All categories

3 years ago

PLEASE HELP I NEED THIS TO PASS THE EIGHTH GRADE AND I ONLY HAVE A COUPLE HOURS LEFT!!!!!!

Physics

1 answer:

zavuch27 [327]3 years ago

7 0

Answer:

the pe at the top of the building: 784 J

the pe halfway through the fall: 392 J

the pe just before hitting the ground: 784 J

Explanation:

Pls brainliest me

I had this question before

You might be interested in

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and

kari74 [83]

Answer:

$a. 4.2057\times 10^-^1^2 F \ or 4.2057\ pF\\b. 2.7344V$

Explanation:

Given the permittivity constant to be $8.854\times 10^-^1^2 F/m$ ,The capacitance of a $cylindrical \ capacitor$ of length, $L$ is given by the equation:

$C=\frac{2\pi \epsilon _o L}{ln(b/a)}$ where $b$ is the radius of the outer cylinder and $a$ the radius of the inner cylinder.

The values are given as: $a=0.550mm(5.5\times 10^-^4m), \ b=4.00mm(4.0\times10^-^3m), \ L=15.0cm(0.150m)$

Substitute in our capacitance equation:

$C=\frac{2\times\pi \times 8.854\times 10^-^1^2 \times 0.15}{In(4.00/0.550)}\\=4.2057\times 10^-^1^2 F$

Hence the capacitance is $4.2057\times 10^-^1^2 F$

b. The charge on the capacitance is related to the potential difference across it. The potential difference is expressed using the equation:

$Q=CV$ , $Q=11.5pC$

From a above, we already have our capacitance value, $C=4.2057\times 10^-^1^2 F$

We substitute $C$ in the pd equation:

$v=>(11.5)/(4.2057)\\=2.7344V$

Hence, the applied potential difference is 2.7344V

4 0

3 years ago

A 25 pF parallel-plate capacitor with an air gap between the plates is connected to a 100 V battery. A Teflon slab (dielectric c

adoni [48]

Answer:

The change in the charge on the positve plate when the Teflon is inserted is +2.5 nC.

Explanation:

It can be showed that the capacitance of a parallel-plate capacitor, can be expresssed as follows:

$C = \frac{\epsilon*A}{d}$

Where ε, is the dielectric constant of the material that fills the space between plates.
When this space is filled with air, ε= ε₀ = 8,85*10⁻¹² F/m.
At the same time, the capacitance of a capacitor, by definition, is as follows:

$C =\frac{Q}{V}$

If we insert a Teflon slab, in such a way that fills completely the gap between the plates, all other parameters being equal, if ε = 2*ε₀, this means that C₂ = 2* C₁. = 50 pF
As V₂=V₁ (due to the capacitor remains connected to the same battery) the charge must be the double, so Q₂ = 2* Q₁ = 5 nC.
So, the change in the charge of the positive plate is +2.5 nC.

3 0

3 years ago

Write down the symbols of nitrogen and neon?

statuscvo [17]

Answer:

Nitrogen is N, neon is Ne

5 0

2 years ago

Read 2 more answers

Why does a magnet attract a steel nail

SVEN [57.7K]

Answer:

The nail will stick to the bar magnet because it will become magnetized, and it's metal. The presence of the nearby north pole rearranges the magnetic domains inside the steel so that their south poles all point toward the north pole of the permanent magnet. As a result, the other end of the nail becomes a north pole.Magnets attract iron due to the influence of their magnetic field upon the iron. ... When exposed to the magnetic field, the atoms begin to align their electrons with the flow of the magnetic field, which makes the iron magnetized as well. This, in turn, creates an attraction between the two magnetized objects.

7 0

2 years ago

Two animals that can be found at the Grand Canyon include the Abert’s and Kaibab squirrels. The Abert’s squirrel is found at the

Tom [10]

D. <span>When placed together in pairs for study, no pair of Abert's squirrel and Kaibab squirrel result in offspring.</span>

6 0

3 years ago