Hazardous waste is ____

The rating of a light bulb at 100 watts indicates

Slav-nsk [51]

That marking means that when the expected voltage is connected
between the two contacts of the bulb, it will dissipate 100 joules of
energy every second, in the form of heat and light.

8 0

4 years ago

Two metal spheres of identical mass m = 4.20 g are suspended by light strings 0.500 m in length. The left-hand sphere carries a

Yanka [14]

Answer:

Explanation:

Answer:

0.632 m

Explanation:

let the equilibrium separation between the charges is d and the angle made by string with the vertical is θ.

According to the diagram,

d = L Sinθ + L Sinθ = 2 L Sinθ .....(1)

Let T be the tension in the string.

Resolve the components of T.

T Sinθ = k q1 q2 / d^2

T Sinθ = k q1 q2 / (2LSinθ)² .....(2)

T Cosθ = mg .....(3)

Dividing equation (2) by equation (3), we get

tanθ = k q1 q2 / (4 L² Sin²θ x mg)

tan θ Sin²θ = k q1 q2 / (4 L² m g)

For small value of θ, tan θ = Sin θ

So,

Sin³θ = k q1 q2 / (4 L² m g)

Sin³θ = (9 x 10^9 x 0.785 x 10^-6 x 1.47 x 10^-6) / (4 x 0.5 x 0.5 x 4.20 x 10^-3 x 9.8)

Sin³θ = 0.2523

Sinθ = 0.632

θ = 39.2 degree

So, the separation between the two charges, d = 2 x L x Sin θ

d = 2 x 0.5 x 0.632 = 0.632 m

6 0

3 years ago

A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =

castortr0y [4]

Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

d) 7.77Nm

Explanation:

(a) The emf of a rotating coil with N turns is given by:

$emf=NBA\omega sin(\omega t)$

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

$emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V$

(b) the maximum rate of change of the magnetic flux is given by:

$\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}$

(c) $emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V$

(d) The torque is given by:

$\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm$