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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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What is an example of kinetic energy

Olenka [21]

An airplane has a large amount of kinetic energy in flight due to its large mass and fast velocity.

6 0

3 years ago

An 800-kHz radio signal is detected at a point 2.7 km distant from a transmitter tower. The electric field amplitude of the sign

dimaraw [331]

Answer:

Option D is correct: 170 µW/m²

Explanation:

Given that,

Frequency f = 800kHz

Distance d = 2.7km = 2700m

Electric field Eo = 0.36V/m

Intensity of radio signal

The intensity of radial signal is given as

I = c•εo•Eo²/2

Where c is speed of light

c = 3×10^8m/s

εo = 8.85 × 10^-12 C²/Nm²

I = 3×10^8 × 8.85×10^-12 × 0.36²/2

I = 1.72 × 10^-4W/m²

I = 172 × 10^-6 W/m²

I = 172 µW/m²

Then, the intensity of the radio wave at that point is approximately 170 µW/m²

7 0

3 years ago

A mass m is tied to an ideal spring with force constant k and rests on a frictionless surface. The mass moves along the x axis.

7nadin3 [17]

Answer: $x=\frac{x_m}{\sqrt{2}}$

Explanation:

Given

initially mass is stretched to $x_m$

Let k be the spring Constant of spring

Therefore Total Mechanical Energy is $\frac{kx_m^2}{2}$

Position at which kinetic Energy is equal to Elastic Potential Energy

$K=\frac{mv^2}{2}$

$U=\frac{kx^2}{2}$

it is given

$k=U$

thus $2U=\frac{kx_m^2}{2}$

$2\times \frac{kx^2}{2}=\frac{kx_m^2}{2}$

$2x^2=x_m^2$

$x=\frac{x_m}{\sqrt{2}}$

3 0

3 years ago

Which of the following best describes the way that scientists make observations

horrorfan [7]

I don’t see any answer choice but the best way is asking questions about the natural phenomenon, making hypothesis, and predicting the consequences in the hypothesis.

6 0

3 years ago

You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped a

AfilCa [17]

Answer:

Explanation:

a Downward acceleration of car A along the slope

= g sinθ - μ g cosθ

= g ( sinθ - μ cosθ)

= 9.8 ( sin 12 - .6 x cos 12 )

= 9.8 x ( .2079 - .5869 )

= - 3.714 m / s²

So there will be deceleration

v² = u² - 2 a s

= 18² - 2 x 3.714 x 24

= 324 - 178

= 146

v = 12 .08 m /s

b )

In the second case , kinetic friction changes

downward acceleration

= g ( sinθ - μ cosθ)

= 9.8 ( sin12 - .1 x cos 12 )

9.8 ( .2079 - .0978 )

= 1.079 m /s

there will be reduced acceleration

v² = u² - 2 a s

= 18² +2 x1.079 x 24

= 324 + 52

= 376

v = 19.4 m /s

7 0

3 years ago