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scoundrel [369]

2 years ago

11

Sophia is planning on going down an 8-m high water slide. Her weight is 50 N. What is Sophia's Gravitational Potential Energy at

the top of the slide?

1 answer:

pogonyaev2 years ago

6 0

Gpe = 50 x 8 = 400 Joules

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A 20 N unbalanced force causes an object to accelerate at 1.5 m/s2. What is the mass of the object?

Ket [755]

<u>Answer:</u>

Force = 20N

acceleration (a) = 1.5 m/s²

Mass of object (m) = ?

<u>From Newtons II law</u>

<em> F = m. a N</em>

m = F/a

m = 20/1.5

<em> m = 13.34 Kg</em>

<em>Mass of an object is 13.34 Kg</em>

8 0

2 years ago

Douglas has a segment with endpoints I(5, 2) and J(9, 10) that is divided by a point K such that IK and KJ form a 2:3 ratio. He

Rufina [12.5K]

For the answer to the question above,
the distance from i to j is 5 parts
(2 parts from i to k and 3 parts from k to j)

The y distance from i to j is
10 - 2 = 8

Each part is 8/5 = 1.6
Therefore the distance between the 2 parts from i to k is 3.2

From the y coordinate of I which is 2 plus the 3.2 to point k
2 + 3.2 = 5.2

Answer y =5.2

Now just convert that to fraction and that will be the answer

3 0

2 years ago

Please help me fill out the chart.??

LekaFEV [45]

Answer:unbalanced: have direction,Change an objects motion, causes object to accelerate

Balanced:Do not change an objects motion, Net forces equal sum of all forces on object, and Does not equal 0 N

Explanation: Thats all I know Hoped I helped sum

5 0

2 years ago

hammer [34]

We can define power as the rate of doing work, it is the work done in unit time. The SI unit of power is Watt (W) which is joules per second (J/s). Sometimes the power of motor vehicles and other machines are given in terms of Horsepower (hp) which is approximately equal to 745.7 watts.

Power is the rate at which a force is applied to an object for example.current wire

4 0

2 years ago

Read 2 more answers

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