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natali 33 [55]
3 years ago
15

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

Physics
1 answer:
chubhunter [2.5K]3 years ago
6 0
Desired operation: A + B = C; {A,B,C) are vector quantities. 

<span>Issue: {A,B} contain error (measurement or otherwise) </span>

<span>Objective: estimate the error in the vector sum. </span>

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

<span>C = A + B </span>

<span>w + dw = (u + du) + (v + dv) </span>

<span>w + dw = (u + v) + (du + dv) </span>

<span>w = u+v; dw = du + dv </span>

<span>The error associated with w is the vector sum of the errors associated with the measured quantities (u,v)</span>
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How much energy is needed to melt 150 g of ice at 0°C to water? (1)(Lf =3.34˟ 10⁵ J/Kg)
NARA [144]

Answer:

5.01×10⁴ J.

Explanation:

Applying,

q = Cm....................... Equation 1

Where q = amount of heat needed to melt the ice, m = mass of the ice, C = specific latent heat of ice.

From the question,

Given: m = 150 g = (150/1000) kg = 0.15 kg, C = 3.34×10⁵ J/kg

Substitute these values into equation 1

q = (0.15×3.34×10⁵)

q = 0.501×10⁵ J

q = 5.01×10⁴ J.

5 0
2 years ago
Un movil aumenta su velocidad de 10m/s a 20m/s acelerando uniformemente a razon de 5m/s2 ¿que distancia logro en dicha operacion
Kruka [31]

v₀ = initial velocity of the mobile = 10 m/s

v = final velocity of the mobile = 20 m/s

a = acceleration of the mobile = 5 m/s²

d = distance traveled during this operation = ?

Using the kinematics equation

v² = v²₀ + 2 a d

inserting the above values in the equation

20² = 10² + 2 (5) d

400 = 100 + 10 d

subtracting 100 both side

400 - 100 = 100 - 100 + 10 d

300 = 10 d

dividing both side by 10

300/10 = 10 d/10

d = 30 m

hence mobile travels 30 m.

8 0
3 years ago
A solar cell generates a potential difference of 0.25 V when a 550 Ω resistor is connected across it, and a potential difference
Andre45 [30]

a) 400 \Omega

b) 0.43 V

c) 0.44 %

Explanation:

a)

For a battery with internal resistance, the relationship between emf of the battery and the terminal voltage (the voltage provided) is

V=E-Ir (1)

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

In this problem, we have two situations:

1) when R_1=550 \Omega, V_1=0.25 V

Using Ohm's Law, the current is:

I_1=\frac{V_1}{R_1}=\frac{0.25}{550}=4.5\cdot 10^{-4} A

2) when R_2=1000 \Omega, V_2=0.31 V

Using Ohm's Law, the current is:

I_2=\frac{V_2}{R_2}=\frac{0.31}{1000}=3.1\cdot 10^{-4} A

Now we can rewrite eq.(1) in two forms:

V_1 = E-I_1 r

V_2=E-I_2 r

And we can solve this system of equations to find r, the internal resistance. We do it by substracting eq.(2) from eq(1), we find:

V_1-V_2=r(I_2-I_1)\\r=\frac{V_1-V_2}{I_2-I_1}=\frac{0.25-0.31}{3.1\cdot 10^{-4}-4.5\cdot 10^{-4}}=400 \Omega

b)

To find the electromotive force (emf) of the solar cell, we simply use the equation used in part a)

V=E-Ir

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

Using the first set of data,

V=0.25 V is the voltage

I=4.5\cdot 10^{-4}A is the current

r=400\Omega is the internal resistance

Solving for E,

E=V+Ir=0.25+(4.5\cdot 10^{-4})(400)=0.43 V

c)

In this part, we are told that the area of the cell is

A=4.0 cm^2

While the intensity of incoming radiation (the energy received per unit area) is

Int.=5.5 mW/cm^2

This means that the power of the incoming radiation is:

P=Int.\cdot A=(5.5)(4.0)=22 mW = 0.022 W

This is the power in input to the resistor.

The power in output to the resistor can be found by using

P'=I^2R

where:

R=1000 \Omega is the resistance of the resistor

I=3.1\cdot 10^{-4} A is the current on the resistor (found in part A)

Susbtituting,

P'=(3.1\cdot 10^{-4})^2(1000)=9.61\cdot 10^{-5} W

Therefore, the efficiency of the cell in converting light energy to thermal energy is:

\epsilon = \frac{P'}{P}\cdot 100 = \frac{9.6\cdot 10^{-5}}{0.022}=0.0044\cdot 100 = 0.44\%

7 0
3 years ago
A 2 kg mass is free falling in the negative Y direction when a 10 N force is exerted in the minus X direction. What is the accel
lara31 [8.8K]

Answer:

The mass's acceleration is 5 m/s^2 in the minus X direction and 9,8 m/s^2 in the minus Y direction.

Explanation:

By applying the second Newton's law in the X and Y direction we found that in the minus X direction an external force of 10 N is exerted, while in the minus Y direction the gravity acceleration is acting:

X-direction balance force: -10 [N] = m.ax

Y-direction balance force: -m*9,8 \frac{m}{s^2} = m.ay

Where ax and ay are the components of the respective acceleration and m is the mass. By solving for each acceleration:

ax=(-10 [N]) / m

ay=-m*9,8\frac{m}{s^2} / m

Note that for the second equation above the mass is cancelled and, the Y direction acceleration is minus the gravity acceleration:

ay=-9,8\frac{m}{s^2}

For the x component aceleration we must replace the Newton unit:

N =\frac{kg.m}{s^2}

ax= -10 \frac{kg.m}{s^2} / (2 kg)

ax= - 5 \frac{m}{s^2}

6 0
3 years ago
Calculate the force of an object that has a mass of 10kg and an acceleration of 4m/s²
ICE Princess25 [194]
The answer is A-40N.
8 0
2 years ago
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