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3 years ago

Which equation shows the relationship between the Kelvin and Celsius temperature scales?

Physics

2 answers:

jenyasd209 [6]3 years ago

5 0

K = C + 273.15

and

C = K - 273.15

Gnom [1K]3 years ago

4 0

Answer:

T(C) = T(K) + 273.15

T(K) = T(C) - 273.15

Explanation:

Kelvin (K) is the scale used to measure the absolute temperature, and it is the SI unit for the temperature.

Another scale used is the Celsius degree (C), which is defined by using the freezing point and boiling point of water. In fact:

- The zero (0 C) in the Celsius scale is defined as the temperature at which the water freezes, and

- 100 C in the Celsius scale is defined as the temperature at which the water boils

The Celsius scale involves the possibility of having negative temperatures; instead, the zero in the Kelvin scale (0 K) is defined as the lowest possible temperature in the Universe. The conversion factors between the two scales are the following:

T(C) = T(K) - 273.15

T(K) = T(C) + 273.15

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The first wave that forms when the wind begins to blow across the ocean surface is a ________.

seraphim [82]

The first wave that forms when the wind begins to blow across the ocean surface is a <u>capillary wave</u>

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<h3>What is a capillary wave?</h3>

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2 years ago

Factor 72x^2 - 8/9<br><br> If you can, please show the steps to solve!

ludmilkaskok [199]

Factor out 8 and then facotr and u get

8/9(9x+1)(9x-1

7 0

3 years ago

A rock is thrown off a cliff at an angle of 53° with respect to the horizontal. The cliff is 100 m high. The initial speed of th

Nadusha1986 [10]

(a) 129.3 m

The motion of the rock is a projectile motion, consisting of two indipendent motions along the x- direction and the y-direction. In particular, the motion along the x- (horizontal) direction is a uniform motion with constant speed, while the motion along the y- (vertical) direction is an accelerated motion with constant acceleration $g=-9.8 m/s^2$ downward.

The maximum height of the rock is reached when the vertical component of the velocity becomes zero. The vertical velocity at time t is given by

$v(t) = v_0 sin \theta +gt$

where

$v_0 = 30 m/s$ is the initial velocity of the rock

$\theta=53^{\circ}$ is the angle

t is the time

Requiring $v(t)=0$ , we find the time at which the heigth is maximum:

$0=v_0 sin \theta + gt\\t=\frac{-v_0 sin \theta}{g}=-\frac{(30)(sin 53^{\circ})}{(-9.8)}=2.44 s$

The heigth of the rock at time t is given by

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

Where h=100 m is the initial heigth. Substituting t = 2.44 s, we find the maximum height of the rock:

$y=100+(30)(sin 53^{\circ})(2.44)+\frac{1}{2}(-9.8)(2.44)^2=129.3 m$

(b) 44.1 m

For this part of the problem, we just need to consider the horizontal motion of the rock. The horizontal displacement of the rock at time t is given by

$x(t) = (v_0 cos \theta) t$

where

$v_0 cos \theta$ is the horizontal component of the velocity, which remains constant during the entire motion

t is the time

If we substitute

t = 2.44 s

Which is the time at which the rock reaches the maximum height, we find how far the rock has moved at that time:

$x=(30)(cos 53^{\circ})(2.44)=44.1 m$

(c) 7.58 s

For this part, we need to consider the vertical motion again.

We said that the vertical position of the rock at time t is

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

By substituting

y(t)=0

We find the time t at which the rock reaches the heigth y=0, so the time at which the rock reaches the ground:

$0=100+(30)(sin 53^{\circ})t+\frac{1}{2}(-9.8)t^2\\0=100+23.96t-4.9t^2$

which gives two solutions:

t = -2.69 s (negative, we discard it)

t = 7.58 s --> this is our solution

(d) 136.8 m

The range of the rock can be simply calculated by calculating the horizontal distance travelled by the rock when it reaches the ground, so when

t = 7.58 s

Since the horizontal position of the rock is given by

$x(t) = (v_0 cos \theta) t$

Substituting

$v_0 = 30 m/s\\\theta=53^{\circ}$

and t = 7.58 s we find:

$x=(30)(cos 53^{\circ})(7.58)=136.8 m$

(e) (36.1 m, 128.3 m), (72.2 m, 117.4 m), (108.3 m, 67.4 m)

Using the equations of motions along the two directions:

$x(t) = (v_0 cos \theta) t$

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

And substituting the different times, we find:

$x(2.0 s)=(30)(cos 53^{\circ})(2.0)=36.1 m$

$y(2.0 s)= 100+(23.96)(2.0)-4.9(2.0)^2=128.3 m$

$x(4.0 s)=(30)(cos 53^{\circ})(4.0)=72.2 m$

$y(4.0 s)= 100+(23.96)(4.0)-4.9(4.0)^2=117.4 m$

$x(6.0 s)=(30)(cos 53^{\circ})(6.0)=108.3 m$

$y(6.0 s)= 100+(23.96)(6.0)-4.9(6.0)^2=67.4 m$

3 0

3 years ago

What is solid, geometric shape formed by a minerals?

77julia77 [94]

crystal hope this helps

6 0

4 years ago

A wave has a wavelength of 8 mm and a frequency of 14 Hertz what is its speed?

iren [92.7K]

To find the speed you multiply the wavelength and the frequency together.

wavelength = 8mm = $8x10^{-3}$ m = 0.008 m
frequency = 14 Hz

speed = 0.008m * 14 Hz = 0.112 m/s

6 0

4 years ago