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fgiga [73]
2 years ago
14

The mass of the skier, including his equipment, is 75kg. In the ski race, the total vertical

Physics
1 answer:
diamong [38]2 years ago
3 0

The gravitational potential energy

gpe = mgh

\tt gpe=75\times 9.8\times880=646,800~J

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1.Convert 340 cm into m *(answer=0.34m)
Nataly [62]

Answer:

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<em>100c</em><em>m</em><em>-</em><em>1</em><em>m</em>

<em>3</em><em>4</em><em>0</em><em>c</em><em>m</em><em>-</em><em>x</em>

<em>3</em><em>4</em><em>0</em><em>/</em><em>100</em>

<em>=</em><em>3</em><em>.</em><em>4</em>

<em>the </em><em>answer </em><em>is </em><em>supposed</em><em> to</em><em> be</em><em> </em><em>3</em><em>.</em><em>4</em><em>,</em><em> maybe</em><em> </em><em>there's</em><em> </em><em>a </em><em>mistake</em><em> </em><em>with </em><em>the </em><em>question</em><em> </em><em>or </em><em>the </em><em>answer</em>

<em>2</em><em>.</em><em>t</em><em>h</em><em>e</em><em> </em><em>weight</em><em> </em><em>of </em><em>a </em><em>body </em><em>is </em><em>given </em><em>by </em><em>the </em><em>formula</em>

<em>mass×</em><em>g</em><em>r</em><em>a</em><em>v</em><em>i</em><em>t</em><em>y</em><em>,</em><em>in </em><em>this </em><em>case </em><em>the </em><em>mass </em><em>is </em><em>7</em><em>5</em><em>k</em><em>g</em><em> </em><em>and </em><em>the </em><em>gravity </em><em>is </em><em>9</em><em>.</em><em>8</em>

<em>weight</em><em>=</em><em>7</em><em>5</em><em>×</em><em>9</em><em>.</em><em>8</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>7</em><em>3</em><em>5</em><em>N</em>

<em>3</em><em>.</em><em>f</em><em>o</em><em>r</em><em> </em><em>this </em><em>one </em><em>the </em><em>mass </em><em>of </em><em>a </em><em>body </em><em>is </em><em>given</em><em> by</em><em> the</em><em> formula</em>

<em>mass=</em><em>weight/</em><em>gravity</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>4</em><em>2</em><em>0</em><em>/</em><em>9</em><em>.</em><em>8</em>

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4 0
3 years ago
Why are some consolation visible to New York State observers at midnight during April , but not visible at midnight during Octob
Svetach [21]
"Midnight" means looking away from the Sun. But in 6 months from April to October the earth goes halfway around the Sun. So midnight in April and midnight in October are exactly opposite directions.
3 0
3 years ago
Consider two ideal gases, A &amp; B, at the same temperature. The rms speed of the molecules of gas A is twice that of gas B. Ho
uysha [10]

Answer:

option (d)

Explanation:

The relation between the rms velocity and the molecular mass is given by

v   proportional to  \frac{1}{\sqrt{M}} keeping the temperature constant

So for two gases

\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}

\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}

{\frac{M_{B}}{M_{A}}} = 4

{\frac{M_{B}}{4}} = M_{A}

7 0
3 years ago
Calculate the change in the energy of an electron that moves from the n = 3 level to the n = 2 level. What type of light is emit
marissa [1.9K]

Answer:

Red light

Explanation:

The energy emitted during an electron transition in an atom of hydrogen is given by

E=E_0 (\frac{1}{n_2^2}-\frac{1}{n_1^2})

where

E_0 = 13.6 eV is the energy of the lowest level

n1 and n2 are the numbers corresponding to the two levels

Here we have

n1 = 3

n2 = 2

So the energy of the emitted photon is

E=(13.6) (\frac{1}{2^2}-\frac{1}{3^2})=1.9 eV

Converting into Joules,

E=(1.9 eV)(1.6\cdot 10^{-19} J/eV)=3.0\cdot 10^{-19} J

And now we can find the wavelength of the emitted photon by using the equation

E=\frac{hc}{\lambda}

where h is the Planck constant and c is the speed of light. Solving for \lambda,

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{3.0\cdot 10^{-19}}=6.63\cdot 10^{-7} m = 663 nm

And this wavelength corresponds to red light.

5 0
3 years ago
A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no apprecia
masha68 [24]

Answer: R = 394.36ohm

Explanation: In a LR circuit, voltage for a resistor in function of time is given by:

V(t) = \epsilon. e^{-t.\frac{L}{R} }

ε is emf

L is indutance of inductor

R is resistance of resistor

After 4s, emf = 0.8*19, so:

0.8*19 = 19. e^{-4.\frac{22}{R} }

0.8 = e^{-\frac{88}{R} }

ln(0.8) = ln(e^{-\frac{88}{R} })

ln(0.8) = -\frac{88}{R}

R = -\frac{88}{ln(0.8)}

R = 394.36

In this LR circuit, the resistance of the resistor is 394.36ohms.

7 0
3 years ago
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