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salantis [7]
3 years ago
7

Which statement accurately describes the speed of the object in the graph above over the four seconds? (2 points)

Physics
1 answer:
Mumz [18]3 years ago
5 0

Answer:

I honestly have no clue so yeah

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Joe noted that the temperature in his aquarium dropped overnight. He concluded that thermal energy moved from the aquarium to th
alukav5142 [94]

Answer:

 validate                                            

Explanation:

<u>Law of Conservation of energy</u>

According to the law of conservation of energy, energy of a body or a system gets transferred to another system as different energy. We cannot create energy nor we can destroy energy. Energy is always transformed into other forms of energy.

In the context, Joe noticed that the aquarium's temperature dropped during the night and he concluded that according to the law of conservation of energy, the energy of the aquarium in terms of thermal energy is being transferred to the surrounding air.

This is because the aquarium is considered to be an open system where energy can come in and move out.

Hence Joe's statement is valid.

5 0
3 years ago
Read 2 more answers
When is it easiest for a person to build up static electricity?
Elena-2011 [213]
Idkk look it up on quizlet, they flash cards to tell you lots of stuff
3 0
4 years ago
Physical Science AIS - AE 20-21-Kotal / Unit 2: Reactions and Radioactivity / The Environment
astraxan [27]

Answer:19

Explanation:

6 0
3 years ago
If a load of 5 kg covers a distance of 50m in 2 min, what is the power? (g=10m/s)​
eduard

Explanation:

the answer is above with its si unit

3 0
2 years ago
A circular rod has a radius of curvature R = 9.09 cm and a uniformly distributed positive charge Q = 6.49 pC and subtends an ang
Digiron [165]

Answer:

E = 1.19 N/C

Explanation:

Let's first determine the length of the arc which can be given as:

L= Rθ

where:

L = length of the arc

R = radius of curvature

θ = angle in radius

L = (9.09×10⁻²m)(2.59)

L = (0.0909)(2.59)

L = 0.235431 m

Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:

E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]

E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]

E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]

Since \lambda = \frac{Q}{L}

where;

L = length

Q = charge

λ =  density of the charge;

then substituting \frac{Q}{L} for λ, we have :

E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]

E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}

substituting our given parameter; we have:

E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}

E = 1.1889 N/C

E = 1.19 N/C

∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C

4 0
3 years ago
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