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HACTEHA [7]
2 years ago
4

A 0.800-kg ball is tied to the end of a string 1.60 m long and swung in a vertical circle. (a)During one complete circle, starti

ng anywhere, calculate the total work done on the ballby (i) the tension in the string and (ii) gravity. (b) Repeat part (a) for motion along thesemicircle from the lowest to the highest point on the path.
Physics
1 answer:
kipiarov [429]2 years ago
5 0

Answer: a) i) 0 ii) 0  b) i) 0 ii) -25.1 J

Explanation:

a) For the first part, as work is defined as the process through an applied force produces a displacement in the direction of the movement, if the start and the end points are the same (as for a complete circle), this means that there is no displacement, i.e. total work (for any applied force) is 0.

b) Taking only a semicircle from the lowest to the highest point on the path, as tension force is always perpendicular to the displacement, we can conclude that the tension force does no work.

Now, for gravity, the displacement, is just the difference in height between the highest and lowest point, which is equal to twice the length of the string, and is a vector with only a vertical component, as follows:

Δy = y₂ - y₁ (by definition of displacement)

Now, the work done by gravity is just the product of the weight of  the ball, times the vertical displacement (as the work done by gravity is independent of the trajectory) , as follows :

W = m. g. (y₂-y₁) = 0.8 Kg. 9.8 m/s². (-2. 1.6 m) = -25.1 J

(The minus sign is due to gravity and the displacement point in opposite directions, in this case we take as positive the downward direction).

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weqwewe [10]

Answer:

a) v =  1.19 m / s , b)   P₁ = 0.922 10⁵ Pa

Explanation:

1) Let's use the fluid continuity equation

       Q = A v

The area of ​​a circle is

      A = π r2 = π d²/4

     

     v = Q / A = Q 4 / pi d²

     v = 0.006 4/π 0.08²

     v =  1.19 m / s

2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point

     P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rho g y₂

The exercise tell us

P₂ = 1.0013 105 Pa

v₁ = 0

y₁ = 1 m

y₂=0  

Rho (water) = 1000 kg / m³

      P₁ + rho y₁ = P₂ + ½ rho v₂²

      P₁ = P₂ + ½ rho v₂² - rho g y₁

      P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1

      P₁ = 1.013 10⁵ +708.5  - 9800

      P₁ =  92208.5Pa

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8 0
2 years ago
A curtain hangs straight down in front of an open window. A sudden gust of wind blows past the window; and the curtain is pulled
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Answer:

option B.

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The correct answer is option B.

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According to Bernoulli's Principle when the speed of the moving fluid increases the pressure within the fluid decrease.

When wind flows in the outside window the pressure outside window decreases and pressure inside the room is more so, the curtain moves outside because of low pressure.

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Explanation:

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2 years ago
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A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15° below the
Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

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Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ F_x=Fcos(\theta)          and             ⇒ F_y=Fsin (\theta)

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⇒ \sum F_x=F_n_e_t =F-f

⇒ F_n_e_t =F-f

⇒ ma =F-f       <em>  ...Newtons second law Fnet = ma</em>

⇒ a =\frac{F-f}{m}              

⇒ Plugging the values.

⇒ a =\frac{15cos(15)-0}{40}     <em>...f is the friction which is zero here.</em>

⇒ a =\frac{14.48}{40}

⇒ a=0.362\ ms^-^2

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0
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