What two things does light provide?

Coiling wire around an iron core and applying an electric current through the wire creates a temporary magnet called an electrom

Serggg [28]

Answer:

C) Use a battery with more voltage.

Explanation:

The equation for the magnetic field around a coil is given by,

B = μ₀NI

where,

B = Magnetic flux density

μ₀ = permeability

N = number of turns per meter

I = Current in the wire

So when using a higher voltage battery, more current passes through the battery as resistance of the wire remains the same.

5 0

3 years ago

Read 2 more answers

A tiny 0.0250 -microgram oil drop containing 15 excess electrons is suspended between to horizontally closely-spaced metal plate

qaws [65]

Answer:

(a) 12 × 10⁻³ C = 12 mC (b) The lower plate

Explanation:

Given

mass of oil drop, m = 0.025 μg = 0.025 × 10⁻⁶

radius of plates, r = 6.50 cm = 6.5 × 10⁻² m

k = 1/4πε₀ = 9.0 × 10⁹ Nm²/C²

electric charge, e= 1.6 × 10⁻¹⁹ C

charge on oil drop, q = 15e

charge on plates, Q = ?

First, we find the charge density of the plates, D = Q/A where Q = charge on plates and A = area of plates. Since the plates are circular, the area is given by A=πr² where r = radius of plates. D=Q/πr²

Also, the electric field, E between the plates is given by E = D/ε =Q/Aε = Q/ε₀πr².

The force on the oil drop due to the electric field between the plates is given by F = qE = qQ/ε₀πr².

Since the oil drop is suspended between the plates, it means that the electric force due to the field on the oil drop balances the weight of the oil drop. So, since weight of oil drop W = mg where g = 9.8 m/s². F =W (for oil drop suspension).

So, qQ/ε₀πr²=mg

So, Q=mgε₀πr²/q

From k = 1/4πε₀, ε₀=1/4πk

So, Q = mgπr²/4πkq = mgr²/4kq = (0.025 × 10⁻⁶ × 9.8 × (6.5 × 10⁻²)²)÷(4 × 9 × 10⁹ × 15 × 1.6 × 10⁻¹⁹)= 0.012 C = 12 mC

(b) The lower plate must be positive because, the direction of the electric field must be upwards, so as to balance out the weight of the oil drop so as to suspend it.

6 0

3 years ago

Mangrove forests play an important role in _______.

AysviL [449]

D. All of the above

At high tide fish will feed among the mangrove roots - rich fishing ground

The trees trap sediment and soil in the river that would flow out to sea which also helps stop erosion

Wildlife utilise almost every part of the tree, with insects and birds, monkeys and lizards in the branches, shrimps and fish in the roots, and snails and clams in the soil

4 0

2 years ago

A sample of copper with a mass of 1.80 kg, initially at a temperature of 150.0°C, is in a well-insulated container. Water at a t

user100 [1]

Answer:

the mass of water is 0.3 Kg

Explanation:

since the container is well-insulated, the heat released by the copper is absorbed by the water , therefore:

Q water + Q copper = Q surroundings =0 (insulated)

Q water = - Q copper

since Q = m * c * ( T eq - Ti ) , where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature

and denoting w as water and co as copper :

m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) = m co * c co * (T co - Ti eq)

m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]

We take the specific heat of water as c= 1 cal/g °C = 4.186 J/g °C . Also the specific heat of copper can be found in tables → at 25°C c co = 0.385 J/g°C

if we assume that both specific heats do not change during the process (or the change is insignificant)

m w = m co * c co * (T eq - Ti co) / [ c w * (T eq - Tiw) ]

m w= 1.80 kg * 0.385 J/g°C ( 150°C - 70°C) /( 4.186 J/g°C ( 70°C- 27°C))

m w= 0.3 kg

7 0

4 years ago

A proton moving in the positive x direction with a speed of 9.9 105 m/s experiences zero magnetic force. When it moves in the po

Alex

Answer:

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

Explanation:

In order to calculate the magnitude and direction of the magnetic field, you take into account the following equation for the magnetic force on the proton:

$\vec{F_B}=q\vec{v}\ X\ \vec{B}$ (1)

v: speed of the proton = 9.9*10^5 m/s

q: charge of the proton = 1.6*10^-19C

B: magnetic field = ?

FB: magnetic force on the proton = 1.6*10^-13N

When the proton travels in the positive y direction (^j), you have that the proton experiences a force in the positive z direction (+^k). To obtain this direction of the magnetic force on the proton, it is necessary that the magnetic field points in the negative x direction, in fact, you have:

^j X (-^i) = -(-^k)=^k

To obtain the magnitude of the magnetic field you use:

$F_B=qvBsin90\°=qvB\\\\B=\frac{F_B}{qv}=\frac{1.6*10^{-13}N}{(1.6*10^{-19}C)(9.9*10^5m/s)}\\\\B=1.01T$

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

8 0

3 years ago