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3 years ago

14

Is work required to pull a nucleon out of an atomic nucleus? Does the nucleon, once outside the nucleus, hove more mass than it

had inside the nucleus?

1 answer:

olga2289 [7]3 years ago

6 0

<span>Work is required to pull a nucleon out of an atomic nucleus. It has more mass outside the nucleus.</span>

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Tory has a mass of 40kg. She sleds down a hill that has a slope of 25 degrees. what is the component of her weight that is along

Fudgin [204]

1.7 x 10^2 N

or 166 N

First you find the vertical component of the weight, which is 9.8*40, (g*m), which is 392 N. You then find the angle between that and the slope, which is 90-25, which is 65. You then multiply the vertical weight by cos(65), to find the component of that that is parallel to the slope. You get 165.666 N

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3 years ago

The field used in the Canadian football League (CFL) has the midfield marker at the 55 yard line.how long is the fiend from goal

kogti [31]

Answer:

110 yds

Explanation:

Well if 55 yards is 1/2 of the field then 2 x 55 = 110 yards is total field length

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1 year ago

A light wave travels through air at a speed of 3.0x108 m/s. Green light has a wavelength of about 5.76x1014Hz. What is the wavel

solong [7]

Answer:

521 nm

Explanation:

Given the values and units we are given, I'm assuming 5.76*10^14 Hz is frequency.

The formula to use here is λ * υ = c, where λ is wavelength, υ is frequency, and c is the speed of light.

λ = $\frac{3*10^8\frac{m}{s} }{5.76*10^{14}Hz} = {5.20833*10^{-7} m}\approx{521 *10^{-9}m}={521 nm}$

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2 years ago

A horizontal net force F is exerted on an object at rest. The object starts at x=0 m and has a speed of 4.0 m/s after moving 4.0

Alex_Xolod [135]

Explanation:

The solution is be found in the attachment.

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3 years ago

. Consider a fully extended arm that is rotating about the shoulder such that with a shoulder-to-hand length of 30 cm. If the ar

Rainbow [258]

Answer:

13.309 m/s²

Explanation:

Length from shoulder to hand, l = 30 cm = 0.3 m

initial velocity, u = 1 m/s

final velocity, v = 2.5 m/s

time, t = 3 s

Let the tangential acceleration is a.

by using first equation of motion

v = u + at

2.5 = 1 + 3 a

a = 0.5 m/s²

Let the centripetal acceleration is a'.

a' = v'²/l

a' = 2 x 2 / 0.3

a' = 13.3 m/s²

The tangential acceleration and the centripetal acceleration are both perpendicular to each other. So, the net acceleration is given by

$A=\sqrt{a^{2}+a'^{2}}$

$A=\sqrt{0.5^{2}+13.3^{2}}$

A = 13.309 m/s²

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3 years ago