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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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~~~NEED HELP ASAP~~~<br>Please solve each section and show all work for each section.
anastassius [24]

Explanation:

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass m_A as

x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)

y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)

Substituting (2) into (1), we get

\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)

where f_N= \mu_kN, the frictional force on m_A. Set this aside for now and let's look at the forces on m_B

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>

Let the x-axis be (+) up along the inclined plane. We can write the forces on m_B as

x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)

y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)

From (5), we can solve for <em>N</em> as

N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)

Substituting (7) into (4) we get

m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba

Collecting similar terms together, we get

(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag

or

a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)

Putting in the numbers, we find that a = 1.4\:\text{m/s}. To find the tension <em>T</em>, put the value for the acceleration into (7) and we'll get T = 21.3\:\text{N}. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get N = 50.9\:\text{N}

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The properties of each state of a substance are due largely to the degree (amount) of intermolecular bonding. True False
marysya [2.9K]

Answer:

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Explanation:

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1. What is work done in holding a 15kg suitcase while waiting for a bus for 15 minutes?
Tems11 [23]
Work is (force applied) x (distance through which the force moves).

Since the suitcase doesn't move up or down during the 15 minutes,
no work is done ... zero, zip, nada ... according to the real Physics
definition of 'work'.
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