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givi [52]
3 years ago
14

The specialty of an athlete on the women's track team is the pole vault. She has a mass of 48 kg and her approach speed is 8.9 m

/s. When she is directly above the bar, her speed is 1.6 m/s. Neglecting air resistance and any energy absorbed by the pole, determine the amount(in m) she has raised herself as she crosses the bar.
Physics
1 answer:
Lyrx [107]3 years ago
4 0

Answer:

H = 3.9 m

Explanation:

mass (m) = 48 kg

initial velocity (initial speed) (U) = 8.9 m/s

final velocity (V) = 1.6 m/s

acceleration due to gravity (g) = 9.8 m/s^{2}

find the height she raised her self to as she crosses the bar (H)

from energy conservation, the change in kinetic energy = change in potential energy

0.5m(V^{2} - [test]U^{2}[/tex]) = mg(H-h)

where h = initial height = 0 since she was on the ground

the equation becomes

0.5m(V^{2} - [test]U^{2}[/tex]) = mgH

0.5 x 48 x (1.6^{2} - [test]8.9^{2}[/tex]) = 48 x 9.8 x H

-1839.6 = 470.4 H  (the negative sign indicates a decrease in kinetic energy so we would not be making use of it further)

H = 3.9 m

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On a nice winter day at the South Pole, the temperature rises to −54°F. What is the approximate temperature in degrees Celsius?
Evgesh-ka [11]

Answer:

Its going to be about -47.78°C

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3 years ago
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As the concentration of a solute in a solution increases, the freezing point of the solution ________ and the vapor pressure of
kykrilka [37]

Answer:

As the concentration of a solute in a solution increases, the freezing point of the solution <u><em>decrease </em></u>and the vapor pressure of the solution <em><u>decrease </u></em>.

Explanation:

Depression in freezing point :

\Delta T_f=K_f\times m

where,

\Delta T_f =depression in freezing point =  

K_f = freezing point constant  

m = molality  ( moles per kg of solvent) of the solution

As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:

  1. If molality of the solution in high the depression in freezing point of the solution will be more.
  2. If molality of the solution in low the depression in freezing point of teh solution will be lower .

Relative lowering in vapor pressure of the solution is given by :

\frac{p_o-p_s}{p_o}=\chi_{solute}

p_o = Vapor pressure of pure solvent

p_s  = Vapor pressure of solution

\chi_{solute} = Mole fraction of solute

p_s\propto \frac{1}{\chi_{solute}}

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.

  1. Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
  2. lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.
8 0
2 years ago
On January 22, 1943, the temperature in Spearfish, South Dakota, rose from -4.0∘F∘F to 45.0∘F∘F in just 2 minutes. What was the
Margaret [11]

Answer:

The change in temperature, \Delta T=9.45^{\circ} C

Explanation:

Given that,

The temperature in Spearfish, South Dakota, rose from -4^{\circ} F\ to\ 45^{\circ} F in just 2 minutes. We need to find the temperature change in Celsius degrees. Change in temperature is given by final temperature minus initial temperature such that,

\Delta T=T_f-T_i\\\\\Delta T=45-(-4)\\\\\Delta T=49^{\circ}F

The relation between degrees Celsius and degrees Fahrenheit is given by :

F=1.8C+32

Here, F = 49 degrees

49=1.8C+32\\\\\Delta T=9.45^{\circ} C

So, the change in temperature is 9.45 degree Celsius. Hence, this is the required solution.

8 0
3 years ago
1. A 700 kg satellite is in orbit 2.4 x 106 meters from the center of the Earth. What is the force of
nevsk [136]

Answer:

4.8 \cdot 10^4 N

Explanation:

The force of gravity acting on the satellite is given by:

F=\frac{GMm}{r^2}

where

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

m is the mass of the satellite

r is the distance of the satellite from the Earth's centre

Here we have

m = 700 kg

r=2.4\cdot 10^6 m

Substituting into the equation, we find:

F=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(700)}{(2.4\cdot 10^6)^2}=4.8 \cdot 10^4 N

<em>Note that the distance mentioned in the problem (2.4 x 10^6 meters) is not realistic, since it is less than the radius of the Earth (6.37 x 10^6 meters).</em>

3 0
2 years ago
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