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3 years ago

Which statement is true for particles of the medium of an earth quake p wave

1 answer:

8 0

I am pretty sure that the only statement which is true for particles of the medium of an earthquake P-wave is being shown in the option : b)vibrate parallel to the wave, forming compressions and rarefactions. As you know, it can be formed in two ways : from alternating compressions and rarefactions or primary wave. I bet you will agree with me.

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Critical angle of glass is 42 .what does it mean?

elena55 [62]

Answer:

i think..its fraction that its have multiple fractions on it..if you minus the 397 000-355 it should be 381+ so i say if you get the 5 multiply it by 9!! so you will get it!

Explanation:

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2 years ago

A teacher wants to perform a classroom demonstration that illustrates both chemical and physical changes. Which would be the bes

Alex17521 [72]

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3 years ago

Read 2 more answers

Ceres is a dwarf planet located in the main asteroid belt.

Pie

Yes that is correct, it <span>lies between the orbits of Mars and Jupiter and is the largest in the belt.

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3 years ago

What is an electric load ?write with example

Ganezh [65]

Explanation:

an electrical load is the part of an electrical circuit in which current is transformed into something useful. examples include a lightbulb, a resistor and a motor. a load converts electricity into heat, light or motion. put another way, the part of a circuit that connects to a well-defined output terminal is considered an electrical load.

4 0

3 years ago

A 30-g bullet is fired with a horizontal velocity of 460 m/s and becomes embedded in block B which has a mass of 3 kg. After the

ICE Princess25 [194]

Answer:

energy loss due to friction and the impacts = 2.97 J; The impact loss due to AB impacting the carrier is =25.72J; The impact loss at first impact is 6,316.64J

Explanation:

First find the velocity of the bullet after the first impact using

M1V1 + 0 = (M1 + M2)v'

Where M1 is the mass of the bullet

M2 is the mass of the block B

M3 is the mass of the carrier

v' is the velocity

v' = M1V1/(M1 + M2)

v'= (30 × 10^-3 kg)(460 m/s) / (30 × 10^-3 kg + 3 kg)

v' = 13.8/3.03

v'= 4.55m/s

Also calculate final velocity of the carrier v2'

v2' = M1V1/(M1 + M2 + M3)

v2'= (30 × 10 kg)(460 m/s) / (30 × 10 kg + 3 kg + 30kg)

v2' =0.42m/s

Now to calculate energy loss due to friction

Normal force

N= W1 + W2 = (m1 + m2)g

Where W1 and W2 is the weight of the bullet and block respectively

g is gravitational acceleration for taken as 9.81m/s

= (0.030 kg + 3 kg)(9.81 m/s) 29.724 N

Friction force = coefficient of kinetic× normal force

Where coefficient of kinetic = 0.2

Ff = (0.2)(29.724)= 5.945 N

Now

Energy loss due to friction = frictional force × distance

Assume distance is 0.5 m.

Energy loss due to friction = 5.945 N × 0.5 m

= 2.97J

Kinetic energy of block with embedded bullet immediately after first impact:

1/2 × (m1 + m2)(v')^2

1/2 × (30 × 10^-3 kg + 3 kg)(4.55m/s)^2

= 1/2 × (3.03kg) × (4.55m/s)^2

= 31.36 J

Final kinetic energy of bullet, Block, and Carrier together

1/2 × (m1 + m2 + m3)(v2')^2

1/2 × (30 × 10^-3 kg + 3 kg + 30kg) (0.42m/s)^2

1/2 × (33.03kg) × (0.42m/s)^2

= 2.91 J

Therefore

Loss due to friction and stopping impact = Kinetic energy of block with embedded bullet immediately after first impact - Final kinetic energy of bullet, Block, and Carrier together

= 31.36 J - 2.91 J

= 28.69 J

Impact loss due to AB impacting the carrier = loss due to friction- energy due to friction

28.69J - 2.97J

=25.72J

Initial kinetic energy of system ABC = 1/2(m1vo)

=1/2(0.030 kg)(460 m/s)^2 = 6,348J

Therefore

Impact loss at first impact = Initial kinetic energy of system ABC - Kinetic energy of block with embedded bullet immediately after first impact:

= 6,348J - 31.36 J

= 6,316.64J

3 0

3 years ago