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Ket [755]
3 years ago
7

A 25.0 kg child slides down a long slide in a playground. She starts from rest at a height h1 of 21.00 m. When she is partway do

wn the slide, at a height h2 of 8.00 m, she is moving at a speed of 7.80 m/s. Calculate the mechanical energy lost due to friction (as heat, etc.).
Physics
1 answer:
____ [38]3 years ago
4 0

Answer:

The mechanical energy lost due to friction is 2,424.5 J

Explanation:

Given;

mass of the child, m = 25 kg

intial velocity of the child, u = 0

final velocity of the child, v = 7.8 m/s

initial position of the child, h₁ = 21 m

final position of the child, h₂ = 8 m

Let the energy lost due to heat = ΔE

ΔE + ΔK.E  + ΔP.E = 0

ΔE  +   ¹/₂m(v² - u²)  +  mg(h₂ - h₁) = 0

ΔE   +   ¹/₂ x 25(7.8² - 0)    +   25 x 9.8(8 - 21) = 0

ΔE   +  760.5 J   - 3185 J =

ΔE   -  2,424.5 J = 0

ΔE = 2,424.5 J

Therefore, the mechanical energy lost due to friction is 2,424.5 J

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a 4050 N sled is being pushed with a force of 575 N. If it encounters a frictional force of 385 N… determine the net force
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According to the solving the net force is:

= 3,860.

<h3>Why is net force important?</h3>

The net force is the total of all the forces operating on an object, according to its definition. Mass can be accelerated by net force. Whether a body is at rest or in motion, another force is at work on it. When a system has a lot of forces acting on it, the phrase "net force" is employed.

<h3>What's the net force equation?</h3>

When a body is moving and multiple forces, such as the normal force, frictional force, and gravitational force, are acting on it, the combined effect is known as net force. As a result, is the net force formula. FNet equals Fa, Fg, Ff, and FN.

<h3>According to the given data:</h3>

4050 N sled

force of 575 N.

encounters a frictional force of 385 N

So the net force will be:

(Net)F = F₁ + F₂ + F₃

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(Net)F = F₁ + F₂ - F₃(frictional force)

(Net)F = 4050 + 385 - 575

= 3,860

According to the solving the net force is:

= 3,860.

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8 0
1 year ago
A right triangle has base b = 100 plusminus 1 ft and adjacent angle theta = 30 degree plusminus 0.5 degree. Calculate the height
aivan3 [116]

Answer:

The height h of this right triangle is 57.74±1.74ft.

Explanation:

Knowing that this is a right triangle, if one of the angles adjacent to the base, θ is 30º, the other angle is 90º. Therefore we can calculate the height h can be calculated with the tangent:

tan(\theta)=\frac{h}{b}\leftrightarrow h=tan(\theta)b=tan(30^\circ)100ft=57.74ft

For the uncertainty we use the  partial derivatives:

\Delta h=|\frac{dh}{db}|\Delta b+ |\frac{dh}{d\theta}|\Delta \theta\\\Delta h=|tan(\theta)|\Delta b+ |\frac{b}{cos^2(\theta)}|\Delta \theta

We have to be careful to use Δθ in radians:

\Delta h=|tan(30^\circ)|1ft+ |\frac{100ft}{cos^2(30^\circ)}|\frac{0.5^\circ2\pi}{360^\circ}=1.74ft

4 0
3 years ago
What does elements in the first two columns of the periodic table have in common​
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The first two columns on the periodic table make the s-block, and the elements in this block have in common that they tend to lose electrons to gain stability.
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3 years ago
Jus answer pls I’m done with physics
dybincka [34]

5. 35 - (-20) = 55

6. v_{av}=\frac{\Delta s}{\Delta t}=\frac{20-(-20)}{6-0} =\frac{40}{6} =6.67

7. D and G

8. A, C, and E

9.

At t = 19,

s=\frac{s_{final}-s_{initial}}{t_{final}-t_{initial}}(t_1-t_{initial})+s_{initial}=\frac{40-(-20)}{23-18}(19-18)-20=12-20= -8

At t = 27,

s=\frac{s_{final}-s_{initial}}{t_{final}-t_{initial}}(t_1-t_{initial})+s_{initial}=\frac{0-40}{29-26}(27-26)+40=\frac{-40}{3}+40=26.67

Therefore, displacement between t = 19 and t = 27 is:

26.67 - (-8) = 34.67

10. v=\frac{\Delta s}{\Delta t}=\frac{0-40}{29-26}=-13.33

11. -20

12. v=\frac{\Delta s}{\Delta t}=\frac{20-(-20)}{6-0}=\frac{40}{6}=6.67

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