Answer:
b)
Explanation:
By convention, the electric field lines (which are tangent to the direction of the electric field at a given point) always begin at positive charges, and finish at negative charges.
This is a consequence of the convention that states that the electric field has the direction of the trajectory of a positive test charge when released from rest in an electric field.
(As the positive charge would move away from positive charges and would be attracted by negative ones).
So, the combination of answers that is true is b) (positive, negative, positive).
Answer:84.672 joules.
Explanation:
1) Data:
m = 7.2 kg
h = 1.2 m
g = 9.8 m / s²
2) Physical principle
Using the law of mechanical energy conservation principle, you have that the kinetic energy of the dog, when it jumps, must be equal to the final gravitational potential energy.
3) Calculations:
The gravitational potential energy, PE, is equal to m × g × h
So, PE = m × g × h = 7.2 kg × 9.8 m/s² × 1.2 m = 84.672 joules.
And that is the kinetic energy that the dog needs.
I'm going to assume that this gripping drama takes place on planet Earth, where the acceleration of gravity is 9.8 m/s². The solutions would be completely different if the same scenario were to play out in other places.
A ball is thrown upward with a speed of 40 m/s. Gravity decreases its upward speed (increases its downward speed) by 9.8 m/s every second.
So, the ball reaches its highest point after (40 m/s)/(9.8 m/s²) = <em>4.08 seconds</em>. At that point, it runs out of upward gas, and begins falling.
Just like so many other aspects of life, the downward fall is an exact "mirror image" of the upward trip. After another 4.08 seconds, the ball has returned to the height of the hand which flung it. In total, the ball is in the air for <em>8.16 seconds</em> up and down.
The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.
<h3>How does test charge affect electric field?</h3>
As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.
Adjusting the amount of charge on the test charge will not change the electric field force.
<h3>What is a test charge used for?</h3>
The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.
To learn more about test charge, refer
brainly.com/question/16737526
#SPJ9
Answer is given below
Explanation:
given data
pressure = double
volume = constant
solution
As we know that an Average velocity and rms velocity is directly proportional to square root of PV ..................1
so if we take P is doubled while keeping V constant
than Velocity increases by a factor 
so that Factor = 1.414 for both the cases
