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AURORKA [14]
3 years ago
7

Where does the majority of the mass of an atom come from

Physics
2 answers:
asambeis [7]3 years ago
5 0

Answer:

The Nucleus

Explanation:

The nucleus contains the majority of an atom's mass because protons and neutrons are much heavier than electrons, whereas electrons occupy almost all of an atom's volume.  I hope this helps you :D

sergey [27]3 years ago
5 0
What the one up there said
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Because you can't physically see the difference.
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The following questions present a twist on the scenario above to test your understanding. Suppose another stone is thrown horizo
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The first part of the text is missing, you can find on google:

"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."

Let's now solve the different parts.

(a) 3.03 s

The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by

y(t) = h -\frac{1}{2}gt^2

where

h = 45 m is the initial height

g = 9.8 m/s^2 is the acceleration of gravity

When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:

0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s

(b) 18.5 m/s

For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.

The horizontal position is given by

x=v_x t

where

v_x is the horizontal speed, which is constant

t is the time

At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:

v_x = \frac{x}{t}=\frac{56}{3.03}=18.5 m/s

The ball was thrown horizontally: this means that its initial vertical speed was zero, so 18.5 m/s was also its initial overall speed.

(c) 35.0 m/s at 58.1 degrees below the horizontal

At the impact, we know that the horizontal speed is still the same:

v_x = 18.5 m/s

we need to find the vertical velocity. This can be done by using the equation

v_y = u_y -gt

where

u_y =0 is the initial vertical velocity

g is the acceleration of gravity

t is the time

Substituting t = 3.03 s, we find the vertical velocity at the time of impact:

v_y = -(9.8)(3.03)=-29.7 m/s

So the magnitude of the velocity at the impact (so, the speed at the impact) is

v=\sqrt{v_x^2+v_y^2}=\sqrt{18.5^2+(-29.7)^2}=35.0 m/s

The angle instead can be found as:

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-29.7}{18.5})=-58.1^{\circ}

so, 58.1 degrees below the horizontal.

4 0
3 years ago
The graph shows how the vertical velocity of a parachutist changes from the moment the parachutist jumps from the aircraft until
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Answer:

BUT WHY

Explanation:

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4 years ago
At what point in a projectile’s trajectory does it experience a minimum velocity?
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Minimum velocity occurs at the top of the trajectory, where
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8 0
3 years ago
Two point charges of +20.0 μC and -8.00 μC are separated by a distance of 20.0 cm. What is the intensity of electric field E mid
algol13

Answer:

The intensity of the net electric field will:

E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C

Explanation:

Here we need first find the electric field due to the first charge at the midway point.

The electric field equation is given by:

|E_{1}|=k\frac{q_{1}}{d^{2}}

Where:

  • k is Coulomb's constant
  • q(1) is 20.00 μC or 20*10⁻⁶ C
  • d is the distance from q1 to the midpoint (d=10.0 cm)

So, we will have:

|E_{1}|=(9*10^{9})\frac{20*10^{-6}}{0.1^{2}}

|E_{1}|=1.8*10^{7}\: N/C

The direction of E1 is to the right of the midpoint.

Now, the second electric field is:

|E_{2}|=k\frac{q_{2}}{d^{2}}

|E_{2}|=(9*10^{9})\frac{8*10^{-6}}{0.1^{2}}

|E_{2}|=7.2*10^{6}\: N/C

The direction of E2 is to the right of the midpoint because the second charge is negative.

Finally, the intensity of the net electric field will:

E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C

I hope it helps you!

5 0
3 years ago
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