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3 years ago

What is the sig fig of 200

1 answer:

8 0

Assuming you don't have a decimal after the zeros, there is only one sig fig. Zeros never count unless they're between other numbers or hold precision like 2.0, because it shows how accurate the measurement was

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If a neighbor pushes a lawnmower six times as far as you do but exerts only one third the force, which one of you does more work

Nat2105 [25]

Answer:

Your neighbor does twice as much work.

Explanation:

Distance(measurement) * exertion(force of pushing mower) = work

6 * 1/3 = 2

Your neighbor does twice as much work as you do.

7 0

2 years ago

Read 2 more answers

Calculate the work done by a 47 n force pushing a 0.025 kg pencil 0.25 m against a force of 23.

inna [77]

Fnet=fa-fr
=47-23
=24n
W=F∆Xcos°
=24n×0.025cos0
=0.6J

8 0

3 years ago

a stone is thrown vertically upwards with a velocity of 20 m per second what will be its velocity when it reaches a height of 10

TEA [102]

Answer:

Explanation:

Here's the info we have:

initial velocity is 20 m/s;

final velocity is our unknown;

displacement is -10.2 m; and

acceleration due to gravity is -9.8 m/s/s. Using the one-dimensional equation

v² = v₀² + 2aΔx and filling in accordingly to solve for v:

$v=\sqrt{(20)^2+2(-9.8)(-10.2)}$ Rounding to the correct number of sig fig's to simplify:

$v=\sqrt{400+2.0*10^2}$ to get

v = $\sqrt{600}=20\frac{m}{s}$ If you don't round like that, the velocity could be 24, or it could also be 24.5 depending on how your class is paying attention to sig figs or if you are at all.

So either 20 m/s or 24 m/s

7 0

3 years ago

A cube made out of wood has a mass of 1.2 kg and the density of the wood is 400kg/m3.What is the length of the side of the cube?

LenaWriter [7]

Answer:

≈ 0.144 m

Explanation:

Density is mass divided by volume:

D = M / V

Solving for volume:

V = M / D

Given M = 1.2 kg and D = 400 kg/m³:

V = 1.2 kg / (400 kg/m³)

V = 0.003 m³

Volume of a cube is the side length cubed:

V = s³

Therefore:

s³ = 0.003 m³

s ≈ 0.144 m

Round as needed.

5 0

3 years ago

Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4

zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

From table A-3E, the specific heat of water is $c_p=1.0\ Btu/lbm.F$ , and the steam properties as, A-4E:

$h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R$

Using the energy balance for the system:

$\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s$

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

$\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s$

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

$\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R$

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

4 0

4 years ago