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posledela
3 years ago
15

The freezing point of water is the same as its

Physics
2 answers:
Setler [38]3 years ago
8 0
(a): melting, ice melts at the same point where water freezes
denis23 [38]3 years ago
8 0

Answer:

A. Melting point

Explanation

Melting point is the point of temperature at which solid changes it's state to liquid while freezing point is the point of temperature at which liquid changes it's state to solid.

Melting and freezing point of water is same i.e. 0 degree centigrade and at this temperature 80 calories of heat per gram of solid ice will convert it into liquid water.

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The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A
kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

3 0
3 years ago
What forces act upon the moon as it orbits the earth
ValentinkaMS [17]

the axis acts against and it would be a contact force



7 0
3 years ago
Read 2 more answers
This is the reduction in the density of a medium
MAXImum [283]
The answer is Rarefaction
3 0
3 years ago
En la Tierra un volcán puede expulsar rocas verticalmente hasta una altura máxima H. A) ¿A qué altura (en términos de H) llegarí
Nonamiya [84]

A) 2.64 H

The maximum height that the expelled rock can reach can be found by using the equation:

v^2-u^2 = 2gd

where

v = 0 is the velocity at the maximum height

u is the initial velocity

g is the acceleration of gravity

d is the maximum height

Solving for d,

d=\frac{-u^2}{2g}

We see that the maximum heigth is inversely proportional to g. On the Earth,

d=H and g=g_e = -9.81 m/s^2

So we can write:

\frac{H}{H'}=\frac{g_m}{g_e}

where H' is the maximum height reached on Mars, and g_m = -3.71 m/s^2 is the acceleration of gravity on Mars. Solving for H',

H' = \frac{g_e}{g_m}H = \frac{-9.81}{3.71}H=2.64 H

B) 2.64T

The time after which the rock reaches the maximum height can be found by using

v=u+gt

where

v = 0 is the velocity at the maximum height

u is the initial velocity

Solving for t,

t=\frac{v-u}{g}

The total time of the motion is twice this value, so:

t=2\frac{v-u}{g}

So we see that it is inversely proportional to g.

On the Earth, t = T. So we can write:

\frac{T}{T'}=\frac{g_m}{g_E}

where T' is the total time of the motion on Mars. Solving for T',

T' = \frac{g_e}{g_m}T=\frac{-9.81}{-3.71}T=2.64T

4 0
3 years ago
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.80×106 N , one at an angle 14.0 ∘ west of north,
Leviafan [203]

Answer:

W = 1,049 10⁹ J

Explanation:

Work is defined by the relation

         W = F. d = F d cos θ

where tea is the angle between the forces and the displacement.

The total work is the sum of the work of each tug.

Tug 1

       W₁ = F d cos θ₁

 

the angle measured from the positive side of the x-axis is

       θ₁ = 14 + 90 = 104º

           

tugboat 2

             W₂ = F d cos θ₂

             θ₂ = 14

we substitute

             W = F d cos θ₁ + F d cos θ₂

             W = F d (cos θ₁ + cos θ₂)

               

let's calculate

             W = 1.80 10⁶  800 (cos 104 + cos 14)

             W = 1,049 10⁹ J

5 0
3 years ago
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