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3 years ago

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A 65.2-kg skier coasts up a snow-covered hill that makes an angle of 33.3 ° with the horizontal. The initial speed of the skier

is 8.05 m/s. After coasting a distance of 1.26 m up the slope, the speed of the skier is 4.54 m/s. (a) Find the work done by the kinetic frictional force that acts on the skis. (b) What is the magnitude of the kinetic frictional force?

1 answer:

Anarel [89]3 years ago

6 0

Answer:

a) -999J

b) -793N

Explanation:

We need to apply the conversation of energy equation, we will take the potencial energy at the bottom as zero.

$u1+k1+W_{friction}=u2+k2$

now rearranging the equation:

$W_{friction}=u2+k2-k1\\W_{friction}=m*g*h+\frac{1}{2}*m*(v_2)^2-\frac{1}{2}*m*(v_1)^2\\W_{friction}=65.2*1.26*9.80*sin(33.3)+\frac{1}{2}*65.2*(4.54)^2-\frac{1}{2}*65.2*(8.05)^2\\W_{friction}=-999J$

we know that work is:

$W=F*d\\F=\frac{W}{d}\\\\F=\frac{-635}{1.26}\\F=-793N$

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A rectangular reflecting pool is 56.5 ft wide and 141 ft long. What is the area of the pool in square meters

konstantin123 [22]

By definition, the area of the pool is 740.112 m².

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The area is the measure of a space delimited by a contour which is called perimeter.

That is, the area gives an idea of how much surface a figure covers and is defined as the measure of the region or surface enclosed by a geometric figure.

<h3>Area of a rectangle</h3>

A rectangle is a flat geometric figure with four sides, of which two sides that are opposite parallel to each other have the same length and the remaining two have another length.

The area of a rectangle is calculated by multiplying the lengths of two contiguous sides of the figure which are different from each other, that is, the length is multiplied by the width:

Area= length× width

<h3>This case</h3>

A rectangular reflecting pool is 56.5 ft wide and 141 ft long. So, the area of the rectangular pool is calculated as:

Area= length× width

Area= 141 ft× 56.5 ft

Solving:

<u><em>Area= 7966.5 ft²= 740.112 m²</em></u> (being 1 ft²= 0.092903 m²)

So, the area of the pool is 740.112 m².

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3 years ago

Calculate the change in internal energy (ΔE) for a system that is giving off 25.0 kJ of heat and is changing from 12.00 L to 6.0

Alexxandr [17]

Explanation:

The given data is as follows.

q = -25.0 kJ, Pressure (P) = 1.50 atm

$\Delta V$ = (12 - 6) L = 6 L

Therefore, product of pressure and change in volume will be as follows.

$P \Delta V = 1.50 atm \times 6 L$

= 7.5 L atm

= $7.5 \times 101.3$

= 759.75 J

Now, we will calculate the change in internal energy as follows.

$\Delta E = q + w$

= $q + P \Delta V$

= -25000 kJ + 759.75 J

= 24240.25 J

or, = 24.240 kJ (as 1 kJ = 1000 J)

Thus, we can conclude that the change in internal energy ( $\Delta E$ ) for a system is 24.240 kJ.

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4 years ago

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

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At x = 1.00 m

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= 18J

Total energy will be =

4J + 18J = 22J

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$= \frac{4-10}{25} = \frac{-6}{25} J$

= -0.24J

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$\frac{1}{2} * 4Vf^2$

= 2Vf²

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2Vf² - 0.024

Using conservation of energy,

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22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

$Vf = \sqrt{frac{22.4}{2}$

= 3.33 m/s

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$\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}$

$= \frac{2}{x^2} - \frac{8}{x^3}$

At x = 5.0 m

$\frac{2}{5^2} - \frac{8}{5^3}$

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