Explanation:
If a positive test charge is placed in an electric field, it will exert the force in the test charge in the direction of electric field vector. We know that the direction of electric field is given by electric field lines. The field lines for a positive charge is outwards. The electric force acting on the charge is given by :
F = q E
Hence, this is the required solution.
Answer:
1) Newton's first law of motion states an object will remain at rest or in uniform will be in uniform motion in a straight line unless a force acts on it
2) Newton's second law states the acceleration of an object is directly proportional to the applied force acting on an object and inversely proportional to the mass of the object
Explanation:
1) With Newton's first law, we are able arrange things within a space and schedule meetings in time knowing that they will remain in place unless an external force changes their positions
2) An example of Newton's second law of motion is that small objects such as a ball are easily accelerated and can be given appreciable acceleration for flight by single, one time contact (such as kicking the ball) while larger objects such as a rock require sustained force application to change their location.
Answer:
configuration of string:
Node - Antinode - Node or N-A-N
This is 1/2 wavelength since a full wavelength is N-A-N-A-N
f (fundamental) = V / wavelength
F0 = 300 m/s / 1 m = 100 / sec
F1 = 300 m/s / .5 m = 600 / sec
Each increase is a multiple of the fundamental since the wavelength
increases by 1/2 wavelength to keep nodes at both ends of the string
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Weight = electric force
<span>mg = qE </span>
<span>6.64x10^-27 x 9.81 = (2 x 1.60x10^-19) E
</span>qE =mg,
<span>E = mg/q = 6.64•10^-27•9.8/2•1.6•10^-19 =2.03•10^-7 V/m</span>
Answer:
0.176m from the flagpole, westward.
Explanation:
Let the Eastward be the positive direction. So initially runner A is at position -6km, running with velocity of 9km/h while runner B is at position 5km running at a velocity of -8km/h. We can conduct the following equation for their distances over the same time t
When A an B meets, they are at the same position and at the same time. So
So where they meet is 0.176m from the flagpole, westward.
