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zubka84 [21]
3 years ago
8

Is the friction of the pendulum (catch mechanism, support axis, etc.) a random or systematic error? Will this source of error ca

use your calculated velocity to be less than or greater than the actual velocity?
Physics
1 answer:
Hatshy [7]3 years ago
5 0

Answer:

l these errors believe that the speed of the system is less than that calculated

Explanation:

When we carry out any measurement in addition to the magnitude, the sources of uncertainty must also be analyzed.

We can have random uncertainties, correspondin

g to momentary errors, for example early warps during medicine, parallax errors, errors in the starting and ending points of the movement; I mean every possible random error. This error is the one that is analyzed and calculated in the statistical equations

There is another source of error, the systematic ones, these are much more complicated, they can be an error in the pendulum length, friction in the pendulum movement mechanism, deformities in the support systems, this errors are not analyzed by the statistic, in general They discover by looking at the results and comparing with the tabulated or real ones.

 

tith the explanation we see that the errors described are systematic.

In general these errors believe that the speed of the system is less than that calculated

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(8 points) Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet
trapecia [35]

Answer:

a ) \dot m = 351.49 kg/s

b)  \dot m_{actual} = 1046.15 kg/s

Explanation:

given data:

pressure ration rp = 12

inlet temperature = 300 K

TURBINE inlet temperature  = 1000 K

AT the end of isentropic process (compression) temperature is

\frac{T_2'}{T_1} = rp ^{\frac{\gamma -1}{\gamma}}

\frac{T_2'}{300} = 12^{\frac{1.4 -1}{1.4}}

T_2' = 610.181 K

AT the end of isentropic process (expansion) temperature is

\frac{T_3}{T_4'} = rp ^{\frac{\gamma -1}{\gamma}}

\frac{1000'}{T_4'} = 12^{\frac{1.4 -1}{1.4}}

T_4' = 491.66 K

isentropic work is given as

w(compressor) = CP (T_2' -T_1)

w = 1.005(610.18 - 300)

w = 311.73 kJ/kg

w(turbine) = 1.005( 1000 - 491.66)

w(turbine) = 510.88 kJ/kg

a) mass flow rate for isentropic process is given as

\dot m = \frac{70000}{510.88 - 311.73}

\dot m = 351.49 kg/s

b) actual mass flow rate uis given as

\dot m_{actual} = \frac{70000}{51.088\times 0.85 - \frac{311.73}{0.85}}

\dot m_{actual} = 1046.15 kg/s

6 0
3 years ago
Difference between rest and motion?
Yuki888 [10]

Answer:

Rest and motion are relative terms. In simple terms, an object that changes its position is said to be in motion while the opposite action causes an object to be at rest.

Explanation:

6 0
2 years ago
The relationship between sharks and remoras is characterized<br><br>TRUE <br>   OR<br>FALSE
Damm [24]
False, sharks and remoras have a symbiotic relationship. The remora removes parasites from the sharks scales, and the shark provides protection for the remora
8 0
3 years ago
A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm
I am Lyosha [343]

Answer:

Rate of change of magnetic field is 3.466\times 10^3T/sec        

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop r=\frac{0.13}{2}=0.065m

Length of the circular loop L=2\pi r=2\times 3.14\times 0.065=0.4082m

Wire is made up of diameter of 2.6 mm

So radius r=\frac{2.6}{2}=1.3mm=0.0013m

Cross sectional area of wire A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2

Resistivity of wire \rho =2.18\times 10^{-8}m

Resistance of wire R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm

Current is given i = 11 A

So emf  e=11\times 1.67\times 10^{-3}=0.0183volt

Emf induced in the coil is e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}

0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}

\frac{dB}{dt}=3.466\times 10^3=T/sec

8 0
3 years ago
A 5 kg ball is sitting on the floor without moving. What is the momentum of the ball?
natita [175]

Answer:

The ball has no momentum

Explanation:

The given parameters are;

The mass of the ball = 5 kg

The velocity of the ball = 0 (The ball is sitting on the floor without moving)

The momentum of the ball = The mass of the ball × Velocity of the ball

Therefore, the momentum of the ball = 5 kg × 0 m/s = 0

The momentum of the ball is zero, the ball has no momentum.

5 0
2 years ago
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