All categories

3 years ago

2. How can we make the percent error value as minimal as possible?

Physics

2 answers:

andriy [413]3 years ago

8 0

Answer:

Percent error can be reduced by improving both your accuracy and precision.

Molodets [167]3 years ago

3 0

Answer:

Percent error can be reduced by improving both your accuracy and precision.

Explanation:

You might be interested in

Define force and types of force

Ratling [72]

Answer:

Force is the strength or weight of things that depends on movement. The types of forces are conteact force, spring force, applied force, air resistance force, normal force, frictional force, tension force, and non-contact force.

3 0

3 years ago

(03.05 LC)

Lynna [10]

Honed I don’t know where the question is

7 0

3 years ago

Read 2 more answers

Two balls are at the same height and released at the same time. One ball is dropped and hits the ground 5s later. The 2nd ball i

ASHA 777 [7]

Answer:

The second ball hits at the same time.

Distance travelled is 65 meters

Explanation:

5 0

2 years ago

On a straight road (taken to be in the x direction) you drive for an hour at 60 km per hour, then quickly speed up to 120 km per

Luda [366]

Answer:

The average velocity is 180 km/hr

Explanation:

Given;

initial velocity, u = 60 km per hour

final velocity, v = 120 km per hour

initial time = 1 hour

final time = 2 hour

Initial position = 60 km/h x 1 hour = 60 km

final position = 120 km/h x 2 hour = 240 km

The average velocity is given by;

$V_{avg} = \frac{Final \ position\ - \ Initial \ position}{final \ time\ - \ initial \ time}\\\\V_{avg} = \frac{240km \ - \ 60km}{2hr\ - \ 1hr} \\\\V_{avg} = \frac{180 \ km}{1hr} \\\\V_{avg}= 180 \ km/hr$

Therefore, the average velocity is 180 km/hr

3 0

3 years ago

Ammonia enters the compressor of an industrial refrigeration plant at 2 bar, −10°C with a mass flow rate of 15 kg/min and is com

Juli2301 [7.4K]

Answer:

a) W=85.225 kW

b) $0.02\frac{kW}{K}$

Explanation:

First, consider the energy balance for the compressor: The energy that enters to the system (W and enthalpy of the feed flow) is equal to the energy that goes out from it (Heat Q and enthalpy of the exit flow):

$W+m*h_1=Q+m*h_2\\W=Q+m*(h_2-h_1)$

Consider the enthalpy data from van Wylen 6th edition, Table B.2.2. According to that, $h_1=h(200kPa,-10C)=1440.6\frac{kJ}{kgK}$ , $h_2=h(1200kPa,140C)=1757.5\frac{kJ}{kgK}$

So, the power input to the compressor is:

$W=6kW+15\frac{kg}{min}*\frac{min}{60s}*(1757.5-1440.6)\frac{kJ}{kg}\\W=85.225kW$

b) The differential entropy change dS for a reversible heat transfer dQ at a temperature T is:

$dS=\frac{dQ}{T}$

This equation can be integrated if the heat transfer surface temperature remains constant, which is the case, giving as a result:

$S_2-S_1=\frac{Q}{T}=\frac{6kW}{300K}=0.02\frac{kW}{K}$

6 0

3 years ago