Ask question

Ask question

All categories

2 years ago

8

Metals are used in many products because of the characteristic properties that most metals have. Which product requires the high

luster of metals in order to work?

2 answers:

sashaice [31]2 years ago

7 0

Answer:

The correct answer is...

Explanation:

METALS!

This is correct on e2020, edge-nuity, and pretty much everywhere else. If not, I don't know what you are learning.

*I am confirming the answer above, to be exact. Don't hesitate to trust the <u>quality assurance</u> user above! I also reached level Ace, but my account was deleted, so you can trust me, too!*

I hope this helps!

labwork [276]2 years ago

3 0

<span>The answer is mirrors. Mirrors are made by applying a metal thin layer on the back surface of a transparent substrat, typically glass. The metal layer in the antiquity was bronze, mercury and later silver whose luster gave the reflective property to the mirror.</span>

You might be interested in

A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a hor

nalin [4]

Answer:

$u_x=38.13\ m/s$

Explanation:

Given that initially ball moves in the horizontal direction ,it means that the velocity in the vertical direction is zero.

Horizontal distance = 13 m

Vertical distance = 57 cm

Lets take time to cover 57 cm distance in vertical direction is t.

We know that g is the constant acceleration in the vertical direction so we can apply the equation of motion in the vertical direction.

$S=u_yt+\dfrac{1}{2}gt^2$

Here $u_y=0$

S= 57 cm

$0.57=0\times t+\dfrac{1}{2}\times 9.81\times t^2$

t=0.34 s

Now in the horizontal direction

$x=u_xt$

Here x=13 m

t= 0.34 s

So

$13=u_x\times 0.34$

$u_x=38.13\ m/s$

So the initial speed of ball is 38.13 m/s.

7 0

1 year ago

Your ear is capable of differentiating sounds that arrive at the ear just 1.00 ms apart. What is the minimum distance (in centim

Andrej [43]

Answer:Your ear is capable of differentiating sounds that arrive at the ear just 1.00 ms apart. What is the minimum distance between two speakers that produce sounds that arrive at noticeably different times on a day when the speed of sound is 340 m/s

Explanation:

5 0

2 years ago

Automobile A starts from O and accelerates at the constant rate of 0.75 m/s2. A short time later it is passed by bus B which is

boyakko [2]

Answer:

they meet from point o at distance 50.46 m and time taken is 11.6 seconds

Explanation:

given data

acceleration = 0.75 m/s²

speed B = 6 m/s

time B = 20 s

to find out

when and where the vehicles passed each other

solution

we consider here distance = x , when they meet after o point

and time = t for meet point z

we find first Bus B distance for 20 s ec

distance B = velocity × time

distance B = 6 × 20

distance B = 120 m

so

B take time to meet is calculate by distance formula

distance = velocity × time

120 - x = 6 × t

x = 120 - 6t .................1

and

distance of A when they meet by distance formula

distance = ut + 1/2 × at²

here u is initial speed = 0 and t is time

x = 0 + 1/2 × 0.75 × t²

x = 0.375 × t² .............2

so from equation 1 and 2

0.375 × t² = 120 - 6t

t = 11.6

so time is 11.6 second

and

distance from point o from equation 2

x = 0.375 (11.6)²

x = 50.46

so distance from point o is 50.46 m

6 0

1 year ago

Terminal velocity. A rider on a bike with the combined mass of 100kg attains a terminal speed of 15m/s on a 12% slope. Assuming

Firlakuza [10]

Answer:

0.9378

Explanation:

Weight (W) of the rider = 100 kg;

since 1 kg = 9.8067 N

100 kg will be = 980.67 N

W = 980.67 N

At the slope of 12%, the angle θ is calculated as:

$tan \ \theta = \dfrac{12}{100} \\ \\ tan \ \theta = 0.12 \\ \\ \theta = tan^{-1}(0.12) \\\\ \theta = 6.84^0$

The drag force D = Wsinθ

$\dfrac{1}{2}C_v \rho AV^2 = W sin \theta$

where;

$\rho = 1.23 \ kg/m^3$

A = 0.9 m²

V = 15 m/s

∴

Drag coefficient $C_D = \dfrac{2 *W*sin \theta}{\rho *A *V^2}$

$C_D =\dfrac{2 *980.67*sin 6.84}{1.23 *0.9 *15^2}$

$C_D =0.9378$

8 0

1 year ago

Coffee is poured at a uniform rate at 20 cm3 / s into a cup whose inside is shaped like a truncated cone. if the upper and lower

Vladimir [108]

Let the cup is filled to height h after some time

now the total volume of coffee filled in the cup is given as

$\frac{2}{y} = \frac{4}{6+y}$

$2y = 6 + y$

$y = 6 cm$

now volume of the coffee will be

$V = \frac{1}{3}\pi r^2(y + 6) - \frac{1}{3}\pi 2^2 (6)$

here we know that

$\frac{r}{y+6} = \frac{2}{6}$

$r = \frac{y+6}{3}$

$V = \frac{1}{3}\pi (\frac{y+6}{3})^2(y+6) - \frac{1}{3}\pi 2^2(6)$

now we know that volume flow rate is given as

$Q = \frac{dV}{dt}$

$20 cm^3/s = \frac{1}{3}\pi (\frac{1}{9})(3(y+6)^2)\frac{dy}{dt}$

$20 \times 9 = \pi (y + 6)^2 v$

here y = 3 cm

$180 = \pi (9)^2 v$

$v = 0.71 cm/s$

so water will rise up with speed 0.71 cm/s

5 0

2 years ago