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3 years ago

Magnetic field lines surrounding a magnet are conventionally drawn

Physics

2 answers:

Lady bird [3.3K]3 years ago

7 0

North to South. Cannot switch around.

satela [25.4K]3 years ago

5 0

B) from north to south

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On what factors does critical velocity depend on

ASHA 777 [7]

Explanation:

The critical velocity is that velocity of liquid flow, up to which its flow is streamlined (laminar)& above which its flow becomes turbulent. It's denoted by Vc & it depends upon: Coefficient of viscosity of liquid (η) Density of liquid. Radius of the tube.

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3 years ago

Submarines need to be extremely strong to withstand the extremely high pressure of water pushing down on them. An experimental r

Brrunno [24]

1500 I think so but I not sure

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2 years ago

Describe how a battery works

Whitepunk [10]

Magic, Nah im just kidding. A battery has two parts, the anode and the cathode. Which anode is positive and cathode is negative, which they are connected to the electrolyte. Once they are connected to a device they once start working from separate ends. Which is the flow of energy

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3 years ago

A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha

USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

$\Phi_E=\frac{Q}{\epsilon_o}$

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

$\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}$

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

$Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C$

Finally, you obtain for E:

$E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}$

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

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2 years ago

Using hooke's law find the elastic constant of a spring that stretches 2 cm when 4newton force is applied to it

Kay [80]

<u>Answer:</u>

2N/cm

<u>Step-by-step explanation:</u>

According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

$F=kx$