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NeX [460]
3 years ago
13

Difference between fundamental unit and derived unit (any two)grade 8 question​

Physics
2 answers:
lara31 [8.8K]3 years ago
8 0

Answer:

\boxed{ \sf{see \: below}}

Explanation:

\underline{ \sf{fundamental \: unit}}

\star This is the unit of a fundamental quantity.

\star This unit is independent of other units.

\star There are only seven fundamental units. They are metre , kilogram , second , Kelvin , ampere , candela and mole.

\underline{ \sf{derived \: unit}}

\star This is the unit of a derived Quantity.

\star This unit is obtained from fundamental units.

\star It is formed in many types using seven fundamental units. They are Newton , Pascal , Joule , Watt , Hertz , Ohm etc.

Hope I helped!

Best regards! :D

mash [69]3 years ago
4 0

Answer:

fundamental units are those units that are independent to other units where as derived units are derived from fundamental units . there are altogether 7 fundamental units where as there are more than seven derived units..

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I=mr^2

reducing r reduces I. However, the angular moment of the system remains always conserved. So, to conserve the angular momentum the angular velocity of the planet increases and so did the  otational kinetic energy

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A rotating light is located 13 feet from a wall. The light completes one rotation every 3 seconds. Find the rate at which the li
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Answer:

29.2 ft/s

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A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to th
kari74 [83]

Given that,

Mass = 9.2 kg

Force = 110 N

Angle = 20.2°

Distance = 5.10 m

Speed = 1.58 m/s

(A). We need to calculate the work done by the gravitational force

Using formula of work done

W_{g}=mgd\sin\theta

Where, w = work

m = mass

g = acceleration due to gravity

d = distance

Put the value into the formula

W_{g}=9.2\times(-9.8)\times5.10\sin20.2

W_{g}=-158.8\ J

(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction

Using formula of potential energy

\Delta U=-W

Put the value into the formula

\Delta U=-(-158.8)\ J

\Delta U=158.8\ J

(C). We need to calculate the work done by 100 N force on the crate

Using formula of work done

W=F\times d

Put the value into the formula

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We need to calculate the work done by frictional force

Using formula of work done

W=-f\times d

W=-\mu mg\cos\theta\times d

Put the value into the formula

W=-0.4\times9.2\times9.8\cos20.2\times5.10

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We need to calculate the change in kinetic energy of the crate

Using formula for change in kinetic energy

\Delta k=W_{g}+W_{f}+W_{F}

Put the value into the formula

\Delta k=-158.8-172.5+510

\Delta k=178.7\ J

(E). We need to calculate the speed of the crate after being pulled 5.00m

Using formula of change in kinetic energy

\Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})

v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2

Put the value into the formula

v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58

v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}

v_{2}=6.35\ m/s

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(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.

(C). The work done by 100 N force on the crate is 510 J.

(D). The change in kinetic energy of the crate is 178.7 J.

(E). The speed of the crate after being pulled 5.00m is 6.35 m/s

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