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3 years ago

Help pleaseee SOS!! Again

Physics

1 answer:

wlad13 [49]3 years ago

8 0

Answer:

A) 7500[N]

Explanation:

This is a problem where we have to apply the pascal principle, in this case replace the given values in the given equation:

F1 = 2500 [N]

A1 = 0.06 [m^2]

A2 = 0.18 [m^2]

$F_{2}=A_{2}*\frac{F_{1}}{A_{1}} \\F_{2}=0.18*\frac{2500}{0.06}\\ F_{2}= 7500[N]$

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Using Mirror equation A, Calculate The Frequency Of The Long Stand And The Shortest Wave Length Of That An Object Is Placed Of A

Leya [2.2K]

The Image distance and Magnification of The Image will be 30 cm and 3.

<h3>What is focal length?</h3>

The focal length of the lens, which is often expressed in millimeters, is the distance between the lens and the image sensor when the subject is in focus.

Given data;

Focal length,f=?

Image distance,v=?

Object distance,u= 10 cm

Magnification,m= 2.85

The focal length is half of the radius;

f=R/2

f=30 Cm/2

f= 15 Cm

The mirror equation is found as;

$\rm \frac{1}{f} =\frac{1}{v} +\frac{1}{u} \\\\\ \frac{1}{15}=\frac{1}{v}+\frac{1}{10} } \\\\\ v= -30 \ cm$

The magnification of the lens is found as;

$\rm m=\frac{30}{10}\\\\ m=3$

Hence, the image distance and magnification of The image will be 30 cm and 3.

To learn more about the focal length refer;

brainly.com/question/16188698

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2 years ago

A greyhound's velocity changes from 10 meters per second to 15 meters per second in 0.5 second. What is the greyhound's average

alexgriva [62]

I think it is 5 m/s

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3 years ago

Read 2 more answers

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

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3 years ago

A graduated cylinder contains 17.5 ml of water. When a metal cube is placed onto the cylinder, its water level rises to 20.3 ml.

igomit [66]

Answer:

V=2.8 ml

Explanation:

volume of the cube is it would be 20.3 - 17.5 ml so 2.8 ml.

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4 years ago

Question 2 of 34

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You can find the magnitude of the resultant vector : (B). By adding the magnitudes of the two vectors

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A vector can be defined as any quantity which possesses magnitude and also has direction.

A Vector quantity is very useful because This type of quantity gives more details to the student or teacher analyzing it.

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2 years ago