Ask question

Ask question

All categories

3 years ago

14

Which particles in an atom are acted upon by the strong force? A. Electrons and neutrons B. Neutrons and protons O C. Protons an

d electrons D. Protons, neutrons, and electrons SUE w

1 answer:

BARSIC [14]3 years ago

6 0

$\large \sf \pmb{B) \: Neutrons \: and \: Protons}$

<em>Neutrons And Protons</em> is a particles in an atom and are acted upon by the strong force.

You might be interested in

A roundabout is a type of playground equipment involving a large flat metal disk that is able to spin about its center axis. A r

Zina [86]

Answer:

116.1 kgm²/s

1.12718 rad/s

Decreases

Explanation:

m = Mass of girl = 43 kg

M = Mass of roundabout = 120 kg

v = Velocity of roundabout = 2.7 m/s

r = Radius of roundabout = 1 m = R

I = Moment of inertia

Her angular momentum

$L_i=mvr\\\Rightarrow L_i=43\times 2.7\times 1\\\Rightarrow L_i=116.1\ kgm^2/s$

Magnitude of angular momentum is 116.1 kgm²/s

Here the angular momentum is conserved

$L_f=L_i\\\Rightarrow I\omega=L_i\\\Rightarrow (\frac{1}{2}MR^2+mr^2)\omega=116.1\\\Rightarrow \omega=\frac{116.1}{\frac{1}{2}\times 120\times 1^2+43\times 1^2}\\\Rightarrow \omega=1.12718\ rad/s$

Angular speed of the roundabout is 1.12718 rad/s

Initial kinetic energy

$K_i=\frac{1}{2}mv^2\\\Rightarrow K_i=\frac{1}{2}43\times 2.7^2\\\Rightarrow K_i=156.735\ J$

Final kinetic energy

$K_f=\frac{1}{2}I\omega^2\\\Rightarrow K_f=\frac{1}{2}\times (\frac{1}{2}\times 120\times 1^2+43\times 1^2)\times 1.12718^2\\\Rightarrow K_f=65.43253\ J$

The overall kinetic energy decreases as can be seen. This loss is converted to heat.

5 0

3 years ago

According to the nebular theory, what early event eventually led to the formation of our solar system?

qaws [65]

Answer:Solar system formed about 4.6 billion year ago, when gravity pulled together low-density cloud of interstellar gas and dust (called a nebula)(movie). The Orion Nebula, an interstellar cloud in which star systems and possibly planets are forming. Initially the cloud was about several light years across.

Explanation:

4 0

3 years ago

A world class sprinter is travelling with speed 12.0 m/s at the end of a 100 meter race. Suppose he decelerates at the rate of 2

Solnce55 [7]

Answer:

after 6 second it will stop

he travel 36 m to stop

Explanation:

given data

speed = 12 m/s

distance = 100 m

decelerates rate = 2.00 m/s²

so acceleration a = - 2.00 m/s²

to find out

how long does it take to stop and how far does he travel

solution

we will apply here first equation of motion that is

v = u + at ......1

here u is speed 12 and v is 0 because we stop finally

put here all value in equation 1

0 = 12 + (-2) t

t = 6 s

so after 6 second it will stop

and

for distance we apply equation of motion

v²-u² = 2×a×s ..........2

here v is 0 u is 12 and a is -2 and find distance s

put all value in equation 2

0-12² = 2×(-2)×s

s = 36 m

so he travel 36 m to stop

3 0

3 years ago

A traveling electromagnetic wave in a vacuum has an electric field amplitude of 81.1 V/m. Calculate the intensity S of this wave

ICE Princess25 [194]

Answer:

U = 4.39 J

Explanation:

Electric field energy stored in the medium or vacuum is given as

$U = \frac{1}{2}\epsilon_0 E^2 V$

here we know that

$\epsilon_0 = 8.85 \times 10^{-12}$

E = 81.1 V/m

V = volume

$V = (0.0253)(speed \times time)$

$V = (0.0253)(3\times 10^8 \times 19.9)$

$V = 1.51 \times 10^8 m^3$

now from above formula we have

$U = \frac{1}{2}(8.85 \times 10^{-12})(81.1)^2(1.51 \times 10^8)$

$U = 4.39 J$

6 0

4 years ago

A closed box has two metal terminals a and b. The inside of the box contains an unknown emf 8 in series with a resistance R. Whe

Lelu [443]

Answer:

The value of $\epsilon$ and R are 12.9 ohm and 2.6 V.

Explanation:

Given that,

Potential difference = 22.0 V

Current = 1.50 A

If this potential difference is reversed,

Then the current = 1.90 A

In forward direction

The net emf is

$\epsilon=(\epsilon+22.0)\ V$

The net current is

$I=\dfrac{V}{R}$

$1.50=\dfrac{\epsilon+22.0}{R}$ ...(I)

In reverse direction

$1.90=\dfrac{22.0-\epsilon}{R}$ ...(II)

From equation (I) and (II)

$\epsilon =2.6\ V$

$R = 12.9\ Omega$

Hence, The value of $\epsilon$ and R are 12.9 ohm and 2.6 V.

8 0

4 years ago