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3 years ago

How would you determine how much error there is between a vector addition and the real results

1 answer:

6 0

Desired operation: A + B = C; {A,B,C) are vector quantities.

Issue: {A,B} contain error (measurement or otherwise) 

Objective: estimate the error in the vector sum. 

Let A = u + du; where u is the nominal value of A and du is the error in A 
Let B = v + dv; where v is the nominal value of B and dv is the error in B 
Let C = w + dw; where w is the nominal value of C and dw is the error in C [the objective] 

C = A + B 

w + dw = (u + du) + (v + dv) 

w + dw = (u + v) + (du + dv) 

w = u+v; dw = du + dv 

The error associated with w is the vector sum of the errors associated with the measured quantities (u,v)

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0.3-L glass of water at 20C is to be cooled with ice to 5C. Determine how much ice needs to be added to the water, in grams, i

abruzzese [7]

Answer:

$a. m_i_c_e=54.6g\\b. m_i_c_e=48.7g\\m_c_o_l_d_w_a_t_e_r=900g$

Explanation:

First we need to state our assumptions:

Thermal properties of ice and water are constant, heat transfer to the glass is negligible, Heat of ice $h_i_f=333.7KJkg$

Mass of water, $m_w=\rho V =1\times0.3=0.3Kg$ .

Energy balance for the ice-water system is defined as

$E_i_n-E_o_u_t=\bigtriangleup E_s_y_s\\0=\bigtriangleup U=\bigtriangleup U_i_c_e+\bigtriangleup U_w$

a.The mass of ice at $0\textdegree C$ is defined as:

$[mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[0+333.7+418\times(5-0)]+0.3\times4.18\times(5-20)=0\\m_i_c_e=0.0546Kg=54.6g$

b.Mass of ice at $20\textdegree C$ is defined as:

$[mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[2.11\times(0-(20))+333.7+4.18\times(5-0)]+0.3\times4.18\times(5-20)=0\\\\m_i_c_e=0.0487Kg=48.7g$

c.Mass of cooled water at $T_c_w=0\textdegree C$

$\bigtriangleup U_c_w+\bigtriangleup U_w=0$

$[mc(T_2-T_1)]_c_w+[mc(T_2-T_1)]_w=0\\m_c_w\times4.18\times(5-0)+0.3\times4.18\times(5-20)\\m_c_w=0.9kg=900g$

8 0

3 years ago

What is the value of work done on an object when a 50-newton force moves it 15 meters in the same direction as the force?

Volgvan

Answer:

W = 750 Joules.

Explanation:

Given that,

Force acting on the object, F = 50 N

Distance moved by the object, d = 15 m

To find,

Work done by the object

Solution,

Let W is the work done by the object. It is given by the product of force and distance. Its expression is given by :

$W=F\times d\ cos\theta$

Here, $\theta=0$ (force and distance are in same direction)

$W=F\times d$

$W=50\ N\times 15\ m$

W = 750 Joules

So, the work done by the force is 750 Joules.

4 0

3 years ago

Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons p

boyakko [2]

Answer:

The number of free electrons per cubic meter is $7.61\times 10^{28}\ m^{-3}$

Explanation:

It is given that,

The number of free electrons per cubic meter is, 1.3

Electrical conductivity of metal, $\sigma=6.8\times 10^7\ \Omega^{-1}m^{-1}$

Density of metal, $\rho=10.5\ g/cm^3$

Atomic weight, A = 107.87 g/mol

Let n is the number of free electrons per cubic meter such that,

$n=1.3\ N$

$n=1.3(\dfrac{\rho N_A}{A})$

Where

$\rho$ is the density of silver atom

$N_A$ is the Avogadro number

A is the atomic weight of silver

$n=1.3\times (\dfrac{10.5\ g/cm^3\times 6.02\times 10^{23}\ atoms/mol}{107.87\ g/mol})$

$n=7.61\times 10^{22}\ cm^{-3}$

$n=7.61\times 10^{28}\ m^{-3}$

Hence, this is the required solution.

6 0

3 years ago

What are the difference between convergent, divergent, and transform?

pashok25 [27]

When two plates come together, it is known as a convergent boundary.Magma from the mantle solidifies into basalt, a dark, dense rock that underlies the ocean floor. Thus at divergent boundaries, oceanic crust, made of basalt, is created. Two plates sliding past each other forms a transform plate boundary.

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4 years ago

. The artificial sweetener NutraSweet is a chemical called aspartame (C14H18N2O5). What is (a) its molecular mass (in atomic mas

loris [4]

The molecular mass, in atomic mass unit, of aspartame would be 294 amu while the mass, in kg, of an aspartame molecule would be 0.294 kg

Aspartame has the chemical formula C14H18N2O5

C = 12, H = 1, N = 14, O = 16

(a) Molecular weight = (12x14) + (1x18) + (14x2) + (16x5)

= 168 + 18 + 28 + 80

= 294 amu

(b) Mass of 1 molecule of aspartame = mole x molar mass

= 1 x 294

= 294 g

Converting 294 g to kg = 294/1000

= 0.294 kg

More on mole and molar mass can be found here: brainly.com/question/6613610

7 0

3 years ago