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pishuonlain [190]
2 years ago
12

Kahalagahan ng diagram​

Physics
2 answers:
timama [110]2 years ago
5 0

  • <em>m</em><em>ahalaga ito upang mas mapaliwanag mo ng maayos at klaro ang isang bagay o pangyayari.</em>

<em>sana </em><em>makatulong</em><em> </em><em>hehe</em>

12345 [234]2 years ago
4 0

Answer:

Diagrams are essential to provide a clear, unambiguous picture. They can be put together in meetings, can make discussions easier than lots of text and are quicker and easier to do iterations and re-work.

Explanation:

Diagrams play an important role in statistical data presentation. Diagrams are nothing but geometrical figures like lines, bars, circles, squares, etc. Diagrammatic data presentation allows us to understand the data in an easier manner.

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Can i eat air? im hungry
Lerok [7]

Answer:

Yes

Explanation:

8 0
2 years ago
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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
What's the formula for power<br>​
poizon [28]

Answer:

What's the fórmula For power

Explanation:

Power=Work

Time

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3 years ago
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A car travels from City A to City B to the north. It takes 4 hours, and the road mileage is 200 miles. The map shows that
ikadub [295]

Answer:

The average speed would be the 200 miles/4 hours.

The average velocity would be the net displacement to the north at 80 miles/4 hrs = mph

Explanation:

4 0
2 years ago
A long solenoid has 1400 turns per meter of length, and it carries a current of 3.0 A. A small circular coil of wire is placed i
steposvetlana [31]

Answer:

Torque = 2.48×10-³ Nm

Explanation:

See attachment below.

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3 years ago
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