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arlik [135]
3 years ago
10

D

Physics
2 answers:
lakkis [162]3 years ago
8 0
The right answer is “Atoms “
SpyIntel [72]3 years ago
7 0

Answer:

Atoms

Explanation:

That's what everything is made up of.

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A transformer connected to a 110–(V) (rms) ac line is to supply 10.0 (V) (rms) to a portable electronic device. The load resista
spayn [35]

Answer:

Explanation:

a ) V( primary ) = 100 V

V( secondary ) = 10 V

No of turns ( secondary ) / No of turns ( primary ) = 10 / 100

= 1 / 10

b ) current in secondary

= volt ( secondary ) / resistance

= 10 /6 = 1.67 A

c )

Average power to secondary

= V  ( secondary ) x current ( secondary )

= 10 x 10 / 6

= 16.67 W

d )

Power in primary = power in secondary = 16.67 W

e ) current drawn by ac line ( primary )

Volt ( primary ) x current ( primary ) = power in primary

= 16.67

current ( primary )

= 16.67 / 100

= 0.167 A

5 0
2 years ago
Fill in the blank.
Fofino [41]
It would decrease, hope it helps
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2 years ago
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What is the chemical formula for mercury(I) nitrate? Hgmc021-1.jpg(NOmc021-2.jpg) Hg(NOmc021-3.jpg)mc021-4.jpg Hgmc021-5.jpg(NOm
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If you just type "<span>What is the chemical formula for mercury(I) nitrate?" into google you get the answer but HG(NO3)2 is the correct one.
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Based on the data table,which car won the race?
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3 0
2 years ago
A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
never [62]

The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

8 0
2 years ago
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