All categories

3 years ago

Why are certain metalloids useful in today's technological equipment?

Physics

1 answer:

fgiga [73]3 years ago

8 0

The answer is B.

Metalloids are elements that posses some of the characteristics of non-metals and also some of the characteristics of metals.

The semiconductivity property of metalloids such as Siliconium, Geranium have been leveraged to the maximum and these are used in LCD screens, televisions and many other technological appliances.

You might be interested in

How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations (grav

JulsSmile [24]

Answer:

maximum height on moon is 6 times more than the maximum height on Earth

Explanation:

Let the Astronaut has its maximum speed by which he can jump is "v"

now for the maximum height that it can jump is given as

$v_f^2 - v_i^2 = 2 aH$

now from above equation we will have

$0 - v^2 = 2(-g)H$

now we have

$H = \frac{v^2}{2g}$

now if Astronaut jump on the surface of moon with same speed

then we know that the acceleration of gravity on surface of moon is 1/6 times the gravity on earth

so at surface of moon we have

$0 - v^2 = 2(-g/6) H$

now we have

$H = \frac{6v^2}{2g}$

so maximum height on moon is 6 times more than the maximum height on Earth

8 0

3 years ago

A certain first-order reaction is 58% complete in 95 s. What are the values of the rate constant and the half-life for this proc

guajiro [1.7K]

Answer:

$0.005734 s^{-1}$ and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

Explanation:

Expression for rate law for first order kinetics is given by:

$a_o=a\times e^{-kt}$

where,

k = rate constant

t = age of sample

$a_o$ = let initial amount of the reactant

a = amount left after decay process

We have :

$a_o=x$

$a=58\%\times x=0.58x$

t = 95 s

$0.58x=x\times e^{-k\times 95 s}$

$\k= 0.005734 s^{-1}$

Half life is given by for first order kinetics::

$t_{1/2}=\frac{0.693}{k}$

$=\frac{0.693}{0.005734 s^{-1}}=120.86 s$

$0.005734 s^{-1}$ and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

3 0

3 years ago

The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 54/(7 + x2 + y2), where T is measured in °C and x,

ANTONII [103]

Answer:

dt/dx = -0.373702

dt/dy = -1.121107

Explanation:

Given data

T(x, y) = 54/(7 + x² + y²)

to find out

rate of change of temperature with respect to distance

solution

we know function

T(x, y) = 54 /( 7 + x² + y²)

so derivative it x and y direction i.e

dt/dx = -54× 2x / (7 +x² + y²)² .........................1

dt/dy = -54× 2y / (7 + x² + y²)² .........................2

now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2

dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²

dt/dx = -0.373702

and

dt/dy = -54× 2(3) / (7 + (1)² + (3)²)²

dt/dy = -1.121107

7 0

3 years ago

What is the name and symbol of the element in the second row and fourteenth column of the periodic table? Hint: Review your peri

Papessa [141]

The answer is Carbon. Beyond being the only element listed here that is located in the second row, it is also in the fourteenth column of the table if you count from left to right. Hope this helps!

4 0

3 years ago

A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor

Nata [24]

Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

the magnetic field is oriented at 14.42°

we wil get the area due to rotation of radius of baton is

$\Delta A = \frac{1}{2} \Delta \theta R^2$

The formula for the induced emf is

$E = \frac{\Delta \phi}{\Delta t}$

$\phi = \texttt {magnetic flux}$

$E=\frac{\Delta (BA) }{\Delta t}$

$=B\frac{\Delta A}{\Delta t}$

B is the magnetic field strength

substitute

$\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A$

$E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega$

The magnetic field of the earth is oriented at 14.42

$\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5$

we plug in the values in the equation above

so, the induce EMF will be

$E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega$

$=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V$

6 0

3 years ago