1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Pani-rosa [81]
3 years ago
6

Think about it: suppose a meteorite collided head-on with mars and becomes buried under mars's surface. what would be the elasti

city of this collision? explain your answer.
Physics
1 answer:
Ket [755]3 years ago
4 0

Answer:

Perfectly inelastic collision

Explanation:

There are two types of collision.

1. Elastic collision : When the momentum of the system and the kinetic energy of the system is conserved, the collision is said to be elastic. For example, the collision of two atoms or molecules are considered to be elastic collision.

2. Inelastic collision: When the momentum  the system is conserved but the kinetic energy is not conserved, the collision is said to be inelastic. For example, collision of a ball with the mud.

For a perfectly elastic collision, the two bodies stick together after collision.

Here, the meteorite collide with the Mars and buried inside it, the collision is said to be perfectly inelastic. here the kinetic energy of a body lost completely during the collision.

You might be interested in
Suppose you're watching your favorite soap opera. After one of the main characters is thrown from a galloping horse, she suffers
Viktor [21]
She suffers from <span>Retrograde </span>amnesia 
5 0
2 years ago
Read 2 more answers
A projectile is launched
Flauer [41]

The vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.

The given parameters;

  • initial horizontal velocity, vₓ = 16 m/s
  • initial vertical velocity, v_y =0
  • time interval 1 seconds

The components of the velocity can be horizontal or vertical velocity.

The vertical component of the velocity is affected by acceleration due to gravity while the horizontal component of the velocity is not affected by gravity.

The vertical component of the velocity is calculated as;

v_y = v_0_y -gt\\\\v_y = 0 - (1\times 9.8)\\\\v_y = -9.8 \ m/s

The horizontal component of the velocity is constant since it is not affected by gravity.

The horizontal component of the velocity = 16 m/s

Thus, the vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.

Learn more here:brainly.com/question/20349275

4 0
2 years ago
FILL IN THE BLANKS!
Aleksandr [31]
1. First blank is A. Conductors
Second blank is D. Insulators

2. C. Heat
8 0
2 years ago
SO MANY POINTS! WILL MARK BRAINIEST!!!! PLZ BE FASSTT!
lora16 [44]

Answer:  Point A is the answer for the potential energy. Point D is the answer for the kinetic energy.

Explanation:

4 0
3 years ago
(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50
Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

Substituting values we obtain

m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg

For the transition between parts B and C, we can write:

m_Bv_B=m_Cv_C

Which means

(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

Since we want the new final speed of the car (v_C) we do:

v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}

Substituting values we obtain

v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s

5 0
3 years ago
Other questions:
  • Why are certain metalloids useful in today's technological equipment?
    6·1 answer
  • If a baseball pitch leaves the pitcher's hand horizontally at a velocity of 150 km/h by what percent will the pull of gravity ch
    7·1 answer
  • Describe how can two or more velocities be combined
    9·1 answer
  • A hypothetical planet has a mass of 1.66 times that of Earth, but the same radius. What is gravitiy near its surface?
    10·2 answers
  • A truck moving at 45 m/s passes a police car moving at 36 m/s in the opposite direction. if the frequency of the siren is 500 Hz
    14·1 answer
  • What is an advantage of using an electromagnet rather than a regular magnet
    10·2 answers
  • Graph the following data tables on different graphs.
    13·1 answer
  • So I kicked my leg in the air trying to do something but I slipped cause I had socks on. I hit the ground hard with my left hand
    10·1 answer
  • What is projectile motion​
    7·2 answers
  • Hey can someone please help answer this question?
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!