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3 years ago

A penny is dropped from the top of a building 290 m high. Ignoring air resistance, if

Physics

1 answer:

Kryger [21]3 years ago

5 0

Answer:

1. Velocity = 75.39 m/s

2. Air resistance

3. Mass of the coin

Explanation:

1. Given that the height of the building is 290 m. From Newton's third law of motion,

$V^{2}$ = $U^{2}$ + 2gh

where: V is the final velocity of the coin, U is the initial velocity, g is the force of gravity and h is the height.

Since the coin was dropped from a height, its initial velocity is zero. The force of gravity is taken as 9.8 m/ $s^{2}$ , so that:

$V^{2}$ = 0 + 2 x 9.8 x 290

= 5684

V = $\sqrt{5684}$

= 75.3923

The velocity with which the coin hits the ground is 75.39 m/s.

2. Air resistance: During the free fall of the coin, the direction of wind flow could either cause an increase or decrease the value predicted.

ii. Mass of the coin: This can also affect the predicted value.

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Difference between output work and input work

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3 years ago

In a physics experiment, a ball is released from rest, and it falls toward the ground. The timer was not paying attention but es

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Answer:

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(B) –42.0m

Explanation:

The complete solution can be found in the attachment below.

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3 years ago

A transverse wave on a string is described by the wave functiony(x, t) = 0.350 sin (1.25x + 99.6t)where x and y are in meters an

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The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

<h3>How to solve for the time interval</h3>

We have y = 0.175

y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.5

99.62 = pi/6

t1 = 5.257 x 10⁻³

99.6t = pi/6 + 2pi

= 0.0683

The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

b. we have k = 1.25, w = 99.6t

v = w/k

99.6/1.25 = 79.68

s = vt

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complete question

A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?

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2 years ago

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2 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:

If molality of the solution in high the depression in freezing point of the solution will be more.
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 :

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

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$p_s$ = 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.

Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
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