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Rudiy27
3 years ago
13

An athlete stretches a spring an extra 28.6 cm beyond its initial length. how much energy has he transferred to the spring, if t

he spring constant is 52.9 n/cm?
Physics
1 answer:
g100num [7]3 years ago
6 0
PE = 0.5 × k × x²

PE potential Energy
k spring constant
x stretch/compression of the spring
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The lons entering the mass spectrometer have the same charges. After being accelerated through a potential difference of 8.20 kV
Ratling [72]

The calculated magnitude is  6.73 x 10³ V/m.

AMU is described as being one-twelfth the mass of a carbon-12 atom (12C). C makes up more than 98% of the carbon that can be found in nature, making it the most prevalent isotope. The magnitude of the field is the change in potential across a small distance in the indicated direction divided by that distance.

Potential difference = 8.20 kV= 8.20 x 10³ V

radius= 19.4/100=0.194 m

total distance that is circumference of the circle= 2πr =2 x 3.14 x 0.194

                                                                               = 1.218 m

therefore Magnitude= 8.20 x 10³ / 1.218

                                  =6.73 x 10³ V/m

Learn more about Magnitude here-

brainly.com/question/15681399

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4 0
1 year ago
1- ¿Cuál sería la energía de un objeto de 50 newton de peso que se encuentra sobre una estantería de 3 metros de altura? ¿Qué ti
Liula [17]

Responder:

<h3>150 Nm </h3><h3>Energía potencial </h3>

Explicación:

El tipo de energía que posee el objeto se conoce como energía potencial. <u>La energía potencial es la energía que posee un objeto, mi virtud de su posición. </u>

Energía potencial = masa * aceleración debido a la gravedad * altura

Dado que Force = masa * aceleración debido a la gravedad

Energía potencial = Fuerza * altura

Fuerza dada = 50N y altura = 3 m

Energía potencial = 50 * 3

Energía potencial = 150 Nm

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3 years ago
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3 years ago
In which mechanical test is a specimen deformed with a gradually increasing load that is applied uniaxially along the long axis
IRINA_888 [86]

Answer:

Compression Test

Explanation:

The Specimen is undergoing a compression test. It is similar to tensile test with the difference that the force is compressive and applied along the direction of stress. Both Tensile and compression tests are performed on Universal Testing machine. Compression test is done to determine the product's reaction when it is compressed, squashed and crushed.

7 0
3 years ago
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