An a.m. station has a frequency of 967 MHz what is the corresponding wavelength to this radio station

Solve for the unknown in each of these circuits

Tamiku [17]

Ohms law: This law relates voltage difference between two points. Mathematically, the law states that V=IR;

Where

V = voltage difference ; in volts

I = Current ; in Amperes

R = Resistance ; in ohms

1. Answer : given that R = 10 ; V= 12 V ; I = ?

From ohms law, I = V/R

= 12/10

= 1.2 Amp.

2. Answer: given that R = 10 ; V= ? ; I = 5

From ohms law, V = IR

= 10×5 = 50 V

3 . Answer: given that R = ? ; V= 120 ; I = 5

From ohms law, R = V/I

= 120/5

= 24 Ω

4 . Answer: given that R = ? ; V= 10 ; I = 20

From ohms law, R = V/I

= 10/20

= 0.5 Ω

5 . Answer: given that R = 480 ; V= 24 ; I = ?

From ohms law, I = V/R

= 24/480

= 0.05 A

6. Answer: given that R = 150 ; V= ? ; I = 1

From ohms law, V = IR

= 1 × 150

= 150 V

6 0

4 years ago

How is the earth a cracked hard boiled egg

xz_007 [3.2K]

Answer:

the earth isn't a cracked hard boiled egg, but I'd guess they mean layers. A thin shell/crust and the yolk and whites of the egg represent layers of the earth

Explanation:

6 0

3 years ago

To enhance the effective surface, and hence the chemical reaction rate, catalytic surfaces often take the form of porous solids.

marysya [2.9K]

Answer:

See solution with all the conditions considered. A gaseous mixture of A and B for which species A is chemically consumed at the catalytic surface.

The total pore reaction rate is stated below and it can be inferred by applying the bellow analogy.

4 0

3 years ago

A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part

mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

$v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t$

$v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}$

At t = 5.0 s, the particle has velocity

$v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}$

$v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}$

The speed at this time is the magnitude of the velocity :

$\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}$

The direction of motion at this time is the angle $\theta$ that the velocity vector makes with the positive x-axis, such that

$\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}$

4 0

3 years ago

Intersystem crossing is:

Semenov [28]

An intersystem crossing (ISC) is a non-radiative process that involves the transition between two electronic states with different spin multiplicity. That is, when an electron is excited in a molecule in a basal singlet state (either by absorption or radiation) into a state of greater energy, an excited singlet or triplet state can be obtained.

Therefore, ISC is understood as an a non radio active transition between states with different spin multiplicity.

Correct answer is C: a radiative transition between states with the same spin.

8 0

4 years ago