cm² to m²

$A = 4[cm^{2}]*\frac{1^{2} m^{2} }{100^{2} cm^{2} }$

A = 0.0004 [m²]

Also, the weight should be calculated as follows:

$w = m*g$

where:

m = mass = 60 [kg]

g = gravity acceleration = 9.81 [m/s²]

Now replacing:

$w = 60*9.81\\w = 588.6[N]$

And the pressure:

$P=588.6/0.0004\\P=1471500 [Pa]$

Because 1 [Pa] = 1 [N/m²]

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An AC generator consists of 20 circular loops of wire with an area of 75 cm2. It has a maximum induced voltage of 24 V. If its a

Monica [59]

Faraday's law allows us to find the magnetic field that produces the emf in the rotating system is:

The magnetic field is: B = 0.424 T

Faraday's law of induction states that when the magnetic flux changes in time, an induced electromotive force is produced.

fem = $- \frac{d \Phi_B }{dt}$

where fem is the induced electromotive force and Ф the flux,

The magnetic flux is the scalar product of the field and the area.

$\Phi_B = B . A = B A \ cos \theta$

In this case we have several turns, so the expression remains.

fem = $- N B A \ \frac{d cos \theta}{dt}$

Indicate that the turns rotate at a constant frequency, therefore we can use the uniform rotational motion ratio.

θ = w t

We substitute

$fem = - N B A \ \frac{d \ cos \ wt}{dt}\\fem = N B A w sin \ wt$

the maximum induced electromotive force occurs when the sine function is ±1

fem = N B A w

They indicate that the fem = 24 V, the number of the turn is N = 20, the area is A = 75 cm² = 75 10⁻⁴ m² and the frequency f = 60 Hz

Frequency and angular velocity are related.

w = 2π f

We substitute.

fem = N B A 2π f

$B = \frac{fem }{2 \pi \ NA \ f}$

Let's calculate.

$B= \frac{24 }{2\pi \ 20 \ 75 \ 10^{-4} 60}$ B = 24 / 2pi 20 75 10-4 60

B = 0.424 T

In conclusion, using Faraday's law we can find the magnetic field that produces the emf in the rotating system is:

The magnetic field is; B = 0.424 T

Learn more about Faraday's law here: brainly.com/question/24617581

8 0

2 years ago

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,

$t_{tot}=2\dfrac{v_0}{g}+t_0$

where $t_0$ is the time it takes the rock to go from the roof of the building to the ground, and it is given by

$H=v_0t_0+\dfrac{1}{2}gt_0^2$

we solve for $t_0$ using the quadratic formula and take the positive value to get:

$t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

Therefore the total time is

$t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

$\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH} }{g}}$

Now the average velocity is

$v_{avg}=\dfrac{y_{tot}}{t}$

$v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }$

$\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }$

5 0

3 years ago

You are performing a knee extension exercise. You hold a 20kg weight at full knee extension. The weight is 0.4m from your knee j

dmitriy555 [2]

Answer:

The moment is -78.4 N-m (clockwise).

Explanation:

Given:

Mass of the object (m) = 20 kg

Distance of the object from the knee joint (d) = 0.4 m

Weight of leg is not considered.

Acceleration due to gravity (g) = 9.8 m/s²

Now, weight of the object is equal to the product of its mass and acceleration due to gravity. So,

Weight = Mass × Acceleration due to gravity

= $mg=20\times 9.8 =196\ N$

We know that, moment of a force about a point is defined as the product of force applied and the perpendicular distance between the point and the line of application of force.

Moment of the given weight about the knee joint is given as:

Moment about knee joint = Weight × Distance from knee joint to weight

Moment about knee joint = 196 × 0.4 = 78.4 Nm

Now, from the diagram below, we can observe that, the weight acts vertically down and thus the sense of rotation about the knee joint at point O is clockwise. So, moment is negative.

Therefore, the moment is -78.4 N-m (clockwise).

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3 years ago

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Morgarella [4.7K]

Answer:

N / NEWTONS

Explanation:

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