The resistance of 3A and 12V

Two generators use the same magnetic field and operate at the same frequency. Each has a single-turn circular coil. One generato

Volgvan

Answer:

The coil radius of other generator is 5.15 cm

Explanation:

Consider the equation for induced emf in a generator coil:

EMF = NBAω Sin(ωt)

where,

N = No. of turns in coil

B = magnetic field

A = Cross-sectional area of coil = π r²

ω = angular velocity

t = time

It is given that for both the coils magnetic field, no. of turn and frequency is same. Since, the frequency is same, therefore, the angular velocity, will also be same. As, ω = 2πft.

Therefore, EMF for both coils or generators will be:

EMF₁ = NBπr₁²ω Sin(ωt)

EMF₂ = NBπr₂²ω Sin(ωt)

dividing both the equations:

EMF₁/EMF₂ = (r₁/r₂)²

r₂ = r₁ √(EMF₂/EMF₁)

where,

EMF₁ = 1.8 V

EMF₂ = 3.9 V

r₁ = 3.5 cm

r₂ = ?

Therefore,

r₂ = (3.5 cm)√(3.9 V/1.8 V)

3 0

3 years ago

Từ độ cao 100 m người ta thả một vật thẳng đứng xuống với v = 10 m/s, g = 10 m/s2 . a. Sau bao lâu vật chạm đất. b. Tính vận tốc

Bas_tet [7]

Answer:

10 points dapat yawa aman daw gud

3 0

3 years ago

Read 2 more answers

In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr

spin [16.1K]

Answer:

a) $v_p=9.35m/s$

Explanation:

From the question we are told that:

Open cart of mass $M_o=50.0 kg$

Speed of cart $V=5.00m/s$

Mass of package $M_p=15.0kg$

Speed of package at end of chute $V_c=3.00m/s$

Angle of inclination $\angle =37$

Distance of chute from bottom of cart $d_x=4.00m$

a)

Generally the equation for work energy theory is mathematically given by

$\frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2$

Therefore

$\frac{1}{2}u^2+gh=\frac{1}{2}v_p^2$

$v_p=\sqrt{2(\frac{1}{2}u^2+gh)}$

$v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}$

$v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}$

$v_p=9.35m/s$

4 0

2 years ago

A 5-m steel beam is lowered by means of two cables unwinding at the same speed from overhead cranes. As the beam approaches the

ArbitrLikvidat [17]

Answer:

a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)

b) The acceleration of point A is 3.25 m/s²

The acceleration of point E is 0.75 m/s²

Explanation:

a) The relative acceleration of B with respect to D is equal:

$a_{B} =a_{D} +(a_{B/D} )_{n} +(a_{B/D} )_{t}$

Where

aB = absolute acceleration of point B = 2.5 j (m/s²)

aD = absolute acceleration of point D = 1.5 j (m/s²)

(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam

(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)

$a_{B} =a_{D} +(a_{B/D} )_{t}$

$2.5j=1.5j +(a_{B/D} )_{t}\\(a_{B/D} )_{t}=j=1m/s^{2}$

We have that

(aB/D)t = BDα

Where α = acceleration of the beam

BDα = 1 m/s²

Where

BD = 2

$2\alpha =1\\\alpha =0.5rad/s^{2}CW$

b) The acceleration of point A is:

$a_{A} =a_{D} +(a_{A/D} )_{t}$

(aA/D)t = ADαj

$a_{A} =a_{D} +AD\alpha j\\a_{A}=1.5j+(3.5*0.5)j\\a_{A}=3.25jm/s^{2}$

The acceleration of point E is:

(aE/D)t = -EDαj

$a_{E} =a_{D} -ED\alpha j\\a_{E}=1.5j-(1.5*0.5)j\\a_{E}=0.75jm/s^{2}$

7 0

3 years ago

A mxiture of n2 and H2 has mole fraction of 0.4 and 0.6 respectively. Determine the density of the mixture at one bar and 0 c.

SOVA2 [1]

Answer:

The density of the mixture is 0.55kg/m^3

Explanation:

P = 1bar = 100kN/m^2, T = 0°C = 273K, n = 0.4+0.6 = 1mole

PV = nRT

V = nRT/P = 1×8.314×273/100 = 22.70m^3

Mass of N2 = 0.4×28 = 11.2kg

Mass of H2 = 0.6×2 = 1.2kg

Mass of mixture = 11.2 + 1.2 = 12.4kg

Density of mixture = mass/volume = 12.4/22.7 = 0.55kg/m^3

3 0

3 years ago