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A 5-turn square loop (10 cm along aside, resistance = 4.0 ) is placed in a magnetic field that makes an angle of30o with the pla

andreev551 [17]

Answer:

Explanation:

Area A of the coil = .1 x .1 = .01 m²

no of turns n = 5

magnetic field B = .5 t²

Flux Φ perpendicular to plane passing through it.= nBA sin30

rate of change of flux

dΦ/dt = nAdBsin30 / dt

= nA d/dt (.5t²x .5 )

= nA x 2 x .25 x t

At t = 4s

dΦ/dt = nA x 2

= 5x .01 x 2

= .1

current = induced emf / resistance

= .1 / 4

= .025 A

= 25 mA.

4 0

3 years ago

Determine the magnitude of the resultant force FR=F1+F2FR=F1+F2. Assume that F1F1F_1 = 235 lblb and F2F2F_2 = 350 lb

Svetlanka [38]

Answer:

585lb

Explanation:

Given the formula for calculating the magnitude of the resultant force as;

$F_R = F_1 + F_2$

Given

$F_1 = 235 lb \ and \ F_2 = 350 lb\\F_R = 235lb + 350lb\\F_R = 585lb\\$

Hence the magnitude of the resultant force is 585lb

4 0

3 years ago

A temperature of a 0.349 kg sample of copper decreases from 87.0 °C to 21.0 °C. How much heat

Arte-miy333 [17]

We have that heat flow out of the copper sample is mathematically given as

Q=8.86809J

Question Parameters

A temperature of a 0.349 kg sample of copper
decreases from 87.0 °C to 21.0 °C.

Endothermic Reaction

It is common knowledge that an endothermic reaction is one where the system absorbs heat energy from the environs.

Exothermic Reaction

An exothermic reaction is one where the system releases heat energy to the environs.

Generally the equation for Heat is mathematically given as

$Q=mCdt\\\\Therefore\\\\Q=0.349*0.385*(87-21)$

Q=8.86809J

For more information on temperature visit

brainly.com/question/15267055

7 0

2 years ago

If you add more hot water than cold water, will the temperature of the cold water change more?

ycow [4]

Answer:

Yes because of the temputer

Explanation:

3 0

3 years ago

Read 2 more answers

A simple pendulum is used to measure gravity using the following theoretical equation,TT=2ππ�LL/gg ,where L is the length of the

Elina [12.6K]

Answer:

g ±Δg = (9.8 ± 0.2) m / s²

Explanation:

For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use

T = $2\pi \sqrt{ \frac{L}{g} }$

T² = $4\pi ^2 \frac{L}{g}$ 4pi2 L / g

g = $4\pi ^2 \frac{L}{T^2}$

They indicate the average time of 20 measurements 1,823 s, each with an oscillation

let's calculate the magnitude

g = $4\pi ^2 \frac{0.823}{1.823^2}$ 4 pi2 0.823 / 1.823 2

g = 9.7766 m / s²

now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation

for the period

T = t / n

ΔT = $\frac{dT}{dt}$ Δt + $\frac{dT}{dn}$ ΔDn

In general, the number of oscillations is small, so we can assume that there are no errors, in this case the number of oscillations of n = 1, consequently

ΔT = Δt / n

ΔT = Δt

now let's look for the uncertainty of g

Δg = $\frac{dg}{dL}$ ΔL + $\frac{dg}{dT}$ ΔT

Δg = $4\pi ^2 \frac{1}{T2}$ ΔL + 4π²L (-2 T⁻³) ΔT

a more manageable way is with the relative error

$\frac{\Delta g}{g} = \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}$

we substitute

Δg = g ( \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}DL / L + ½ Dt / T)

the error in time give us the stanndard deviation

let's calculate

Δg = 9.7766 ( $\frac{0.001}{0.823} + \frac{1}{2} \ \frac{0.671}{1.823}$ )

Δg = 9.7766 (0.001215 + 0.0184)

Δg = 0.19 m / s²

the absolute uncertainty must be true to a significant figure

Δg = 0.2 m / s2

therefore the correct result is

g ±Δg = (9.8 ± 0.2) m / s²

5 0

3 years ago