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Stels [109]
3 years ago
15

A small object is attached to the end of a relaxed, horizontal spring whose opposite end is fixed. The spring rests on a frictio

nless surface. Let the initial position of the object be defined as x-0. The object is pulled to position x = A and then released, after which it undergoes simple harmonic motion.
In one full cycle of its motion, the total distance traveled by the object is:

1) A
2) A/2
3) 2A
4) 4A
5) A/4
Physics
1 answer:
eduard3 years ago
7 0

Answer:4A

Explanation:

Given

Mass is displace x= A units from its mean position x=0'

When it is set to free it will oscillate about its mean position with maximum amplitude A i.e. from x=-A to x=A

One cycle is completed when block returns to its original position

so first block will go equilibrium position x=0 and then to x=-A

from x=-A it again moves back to x=0 and finally back to its starting position x=A

so it travels a distance of A+A+A+A=4A    

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Describe how one plays Dr.Dogeball​
JulsSmile [24]
•To play Dr. Dodgeball you need to have 2 teams to verse each other.
•Next, select one person from each team to be the doctor (depending on the size of the teams you can have varying amounts of doctors)
•Continue to play dodgeball how you normally would
•When a player gets hit and is “out” they have to sit on the ground and wait for the doctor to “revive them” (this usually requires the doctor dragging,touching, or moving the player that is out to a “revival place” which is usually decided on by the advisor or person in charge.
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4 0
3 years ago
Real life examples of atomic physics
Flauer [41]

Answer:

magnets

static electricity

heat

7 0
2 years ago
An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 1017 J. Calculate
riadik2000 [5.3K]

Answer: 6.408(10)^{-19} C

Explanation:

This problem can be solved by the following equation:

\Delta K=q V

Where:

\Delta K=7.37(10)^{-17} J is the change in kinetic energy

V=115 V is the electric potential difference

q is the electric charge

Finding q:

q=\frac{\Delta K}{V}

q=\frac{7.37(10)^{-17} J}{115 V}

Finally:

q=6.408(10)^{-19} C

4 0
3 years ago
A race car makes one lap around a track of radius 50 m in 9.0 s. What is the average velocity? *
Oksi-84 [34.3K]

Given that,

Radius of track, r = 50 m

time , t = 9 s

velocity, v = ?

Distance covered by car in one lap around a track is equal to the circumference of the track.

C = 2 π r = 2 * 3.14 * 50

C = 314.159 m

Distance covered by car, s = 314.159 m

Velocity = distance/ time

V = 314.159 / 9

V = 34.9 m/s

The average velocity of car is 34.9 m/s.

7 0
2 years ago
Calcular la aceleración que produce una fuerza de 40 N sobre un cuerpo con 88 Kg de masa. Expresar el resultado en m⁄s^2 *
tigry1 [53]

Answer:

a = 0.45 m/s²

Explanation:

The given question is ''Calculate the acceleration that produces a force of 40 N on a body with 88 kg of mass".

Given that,

Force, F = 40 N

Mass of the body, m = 88 kg

The net force acting on the body is given by :

F = ma

Where

a is the acceleration of the body

a=\dfrac{F}{m}\\\\a=\dfrac{40\ N}{88\ kg}\\\\a=0.45\ m/s^2

So, the required acceleration is 0.45 m/s².

4 0
3 years ago
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