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Lera25 [3.4K]
2 years ago
11

What is the pressure 100m below the surface of the sea , if the density of sea water is 1150kg/m3

Physics
1 answer:
Olin [163]2 years ago
4 0
The pressure at 100 meters below the surface of sea water with a density of 1150kg is 145.96 psi.
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A 6,000 kg train car is moving to the right at 10 m/s and connects to a 4,000-kg train car that wasn't moving. What is the veloc
vredina [299]

1) The final velocity of the two trains is 6 m/s to the right

2) It is an inelastic collision

Explanation:

1)

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two trains must be conserved before and after the collision.

Therefore we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

m_1 = 6,000 kg is the mass of the first train

u_1 = 10 m/s is the initial velocity of the first train

m_2 = 4,000 kg is the mass of the second train

u_2 = 0 is the initial velocity of the second train  (initially at rest)

v is the final combined velocity of the two trains

Solving the equation for v, we find the final velocity of the two trains:

v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(6000)(10)+0}{6000+4000}=6 m/s

2)

There are two types of collisions:

  • Elastic collision: in an elastic collision, both the total momentum and the total kinetic energy of the system are conserved
  • Inelastic collision: in an inelastic collision, only the total momentum is conserved, while the total kinetic energy is not (part of it is converted into thermal energy due to internal frictions)

To verify what type of collision is this, we can compare the total kinetic energy before and after the collision:

Before:

K=\frac{1}{2}m_1 u_1^2 = \frac{1}{2}(6000)(10)^2=300,000 J

After:

K=\frac{1}{2}(m_1 +m_2)v^2 = \frac{1}{2}(6000+4000)(6)^2=180,000 J

As we can see, the kinetic energy is not conserved, so this is an inelastic collision.

Learn more about momentum here:

brainly.com/question/7973509  

brainly.com/question/6573742  

brainly.com/question/2370982  

brainly.com/question/9484203  

#LearnwithBrainly

3 0
3 years ago
Lien uses a spring scale to pull a block toward the right across the lab table. The scale reads 8 N. Which force should Lien con
Alja [10]

Answer

The force on the left across the lab table.

Explanation

The Newton's third law of motion states that; <em>action and reaction are equal but a ct in opposite direction. </em>

When the block of is pulled on the right with a force of X Newtons then there is a force -X Newtons applying and equal force on the left. For every action there must be a reaction with is equal and applying in the opposite direction.

So, if the block is pulled on the right by a force of 8 N there is another equal force applying on the left.


5 0
3 years ago
Read 2 more answers
How do physicists define velocity?
Natalija [7]

Answer:

Velocity, quantity that designates how fast and in what direction a point is moving.

Explanation:

7 0
3 years ago
Read 2 more answers
1.How far is Object Z from the origin at t = 3 seconds?
GenaCL600 [577]

Answer:

Please find the answer in the explanation

Explanation:

1.) How far is Object Z from the origin at t = 3 seconds

The distance of the object Z from the origin will be the slope of the graph.

Slope = 4/2 = 2m

2.) Which object takes the least time to reach a position 4 meters from the origin ?

According to the graph given to the question above, object Z has the list time which is 2 seconds since object X does not start from the origin.

3.) Which object is farthest from the origin at t = 2 seconds?

The correct answer is still object Z because it has the highest slope.

4 0
3 years ago
Light of wavelength 623.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 76.5 cm from the slit.
Dvinal [7]

Answer:

Explanation:

wave length of light λ = 623 x 10⁻⁹ m .

Distance of screen D = 76.5 x 10⁻² m

width of slit        =      d

Distance on the screen between the second order minimum and the central maximum       =  2  λ D / d

1.11 x 10⁻²  = (2 x 623 x 10⁻⁹ x 76.5 x 10⁻² )/ d

d =  ( 2 x 623 x 10⁻⁹ x 76.5 x 10⁻²) / 1.11 x 10⁻²

= 85872.97 x  10⁻⁹

=  85.87297 x  10⁻⁶

= 85.87 μm

width a of the slit is = 85.87 μm

6 0
3 years ago
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