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2 years ago

8. List three temperature scales. Then write the boiling and freezing

Physics

1 answer:

Tomtit [17]2 years ago

5 0

Explanation:

For most temperature scales, the boiling point of water and the freezing point is used to calibrate it.

Three known temperature scales;

Kelvin scale
Celcius scale
Fahrenheit scale

Kelvin scale Celcius scale Fahrenheit scale

Freezing point 273K 0°C 32°F

Melting point 373K 100°C 212°F

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A frictionless toy car is placed on a ramp, which is inclined at an unknown angle with respect to the horizontal. Starting from

NikAS [45]

The final speed of the toy car at the end of the given time period is 3.58 m/s.

The given parameters;

distance traveled by the car, s = 1.2 m
time of motion of the car, t = 0.67 s
initial velocity of the car, u = 0

The acceleration of the car is calculated as;

$s = ut + \frac{1}{2} at^2\\\\1.2 = 0 + 0.5\times a\times (0.67)^2\\\\1.2 = 0.225a\\\\a = \frac{1.2}{0.225} \\\\a = 5.33 \ m/s^2$

The final velocity of the toy car is calculated as;

$v_f^2 = u^2 + 2as\\\\v_f^2 = 0 + 2\times 5.33 \times 1.2\\\\v_f^2 = 12.792\\\\v_f = \sqrt{12.792} \\\\v_f = 3.58 \ m/s$

Thus, the final speed of the toy car at the end of the given time period is 3.58 m/s.

Learn more here: brainly.com/question/20352766

6 0

2 years ago

On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole

ohaa [14]

The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
$F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\$
Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation. $\alpha$ is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
$g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\
g_0=g'\\
g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\
ag_0=g_0-w^2r\\
w^2r=g_0(a-1)
$
Our final equation is:
$g=g_0-g_0(a-1)cos^2(\alpha)$
Colatitude is:
$\alpha_c=90^\circ-\alpha$
The answer is:
$g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)$

5 0

3 years ago

Difine scalar quantity

vfiekz [6]

Scalar quantity are physical quantities that have just magnitude, not direction.

It is always positive.
Examples: Speed, distance

8 0

2 years ago

Yall have helped a lot i just need help on this then ill be done for a while

Lubov Fominskaja [6]

Answer:

f(x)=a(x - h)2 + k

Much like a linear function, k works like b in the slope-intercept formula. Like where add or subtract b would determine where the line crosses, in the linear, k determines the vertex of the parabola. If you're going to go up 2, then you need to add 2.

The h determines the movement horizontally. what you put in h determines if it moves left or right. To adjust this, you need to find the number to make the parentheses equal 0 when x equals -2 (because moving the vertex point to the left means subtraction/negatives):

x - h = 0

-2 - h = 0

-h = 2

h = -2

So the function ends up looking like:

f(x)=a(x - (-2))2 + 2

Subtracting a negative cancels the signs out to make a positive:

f(x)=a(x + 2)2 + 2Explanation:

6 0

3 years ago

A race car starts from rest and travels east along a straight and level track. For the first 5.0 s of the car’s motion, the east

Marrrta [24]

Answer:

ax = 6.43m/s²

Explanation:

The acceleration is the time derivative of the velocity function ax = dvx(t)/dt

We have been given the velocity function v(t) and also the velocity v = 12.0m/s and we are requested to calculate the acceleration at this time which we don't know.

So the first step is to calculate the time at which the velocity =12.0m/s and with this time calculate the acceleration. Detailed solution can be found in the attachment below.

7 0

3 years ago