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AURORKA [14]
3 years ago
6

A 1000-kg whale swims horizontally to the right at a speed of 6.0 m/s. It suddenly collides directly with a stationary seal of m

ass 200 kg. The seal grabs onto the whale and holds fast. What is the momentum of these two sea creatures just after their collision
Physics
1 answer:
anzhelika [568]3 years ago
5 0

Answer:

Momentum after collision will be 6000 kgm/sec

Explanation:

We have given mass of the whale = 1000

Initial velocity v = 6 m/sec

It collides with other mass of 200 kg which is at stationary

Initial momentum of the whale = 1000×6 = 6000 kgm/sec

We have to find the momentum after collision

From conservation of momentum

Initial momentum = final momentum

So final momentum = 6000 kgm/sec

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A generator converts mechanical energy into electrical energy, while a motor does the opposite - it converts electrical energy into mechanical energy
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Joan stirred a pot of hot soup with a metal spoon. after a minute or two, the handle of the spoon became too hot to hold. what k
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When the metal spoon comes in contact with the hot soup, heat transfer from hot soup to the spoon because of difference in temperature of the two. The kind of heat transfer represented here is conduction. In conduction, heat is transferred from a region of higher temperature to a region of lower temperature when there is a physical contact between the two.

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A conducting loop has an area of 0.065 m2 and is positioned such that a uniform magnetic field is perpendicular to the plane of
aev [14]

Answer:

initial magnetic field  1.306 T

Explanation:

We have given area of the conducting loop A=0.065m^2

Emf induced = 1.2 volt

Initial magnetic field B = 0.3 T

Time dt = 0.087 sec

We know that induced emf is given by e=\frac{d\Phi }{dt}=-A\frac{db}{dt}

1.2=0.065\times \frac{db}{0.087}

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So initial magnetic field = 1.606-0.3= 1.306 T

5 0
2 years ago
How are series and parallel circuits simililar?
Ainat [17]
They both have currents that don't change in the beginning 
6 0
3 years ago
Read 2 more answers
A student releases a block of mass m at the top of a slide of height h1. the block moves down the slide and off the end of the t
Virty [35]

Answer:

B)   d = √  ( 4 h₂ ( H - 2h₂))

Explanation:

A) 1) If the height of the slide is very small, there is no speed to leave the table, therefore do not recreate almost any horizontal distance

2) If the height is very small downwards, it touches the earth a little and the horizon is small,

B) to find an equation for horizontal distance (d)

We must maximize the speed at the bottom of the slide let's use energy

Starting point Higher

          Em₀ = U = m g h₁

Final point. Lower (slide bottom)

            Emf = K + U = ½ m v² + m gh₂

As there is no friction the energy is conserved

             mgh₁ = ½ m v² + mgh₂

             v² = 2 g (h₁-h₂)

This is the speed with which the block leaves the table, bone is the horizontal speed (vₓ)

The distance traveled when leaving the table can be searched with kinematics, projectile launch

           x = v₀ₓ t

          y = v_{oy} t - ½ g t²

The height is the height of the table (y = h₂), as it comes out horizontally the vertical speed is zero

         t = √ 2h₂ / g

We substitute in the other equation

         d = √ (2g (h₁-h₂))  √ 2h₂ / g

         d = √ (4 h₂ (h₁-h₂))

         H = h₁ + h₂

         h₁ = H -h₂

         d = √  ( 4 h₂ ( H - 2h₂))

3 0
3 years ago
Read 2 more answers
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