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Rzqust [24]
3 years ago
15

1).which of the following describes the interaction between a south pole and a north pole of a magnet

Physics
1 answer:
kumpel [21]3 years ago
6 0

1) a) attract

The magnetic force between two magnetic poles is attractive for two unlike poles and repulsive for two like poles. Therefore we have:

1- For two north poles, the force between them is repulsive

2- For two south poles, the force between them is repulsive

3- For a north pole and a south pole, the force between them is attractive

In this problem, we are in the situation described in 3), so the force between the poles is attractive.

2) a) motion of electrons

While electric fields are produced by static electric charges, magnetic fields are produced by charges in motion (currents). In particular, a current in a wire (where a current is simply the motion of electrons inside the wire) produces a magnetic field whose intensity is

B=\frac{\mu_0 I}{2 \pi r}

where

I is the current in the wire

r is the radial distance from the wire

And the direction of the field lines are such that the field form concentric circles around the wire.

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Which objects in space formed from the huge disk cisce and debris beyond the outer planets? Select two option
Dennis_Churaev [7]

Answer:

Comets

Explanation:

The Kuiper Belt is a collection of trans-Neptunian objects that consist of comets and other dwarf planets, including Pluto.

4 0
3 years ago
A truck initially traveling at a speed of 22 m/s increases at a constant rate of 2.4 m/s^2 for 3.2s. What is the total distance
FinnZ [79.3K]

Answer:

82.7 m

Explanation:

u= 22m/s

a= 2.4 m/s^2.

t= 3.2 secs

Therefore the distance travelled can be calculated as follows

S= ut + 1/2at^2

= 22 × 3.2 + 1/2 × 2.4 × 3.2^2

= 70.4 + 1/2×24.58

= 70.4 + 12.29

= 82.7 m

Hence the distance travelled by the truck is 82.7 m

6 0
3 years ago
ONLY TWO ANSWER CHOICES!!
Otrada [13]
A: Buoyant force is equal to the weight
3 0
3 years ago
A switch can
oee [108]

Answer: a switch can do A, B and E

Explanation:

8 0
3 years ago
A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subject
kow [346]

Answer:

The deflection of the spring is 34.56 mm.

Explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

Radius_{mean} = 60\ mm

Diameter_{mean} = 120\ mm

Load = 200 N

We need to calculate the deflection

Using formula of deflection

\delta=\dfrac{8pD^3n}{Cd^4}

Put the value into the formula

\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}

\delta =34.56\ mm

Hence, The deflection of the spring is 34.56 mm.

4 0
3 years ago
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