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3 years ago

14

how is the motion of the particles in the soup different after the temperature of the soup increases?

1 answer:

eduard3 years ago

6 0

The particles break down and liquify as the tempeture increases.

You might be interested in

A planet in elliptical orbit around a star moves from the point in its orbit furthest from the star (A) to the closest point (P)

Virty [35]

Answer:

Zero work done,since the body isn't acting against or by gravity.

Explanation:

Gravitational force is usually considered as work done against gravity (-ve) and work by gravity ( +ve ) and also When work isn't done by or against gravity work done in this case is zero.

Gravitational force can be define as that force that attracts a body to any other phyical body or system that have mass.

The planet been considered as our system in this case is assumed to have mass, and ought to demonstrate such properties associated with gravitational force in such system. Such properties include the return of every object been thrown up as a result of gravity acting downwards. The orbiting nature of object along an elliptical part when gravitational force isn't acting on the body and it is assumed to be zero.

6 0

3 years ago

Energy producing technologies can positively impact soil fertility.

xenn [34]

The Answer is TRUE. Hope it helps

8 0

2 years ago

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

6 0

3 years ago

jeka94

Answer:

v = 5.15 m/s

Explanation:

At constant velocity, the cable tension will equal the car weight of 984(9.81) = 9,653 N

As the cable tension is less than this value, the car must be accelerating downward.

7730 = 984(9.81 - a)

a = 1.95 m/s²

kinematic equations s = ut + ½at² and v = u + at

-5.00 = u(4.00) + ½(-1.95)4.00²

u = 2.65 m/s the car's initial velocity was upward at 2.65 m/s

v = 2.65 + (-1.95)(4.00)

v = -5.15 m/s

3 0

2 years ago

You send a beam of light from a material with index of refraction 1.21 into an unknown material. In order to help identify this

White raven [17]

Answer:

refractive index of the unknown material is 1.33.

Explanation:

μ₁ = 1.21

incidence angle (i) = 41.9°

refraction angle (r) = 37.3°

Let us assume μ be the refractive index of the unknown material

according to snell's law of refraction.

μ₁ sin i = μ₂ sin r

1.21 × sin 41.9° = μ × sin 37.3°

μ = 1.33

hence the refractive index of the unknown material comes out top be 1.33

6 0

3 years ago