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Elena-2011 [213]
3 years ago
15

Circle the larger unit:

Physics
1 answer:
DerKrebs [107]3 years ago
4 0
1. Centimeter
2. Kilogram
3. Millisecond
4. DL
5. Kg
6. Mm
7. S
8. Mm
9. Us
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Calculate the pressure exerted on the floor by the boy standing on both feet if the weight of the boy is 40kg. Assume that the a
Jlenok [28]

Answer:

P = 1333.33 N

Explanation:

The pressure exerted by the boy on the floor can be calculated by the following equation:

P = \frac{F}{A}

where,

P = Pressure exerted by the boy = ?

F = Force Applied = Weight of Boy = 40 kg = 40 N (since 1 kg = 1N)

A = Area of application of force = 2(Area of one show) = 2(6 cm x 25 cm)

A = 2(0.06 m x 0.25 m) = 0.03 m²

Therefore,

P = \frac{40\ N}{0.03\ m^2}\\\\

<u>P = 1333.33 N</u>

4 0
3 years ago
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nirvana33 [79]

Answer:

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Explanation:

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An engineer weighs a sample of mercury (ρ = 13.6 × 103 kg/m3 ) and finds that the weight of the sample is 6.0 n. what is the sam
Amanda [17]
Given:
ρ = 13.6 x 10³ kg/m³, density of mercury
W = 6.0 N, weight of the mercury sample
g = 9.81 m/s², acceleration due to gravity.

Let V =  the volume of the sample.
Then
W = ρVg
or
V =  W/(ρg)
   = (6.0 N)/[(13.6 x 10³ kg/m³)*(9.81 m/s²)]
   = 4.4972 x 10⁻⁵ m³

Answer: The volume is 44.972 x 10⁻⁶ m³
5 0
3 years ago
Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.500 A when R2 is removed,
KATRIN_1 [288]

Answer:

a)   R₁ = 14.1 Ω,   b)  R₂ =  19.9 Ω

Explanation:

For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances

all resistors connected

           V = i (R₁ + R₂)

with R₁ connected

           V = (i + 0.5) R₁

with R₂ connected

           V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

           V = i ( \frac{V}{ i+0.5} + \frac{V}{i+0.25} )

           1 = i ( \frac{1}{i+0.5} + \frac{1}{i+0.25} )

           1 = i ( \frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) } ) =  \frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}

           i² + 0.75 i + 0.125 = 2i² + 0.75 i

           i² - 0.125 = 0

           i = √0.125

           i = 0.35355 A

with the second equation we look for R1

          R₁ = \frac{V}{i+0.5}

          R₁ = 12 /( 0.35355 +0.5)

          R₁ = 14.1 Ω

with the third equation we look for R2

          R₂ = \frac{V}{i+0.25}

          R₂ =\frac{12}{0.35355+0.25}

          R₂ =  19.9 Ω

3 0
3 years ago
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