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DaniilM [7]
3 years ago
6

Bodies A and B have equal mass. Body B is initially at rest. Body A collides with body B in a one-dimensional elastic collision.

Which statement is true?
A. BODY A COMES TO REST BODY B STARTS MOVING WITH INITIAL VELOCITY OF BODY A

B. THE MOMENTUM OF THE SYSTEM IS NOT CONSERVED

C.THE KINETIC ENERGY OF THE SYSTEM IS NOT CONSERVED

D.THE FINAL VELOCITY
Physics
2 answers:
inn [45]3 years ago
6 0
After looking on the scaleattached with the question I recognized the right answer in the first option represented above. A. BODY A COMES TO REST BODY B STARTS MOVING WITH INITIAL VELOCITY OF BODY A - is definitely the only correct answer which perfectly concides with the given task.Hope you still need the answer it because this one I gave you is really useful.
jek_recluse [69]3 years ago
4 0
According to the statement " Collision <span>between two bodies in which the total kinetic energy of the two bodies after the collision is equal to their total kinetic energy before the collision."
The best answer is :
Option A " </span><span>BODY A COMES TO REST BODY B STARTS MOVING WITH INITIAL VELOCITY OF BODY A "</span>
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Which phrase describes an electromagnetic wave
Viefleur [7K]

Answer:

<u>a transverse wave consisting of changing electric fields and changing magnetic fields.</u>

Explanation:

An electromagnetic wave is a wave generated by the vibration of perpendicular electric and magnetic fields, which may progate through vacuum (empty space) or a material medium.

All electromagnetic waves propagate at the same speed in vacuum. This speed is approximately 3.0 × 10⁸ m/s. Which is generally referred as the speed of light, but it is the same constant speed of any electromagnetic wave in the vacuum, c.

In general, waves transfer energy when they travel, but only electromagnetic waves can travel in vacuum. The waves that cannot travel in vacuum are named mechanical waves (they need a medium to travel).

There are two types of waves depending on how they propagate: transverse waves and longitudinal waves. The transverse waves travel perperdiculary to the direcction of the vibration, while longitudinal waves travel parallel to the direction of the vibration.

The classical example of transverse waves is a rope that oscilates up and down. The classical example of  longitudinal waves is a spring that you pull and push by an end and so it  moves forward and back. Sound is also a longitudinal wave.

4 0
2 years ago
SOMEBODY PLEASE HELP!! 20 PTS!!!
aliya0001 [1]

Answer:

1. Reflection

2. travel from one medium to another

3. Same waves to travel in opposite direction.  

Explanation:

1. When a wave strikes a solid barrier, it bounces back in the same medium. This wave behavior of bouncing back is known as reflection. Its like a basketball hitting a backboard. The ball bounces back at the same angle as it was incident. ∠i = ∠r

2. For refraction to occur in a wave, the wave must travel from one medium to another. When light travels from through mediums of different optical densities, it bends. The wave bends away normal when it enters from denser medium to rarer medium. The wave bends towards the normal when it enters from rarer to denser medium. The angle of refraction and angle of incidence are related by Snell's law.

\frac{sin(i)}{sin(r)} = \frac{\mu_2}{\mu_1}

3. The formation of standing wave requires two same waves to travel in the opposite direction and interfere. The incident wave and reflected wave when interfere, form standing waves. There waves are also resonances or harmonics. A standing wave oscillates at one place and does not transfers any energy.

3 0
2 years ago
Read 2 more answers
two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i
NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 - 14.7t - 4.9t^2

4.9t^2 + 14.7t - 19.6 = 0

t^2 + 3t - 4 = 0

(t + 4)(t - 1) = 0

(t - 1) = 0

\boxed {t = 1 ~ second}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 + 14.7t - 4.9t^2

4.9t^2 - 14.7t - 19.6 = 0

t^2 - 3t - 4 = 0

(t - 4)(t + 1) = 0

(t - 4) = 0

\boxed {t = 4 ~ seconds}

The difference in the two ball's time in the air is:

\Delta t = 4 ~ seconds - 1 ~ second

\large {\boxed {\Delta t = 3 ~ seconds} }

<h2>Question B:</h2><h3>First Ball</h3>

v^2 = u^2 - 2gH

v^2 = (-14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

<h3>Second Ball</h3>

v^2 = u^2 - 2gH

v^2 = (14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 11.025 ~ m}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 25.725 ~ m}

The difference in the two ball's height after 0.500 s is:

\Delta h = 25.725 ~ m - 11.025 ~ m

\large {\boxed {\Delta h = 14.7 ~ m} }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0
2 years ago
The sprinter ran 110 m in 11 seconds. What was her average speed in m/s?
Tanya [424]

Answer:

10 m/s

Explanation:

110/11=10

5 0
2 years ago
Read 2 more answers
A golf club hits a stationary 0.05kg golf ball with and average force of 5.0 x 10^3 newtons accelerating the ball at 44 meters p
maxonik [38]

Answer: The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

Explanation:

Force applied on the golf ball = 5.0\times 10^3 N

Mass of the ball = 0.05 kg

Velocity with which ball is accelerating = 44 m/s

Time period over which forece applied = t

f=ma=\frac{m\times v}{t}

t=\frac{0.05 kg\times 44m/s}{5.0\times 10^3 N}=4.4\times 10^{-4} seconds

Impulse=(force)\times (time)=f\times t = 5.0\times 10^3\times 4.4\times 10^{-4} seconds=2.2 Newton seconds

The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

7 0
3 years ago
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