Pahelp po ako.

A demented scientist creates a new temperature scale, the "Z scale." He decides to call the boiling point of nitrogen 0°Z and th

slavikrds [6]

Answer:

A) $Z = 0.577 C +112.931$

$Z = 0.577*(100) +112.931=170.631 Z$

B) $C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C$

C) $K = C +273.15$

$K = -22.41 +273.15 =250.739 K$

Explanation:

For this case we want to create a function like this:

$Z = a C + b$

Where Z represent the degrees for the Z scale C the Celsius grades and tha valus a and b parameters for the model.

The boiling point of nitrogen is -195,8 °C

The melting point of iron is 1538 °C

We know the following equivalences:

-195.8 °C = 0 °Z

1538 °C = 1000 °Z

Let's say that one point its (1538C, 1000 Z) and other one is (-195.8 C, 0Z)

So then we can calculate the slope for the linear model like this:

$a = \frac{z_2 -z_1}{c_2 -c_1}= \frac{1000 Z- 0Z}{1538C -(-195.8 C)}=\frac{1000 Z}{1733.8 C}=0.577 \frac{Z}{C}$

And now for the slope we can use one point let's use for example (-195.8C, 0Z), and we have this:

$0 = 0.577 (-195.8) + b$

And if we solve for b we got:

$b = 0.577*195.8 =112.931 Z$

So then our lineal model would be:

$Z = 0.577 C +112.931$

Part A

The boiling point of water is 100C so we just need to replace in the model and see what we got:

$Z = 0.577*(100) +112.931=170.631 Z$

Part B

For this case we have Z =100 and we want to solve for C, so we can do this:

$Z-112.931 = 0.577 C$

$C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C$

Part C

For this case we know that $K = C +273.15$

And we can use the result from part B to solve for K like this:

$K = -22.41 +273.15 =250.739 K$

6 0

2 years ago

Please help on this one?

olganol [36]

the object distance of both lenses are positive.

6 0

2 years ago

Light of wavelength 623.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 76.5 cm from the slit.

Dvinal [7]

Answer:

Explanation:

wave length of light λ = 623 x 10⁻⁹ m .

Distance of screen D = 76.5 x 10⁻² m

width of slit = d

Distance on the screen between the second order minimum and the central maximum = 2 λ D / d

1.11 x 10⁻² = (2 x 623 x 10⁻⁹ x 76.5 x 10⁻² )/ d

d = ( 2 x 623 x 10⁻⁹ x 76.5 x 10⁻²) / 1.11 x 10⁻²

= 85872.97 x 10⁻⁹

= 85.87297 x 10⁻⁶

= 85.87 μm

width a of the slit is = 85.87 μm

6 0

3 years ago

Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t

salantis [7]

Explanation:

Mass of bumper cars, $m_1=m_2=110\ kg$

Initial speed of car A, $u_1=8\ m/s$

Initial speed of car Z, $u_2=-10\ m/s$

Final speed of car A after the collision, $v_1=-10\ m/s$

We need to find the velocity of car Z after the collision. Let it is equal to $v_2$ . Using the conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$110\times 8+110\times (-10)=110\times (-10)+110v_2$

$v_2=\dfrac{-1320}{110}\ m/s$

$v_2=-12\ m/s$

So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.

5 0

2 years ago

Which component of a galaxy is often found between the stars and looks like a cloud or smoke?

slava [35]

Answer:

Hey I would say Dust

hope this helped

5 0

2 years ago