All categories

2 years ago

Write an expression to evaluate the relative error in g (given that g= 2h/t^2) in terms of h,Δh,t,Δt

1 answer:

6 0

An expression which can be used to evaluate the relative error in g (in terms of h, Δh, t, and Δt) is: $\delta = \frac{t^2(h\;+\Delta \;h) }{h(t\;+\;\Delta t)^2} - 1$

<h3>What is relative error?</h3>

Relative error can be defined as a measure of the ratio of an absolute (real) value of a measurement to an expected (theoretical) value. Also, it's independent of the magnitude of its values.

<h3>How to evaluate the relative error in g?</h3>

In order to write this expression, we would divide the absolute (real) value by the expected (theoretical) value as follows:

$\delta = \frac{g(h\;+\;\Delta h,t \;+ \;\Delta t) - \;g(h,t)}{g(h,t)} \\\\\delta = \frac{g(h\;+\;\Delta h,t \;+ \;\Delta t) }{g(h,t)} - 1\\\\\delta = \frac{t^2(h\;+\Delta \;h) }{h(t\;+\;\Delta t)^2} - 1$

<u>Note:</u> g = 2h/t²

Read more on relative error here: brainly.com/question/13370015

#SPJ1

You might be interested in

You measure a watch's hour and minute hands to be 7.4 mm and 12.1 mm long, respectively. Part A In one day, by how much does the

pychu [463]

Answer:

109.385m

Explanation:

In 1 day, the hour hand travels 2 circles, or 4π rad in angular. The distance it travels is its angle times the radius

7.4 * 4π = 93 mm

In 1 day, the minute hand travels 24*60 = 1440 circles, or 1440 * 2π = 2880π rad in angular. The distance it travels is

12.1 * 2880π = 109478 mm

So the distance traveled by the tip of the minute hand that exceed the distance traveled by the tip of the hour hand is

109478 - 93 = 109385 mm or 109.385 m

3 0

4 years ago

A real heat engine operates between temperatures Tc and Th. During a certain time, an amount Qc of heat is released to the cold

Vlad [161]

Answer:

$W_{max} =Q_C(\dfrac{T_H}{T_C} - 1)$

Explanation:

given,

Temperature of heat engine operate between

Th (temperature in hot reservoir) and Tc(temperature in cold reservoir)

amount of heat released to = Qc

to find maximum amount of work = ?

now,

efficiency of heat engine

$\eta = 1 - \dfrac{T_C}{T_H} = 1 -\dfrac{Q_C}{Q_H}$

now,

$\dfrac{T_C}{T_H} = \dfrac{Q_C}{Q_H}$

$Q_H= \dfrac{T_H}{T_C} Q_C$

maximum work =

$W_{max} = Q_H - Q_C$

$W_{max} =\dfrac{T_H}{T_C} Q_C - Q_C$

$W_{max} =Q_C(\dfrac{T_H}{T_C} - 1)$

above expression gives the expression of maximum amount of work.

5 0

3 years ago

What does critical mean?

iren2701 [21]

expressing adverse or disapproving comments or judgments.

5 0

3 years ago

Read 2 more answers

The stopping distance of a vehicle is an important safety feature. Assuming a constant braking force is applied, use the work-en

elena-14-01-66 [18.8K]

Answer:

The stopping distance would be 200 m.

Explanation:

Hi there!

The work done to stop the vehicle is equal to its change in kinetic energy.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass.

v = velocity.

The change in kinetic energy is calculated as follows:

ΔKE = final kinetic energy - initial kinetic energy

In this case, the vehicle is brought to stop, so, the final kinetic energy will be zero.

ΔKE = 0 - 1/2 · m · v²

The work done is calculated as follows:

W = F · d

Where:

W = work done

F = applied force

d = traveled distance (stopping distance in this case)

The force F is calculated as follows:

F = m · a

Where:

m = mass

a = acceleration

Then:

W = ΔKE

F · d = -1/2 · m · v²

m · a · d = -1/2 · m · v²

a · d = -1/2 · v²

d = -1/2 · v² / a

Let´s find the acceleration of the vehicle that is brought to stop in 50 m with an initial velocity of 45 km/h.

Let´s convert 45 and 90 km/h into m/s

45 km/h · 1000 m/ 1 km · 1 h /3600 s = 12.5 m/s

90 km/h · 1000 m/ 1 km · 1 h /3600 s = 25 m/s

The distance and velocity of the vehicle is calculated using the following equations:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = traveled distance at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity at time t.

