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3 years ago

How is potential energy unique?

Physics

1 answer:

ahrayia [7]3 years ago

3 0

Answer:

is uniquely defined.

Explanation:

The gravitational potential energy gained by an object being lifted is equal to the work done on the object by an applied force, which exactly cancels gravity. The applied force does positive work. When the object is being lifted, the gravitational force does negative work.

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Sammy squirrel is steering his boat at a heading of 327 degree at 18mph. The current is flowing at 4mph at a heading of 60 degre

Tanya [424]

Answer:

59.97 º at 18.23 mph

Explanation:

To find Sammy's course you have to add the two velocities (vectors), 18 mph 327º and 4 mph 60º.

To add the two vectors analytically you decompose each vector into their vertical and horizontal components.

1. 18 mph 327º

Horizontal component: 18 mph × cos (327º) = 15.10 mph

Vertical component: 18 mph × sin (327º) = - 9.80 mph

Vector notation:

$15.10\hat i-9.80\hat j$

2. 4 mph 60º

Horizontal component: 4 mph × cos (60º) = 2.00 mph

Vertical component: 4 mph × sin (60º) = 3.46 mph

Vector notation:

$2.00\hat i+3.46\hat j$

3. Addition:

You add the corresponding components:

$15.10\hat i-9.80\hat j+2.00\hat i+3.46\hat j\\ \\ 17.10\hat i-6.34\hat j$

To find the magnitude use Pythagorean theorem:

$\sqrt{17.1^2+6.34^2}= 18.23$

4. Direction:

Use the tangent ratio:

$tan(\alpha )=opposite/adjacent=3.46/2.00=1.73$

Find the inverse:

arctan (1.73) ≈ 59.97º

5 0

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

2 years ago

How can we show the magnetic force of a magnet illustrate with an example

morpeh [17]

Answer:

take some sharp mixture of iron spread it from up the magnet you will see the magnetic field of magnet

3 0

3 years ago

What is done when an object is moved by a force

Flauer [41]

1). The object winds up in a different location.

2). Work is done.

7 0

3 years ago

a 10-N force is exerted on a box, moving it 20 m in the same direction. What is the magnitude of work done on the box?

-Dominant- [34]

F = 10 N

d = 20 m

θ = 0°

W = F (dot product) D = F * D * cos(angle between them)

W = FDcosθ

W = 10 * 20 * cos0 = 200 J

8 0

2 years ago