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3 years ago

Part 1

Physics

1 answer:

laila [671]3 years ago

7 0

Answer:

pressure is applied to the ground

by a 79 kg man who is standing on square

stilts that measure 0.04 m on each edge?

Answer in units of Pa.

t 2

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MANY POINTS!!! PLZ! I NEED HELP!!!

liq [111]

Cetane rating of the diesel ensures the low temperature and uniform combustion. Centane number is just opposite to the octane number of petrol engine, cetane number of diesel determines at which temperature the diesel will going to autoignite.

8 0

4 years ago

Three arrows are shot horizontally. They have left the bow and are traveling parallel to the ground. Air resistance is negligibl

timurjin [86]

Answer:

F₁ = F₂ = F₃ = 0 N

Explanation:

given,

Arrow 1 mass = 80 g speed = 10 m/s

Arrow 2 mass = 80 g speed = 9 m/s

Arrow 3 mass = 90 g speed = 9 m/s

Horizontal Force:- F₁ , F₂ and F₃

There is no air resistance.

If Air resistance is zero then the horizontal acceleration of the arrow also equal to zero.

We know,

According to newton's second law

F = m a

If Acceleration is equal to zero

Then Force is also equal to zero.

Hence, F₁ = F₂ = F₃ = 0 N

4 0

3 years ago

Find the right answer please

Ghella [55]

Answer:

3.78

Explanation:

bc ik it is :3

5 0

3 years ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

3 years ago

What tool is used to measure air pressure?

alukav5142 [94]

A<span> barometer is used to measure air pressure. </span>

8 0

4 years ago