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3 years ago

13

An 85-kg rock climber, when reaching the summit, built up 250,000 J of gravitational potential energy. How high did the climber

go?

2 answers:

Elodia [21]3 years ago

4 0

Answer: 300 m

Explanation:

A body possesses gravitational potential energy due to virtue of its height above the surface. It is measured in Joules. it is given by:

G.P.E = m g h

where, m is the mass, g is the acceleration due to gravity and h is the height.

Given: m = 85 kg

G.P.E = 250,000 J

g = 9.8 m/s²

⇒ h = (G.P.E}/ mg

⇒h = (250,000) / ( 85 × 9.8) = 300 m.

The climber climbed 300 m high.

Vesna [10]3 years ago

3 0

It's 300! I just got it correct:)

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Planet geos in orbit a distance of 1

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Planet Geos in orbit a distance of 1 A.U. (astronomical unit) from the star Astra has an orbital period of 1 "year." If planet Logos is 4 A.U. from Astra, how long does Logos require for a complete orbit?

TB = <span>8</span> years

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Lelechka [254]

Answer:

a = 0.8 [m/s²]

Explanation:

To solve this problem we must use Newton's second law which tells us that the resulting force on a body is equal to the product of mass by acceleration, in this way we come to the following equation:

∑F = m*a

where:

F = forces applied [N]

m = mass = 1000 [kg]

a = acceleration [m/s²]

Now using Newton's second law.

$1200 - 400 = 1000*a\\800 = 1000 *a\\a=0.8[m/s^{2} ]$

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3 years ago

Energy can be transferred from one _______ to another and one __________ to another. *

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Answer:

Source, form

Explanation:

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Given theta = 7pi/6, find (sec theta, cos theta) ...?

uysha [10]

There is an identity of sin(2pi-x) = -sin(x) and cos(2pi-x) = cos(x). This is what we are going to use.

<span>(7pi/6) = (2pi)-(pi/6)

</span>
Therefore:
<span>
1) sec(7pi/6) = 1/cos(2pi-(pi/6)) = 1/cos(pi/6) = 2sqrt(3)/3
</span>
<span>2) cos(7pi/6) = cos(pi/6) = sqrt(3)/2

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3 years ago

At a beach the light is generallypartially polarized owing to reflections off sand and water. At a particular beach on a particu

Ne4ueva [31]

Answer:

a) 0.159

b) 0.84

Explanation:

The Horizontal component is 2.3 times the vertical component

Let the horizontal electric field component = $E_{h}$

Let the vertical electric field component = $E_{v}$

The formula for light intensity is given by:

$I = \frac{E_{m} ^{2} }{2c \mu}$ ..............................(1)

$E_{m}$ is the resolution of the vertical and horizontal components, $E_{h} and E_{v}$

$E_{m} ^{2} = E_{h} ^{2} + E_{v} ^{2}$ ..................(2)

Light intensity before the glasses were put on:

$I_{1} = \frac{E_{m} ^{2} }{2c \mu_{1} }$ .............................(3)

Put equation (2) into equation (3)

$I_{1} = \frac{E_{h} ^{2} + E_{v} ^{2}}{2c \mu_{1} }$ .............................(4)

After the glasses were put on the horizontal component vanishes, i.e. $E_{h} = 0$

$I_{2} = \frac{ E_{v} ^{2}}{2c \mu_{2} }$ ...................................(5)

Divide equation (5) by equation (4)

$\frac{I_{2} }{I_{1} } = \frac{E_{v} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}$ ...............................(6)

But $E_{h} = 2.3E_{v}$ ......................(7)

Insert equation (7) into (6)

$\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{(2.3E_{v})^{2} + E_{v} ^{2} } \\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{5.29E_{v}^{2} + E_{v} ^{2} }\\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } =\frac{1}{6.29} \\$

$\frac{I_{2} }{I_{1} }= 0.159$

b) When the sunbather lies on his side, the vertical component vanishes, i.e $E_{v} = 0$

$\frac{I_{2} }{I_{1} } = \frac{E_{h} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}$

$\frac{I_{2} }{I_{1} } = \frac{(2.3E_{v} )^{2} }{E_{v} ^{2} +(2.3E_{v} )^{2}}$

$\frac{I_{2} }{I_{1} } = \frac{5.29E_{v}^{2} }{E_{v} ^{2} +5.29E_{v}^{2} }$

$\frac{I_{2} }{I_{1} } = \frac{5.29E_{v}^{2} }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } = \frac{5.29}{6.29} \\\frac{I_{2} }{I_{1} } = 0.84$

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3 years ago