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inn [45]
3 years ago
14

Listed below are ACTION forces. Tell the REACTION force.

Physics
1 answer:
Eduardwww [97]3 years ago
3 0

Answer:

a. The desk seat will push back on the bottom (reaction force).

b. baseball hit the bat (reaction force)

c. The phone screen presses on your finger (reaction force).

Explanation:

The third law of Isaac Newton explains that, to every force, there is an equal and opposite force. The initial force is called the ACTION FORCE while the opposite force is called the REACTION FORCE. The reaction force for the following action forces in this question are:

- Your bottom pushing on your desk seat (action force)- The desk seat will push back on the bottom (reaction force).

- A bat hitting a baseball (action force) - baseball hit the bat (reaction force)

- Your finger pressing on your phone screen while texting (action force) - The phone screen presses on your finger (reaction force).

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Boron (B) has an atomic number of 5 and an atomic mass of 11. Boron has _____.
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the answer is 5 electrons

Explanation:

because its the same name as the amount of protons

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MA=\frac{33}{16.5} \\MA=2

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If continuous spectrum light from a star passes through a cool, low-density gas on its way to your telescope and spectroscope, _
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A metal ring 4.60 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend
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Answer:

A. Ein = 8.05*10^-4 V/m

B. Clockwise sense

Explanation:

A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:

\int E_{in} \cdot ds=-\frac{d\Phi_B}{dt}            (1)

Ein: induced electric field

ds: differential of a path of the ring

ФB: magnetic flux in the ring

The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):

\int E_{in}ds=E_{in}(2\pi r)=-A\frac{dB}{dt}\\\\E_{in}=-\frac{A}{2\pi r}\frac{dB}{dt}   (2)

dB/dt = -0.280T/s     (it is decreasing)

A: area of the ring = π(r/2)^2= (π/4) r^2

r: radius of the ring = 4.60/2 = 2.30 cm

Then, you replace the values of all variables in the equation (2):

E_{in}=-\frac{(\pi/4)r^2}{2\pi r}\frac{dB}{dt}=\frac{r}{8}\frac{dB}{dt}\\\\E_{in}=-\frac{0.0230m}{8}(-0.280T)=8.05*10^{-4}\frac{V}{m}

hence, the induced electric field is 8.05*10^-4 V/m

B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.

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