How many sig figs does 0.009 have?

ivann1987 [24]

Answer:

B is right at first I thought I was wrong

Explanation:

Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This tendency to resist changes in a state of motion is inertia.

7 0

2 years ago

Is position a base or derived quantity?

amid [387]

Position is measured in meters (m), so it is a base quantity.

<h3>What is base quantity?</h3>

A base or fundamental quantity is a physical quantity, in which other quantities are derived from.

Example of fundamental quantities;

Mass
Length (position)
Time
Temperature
Amount of substance

<h3>What is a derived quantity?</h3>

Derived quantities are those quantities obtained or expressed from fundamental quantities.

Example of derived quantities;

Speed
Acceleration
Volume
Area
Density, etc

Thus, we can conclude that position measured in meters (m) is a base quantity.

Learn more about base quantities here: brainly.com/question/14480063

#SPJ1

8 0

2 years ago

Helppppppppp plzzzzzz

seropon [69]

I think number 1 is incorrect I believe that answer is D. Number 6 I believe would be B. The rest seem to be correct.

4 0

3 years ago

Read 2 more answers

A train whose proper length is 1200 m passes at a high speed through a station whose platform measures 900 m, and the station ma

TiliK225 [7]

Answer:

0.66c

Explanation:

Use length contraction equation:

L = L₀ √(1 − (v²/c²))

where L is the contracted length,

L₀ is the length at 0 velocity,

v is the velocity,

and c is the speed of light.

900 = 1200 √(1 − (v²/c²))

3/4 = √(1 − (v²/c²))

9/16 = 1 − (v²/c²)

v²/c² = 7/16

v = ¼√7 c

v ≈ 0.66 c

6 0

3 years ago

A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com

bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature $=20^{\circ}C$

Final Temperature $=80^{\circ}C$

mass flow rate of cold fluid $\dot{m_c}=1.2 kg/s$

Initial Geothermal water temperature $T_h_i=160^{\circ}C$

Let final Temperature be T

mass flow rate of geothermal water $\dot{m_h}=2 kg/s$

diameter of inner wall $d_i=1.5 cm$

$U_{overall}=640 W/m^2K$

specific heat of water $c=4.18 kJ/kg-K$

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

$\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)$

$2\times (160-T)=1.2\times (80-20)$

$160-T=36$

$T=124^{\circ}C$

As heat exchanger is counter flow therefore

$\Delta T_1=160-80=80^{\circ}C$

$\Delta T_2=124-20=104^{\circ}C$

$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}$

$LMTD=\frac{80-104}{\ln \frac{80}{104}}$

$LMTD=91.49^{\circ}C$

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

$\dot{m_c}c(80-20)=U\cdot A\cdot$ (LMTD)

$A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2$

$A=\pi DL=5.144$

$L=\frac{5.144}{\pi \times 0.015}$

$L=109.16 m$

6 0

3 years ago