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4 years ago

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Ben wants to model how the motion of particles changes with temperature. He considers water in an iron pot. Ben draws a model to

show how the water particles and iron particles move at room temperature. Then, he draws a second model to show how the water particles and iron particles move as the pot of water is heated. How should Ben’s second model differ from the first?

1 answer:

Lilit [14]4 years ago

7 0

Answer:

The particles should move faster and have more space between them.

Explanation:

As the molecules heat, they should start to vibrate with the energy. When they vibrate, the space between them increases.

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Anything made of matter must

prohojiy [21]

Answer:

positively charged protons

Explanation:

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3 years ago

A 5.00-kg object is initially at rest. The object is acted on by a 9.00-N force toward the east for 3.00 s. No force acts on the

olya-2409 [2.1K]

Answer:

The total distance at 7 s is:

$x_{tot}=27\: m$

Explanation:

<u>Distance due to the force</u>

We can use second Newton's law to find the acceleration.

$F=ma$

$a=\frac{F}{m}=\frac{9}{5}=1.8\: m/s^{2}$

Now, using the kinematic equation we will find the distance during this interval of time. Let's recall that the initial velocity is zero.

$x_{1}=0.5at_{1}^{2}$

$x_{1}=0.5(1.8)(3)^{2}$

$x_{1}=8.1\: m$

In the second part of the motion, the object moves at a constant velocity, as long as there is no friction between the object and the floor.

First, we need to find the final velocity of the first interval

$v=v_{i}+at_{1}=0+(1.8)3=5.4\: m/s$

So the second distance will be:

$x_{2}=vt_{2}=5.4*4=21.6\: m$

Therefore, the total distance is:

$x_{tot}=x_{1}+x_{2}=5.4+21.6=27\: m$

I hope it helps you!

7 0

3 years ago

A ball is kicked from a location <8, 0, -7> (on the ground) with initial velocity <-8, 19, -3> m/s. The ball's speed

wel

Answer:

a) v(0.3) = ( 8, 0 , 9.943 )

b) 21.1 m/s

c) S(0.5) = (4, 9.5 , 9.72625)

Explanation:

s(0) = 8i + 0j -7k

v(0) = -8i +19j -3k

a(t) = (0i+0j-9.81k)

part a

v(t) = v(0) + a(t)*t

v(0.3) = 8i + 0j -7k + (0i+0j-9.81k)* 0.3

v(0.3) = 8i + 0j - 9.943k

part b

S(t) = s(0) + v(0)*t + 0.5*a(t)*t^2

S(0.3) - S (0) = v(0)*t + 0.5*a(t)*t^2

= (-8i +19j -3k)*0.3 + 0.5*(0i+0j-9.81k)*(0.3)^2

= (-2.4i + 5.7j - .9k) + (0i+0j-0.44145k)

Displacement = (-2.4i + 5.7j - 1.34145k)

$distance = \sqrt{2.4^2+5.7^2+1.34145^2}\\ = 6.3285 m$

Average velocity = 6.3285 / 0.3 = 21.1 m/s

part c

S(t) = s(0) + v(0)*t + 0.5*a(t)*t^2

S(0.5) = v(0)*t + 0.5*a(t)*t^2 + S(0)

=(-8i +19j -3k)*0.5 + 0.5*(0i+0j-9.81k)*(0.5)^2 + (8i + 0j -7k)

=(-4i + 9.5j - 1.5k) + (0i+0j-1.22625k) + (8i + 0j -7k)

=(4i + 9.5j - 9.72625k)

7 0

4 years ago

You work for the city water department and need to pump 3400 liters/minute of water from a tank at ground level into a vented (i

umka21 [38]

Answer:

$P_E=46.2778\ kW$

$v=1.804\ m.s^{-1}$

Explanation:

Given:

flow rate of water, $\dot{V}=3400\ L.min^{-1}=3.4\ m^3.min^{-1}$

<em>∵Density of water is 1 kg per liter</em>

∴mass flow rate of water, $\dot{m}=3400\ kg.min^{-1}$

height of pumping, $h=75\ m$

efficiency of motor drive, $\eta=0.9$

diameter of pipe, $D =0.2\ m$

<u>Now the power required for pumping the water at given conditions:</u>

$P=\dot{m}.g.h$

$P=\frac{3400}{60} \times 9.8\times 75$

$P=41650\ W$

<u>Hence the electric power required:</u>

$P_E \times \eta=P$

$P_E \times 0.9=41650$

$P_E=46.2778\ kW$

<u>Flow velocity is given as:</u>

$v=\dot{V}\div a$

where: a = cross sectional area of flow through the pipe

$v=\frac{3.4}{60}\div (\pi.\frac{0.2^2}{4} )$

$v=1.804\ m.s^{-1}$

3 0

3 years ago

Part E<br> Check the Show planet Earth box. Describe what you see.

Leona [35]

Answer:

Earth is shown with a vertical bar magnet inside of it. The south end of the magnet points up, and the north end points down. If the polarity is still flipped, the north end of the magnet points up and the south end points down.

Explanation:

8 0

3 years ago