Answer:
Perpendicular to the surface
Explanation:
- Electric field lines represent the direction of the electric field. The electric field lines also correspond to the direction along which the gradient of the electric potential is maximum.
- Equipotentials are lines or surfaces along which the electric potential is constant: the electric potential does not change moving along an equipotential surface.
Given the two definitions, equipotential lines are always perpendicular to the electric field lines. Therefore, in this problem, the direction of the electric field is perpendicular to the spherical equipotential surface.
Answer:
The answer is mutualism because they are both on the receiving and giving ends
I'll be happy to solve the problem using the information that
you gave in the question, but I have to tell you that this wave
is not infrared light.
If it was a wave of infrared, then its speed would be close
to 300,000,000 m/s, not 6 m/s, and its wavelength would be
less than 0.001 meter, not 12 meters.
For the wave you described . . .
Frequency = (speed) / (wavelength)
= (6 m/s) / (12 m)
= 0.5 / sec
= 0.5 Hz .
(If it were an infrared wave, then its frequency would be
greater than 300,000,000,000 Hz.)
Answer:
3.1 m/s
Explanation:
The total distance she has to run is the addition of the three lengths:
47 + 63 + 76 = 186 meters.
She needs to cover it one minute (60 seconds). Therefore her speed must be:
186 m / 60 s = 3.1 m/s
The force required to slow the truck was -5020 N
Explanation:
First of all, we find the acceleration of the truck, which is given by
where
v is the final velocity
u is the initial velocity
t is the time
For the truck in this problem,
v = 11.5 m/s
u = 21.9 m/s
t = 2.88 s
So the acceleration is
where the negative sign means that this is a deceleration.
Now we can find the force exerted on the truck, which is given by Newton's second law:
where
m = 1390 kg is the mass of the truck
is the acceleration
And substituting,
So the closest answer among the option is -5020 N.
Learn more about acceleration and forces:
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