Which features do sound waves have the ocean and light waves do not

Near the surface of Earth an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (

pentagon [3]

Answer:

$q=2.997\times 10^{-4}$ C

Sign-Negative

Explanation:

We are given that

Electric field = $E=100NC^{-1}$ (Radially downward)

Acceleration= $0.19 ms^{-2}$ (Upward)

Mass of charge=3 g= $3\times 10^{-3}$ kg

1kg=1000g

We have to find the magnitude and sign of charge would have to be placed on a penny .

By newton's second law

$\sum F_y=ma$

$\sum F_y=qE-mg$

Substitute the values then we get

$qE-mg=ma$

Substitute the values then we get

$q(100)-3\times 10^{-3}(9.8)=3\times 10^{-3}(0.19)$

$100q-29.4\times 10^{-3}=0.57\times 10^{-3}$

$100q=0.57\times 10^{-3}+29.4\times 10^{-3}=29.97\times 10^{-3}$

$q=\frac{29.97\times 10^{-3}}{100}$

$q=2.997\times 10^{-4}$ C

Sign of charge =Negative

Because electric force acting in opposite direction of electric field therefore,charge on penny will be negative.

4 0

3 years ago

Jennifer and Katie are leaning on each other. Jennifer weighs 150 and Katie weighs 120. Which one is pushing harder on the other

murzikaleks [220]

Jennifer

Explanation:

she has more mass which means she is using more force

3 0

3 years ago

Read 2 more answers

Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plate

vichka [17]

Answer:

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Explanation:

The capacitance of a parallel plate capacitor is given by the following formula:

C = ε₀A/d

where,

C = Capacitance

ε₀ = Permeability of free space

A = Area of plates

d = Distance between plates

FOR CAPACITOR A:

C = CA = 17.8 nF = 17.8 x 10⁻⁹ F

A = A₁

d = d₁

Therefore,

CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F ----------------- equation 1

FOR CAPACITOR B:

C = CB = ?

A = A₁/2

d = 2 d₁

Therefore,

CB = ε₀(A₁/2)/2d₁

CB = (1/4)(ε₀A₁/d₁)

using equation 1:

CB = (1/4)(17.8 X 10⁻⁹ F)

5 0

2 years ago

Which colors allowed the most electricity to be generated

Naddik [55]

Yellow and red hope that helped

3 0

3 years ago

A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find

FrozenT [24]

Answer:

The time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

Explanation:

Given :

Current $I = 0.106$ A

Area of plate $A = 36 \times 10^{-4}$ $m^{2}$

Plate separation $d = 4 \times 10^{-3}$ m

(A)

First find the capacitance of capacitor,

$C = \frac{\epsilon _{o} A }{d}$

Where $\epsilon _{o} = 8.85 \times 10^{-12}$

$C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4} }{4 \times 10^{-3} }$

$C = 7.9 \times 10^{-12}$ F

But $C = \frac{Q}{V}$

Where $Q = It$

$C = \frac{It}{V}$

$V = \frac{It}{C}$

Now differentiate above equation wrt. time,

$\frac{dV}{dt} = \frac{I}{C}$

$= \frac{0.106}{7.9 \times 10^{-12} }$

$= 1.34 \times 10^{10}$ $\frac{V}{s}$

Therefore, the time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

8 0

2 years ago