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Mnenie [13.5K]
1 year ago
11

Magnet A has twice the magnetic field strength of magnet B and pulls on magnet B with a force of 100 N. The amount of force that

magnet A exerts on magnet B isA) about 50 N.B) exactly 100 N.C) more than 100 N.D) not enough information
Physics
1 answer:
son4ous [18]1 year ago
3 0

The force exerted by the magnetic in terms of the magnetic field is,

F\propto B

Where B is the magnetic fied strength and F is the force.

Thus, if the magnetic A has twice magnetic field strength than the magnet B,

Then,

B_A=2B_B

Thus, the force exerted by the magnet B is,

\begin{gathered} F_B\propto B_B \\ F_B\propto\frac{B_A}{2} \\ F_B=\frac{F_A}{2} \\ F_B=\frac{100}{2} \\ F_B=50\text{ N} \end{gathered}

Thus, the force exerted by the magnet B on magnet A is 50 N.

The force exerted by the magnet A exerts on the magnet B is exactly 100 N as given.

Hence, the option B is the correct answer.

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5 0
3 years ago
A proton is placed in an electric field of intensity 700 n/
kakasveta [241]
F=ma
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8 0
3 years ago
The aluminum atom______electrons to form an ion.<br> The ion that is formed is______.
Alchen [17]

The aluminum atom_loses_____electrons to form an ion.

The ion that is formed is_Al³⁺_____.

aluminium has the electronic configuration as 1s² 2s² 2p⁶ 3s² 3p¹

from the electronic configuration , we see that aluminium can attain stability by losing 3 electrons from outer shell.

after losing 3 electrons , the ion formed is given as Al³⁺

hence the correct options to fill in the blanks are lose  and Al³⁺


3 0
3 years ago
Read 2 more answers
A 75-m-long train begins uniform acceleration from rest. the the train has a speed of 23 m/s when it passes a railway worker tan
Lapatulllka [165]

Answer:

acceleration a = 1.04 m/s2

Explanation:

Assume the train has a speed of 23m/s when the last car passes the railway workers. Once this happens the last car would have traveled a total distance of the 180m distance between the railway worker standing 180 m from where the front of the train started plus the 75m distance from the first car to the last car:

s = 75 + 180 = 255 m

We can use the following equation of motion to find out the distance traveled by the car:

v^2 - v_0^2 = 2aswhere v = 23 m/s is the velocity of the car when it passes the worker, v_0 = 0m/s is the initial velocity of the car when it starts, a m/s2 is the acceleration, which we are looking for.

23^2 - 0^2 = 2*a*255

510a = 529

a = 529 / 510 = 1.04 m/s^2

8 0
3 years ago
A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication
koban [17]

Answer:

r = 4.24x10⁴ km.  

     

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km.                           </em>                              

From equation (1), r₁ is:

r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}                            

r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}      

r_{1} = 4.24 \cdot 10^{4} km      

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

3 0
3 years ago
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