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andreyandreev [35.5K]
3 years ago
6

Which describes the electromagnetic force only? Check all that apply. is attractive is repulsive has an infinite range has a ver

y small range produces light produces electricity
Physics
1 answer:
Oxana [17]3 years ago
7 0

Answer:Electromagnetic force, like all forces, is measured in Newtons. Electrostatic forces are described by Coulomb’s law, and both electric and magnetic forces are covered by the Lorentz force law. However, Maxwell’s four equations provide the most detailed description of electromagnetism.

Explanation:

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A mobile phone is pulled northward by a force of 10 n and at the same time pulled southward by another force of 15 n. the result
makvit [3.9K]
5 Southward would be the correct answer.
6 0
3 years ago
Two obiect accumulated a charge of
tamaranim1 [39]

Answer:

A. 181.24 N

Explanation:

The magnitude of hte electrostatic force between two charged objects is given by the equation

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

r is the separation between the charges

In this problem, we have:

q_1=4.5\mu C=4.5\cdot 10^{-6}C is the magnitude of the 1st charge

q_2=2.8\mu C=2.8\cdot 10^{-6}C is the magnitude of the 2nd charge

r = 2.5 cm = 0.025 m is the separation between the charges

Therefore, the magnitude of the electric force is:

F=\frac{(9\cdot 10^9)(4.5\cdot 10^{-6})(2.8\cdot 10^{-6})}{(0.025)^2}=181.44 N

So, the closest answer is

A) 181.24 N

3 0
4 years ago
A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber
Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}

Now calculate the bouyant force acting on the lead

F_L = Vpg

F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N

This force will act in upward direction

Gravitational force on the lead due to its mass  will act in downward direction

Hence the difference of this two force

T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N

If V is the volume submerged in the water then bouyant force on the bobber is

F_B=V'pg

Equate bouyant force with the tension and gravitational force

F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}

Now Total volume of bobble is

\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\

=\large\boxed{4.52 \%}

7 0
3 years ago
A river flows with a speed of 0.600 m/s. A student first swims upriver 0.500 km, then turns around and returns to his starting p
DerKrebs [107]

Answer:

a) 1111.0 seconds

b) 833.3 s

c) Because of proportions

Explanation:

a) Total time of round trip is the sum of time upriver and time downriver

t_{total}=t_{up}+t_{down}

Time upriver is calculated with the net speed of student and 0.500 km:

t_{up}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=v_{relative to river}+v_{river}=-1.2+0.6=-0.6 m/s\\t_{up}=\frac{500 m}{0.6 m/s}=833.3 s

(Becareful with units 0.5 km= 500m) Similarly of downriver:

t_{down}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=1.2+0.6=1.8 m/s\\t_{down}=\frac{500 m}{1.8 m/s}=277.7 s

So the sum is:

t_{total}=1111.0s

b) Still water does not affect student speed, so total time would be simply:

t_{total}=\frac{1000 m}{1.2 m/s}=833.3 s

c) For the upriver trip, student moved half the distance in half speed of the calculation in b), so it kept the same ratio and therefore, same time. So the aditional time is actually the downriver.  

6 0
3 years ago
A 0.150 kg baseball has 118 j of KE. how fast is the ball moving?(unit=m/s)
MrRissso [65]

Answer:

Explanation 118 = (1/2) * 0.15 * v² 118 = 0.075 * v² v² = 1573.33 m/s ... since KE = m/2*V^2 , then : V = √2KE/m = √20*118/1.5 = 39.67 m//sec ( 142.8 km/h ; 88.75 mph).:

4 0
3 years ago
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