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dimulka [17.4K]
2 years ago
10

State the reason why the magnetic field strength of a magnet is greatest at the poles​

Physics
1 answer:
icang [17]2 years ago
5 0

Answer:

As shown above, the magnetic field is strongest near to the poles of the magnet were the lines of flux are more closely spaced. The general direction for the magnetic flux flow is from the North ( N ) to the South ( S ) pole.

Explanation:

please give me brainlist and follow

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SOMEONE HELP PLEASE ASAP I BELIEVE THE ANSWER IS C OR D
Serhud [2]
You are correct the answer would be C

hope i helps
4 0
3 years ago
In this graph, what is the displacement of the particle in the last two seconds?
nikklg [1K]
In this graph, what is the displacement of the particle in the last two seconds?of the particle in the last two seconds? 

<span>0.2 meters
2 meters
4 meters
6 meters</span>
In this graph, the displacement of the particle in the last two seconds is 2 meters.
7 0
3 years ago
What is the energy released in this B- nuclear reaction 2K-&gt; 2Ca0,e? (The atomic mass of 42 K is 41.962403 u and that of 42Ca
Katyanochek1 [597]

<u>Answer:</u> The energy released in the given nuclear reaction is 3.526 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

_{19}^{42}\textrm{K}\rightarrow _{20}^{42}\textrm{Ca}+_{-1}^{0}\textrm{e}

We are given:

Mass of _{19}^{42}\textrm{K} = 41.962403 u

Mass of _{20}^{42}\textrm{Ca} = 41.958618 u

To calculate the mass defect, we use the equation:

\Delta m=\text{Mass of reactants}-\text{Mass of products}

Putting values in above equation, we get:

\Delta m=(41.962403-41.958618)=0.003785u

To calculate the energy released, we use the equation:

E=\Delta mc^2\\E=(0.003785u)\times c^2

E=(0.003785u)\times (931.5MeV) (Conversion factor: 1u=931.5MeV/c^2 )

E=3.526MeV

Hence, the energy released in the given nuclear reaction is 3.526 MeV.

7 0
3 years ago
2. What is the accelaration of a<br> 4000 kg airplane that has a force<br> of 24,000 N?
Bond [772]

Answer:

c) 6 m/s²

Explanation:

8 0
2 years ago
A hose on the ground projects a water current upwards at an angle 40 to the horizontal at velocity 20 m/s find height at which w
irinina [24]

Answer:

<em>The water hits the wall at a height of 5.38 m</em>

Explanation:

<u>Projectile Motion </u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

The object describes a parabolic path given by the equation:

{\displaystyle y=\tan(\theta )\cdot x-{\frac {g}{2v_{0}^{2}\cos ^{2}\theta }}\cdot x^{2}}

Where:

y   = vertical displacement

x   = horizontal displacement

θ   = Elevation angle

vo = Initial speed

The hose projects a water current upwards at an angle of θ=40° at a speed vo=20 m/s.

The height at which the water hits a wall located at x=8 m from the hose is:

{\displaystyle y=\tan40^\circ\cdot 8-{\frac {9.8}{2*20^{2}\cos ^{2}40^\circ }}\cdot 8^{2}}

Calculating:

y = 5.38 m

The water hits the wall at a height of 5.38 m

5 0
3 years ago
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