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Bad White [126]
3 years ago
9

The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s

= 5 sin(πt) + 5 cos(πt), where t is measured in seconds. (Round your answers to two decimal places.) (a) Find the average velocity during each time period. g
Physics
1 answer:
Veseljchak [2.6K]3 years ago
4 0

Answer:

v_{avg} = 0

Explanation:

The object is moving to and fro about its mean position

and the equation for its displacement is given as

s = 5 sin(\pi t) + 5 cos(\pi t)

now in one complete time period we know that the the object will complete its one oscillation

So the change in the position of the object in one complete time period will be ZERO

so here we can say that

v_{avg} = \frac{displacement}{time}

since change in position is zero

so displacement = 0

v_{avg} = 0

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A constant net torque is applied to a rotating object. Which of the following best describes the object's motion? A constant net
ahrayia [7]

Answer:

The object will rotate with constant angular acceleration

Explanation:

According to the Newton's Second Law for Whenever there is more than one torque acting on a rigid body that posses fixed axis, the moment of inertia as well as the angular acceleration is equals or proportional to the summation of the torques. It gives details on the relationship between rotational kinematics and torque as well as moment of inertia. This can be represented by the below equation.

∑iτi=Iα.

.Therefore when constant net torque is applied to object that is rotating, the object will rotate with constant angular acceleration

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Which of the following is an example of an endothermic reaction?
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D, photosynthesis

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A. two forces

B. P-f

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The best time to get my RESTING HEART RATE is right after I run.<br><br>true or false ​
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Explanation:

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2 years ago
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En un recipiente de cobre de forma rectangular de aproximadamente 3.5 m largo por 4.5m ancho tiene una temperatura en temporada
Darina [25.2K]

Answer:

A_f= 15,769 m²

Explanation:

Este es un ejercicio de dilatación térmica,  

          ΔA = (2α) A₀ ΔT

el arrea de recipiente

          A₀ = L A

          A₀ = 3,5 4,5

          A₀=  15,75 m²

el coeficiente de dilatación térmica es alfa = 16,6 10⁻⁶ C⁻¹

calculemos

            ΔA = 2 16,6 10⁻⁶  15,75 ( 40 -4)

            ΔA = 522,9 (36)  10⁻⁶  

            A= 1,88 10⁻²  m2

el cambio de volumen es

          ΔA = A_f – A₀

          A_f = A₀ +  ΔA

          A_f=  15,75 +1,88 10⁻²  

          A_f= 15,769 m²

3 0
2 years ago
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