Which planet has the GREATEST attraction to the sun?

A 0.35 m2 coil with 50 turns rotates at 5 radians per sec2 in a magnetic field of 0.6 Tesla. What is the value of the rms curren

soldier1979 [14.2K]

Answer:

11.25 amps

Explanation:

For transformers, the magnetic flux

$\Phi _{max} = \beta \times A$

Therefore;

$\Phi _{max} = 0.6 \times 0.35 = 0.21 \ Weber$

Ф = Фmax (cosωt) = 0.21·(cos(5·t))

From Faraday's law of induction, we have;

ε = -N × dΦ/dt

Which gives;

dΦ/dt = -1.05(sin (5t) )

ε = -N × dΦ/dt = -50× -1.05(sin (5t) )

ε = 52.5(sin (5t) )

I = ε/R = 52.5(sin (5t) )/3.3 = 15.9091(sin (5t) ) amps

The peak current is therefore = 15.9091 amps

The rms current = Peak current /√2 = 15.9091/(√2) = 11.25 amps.

5 0

3 years ago

On a Vernier Caliper, how do you know which mark to use on the very top scale?

madreJ [45]

Answer

To know where it starts we look where the zero mark of the vernier scale starts. The make just before reaching where the zero mark is marks the value to use.

Explanation

A vernier caliper is an instrument that is used to measure the diameter of small circular objects such as diameter of a wires, thickness of an iron sheet.

The objects to be measured is place between the jaws of the calipers.

The vernier scale has two scales, the vernier scale and the main scale which is the very top scale. To know where it starts we look where the zero mark of the vernier scale starts. The make just before reaching where the zero mark is marks the value to use.

4 0

3 years ago

Small-plane pilots regularly compete in "message drop" competitions, dropping heavy weights (for which air resistance can be ign

iren [92.7K]

plane is flying at an altitude of 70 m

now if an object is dropped from it then time taken by object to drop on ground will be given as

$y = v_i* t + \frac{1}{2}at^2$

here initial speed in vertical direction must be zero as plane is moving horizontal

given that

y = 70 m

a = 9.8 m/s^2

$70 = 0 + \frac{1}{2}*9.8*t^2$

$t = 3.77 s$

now since the plane is moving horizontally with speed v = 44 m/s

so the horizontal distance moved by the object will be

$d = v_x * t$

$d = 44 * 3.77$

$d = 166.3 m$

so the distance moved by the box is 166.3 m

3 0

3 years ago

Read 2 more answers

A projectile is fired into the air from the top of a 200-m cliff above a valley as shown below. Its initial velocity is 60 m/s a

anastassius [24]

a) y(max) = 337.76 m

b) t₁ = 5.30 s the time for y maximum

c)t₂ = 13.60 s time for y = 0 time when the fly finish

d) vₓ = 30 m/s vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα v₀ₓ = 60*(1/2) v₀ₓ = 30 m/s ( constant )

v₀y = v₀ * sinα v₀y = 60*(√3/2) v₀y = 30*√3 m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y = y₀ + v₀y*t - (1/2)*g*t² (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt = v₀y - g*t

dy/dt = 0 v₀y - g*t₁ = 0 t₁ = v₀y/g

v₀y = 60*sin60° = 60*√3/2 = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s the time for y maximum

And y maximum is obtained from the substitution of t₁ in equation (1)

y (max) = 200 + 30*√3 * (5.30) - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max) = 337.76 m

Total time of flying (t₂) is when coordinate y = 0

y = 0 = y₀ + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂ - 4.9*t₂² 4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for t₂

t = [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t = [51.96 ±√2700 + 3920]/9.8

t = [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8 we dismiss this solution ( negative time)

t₂ = 13.60 s time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60° vₓ = 30 m/s as we said before v₀ₓ is constant

vy = v₀y - g *t vy = 30*√3 - 9.8 * (13.60)

vy = 51.96 - 133.28 vy = - 81.32 m/s

The sign minus means that vy change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60 m

x = 408 m

5 0

3 years ago

A bag of sugar weighs 3.50 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration isone-sixth

Alex73 [517]

Answer:

$F_{Earth}$ = 15.57 N

$F_{Moon}$ = 2.60 N

$F_{Uranus}$ = 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

Explanation:

The weight of the sugar bag on Earth is:

g=9.81 m/s²

m=3.50 lb=1.59 kg

$F_{Earth}$ =m·g=1.59 kg×9.81 m/s²= 15.57 N

The weight of the sugar bag on the Moon is:

g=9.81 m/s²÷6= 1.635 m/s²

$F_{Moon}$ =m·g=1.59 kg× 1.635 m/s²= 2.60 N

The weight of the sugar bag on the Uranus is:

g=9.81 m/s²×1.09=10.69 m/s²

$F_{Uranus}$ =m·g=1.59 kg×10.69 m/s²= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

5 0

3 years ago