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3 years ago

Which is correct? 1.00 m = 100. dam 1.00 m = 100. dm 1.00 m = 100. cm 1.00 m = 100. mm

Physics

2 answers:

weqwewe [10]3 years ago

8 0

Answer:

1.00 m = 100. cm

Explanation:

1m=0.1dam

1m=100cm

1m=1000mm

babunello [35]3 years ago

3 0

Answer:

1.00 meter = 100 cm

(3) is correct.

Explanation:

The SI unit of length is meter.

This is the conversion factor. it is used for lower unit to higher unit and higher unit to lower unit.

We know that,

1.00 meter = 0.1 dam

1.00 meter = 10 dm

1.00 meter = 100 cm

1.00 meter = 1000 mm

So, we can say that 1.00 m = 100 cm is correct.

Hence, This is the required solution.

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What is the magnitude of fs on an object lying on a flat surface without moving, on

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The magnitude of the force acting on the object lying on a flat surface without moving is 10 N.

The given parameters;

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$\Sigma F_{net} = F\sin(\theta ) + F cos(\theta)\\\\\Sigma F_{net} = 10 sin(0) + 10cos(0)\\\\\Sigma F_{net} = 10 \ N$

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Read 2 more answers

The compressor of an air conditioner draws an electric current of 23.7 A when it starts up. If the start-up time is 2.35 s long,

defon

Answer:

Electric charge, Q = 55.69 C

Explanation:

It is given that,

Electric current drawn by the compressor, I = 23.7 A

Time taken, t = 2.35 s

We need to find the electric charge passes through the circuit during this period. The definition of electric current is given by total charge divided by total time taken.

$I=\dfrac{q}{t}$

Where,

q is the electric charge

$q=I\times t$

$q=23.7\ A\times 2.35\ s$

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So, the electric charge passes through the circuit during this period is 55.69 C. Hence, this is the required solution.

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3 years ago

If R = 12 cm, M = 520 g, and m = 20 g (below), find the speed of the block after it has descended 50 cm starting from rest. Solv

timofeeve [1]

Answer:

v = 0.84 m/s

Explanation:

given,

R = 12 cm

M (mass of pulley )= 520 g

m (mass of block)= 20 g

s = 50 cm = 0.5 m

using conservation of energy

Potential energy = Kinetic energy

$m g h = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

$I_{disk}= \dfrac{1}{2}MR^2$ and v = r ω

$m g h = \dfrac{1}{2}mv^2 + \dfrac{1}{2}(\dfrac{1}{2}MR^2)(\dfrac{v}{R})^2$

$m g h = \dfrac{1}{2}mv^2 +\dfrac{1}{4}Mv^2$

$m g h = \dfrac{1}{2}v^2(m +\dfrac{1}{2}M)$

$v=\sqrt{\dfrac{2mgh}{m + 0.5 M}}$

$v=\sqrt{\dfrac{2\times 0.020 \times 9.8 \times 0.5}{0.02 + 0.5\times 0.52}}$

v = √0.7

v = 0.84 m/s

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3 years ago