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2 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]2 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.

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Recalculate 100 km/h to m/s

qwelly [4]

= 27.777

Explanation:

A kilometer has 1,000 meters, and an hour has 3,600 seconds, so 100 kilometers per hour is: 100 x 1,000 / 3,600 = 27.777... m/s.

4 0

3 years ago

We have three identical metallic spheres A, B, C. Initially sphere A is charged with charge Q, while B and C are neutral. First,

larisa [96]

Answer:

The final charges of each sphere are: q_A = 3/8 Q , q_B = 3/8 Q , q_C = 3/4 Q

Explanation:

This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.

Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point

q_A = Q / 2

q_B = Q / 2

Now sphere A touches sphere C, ending with half the charge

q_A = ½ (Q / 2) = ¼ Q

q_B = ¼ Q

Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge

q = Q / 4 + Q / 2 = ¾ Q

This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q

q_A = 3/8 Q

q_B = 3/8 Q

The final charges of each sphere are:

q_A = 3/8 Q

q_B = 3/8 Q

q_C = 3/4 Q

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3 years ago

Imagine that you move a substance from one container to another and its volume changes. What state of matter is that substance?

Naddika [18.5K]

If it's volume changes when you move it to the new container it would be a solid

4 0

3 years ago

Read 2 more answers

A bat emitts a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away

Varvara68 [4.7K]

The time lapse between when the bat emits the sound and when it hears the echo is 0.05 s.

From the question given above, the following data were obtained:

Velocity of sound (v) = 343 m/s

Distance (x) = 8.42 m

Time (t) =?

We can obtain obtained the time as illustrated below:

v = 2x / t

343 = 2 × 8.42 / t

343 = 16.84 / t

Cross multiply

343 × t = 16.84

Divide both side by 343

t = 16.84/343

t = 0.05 s

Thus, the time between when the bat emits the sound and when it hears the echo is 0.05 s.

<h3>How does a bat know how far away something is?</h3>

A bat emits a sound wave and carefully listens to the echoes that return to it. The returning information is processed by the bat's brain in the same way that we processed our shouting sound with a stopwatch and calculator. The bat's brain determines the distance of an object by measuring how long it takes for a noise to return.

Learn more about time elapses between when the bat emits the sound :

<u>brainly.com/question/16931690</u>

#SPJ4

Correction question:

A bat emits a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away. How much time elapses between when the bat emits the sound and when it hears the echo? (Unit = s)

8 0

2 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

7 0

3 years ago