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Sunny_sXe [5.5K]
3 years ago
6

The basic unit for mass is the a. gram. c. cubic meter. b. metric ton. d. meter.

Physics
2 answers:
Zinaida [17]3 years ago
7 0
Gram according to the metric system
Karolina [17]3 years ago
6 0

Answer: The correct answer is option a.

Explanation:

Basic unit to measure mass is defined as the unit which is commonly used to measure the mass of an object.

From the given options,

Option a: Grams is a unit which is used to measure mass and it is the commonly used unit.

Option b: Cubic meter is the unit which is used to measure the volume of an object.

Option c: Metric ton is a unit which is used to measure the mass of an object but this unit is not commonly used. hence, this is not  basic unit to measure mass.

Option d: Meter is the unit which is used to measure length.

From the above information, the correct answer is option a.

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The conventional relatively small unit for work(ignoring time) such as raising one pound one foot is the foot-pound(ft. lb.). Si
Lelechka [254]

Answer:

1 joule = 0.737 foot-pound

Joule is the unit of work.

1 J = 1 N·m

In SI units

1 J = 1 kg· m/s²

0.737 foot-pound is the amount of work to raise 0.737 pounds one foot or raising one pound to 0.737 ft.

6 0
3 years ago
You are given two infinite, parallel wires, each carrying current.The wires are separated by a distance, and the current in the
miv72 [106K]

Answer:

A. Attractive

B. ( μ₀I² ) / ( 2πd )

Explanation:

A. We know that currents in the same direction attract, and currents in the opposite direction repel, according to ampere's law. In this case the current in the two wires are flowing in the same direction, and hence the force between the two wires are attractive.

B. Suppose that two wires of length l_1 and l_2 both carry the current I in the same direction ( given ). In the presence of a magnetic field produced by wire 1, a force of magnitude m say, is experienced by wire 2. The magnitude of the magnetic field produced by wire 1 at distance say d, from it's axis, should thus be the following -

B_1 = μ₀I / 2πd

The force experienced by wire 2 should thus be -

F_2 = I( l_2 * B_1 )

= I * l_2 * B_1 * Sin( 90 )

= I * l_2 ( μ₀I / 2πd )

Therefore the force per unit length experienced by wire 2 toward wire 1 should be ...

( F_2 / l_2 ) = ( μ₀I² ) / ( 2πd ) ... which is our solution

3 0
3 years ago
Venus has an average distance to the sun of 0.723 AU. In two or more complete sentences, explain how to calculate the orbital pe
Mice21 [21]

As per the question the distance of venus from sun is given as 0.723 AU

We have been asked to calculate the time period of the planet venus.

As per kepler's laws of planetary motion the square of time period of planet is directly proportional to the cube of semi major axis. mathematically

                                        T^{2} \alpha R^{3}

                                         ⇒ T^{2} = KR^{3} where is k is the proportionality  constant

We may solve this problem by comparing with the time period of the earth . We know that time period of earth is 365.5 days

Hence T_{1} =365.5 days

The distance of sun from earth is taken as 1 AU i.e the mean distance of earth from sun

Hence R_{1} =1 AU

The distance of venus from sun is 0.723 AU i.eR_{2} =0.723

From keplers law we know that-\frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{3} }

                            ⇒T_{2} ^{2} =T_{1} ^{2} *\frac{R_{2} ^{3} }{R_{1} ^{3} }

Putting the values mentioned above we get-

                                      T_{2} ^{2} =50,350.132851075

                                         ⇒ T_{2} =\sqrt{50,350.132851075}

                                        ⇒T_{2} = 224.388352752710 days.

Hence the time period of venus is 224.388352752710 days

                                         

                     






                           

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