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Leviafan [203]
3 years ago
7

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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A small block with mass 0.0400 kg is moving in the xy-plane. The net force on the block is described by the potentialenergy func
OlgaM077 [116]

Answer:

A= 148.92  m/s²

Explanation:

Given that

U(x,y) = (6.00  )x²  - (3.75  )y ³

m= 0.04 kg

Now force in the x-direction

Fx= - dU/dx

U(x,y) = (6.00  )x²  - (3.75  )y ³

dU/dx= 12 x

When x=0.4 m

dU/dx= 12 x 0.4 = 4.8

So we can say that

Fx= - 4.8 N

From Newtons law

F= m a

- 4.8 = 0.04 x a

a = -120 m/s²

Acceleration in x direction ,a = -120 m/s²

In y -direction

F= - dU/dy

U(x,y) = (6.00  )x²  - (3.75  )y ³

dU/dy = 0 - 3.75 x 3 y²

When y = 0.56 m

dU/dy = - 3.75 x 3 x 0.56 x 0.56

dU/dy = - 3.52

So we can say that force in y -direction

F= 3.52 N

F= m a'

3.52 = 0.04 x a'

a'=88.2 m/s²

acceleration in y direction is 88.2 m/s²

The resultant acceleration

A=\sqrt{a^2+a'^2}

A=\sqrt{120^2+88.2^2}

A= 148.92  m/s²

7 0
4 years ago
Calculate the force of gravity on the 1.2-kg mass if it were 1.9×107 m above earth's surface (that is, if it were four earth rad
Anettt [7]

Answer:

Force of gravity, F = 0.74 N

Mass of an object, m = 1.2 kg

Distance above earth's surface, d=1.9\times 10^7\ m

Mass of Earth, M=5.97\times 10^{24}\ kg

Radius of Earth, r=6.37\times 10^6\ m

We need to find the force of gravity above the surface of Earth. It is given by :

F=G\dfrac{mM}{R^2}

R = r + d

R = 25370000 m

F=6.67\times 10^{-11}\times \dfrac{5.97\times 10^{24}\ kg\times 1.2\ kg}{(25370000\ m)^2}

F = 0.74 N

So, the force of gravity on the object is 0.74 N. Hence, this is the required solution.

5 0
4 years ago
A satellite is put into an orbit at a distance from the center of the Earth equal to twice the distance from the center of the E
Sveta_85 [38]

Answer:

995 N

Explanation:

Weight of surface, w= 4000N

Gravitational constant, g, is taken as 9.81 hence mass, m of surface is W/g where W is weight of surface

m= 4000/9.81= 407.7472

Using radius of orbit of 6371km

The force of gravity of satellite in its orbit, F=\frac {GMm}{(2r)^{2}}=\frac {GMm}{4(r)^{2}}

Where G=6.67*10^{-11} and M=5.94*10^{24}

F=\frac {(6.67*10^{-11}*5.94*10^{24}*407.7472)}{4*({6.371*10^{6}m)}^{2}}

F= 995.01142 then rounded off

F=995N

6 0
3 years ago
Is there any change in the pressure of container filled with water when the volumed is increased
marshall27 [118]
Not really the volume of a container is simply length X width X depth so just how big the container unless the water is pressurized by some sort of weight or if the containers air pressure is lowered
7 0
4 years ago
With the frequency set at the mid-point of the slider and the amplitude set at the mid-point of the slider, approximately how ma
Gre4nikov [31]

Answer:

correct answer is b

Explanation:

The frequency of a wave depends on the properties of medium density and the elasticity properties change the amplitude depends on the energy carried by the wave, that is, the amplitude is proportional to the height of the wave (oscillation).

Consequently the amplitude of independent of the frequency because it depends on different factors.

Therefore when changing the amplitude the wavelength remains constant

the correct answer is b

8 0
3 years ago
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