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Damm [24]
3 years ago
11

Which could be an example of a balanced force?

Physics
2 answers:
Rus_ich [418]3 years ago
4 0

Answer:

An  asteroid moving at a constant speed through space.

Explanation:

AleksandrR [38]3 years ago
4 0
The answer is A sorry if I’m wrong
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A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that
VMariaS [17]

Answer:

E = 9.66\times 10^{-6} N/C

direction is Horizontal

Explanation:

As we know that the string is horizontal here

so the tension force in the string is due to electrostatic force on it

now we will have

F = qE

so here the force is tension force on it

F = 6.57 \times 10^{-2} N

Q = 6.80 \times 10^3 C

now we have

6.57 \times 10^{-2} = (6.80 \times 10^3)E

E = 9.66\times 10^{-6} N/C

direction is Horizontal

4 0
3 years ago
As the boat in which he is riding approaches a dock at 3.0 m/s, Jasper stands up in the boat and jumps toward the dock. Jasper a
Annette [7]
Im pretty sure it’s a because it makes more sense you know?.
5 0
3 years ago
When light is reflected by a mirror, the angle of incidence is always A. equal to the angle of reflection. B. less than the angl
ankoles [38]
When light is reflected by a mirror, the angle of incidence is always <span>A. equal to the angle of reflection. We know this by the Law of Reflection.</span>
6 0
3 years ago
Read 2 more answers
An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if
maxonik [38]
The electron is accelerated through a potential difference of \Delta V=780 V, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
\frac{1}{2}mv^2 =  e \Delta V
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
\Delta V is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s


Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
evB=m \frac{v^2}{r}
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T
3 0
3 years ago
Who is the leader of the party's national committee
cestrela7 [59]
That would be <span>the national chairperson

-I hope this helped.</span>
3 0
3 years ago
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