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galben [10]
3 years ago
8

julia throws a ball vertically upward from the ground with a speed of 5.89m/s. Andrew catches it when it is on its way down at a

height of 1.27m from the ground. After how much time does Andrew catch the ball?
Physics
1 answer:
aliya0001 [1]3 years ago
8 0
Vo = 5.89 m/s Y = 1.27 m g = 9.81 m/s^2 
Time to height 
Tr = Vo / g Tr = (5.89 m/s) / (9.81 m/s^2) Tr = 0.60 s 
Max height achieved is:
H = Vo^2 / [2g] H = (5.89 )^2 / [ 2 * (9.81) ] H = (34.69) / [19.62] H = 1.77 m 
It falls that distance, minus Andrew's catch distance:
h = H - Y h = (1.77 m) - (1.27 m) h = 0.5 m 
Time to descend is therefore:
Tf = √ { [2h] / g ] Tf = √ { [ 2 * (0.5 m) ] / (9.81 m/s^2) } Tf = √ { [ 1.0 m ] / (9.81 m/s^2) } Tf = √ { 0.102 s^2 } Tf = 0.32 s 
Total time is rise plus fall therefore:
Tt = Tr + Tf Tt = (0.60 s) + (0.32 s) Tt = 0.92 s           (ANSWER)
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MariettaO [177]

1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.

<u>Explanation</u>:

We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the (velocity)^2 and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:

\text { Kinetic Energy }=\frac{1}{2} m v^{2}

Better understood from numerical example as given:

If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?

This can be solved as follows:

\text { Kinetic Energy of } \mathrm{A}=\frac{1}{2} 50 \times 5^{2}=625 \mathrm{J}

\text { Kinetic Energy bf } \mathrm{B}=\frac{1}{2} 100 \times 2.5^{2}=312.5 \mathrm{J}

It shows that man A will have more K.E.

Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.

4 0
3 years ago
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.
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3 years ago
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Vladimir [108]

Answer:

La energía interna es el resultado de la contribución de la energía cinética de las moléculas o átomos que lo constituyen, de sus energías de rotación, traslación y vibración, además de la energía potencial intermolecular debida a las fuerzas de tipo gravitatorio, electromagnético y nuclear.

Explanation:

4 0
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Vesna [10]

Answer:

The time is 106.7 minute.

Explanation:

Given that,

Density = 8.4\times10^{28}\ m^3

Current i = 4.6\times10^{18}\ electron/s

Diameter of wire = 1.2 mm

Length = 31 cm

We need to calculate the drift velocity

Using formula of drift velocity

v_{d}=\dfrac{I}{neA}

v_{d}=\dfrac{Ne}{tne\times\pi r^2}

Put the value into the formula

v_{d}=\dfrac{4.6\times10^{18}}{8.4\times10^{28}\times\pi\times(0.6\times10^{-3})^2}

v_{d}=4.842\times10^{-5}\ m/s

We need to calculate the time

Using formula for time

v_{d}=\dfrac{l}{t}

t=\dfrac{l}{v_{d}}

Where, l = length

v_{d} = drift velocity

Put the value into the formula

t=\dfrac{31\times10^{-2}}{4.842\times10^{-5}}

t=6402.31\ sec

t=106.7\ minute

Hence, The time is 106.7 minute.

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