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kompoz [17]
3 years ago
6

The heaviest invertebrate is the giant squid, which can have a weight of about 1.7 tons spread out over its length of 68 feet. w

hat is its weight in newtons?
Physics
1 answer:
DENIUS [597]3 years ago
4 0
In science and engineering, the weight<span> of an object is usually taken to be the force on the object due to gravity. It is calculated by multiplying the mass with the gravitational acceleration.

W = mg = 1700 (9.81) = 16677 N

Hope this answers the question. Have a nice day.</span>
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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:

1)

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})

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}

3)

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
A series circuit has a 12-volt power source and two resistors of 1 ohm and 2 ohms respectively. How many amps will flow when
Andrej [43]

Answer:

ayoooooooo

Explanation:

3 0
3 years ago
If a fuse melts, does it create an open circuit, a closed circuit, or a short circuit?
Andrew [12]

Answer:

short circuit

7 0
3 years ago
Read 2 more answers
Can somebody help me on this mcq question?
andrew11 [14]
A) lighting an electric lamp as it becomes darker
3 0
3 years ago
How much equal charge should be placed on the earth and the moon so that the electrical repulsion balances the gravitational for
kumpel [21]

As we know that electrostatic force between two charges is given as

F = \frac{kq_1q_2}{r^2}

here we know that electrostatic repulsion force is balanced by the gravitational force between them

so here force of attraction due to gravitation is given as

F_g = 1.98 \times 10^{20} N

here we can assume that both will have equal charge of magnitude "q"

now we have

1.98 \times 10^{20} = \frac{kq^2}{r^2}

1.98 \times 10^{20} = \frac{(9\times 10^9)(q^2)}{(3.84 \times 10^8)^2}

1.98 \times 10^{20} = (6.10 \times 10^{-8}) q^2

now we have

q = 5.7 \times 10^{13} C

6 0
3 years ago
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