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4 years ago

A shovel is the third class lever

Physics

1 answer:

Nookie1986 [14]4 years ago

8 0

Answer:

Yes

Explanation:

In a third-class lever, the effort force lies between the resistance force and the fulcrum. Some kinds of garden tools are examples of third-class levers. When you use a shovel, for example, you hold one end steady to act as the fulcrum, and you use your other hand to pull up on a load of dirt.

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Of all the planets, which is the most different from the others? Explain.

Alenkinab [10]

Answer: VENUS

Explanation:

Venus tiene una lenta rotación retrógrada, lo que significa que gira de este a oeste, en lugar de hacerlo de oeste a este como lo hacen la mayoría de los demás planetas mayores (Urano también tiene una rotación retrógrada, aunque el eje de rotación de Urano, inclinado 97.86°, prácticamente descansa sobre el plano.

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3 years ago

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Which statement best describes the isothermal process? A. the temperature remains constant B. the temperature increases at a con

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Answer:

The answer would be A. - the temperature remains constant

Explanation:

An isothermal process is a change of a system, in which the temperature remains constant: ΔT = 0

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3 years ago

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A spaceship is traveling toward Earth while giving off a constant radio signal with a wavelength of 1 meter (m). What will the s

Nadya [2.5K]

Answer:

Less than 1 m

Explanation:

When objects are getting closer to each other there is a slight change in the wavelength that is being transmitted by either objects. This is known as the blue shift of waves. Here, the wavelength reduces.

In the opposite case the when objects are getting farther from each other there is a slight change in the wavelength that is being transmitted by either objects. This is known as the red shift. Here, the wavelength increases.

In this case the spaceship is getting close to Earth hence the wavelength will be lower than 1 m.

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3 years ago

21. A 60 kg student on a scooter is pushed with a constant force of 40 N across a horizontal concrete driveway at a constant spe

DedPeter [7]

The magnitude of the force of friction is 40 N

Explanation:

To solve the problem, we just have to analyze the forces acting on the student and the scooter along the horizontal direction. We have:

- The constant pushing force forward, of magnitude F = 40 N

- The frictional force, acting backward, $F_f$

Since the two forces are in opposite direction, the equation of motion is

$F-F_f = ma$

where

m is the mass of the student+scooter

a is the acceleration

However, here the scooter is moving at constant speed: this means that its acceleration is zero, so

a = 0

And therefore,

$F - F_f = 0\\F_f = F$

which means that the magnitude of the force of friction is also equal to 40 N.

Learn more about force of friction here:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

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3 years ago

An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _

Viefleur [7K]

Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
where
$f=5 cm$ is the focal length
$p=10 cm$ is the distance of the object from the mirror
$q$ is the distance of the image from the mirror.

Rearranging the equation, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}$
so, the distance of the image from the mirror is $q=10 cm$ .

1b) The image height is given by the magnification equation:
$\frac{h_i}{h_o}=- \frac{p}{q}$
where $h_i$ is the heigth of the image and $h_o=1 cm$ is the height of the object. By rearranging the equation and using p and q, we find
$h_i=-h_o \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm$
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

3) As before, we find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

And now we can use the magnification equation to find the image height:
$\frac{h_i}{h_o}=- \frac{p}{q}$
Rearranging it, we find
$h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm$
and the negative sign means the image is inverted.

5 0

3 years ago

A shovel is the third class lever​

A shovel is the third class lever