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2 years ago

2. If you are 5'10" tall, that is, 5 feet 10 inches, what is your height in meters? (2 54 cm = 1.00 in)

Physics

2 answers:

likoan [24]2 years ago

8 0

Answer:D 1.7

Explanation:

Just trust me

liraira [26]2 years ago

5 0

Answer:

5’10” = 1.778 = 1.8m

Explanation:

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A rocket rises vertically, from rest, with an acceleration of 3.2 m/s2 until it runs out of fuel at an altitude of 1300 m. After

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Explanation:

Below is an attachment containing the solution.

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3 years ago

An intravenous (IV) system is supplying saline solution to a patient at the rate of 0.06 cm3/s through a needle of radius 0.2 mm

horsena [70]

Answer:

Pressure applied to the needle is 7528 Pa

Explanation:

As we know by poiseuille's law of flow of liquid through a cylindrical pipe

the rate of flow through the pipe is given as

$Q = \frac{\Delta P \pi r^4}{8\eta L}$

now we know that

$Q = 0.06 \times 10^{-6} m^3/s$

radius = 0.2 mm

Length = 6.32 cm

$\eta = 1\times 10^{-3} Pa s$

now we have

$6 \times 10^{-8} = \frac{\Delta P \pi (0.2 \times 10^{-3})^4}{8(1 \times 10^{-3})6.32 \times 10^{-2}}$

$3.03 \times 10^{-11} = \Delta P 5.02 \times 10^{-15}$

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now we have

$P - 1500 = 6028 Pa$

$P = 7528 Pa$

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3 years ago

A 0.683 kg mass moves in SHM at the end of a spring. It takes 1.41 s to move from the position with the spring fully extended to

dsp73

Answer:

Spring constant of the spring will be equal to 9.255 N /m

Explanation:

We have given mass m = 0.683 kg

Time taken to complete one oscillation is given T = 1.41 sec

We have to find the spring constant of the spring

From spring mass system time period is equal to $T=2\pi \sqrt{\frac{m}{K}}$ , here m is mass and K is spring constant

So $1.41=2\times 3.14\sqrt{\frac{0.683}{K}}$

$0.2245=\sqrt{\frac{0.683}{K}}$

Squaring both side

$0.0504=\frac{0.4664}{K}$

$K=9.255N/m$

So spring constant of the spring will be equal to 9.255 N /m

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3 years ago

Biker A is cruising at a speed of 10.0 m/s when she passes biker B who is at rest at the origin of the coordinate system. Biker

Burka [1]

Answer: attached

Explanation:

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3 years ago

A dumbell has a mass of 95 kg. What force must be applied to accelerate it upward at 2.2 m/s2?

Sveta_85 [38]

A :-) F = ma
Given - m = 95 kg
a = 2.2 m/s^2
Solution -
F = ma
F = 95 x 2.2
F = 209

.:. The force is 209 N

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2 years ago