Answer:
applying 1st eq of motion vf=vi+at we have to find a=vf-vi/t here a=50-30/2=10 so we got a=10m/s²
Answer
The dedicated graphics card is used when performing hardware-intensive tasks so as to ensure efficiency and balanced performance. However, it uses more power and thus produces more heat. When the cooling system is not sufficient or the room is not well ventilated, your PC begins to overheat while playing games. Explanation: How does the second law of thermodynamics relate to the direction of heat flow? Heat of itself never flows from a cold object to a hot object. ... The second law expresses the maximum efficiency of a heat engine in terms of hot and cold temperatures. one of these answers i am not sure
Answer:
False -
F = G M1 M2 / R^2
So F depends on M1 and M2 and as long either is not zero there will be a gravitational force between them.
Answer:
B. Maximum velocity of ejected electrons.
Explanation:
The ejection of electrons form a metal surface when the metal surface is exposed to a monochromatic electromagnetic wave of sufficiently short wavelength or higher frequency (or equivalently, above a threshold frequency), which leads to the enough energy of the wave to incident and get absorbed to the exposed surface emits electrons. This phenomenon is known as the photoelectric effect or photo-emission.
The minimum amount of energy required by a metal surface to eject an electron from its surface is called work function of metal surface.
The electrons thus emitted are called photo-electrons.
The current produced as a result is called photo electricity.
Energy of photon is given by:
where:
h = Planck's constant
frequency of the incident radiation.
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
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