Bilangan 0,000 000 0124 bila di tulis dengan notasi ilmiah adalah

Answers

Answer 1
Answer: Notasi ilmiah adalah cara penulisan nomor yang terlalu besar atau kecil untuk dengan mudah ditulis dalam notasi desimal standar
jawaban :
0,000 000 0124
;1,24×10-8 (1/10 pangkat 8)

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What is the acceleration of a body revolving in a circle at uniform speed? The acceleration is zero. The acceleration is directed tangentially. The acceleration is directed toward the center. The acceleration is directed away from the center.

Answers

A body revolving in circle generally has tangential acceleration if the speed is changing. Since the speed is uniform here, the tangential acceleration is zero.

centripetal acceleration is directed towards the center and is given as

a = v²/r                    where a = centripetal acceleration , v = speed , r = radius

Since both speed and radius are constant here, the centripetal acceleration is constant and is always directed towards the center.

hence the correct choice is

The acceleration is directed toward the center.

An ionic bond form when atoms

Answers

between two ions of opposite charges. In ionic bonding, electrons transfer from one atom to another.

The diagram below shows the location of three stars in space.The light from Star 2 reaches Star 1 in 63 years. If Star 3 is 75 light years from Star 2, which of these conclusions about the stars is correct? (2 points)
Star 2 is 75 miles away from Star 3.
Star 2 is 63 kilometers away from Star 3.
Star 1 is 12 light years away from Star 3.
Star 1 is 138 light years away from Star 3.

Answers

Because Star 1 is 63 light years from Star 2 and Star 2 is 75 light years from Star 3 we can form a simple equation to get our answer.

75 + 63 = 138

Therefore your answer would be...

D. Star 1 is 138 light years away from Star 3.

I just took th test it is letter D for sure!

The headlights are shining on a truck travelling at 100 km/h. The speed of the light from the headlights relative to the road will be:A) c

B) c + 100 km/h

C) c – 100 km/h

D) depends on the temperature, but faster than the speed if the truck was not moving.

E) faster than if the truck was not moving, but impossible to calculate with the given information

Answers

Answer:

A) c

Explanation:

Speed of light is always constant irrespective of the wave source's motion and the observer's inertial frame of reference. So, no matter how fast the car is moving the speed of light will always be constant. The speed of light in air is around 299704644.54 m/s. The meter is also defined by the speed of light as 1 meter is the distance travelled by light in 1/299792458 second.

A, the light generated from the headlights travels at a fixed speed relative to the road

How do you hook an ammeter and a voltmeter in a circuit?

Answers

Answer:

ammeter goes in series and a voltmeter goes in parallel

Explanation:

A container holding 1.2 kg of water at 20.0 °C is placed in a freezer that is kept at –20.0 °C. The water freezes and comes into thermal equilibrium with the interior of the freezer. a) How much heat is extracted from the water in thisprocess?
b) What is the minimum amount of electrical energy required bythe refrigerator to carry out this process if it operates betweenreservoirs at temperatures of 20.0 °C and -20.0 °C?

Answers

Answer:

(a) Q=556464\ J

(b) 556464 joule

Explanation:

Given:

  • mass of water, m_w=1.2\ kg
  • initial temperature of water, T_i=20^(\circ)C
  • final temperature of frozen water, T_f=-20{}^(\circ)C

The conversion of water of 20.0 °C to the ice of –20.0 °C will comprise of three steps:

  1. cooling of water to 0 °C Q_w
  2. formation of ice at 0 °C from the water of 0 °C Q_L
  3. further cooling of ice of 0 °C to -20 °C Q_i

We have,

  • Latent heat of fusion of ice, L=3.4* 10^5\ J.kg^(-1)
  • specific heat of water, c_w=4186\ J.kg^(-1).^(\circ)C^(-1)
  • specific heat of ice, c_i=2000\ J.kg^(-1).^(\circ)C^(-1)

(a)

Now, total heat lost in the process:

Q=Q_w+Q_L+Q_i

Q=m_w(c_w. \Delta T_w+L+c_i.\Delta T_i)

where:

\Delta T_i\ \&\ \Delta T_w = change in temperature of ice and water respectively.

\Rightarrow Q=1.2(4186* 20+3.4* 10^5+2000* 20)

Q=556464\ J is the total heat extracted during the process.

(b)

So, 556464 joule is the minimum electrical energy (by the law of energy conservation under no loss condition) required by refrigerator to carry out this process if it operates between the reservoirs at temperatures of 20.0 °C and -20.0 °C, because for a refrigerator to work in a continuous cycle it is impossible to transfer heat from a low temperature reservoir to a high temperature reservoir without consuming energy in the form of work. Here 556464 joule is the heat of the system to be eliminated.

Final answer:

The amount of heat extracted from the water involves the sum of heat lost as it cools and then freezes. The minimum energy needed by the refrigerator to do this is given by the formula for Carnot efficiency.

Explanation:

To answer these questions, we'll need to understand some fundamental principles of thermodynamics.

a) The heat Q taken from the water will be the sum of the heat released during cooling of the water until 0.0°C, and the heat released during freezing at 0.0°C. The heat loss as the water cools can be calculated using Q = mcΔt where m=mass of water, c=specific heat of water, and Δt=change in temperature. The heat loss as water freezes can be calculated using Q = mlf where lf is the latent heat of fusion. Adding these two quantities gives the total heat extracted.

b) The minimum energy needed by the refrigerator, W, is given by the Carnot efficiency formula, W = Q*(T_hot - T_cold)/T_hot, where T is in Kelvin. This would tell you how much energy the refrigerator needs to remove the heat from the water and cool it down to the freezer temperature.

Learn more about Thermodynamics here:

brainly.com/question/35546325

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