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3 years ago

The diagram shows a person using a piece of gym equipment to lift weights.

Physics

2 answers:

Lelechka [254]3 years ago

5 0

Answer:

C. The lower legs are levers, and the knees are fulcrums. The ankles hold the loads.

Explanation:

As we know that while the person is doing exercise the weight is hold by the ankles as it is placed there.

Now the lower legs are bend while doing the exercise so we can say that lower legs are acting like lever which can rotate about one end and at the other end it is holding the weight

Now the knees are acting like fulcrum here because they are supporting the lower legs in such a way that they can rotate about its one end connected to knees

so correct answer will be

C. The lower legs are levers, and the knees are fulcrums. The ankles hold the loads.

ki77a [65]3 years ago

3 0

Answer:

C. The lower legs are levers, and the knees are fulcrums. The ankles hold the loads.

Explanation:

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A temperature of 20°C is equal to ? °F.

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The answer is 68 F. i hope this helps

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3 years ago

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A 30kg boxed is pushed with a force of 20N. What is the boxes acceleration. Please show work

kozerog [31]

Answer:

<h3>The answer is 0.67 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

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3 years ago

An airplane has wings, each with area A, designed so that air flows over the top of the wing at 265 m/s and underneath the wing

exis [7]

Answer

given,

Pressure on the top wing = 265 m/s

speed of underneath wings = 234 m/s

mass of the airplane = 7.2 × 10³ kg

density of air = 1.29 kg/m³

using Bernoulli's equation

$P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2$

$\Delta P =\dfrac{1}{2}\rho (v_2^2-v_1^2)$

$\Delta P =\dfrac{1}{2}\times 1.29\times (265^2-234^2)$

$\Delta P =9977.5 Pa$

Applying newtons second law

2 Δ P x A - mg = 0

$A =\dfrac{mg}{2\Delta P}$

$A =\dfrac{7.2\times 10^3 \times 9.8}{2\times 9977.5}$

A = 3.53 m²

7 0

3 years ago

Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’2) mm apart and then falls on a screen 7

IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

$sin \theta=\frac{n \lambda}{d}$

where

n is the order of the maximum

$\lambda$ is the wavelength

$a$ is the distance between the slits

In this problem,

n = 5

$\lambda=525 nm =5.25\cdot 10^{-7} m$

$a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m$

So we find

$\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}$

And given the distance of the screen from the slits,

$D=75.0 cm = 0.75 m$

The distance of the 5th bright fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm$

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

$sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}$

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

$\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}$

And the distance of the 8th dark fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm$

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3 years ago

The blood pressure in the human body is greater at the feet than at the brain

GarryVolchara [31]

Blood pressure is greater in feet because of gravity

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