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
kozerog [31]
4 years ago
12

Helppppp!! will mark brainliest!!

Physics
1 answer:
Ne4ueva [31]4 years ago
3 0
Pick which "black" box goes with each rock....
...ex- the bottom right can be in the sedimentary rock...box
You might be interested in
Please look at the diagram and ans that question
TEA [102]
I can see that they are running away like my dad did
3 0
3 years ago
How many atoms are there in 3.4moles if helium ,show the calculation
prohojiy [21]
Answer: 20.4752789138x x  10^23 atoms
To count how many atoms in moles you need to know Avogadro's number. Avogadro's number dictate that for every mole there is 6.022140857 × 10^23 molecule/atoms.
Then 3.4 moles of helium will be 3.4x 6.022140857 x  10^23 atoms= 20.4752789138x x  10^23 atoms
6 0
4 years ago
A sinusoidal wave has period 0.20 s and wavelength 2.0 m. What is the wave speed?
il63 [147K]

Answer:10m/s

Explanation:

Wave speed ,v=for

Where π= wavelength=2m

Period =1/f f=frequency of wave

F=1/period

=1/0.2=5Hz

So speed of waves,v=5×2=10m/s

5 0
4 years ago
La masa de un camión es de
Nutka1998 [239]

Answer:

45000 kg and 45 tons  

Explanation:

The expression in kilograms and tons is shown below;

As we know that

1 gr is 0.001 kg

So, 45000000 = 45,000 kg

And,

1 kg = 0.001 tons  

So, 45000 kg = 45 tons    

Therefore the same would be considered

7 0
3 years ago
Sphere A of mass 0.600 kg is initially moving to the right at 4.00 m/s. sphere B, of mass 1.80 kg is initially to the right of s
anzhelika [568]

A) The velocity of sphere A after the collision is 1.00 m/s to the right

B) The collision is elastic

C) The velocity of sphere C is 2.68 m/s at a direction of -5.2^{\circ}

D) The impulse exerted on C is 4.29 kg m/s at a direction of -5.2^{\circ}

E) The collision is inelastic

F) The velocity of the center of mass of the system is 4.00 m/s to the right

Explanation:

A)

We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:

p_i = p_f\\m_A u_A + m_B u_B = m_A v_A + m_B v_B

m_A = 0.600 kg is the mass of sphere A

u_A = 4.00 m/s is the initial velocity of the sphere A (taking the right as positive direction)

v_A is the final velocity of sphere A

m_B = 1.80 kg is the mass of sphere B

u_B = 2.00 m/s is the initial velocity of the sphere B

v_B = 3.00 m/s is the final velocity of the sphere B

Solving for vA:

v_A = \frac{m_A u_A + m_B u_B - m_B v_B}{m_A}=\frac{(0.600)(4.00)+(1.80)(2.00)-(1.80)(3.00)}{0.600}=1.00 m/s

The sign is positive, so the direction is to the right.

B)

To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

Before the collision:

K_i = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 =\frac{1}{2}(0.600)(4.00)^2 + \frac{1}{2}(1.80)(2.00)^2=8.4 J

After the collision:

K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(0.600)(1.00)^2 + \frac{1}{2}(1.80)(3.00)^2=8.4 J

The total kinetic energy is conserved: therefore, the collision is elastic.

C)

Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.

Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:

p_x = m_B v_B = (1.80)(3.00)=5.40 kg m/s

While the total momentum along the y-axis is zero:

p_y = 0

We can now write the equations of conservation of momentum along the two directions as follows:

p_x = p'_{Bx} + p'_{Cx}\\0 = p'_{By} + p'_{Cy} (1)

We also know the components of the momentum of B after the collision:

p'_{Bx}=(1.20)(cos 19)=1.13 kg m/s\\p'_{By}=(1.20)(sin 19)=0.39 kg m/s

So substituting into (1), we find the components of the momentum of C after the collision:

p'_{Cx}=p_B - p'_{Bx}=5.40 - 1.13=4.27 kg m/s\\p'_{Cy}=p_C - p'_{Cy}=0-0.39 = -0.39 kg m/s

So the magnitude of the momentum of C is

p'_C = \sqrt{p_{Cx}^2+p_{Cy}^2}=\sqrt{4.27^2+(-0.39)^2}=4.29 kg m/s

Dividing by the mass of C (1.60 kg), we find the magnitude of the velocity:

v_c = \frac{p_C}{m_C}=\frac{4.29}{1.60}=2.68 m/s

And the direction is

\theta=tan^{-1}(\frac{p_y}{p_x})=tan^{-1}(\frac{-0.39}{4.27})=-5.2^{\circ}

D)

The impulse imparted by B to C is equal to the change in momentum of C.

The initial momentum of C is zero, since it was at rest:

p_C = 0

While the final momentum is:

p'_C = 4.29 kg m/s

So the magnitude of the impulse exerted on C is

I=p'_C - p_C = 4.29 - 0 = 4.29 kg m/s

And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore -5.2^{\circ}.

E)

To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:

K_i = \frac{1}{2}m_B u_B^2 = \frac{1}{2}(1.80)(3.00)^2=8.1 J

The total kinetic energy after the collision is the sum of the kinetic energies of B and C:

K_f = \frac{1}{2}m_B v_B^2 + \frac{1}{2}m_C v_C^2 = \frac{1}{2}(1.80)(1.20)^2 + \frac{1}{2}(1.60)(2.68)^2=7.0 J

Since the total kinetic energy is not conserved, the collision is inelastic.

F)

Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:

F=Ma

Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.

Therefore, the centre of mass after the 2nd collision must be equal to the velocity of the centre of mass before the 1st collision: which is the velocity of the sphere A before the 1st collision (because the other 2 spheres were at rest), so it is simply 4.00 m/s to the right.

Learn more about momentum and collisions:

brainly.com/question/6439920

brainly.com/question/2990238

brainly.com/question/7973509

brainly.com/question/6573742

#LearnwithBrainly

8 0
3 years ago
Other questions:
  • An operation has a 20 percent scrap rate. As a result, 80 pieces per hour are produced. What is the potential labor productivity
    8·1 answer
  • Which of the following is true about natural selection?
    9·1 answer
  • you measure or observe two events to occur at the same time one nearby and one further away. Which one did you actually see firs
    5·2 answers
  • Which vector best represents the net force acting on +3C charge in the diagram?
    10·2 answers
  • an aeroplane taking off from a field has a run of 500m. What is the acceleration and take off velocity if it leaves the ground 1
    6·1 answer
  • 15. Sandra decided to talk a walk at her neighborhood park. She walks 20 meters
    10·1 answer
  • Suppose of lycerin have a temperature of 30°C. Heat is supplied to a steady rate of 100 W. The
    7·1 answer
  • Describe the importance of conservative forces to conservation of energy.
    7·1 answer
  • Let me know if you need help figuring out how to make something in little alchemy 2.
    8·2 answers
  • if your oven uses a 220.0 volt line and draws a maximum 8.00 A current what is the resistance of the oven when it is fully heate
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!