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EastWind [94]
3 years ago
15

A 0.547 kg pizza is thrown straight up in the air. At a height of 2.30 m above the surface of the earth it has a speed of 5.00 m

/s. Calculate the total mechanical energy of the pizza crust.
Physics
1 answer:
natima [27]3 years ago
6 0

Answer:

The total mechanical energy of the pizza crust is 19.2 J.

Explanation:

Mechanical energy is that which a body or a system obtains as a result of the speed of its movement or its specific position, and which is capable of producing mechanical work. Then:

Potential energy + kinetic energy = total mechanical energy

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a certain mass and in a position of rest, until it reaches a certain speed.

Kinetic energy is represented by the following formula:

Ec = ½ *m*v²

Where Ec is kinetic energy, which is measured in Joules (J), m is mass measured in kilograms (kg), and v is velocity measured in meters over seconds (m / s).

In this case:

  • m=0.547 kg
  • v= 5 m/s

Replacing:

Ec = ½ *0.547 kg*(5 m/s)²

and solving you get:

Ec= 6.8375 J

On the other hand, potential energy is the energy that measures the ability of a system to perform work based on its position. In other words, this is the energy that a body has at a certain height above the ground.

Gravitational potential energy is the energy associated with the gravitational force. This will depend on the relative height of an object to some reference point, the mass, and the force of gravity. Then for an object with mass m, at height h, the expression applied to the gravitational energy of the object is:

Ep = m*g*h

Where Ep is the potential energy in joules (J), m is the mass in kilograms (kg) is h the height in meters (m) and g is the acceleration of fall in m / s² (approximately 9.81 m/s²)

In this case:

  • m= 0.547 kg
  • g= 9.81 m/s²
  • h= 2.30 m

Replacing

Ep= 0.547 kg *9.81 m/s²* 2.30 m

and solving you get:

Ep= 12.342 J

So:

Total mechanical energy= 12.342 J + 6.8375 J

Total mechanical energy= 19.1795 J≅ 19.2 J

<u><em>The total mechanical energy of the pizza crust is 19.2 J.</em></u>

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10. The electron dot diagram for the element Ne would have ​
zmey [24]

Answer:

2,8

Explanation:

The first electron shell would have 2 electrons, the second shell would have 8 electrons. This is because Neon has a relative charge of 10.

7 0
2 years ago
Question 5 (Multiple Choice Worth 3 points) (02.07 MC) Rachel needs to eat fewer carbohydrates to improve her health. Which of t
patriot [66]
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5 0
3 years ago
While finding the spring constant, if X1 = 12 cm, X2 = 15 cm, and hanging mass = 22 grams, the value of spring constant K would
Pavel [41]

Answer:

If x₁=12 cm then k=1.7985 N/m

If x₂=15 cm then k=1.4388 N/m

Explanation:

Hanging mass= 22 g=0.022 kg

Acceleration due to gravity g=9.81 m/s²

If x₁=displacement= 12 cm=0.12 m

k= spring constant

F=ma\\\Rightarrow F=0.022\times 9.81\\\Rightarrow F=0.21582\ N

\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.012\\\Rightarrow k=1.7985\ N/m\\

∴k = 1.7985 N/m

If x₂=15 cm=0.15 m

Force of the hanging mass is same however the spring constant will change

\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.015\\\Rightarrow k=1.4388\ N/m\\

∴k = 1.4388 N/m

As the mass is not changing the spring constant has to change. That means that here there are two spring one with k=1.7985 N/m and the other with k= 1.4388 N/m

4 0
3 years ago
Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o
sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

a_{x} = -\omega^{2}  x

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders  is given as:

                                  I=\frac{1}{2}MR^{2} -----(1)

Newton's second law for angular motion is:

                                             ∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

                                             ∑Fₓ = Maₓ ------ (3)

By definition of torque:

                                               τ  = RF --------(4)        

Put (4) and (1) in (2)

                                       RF=\frac{1}{2}MR^{2}\alpha

                                       RF=\frac{1}{2}MR^{2}\alpha

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

                                   f_{s}=\frac{1}{2}MR\alpha------ (5)

As angular acceleration is related to linear by:

                                          a= R\alpha

Eq (5) becomes

                                    f_{s}=\frac{1}{2}Ma_{x}---- (6)

aₓ shows displacement in horizontal direction

From (3)

                                              ∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

                                  \sum F_{x} = f_{s} - kx ----(7)

Put (7) in (3)

                                  f_{s} - kx  = Ma_{x}[/tex] -----(8)

Using (6) in (8)

                               \frac{1}{2}Ma_{x} - kx =Ma_{x}

                                     a_{x} = \frac{2k}{3M} x --- (9)

For spring mass system

                                  a= -\omega^{2} x ----- (10)

Equating (9) and (10)

                                  \omega^{2} = \frac{2k}{3M}

\omega = \sqrt{ \frac{2k}{3M}}

then (9) becomes

                                a_{x} = - \omega^{2}x

(The minus sign says that x and  aₓ  have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

                                   T=\frac{2\pi }{\omega}

                                  T = 2\pi \sqrt{\frac{3M}{2k}

                           

 

5 0
3 years ago
An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f
sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

f=\frac{v_s}{2L}         (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

f=\frac{343m/s}{2(0.47228m)}=362.36Hz      

Closed tube:

f'=\frac{v_s}{4L'}

L': length of the closed tube = 0.702821m

f'=\frac{343m/s}{4(0.702821m)}=122.00Hz

Next, you use the following formula for the beat frequency:

f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

7 0
3 years ago
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