Let´s place the origin of the frame of reference at the point where the vehicle begins to decelerate so that x0 = 0. When the vehicle stops, its velocity is zero. Let´s use the equation of velocity to find the time it takes the vehicle to stop (and travel a distance of 50 m):

v = v0 + a · t

0 = 12.5 m/s + a · t

-12.5 m/s / t = a

Using the equation of traveled distance, let´s find the time it takes the vehicle to travel 50 m until stop:

x = x0 + v0 · t + 1/2 · a · t²

Replacing a = -12.5 m/s / t

50 m = 12.5 m/s · t + 1/2 · (-12.5 m/s/t) · t²

50 m = 12.5 m/s · t - 6.25 m/s · t

50 m = 6.25 m/s · t

50 m/ 6.25 m/s = t

t = 8.0 s.

Then, the acceleration is the following:

-12.5 m/s / t = a

-12.5 m/s / 8 s = a

a = -1.5625 m/s²

Then, the stopping distance of the vehicle if it travels at an initial speed of 90 km/h would be the following:

d = -1/2 · v² / a

d = -1/2 ·(25 m/s)² / -1.5625 m/s²

d = 200 m

The stopping distance would be 200 m.

3 0

4 years ago

In a mass spectrometer used for commercial purposes, uranium ions of mass 3.76 X 10^(-25) kg and charge 3.5 X 10^(-19) C are sep

Flauer [41]

Answer:

a. 0.394 T b. 0.255 A c. 1.309 × 10⁸ J

Explanation:

Here is the complete question

A certain commercial mass spectrometer (Fig. 28-12) is used to separate uranium ions of mass 3.92 x 10-25 kg and charge 3.20 x 10-19 C from related species. The ions are accelerated through a potential difference of 109 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 1.31 m. After traveling through 180° and passing through a slit of width 0.752 mm and height 0.991 cm, they are collected in a cup. (a) What is the magnitude of the (perpendicular) magnetic field in the separator? If the machine is used to separate out 1.12 mg of material per hour, calculate (b) the current (in A) of the desired ions in the machine and (c) the thermal energy (in J) produced in the cup in 1.31 h.

Solution

a. The magnitude of the (perpendicular) magnetic field in the separator

The kinetic energy of the uranium ions = electric potential energy

¹/₂mv² = qV

v = √(2qV/m) where v = speed of uranium ions, q = uranium ion charge = 3.2 × 10⁻¹⁹ C , m = mass of uranium ions = 3.92 × 10⁻²⁵ kg and V = 109 kV = 1.09 × 10⁵ V

v = √(2qV/m) = √(2 × 3.2 × 10⁻¹⁹ C × 1.09 × 10⁵ V/3.92 × 10⁻²⁵ kg)

v = 4.22 × 10⁵ m/s

Also, the magnetic force on the uranium ions equals the centripetal force when it passes through the magnetic field.

Bqv = mv²/r

B = mv/rq where B = magnetic field strength and r = radius of circle = 1.31 m

B = m(√(2qV/m))/rq

B = √(2mV/q)/r

B = √(2 × 3.92 × 10⁻²⁵ kg × 1.09 × 10⁵ V/3.2 × 10⁻¹⁹ C)/1.31 m

B = √0.26705/1.31

B = 0.394 T

(b) the current (in A) of the desired ions in the machine

Since a mass m of 3.92 × 10⁻²⁵ kg of uranium ions carries a charge q of 3.2 × 10⁻¹⁹ C, then 1.12 mg per hour = 1.12 × 10⁻³ kg/h. In 1.31 h, our mass is M = 1.12 × 10⁻³ kg/h × 1.31 h = 1.47 × 10⁻³ kg carries a charge of Q of

m/q = M/Q

Q = Mq/m

Q = 1.47 × 10⁻³ kg × 3.2 × 10⁻¹⁹ C/3.92 × 10⁻²⁵ kg

Q = 1200 C

The current i = Q/t where t = time = 1.31 h = 1.31 × 60 × 60 s = 4716 s

i = 1200/4716

i = 0.2545 A

i ≅ 0.255 A

(c) the thermal energy (in J) produced in the cup in 1.31 h.

The thermal energy produced in the cup equals the kinetic energy lost by the uranium ions hitting the cup in 1.31 h.

E = ¹/₂Mv² = ¹/₂ × 1.47 × 10⁻³ kg × (4.22 × 10⁵ m/s)²

E = 1.309 × 10⁸ J

3 0

4 years ago