Which group of items all contain a transparent part with at least one curved surface to refract light?

Which of the following usually occurs with a short circuit? a. All parts of the circuit will begin to carry higher amounts of cu

fgiga [73]

Answer:

Hi,

The correct answer option is; D. Most of the current will flow through one part of the circuit.

Explanation:

A short circuit is a low resistance path in an electric connection between two conductors supplying current in a circuit.

It happens when excess amounts of current flow in the power source through a 'short path'.

Short circuits occur at very high temperatures which is of course caused by the heat produced during dissipation.

An example of application of short circuit is arc welding, where heating is achieved through short circuit.

6 0

2 years ago

What happens when a person’s immune system is very weak?

Alexeev081 [22]

The correct answer is C) becuse without certain medicine they will dieeeee

6 0

3 years ago

A 52.5-turn circular coil of radius 5.35 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0

krek1111 [17]

Answer:

5.43 x 10^-3 Nm

Explanation:

N = 52.5, radius, r = 5.35 cm = 0.0535 m, B = 0.455 T, I = 25.3 mA = 0.0253 A

Torque = N I A B Sin theta

Here, theta = 90 degree

Torque = 52.5 x 0.0253 x 3.14 x 0.0535 x 0.0535 x 0.455

Torque = 5.43 x 10^-3 Nm

6 0

2 years ago

A point charge +2Q is at the origin and a point charge −Q is located along the x axis at x = d as in the figure below. Find a sy

Akimi4 [234]

Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis

$F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}$

Explanation:

Let the force on +Q charge y-axis due to +2Q charge be $F_1$ and force on +Q charge y axis due to -Q charge on x-axis be $F_2$ .

Distance between the +2Q charge and +Q charge = d units

Distance between the -Q charge and +Q charge = $\sqrt{2}d$ units

$k_e$ = Coulomb constant

$F_1=k_e\frac{(+2Q)(+Q)}{d^2}=k_e\frac{+2Q^2}{d^2} N$

$F_2=k_e\frac{(-Q)(+Q)}{(\sqrt{2}d)^2}=k_e\frac{-Q^2}{2d^2} N$

Net force on +Q charge on y-axis is:

$F_x=F_2sin 45^o=k_e\frac{-Q^2}{2d^2}\times \frac{1}{\sqrt{2}} N$

$F_y=F_1-F_2cos45^o$

$F_y=(F_1-F_2cos45^o)=(k_e\frac{+2Q^2}{d^2})-(k_e\frac{-Q^2}{2d^2}\frac{1}{\sqrt{2}})$

$F_N=\sqrt{F_x^2+F_y^2}$

$|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|$

The net froce on the +Q charge on y-axis is

$F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}$

4 0

3 years ago

Read 2 more answers

If the force on the tympanic membrane (eardrum) increases by about 1.4 N above the force from the atmospheric pressure, the memb

Sveta_85 [38]

Answer:

10.58 ft

Explanation:

Force, F = 1.4 N

Diameter of membrane = 7.4 mm

radius of membrane, r = 7.4 / 2 = 3.7 mm = 3.7 x 10^-3 m

Area, A = 3.14 x (3.7 x 10^-3)^2 = 4.3 x 10^-5 m^2

Density, d = 1.03 x 10^3 kg/m^3

Pressure at depth, P = h x d x g

Let h be the depth.

Pressure = force / Area

h x 1.03 x 10^3 x 9.8 = 1.4 / (4.3 x 10^-5)

h = 3.225 m = 10.58 ft

Thus, the depth of water is 10.58 ft.

6 0

2 years ago