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3 years ago

on cloudless nighs, heat escapes from our planet and goes into empty space A.conduction B.convection C.radiation

2 answers:

6 0

Convention.

On cloudless nighs, heat escapes from our planet and goes into empty space <span>convection.</span>

PIT_PIT [208]3 years ago

5 0

On cloudless nighs, heat escapes from our planet and goes into empty space is an example of convection.

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Douglas has a segment with endpoints I(5, 2) and J(9, 10) that is divided by a point K such that IK and KJ form a 2:3 ratio. He

Rufina [12.5K]

For the answer to the question above,
the distance from i to j is 5 parts
(2 parts from i to k and 3 parts from k to j)

The y distance from i to j is
10 - 2 = 8

Each part is 8/5 = 1.6
Therefore the distance between the 2 parts from i to k is 3.2

From the y coordinate of I which is 2 plus the 3.2 to point k
2 + 3.2 = 5.2

Answer y =5.2

Now just convert that to fraction and that will be the answer

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2 years ago

Which of the following does NOT represent Newton’s second law? Question 20 options: a = m/Fnet m = Fnet/a Fnet = ma a = Fnet/m

Natali [406]

Answer:

a=m/f is not an equation under newton's second law

Explanation:

newton's second law of motion is represented using: f=ma

where a=v-u/t

therefore it becomes,f=m(v-u)/t

from f=ma,

a will become f/m,

m will become f/a

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2 years ago

Lab: Motion

I am Lyosha [343]

Answer: In this lab we wanted to know how motion can be described. So the hypothesis is if the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will increase because the toy car will have a greater acceleration. My prediction is that cars travel faster on higher tracts. So the heighten the track was intentionally manipulated. So it is the independent variable the speed of the car is the dependent variable. The speed at the first quarter checkpoint is 1.09 m/s. The speed at the second quarter checkpoint is 1.95 m/s. The speed at the third quarter checkpoint is 2.373.36 m/s. The speed at the finish line is 2.803.00 m/s. The average speed increases as the height increases.

The cars on the higher track travel farther than the cars on the lower track, in the same time.

This means that the cars on the higher track have a greater average speed than those on the lower track. This is demonstrated by the

slope of the higher track line being greater than the slope of the lower track line.

Explanation: put it in notes then send it to files to compress it to submit it.

5 0

2 years ago

Read 2 more answers

?????Help please?????

Sindrei [870]

The correct selections are C, C, B, D, A, B, and A .

6 0

3 years ago

Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12

PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

$q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}$

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

$F_e = k\frac{q_1q_2}{r^2}$

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

$F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2} \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o$

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

$F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}$

$(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}$

Solving the quadratic equation:

$q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C$

for this values q_1 wil be:

$q_1_o = -1.325 *10^{-6}C | 4.17*10^{-6}C$

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0

3 years ago