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3 years ago

Are neutrons a lot lower in mass than protons

1 answer:

3 0

The majority of the nucleon masses comes from quark interactions! the differences between the proton (udu) and neutron (udd) is that the has neutron's second down quark is heavier than the proton's second up quark. So the greater mass of this down quark gives the neutron a greater mass than the proton

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Open the attached file. It shows 2 waves (a blue one and an orange one) that are moving through the same medium at the same time

Taya2010 [7]

The diagram is labelled to DRAW the RESULTANT wave and LABEL the areas of interference with CI, DI and/or points of TDI.

<h3>What is refraction?</h3>

When a light ray falls on the interface of two medium like air and water, then the portion of light is reflected and another portion of the light is refracted into the water medium. The angle of incidence is equal to the angle of reflection but the angle of incidence is larger then the angle of refraction.

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Thus, the diagram is labelled.

Learn more about refraction.

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#SPJ1

3 0

2 years ago

The isotope of an atom containing 40 protons and 51 neutrons suddenly has 2 neutrons added to it. What isotope is created?

timama [110]

<span>The isotope of an atom containing 40 protons and 51 neutrons suddenly has 2 neutrons added to it
That is X-93 so it will be
</span><span>Zirconium-93
</span>hope it helps

7 0

3 years ago

Attempting to impress the skeptical patrol officer with your physics knowledge, you claim that you were traveling so fast that t

horsena [70]

Answer:

v_r = 1.268 × 10⁸ mi/hr

Explanation:

We are given;

wavelength of the red light; λr = 693 nm = 693 × 10^(-9) m

wavelength of the yellow light; λy = 582 nm = 582 × 10^(-9) m

Frequency is given by the formula;

f = v/λ

Where v is speed of light = 3 x 10^(8) m

Frequency of red light; f_o = [3 x 10^(8)]/(693 × 10^(-9)) = 4.33 x 10¹⁴ Hz

Similarly, Frequency of yellow light;

f = [3 x 10⁸]/(582 × 10^(-9)) = 5.15 x 10¹⁴ Hz

To find the speed of the car, we will use the formula;

f = f_o[(c + v_r)/c)]

Where c is speed of light and v_r is speed of car.

Making v_r the subject;

cf/f_o = c + v_r

v_r = c(f/f_o - 1)

So, plugging in the relevant values, we have;

v_r = 3 × 10⁸[((5.15 x 10¹⁴)/(4.33 x 10¹⁴)) - 1]

v_r = 3 × 10⁸(0.189)

v_r = 5.67 x 10⁷ m/s

Converting to mi/hr, 1 m/s = 2.23694 mile/hr

So, v_r = 5.67 × 10⁷ × 2.23694

v_r = 1.268 × 10⁸ mi/hr

5 0

4 years ago

DISPLACEMENT WITH CONSTANT ACCELEration. Given information: displacement=64m, acceleration=9.81m\s and time interval=3s. calcula

balandron [24]

HEYEYEYEYYEYEYDYDYDYD

3 0

3 years ago

A 37.5 kg box initially at rest is pushed 4.05 m along a rough, horizontal floor with a constant applied horizontal force of 150

vodomira [7]

Answer:

a) 607.5 J

b) 160.531875 J

c) 0 J

d) 0 J

e) 2.925 m\s

Explanation:

The given data :-

Mass of the box ( m ) = 37.5 kg.

Displacement made by box ( x ) = 4.05 m.

Horizontal force ( F ) = 150 N.
The co-efficient of friction between box and floor ( μ ) = 0.3
Gravitational force ( N ) = m × g = 37.5 × 9.81 = 367.875

Solution:-

a) The work done by applied force ( W )

W = force applied × displacement = 150 × 4.05 = 607.5 J

b) The increase in internal energy in the box-floor system due to friction.

Frictional force ( f ) = μ × N = 0.3 × 367.875 = 110.3625 N

Change in internal energy = change in kinetic energy.

ΔU = ( K.E )₂ - ( K.E )₁

Since the initial velocity is zero so the ( K.E )₁ = 0

ΔU = ( K.E )₂ = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

c) The work done by the normal force .

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

d) The work done by the gravitational force.

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

e) The change in kinetic energy of the box

( K.E )₂ - ( K.E )₁ = ( K.E )₂ - 0 = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

f) The final speed of the box

( K.E )₂ = 160.531875 J = 0.5 × 37.5 × v²

v² = 8.56

v = 2.925 m\s.

5 0

4 years ago