Answer:
(a) The distance-time graph for an object with uniform speed is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram
(b) The distance-time graph for an object with non-uniform speed is giving by a curved line sloped graph with varying gradient as shown in the attached diagram
(c) The velocity-time graph for a car with uniform motion is giving by a horizontal line graph at the speed of constant motion with a zero gradient as shown in the attached diagram
(d) The velocity-time graph for a car moving with uniform acceleration is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram
(e) The velocity-time graph for a car moving with non-uniform acceleration is giving by a curved line sloped graph with varying gradient as shown in the attached diagram
(f) According to Newton's first law of motion, an object at rest will remain at rest with no motion unless acted by a force, an therefore, will have no motion with time
Explanation:
The resistance of a single light bulb is 220 ohms per bulb.
<h3>What is Ohm's Law?</h3>
Ohm's Law is a formula used to determine how voltage, current, and resistance in an electrical circuit relate to one another.
Ohm's Law (E = IR) is as basic to students of electronics as Einstein's Relativity equation (E = mc2) is to physicists.
E = I x R
The formula reads voltage = current x resistance, or V = A xΩ., or volts = amps x ohms.
110volts divided by .25amps = 440 ohms. 440 divided by 2 =220 ohms per bulb.
R = 110/(2*0.25) = 220 ohms
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Answer:
a. Potential energy decreases and Kinetic energy increases
Explanation:
Because as he comes down due to its steepness the speed of the boy or you can say his KE increases and since he comes from a high position (hill) to the lower ground his potential energy decreases simultaneously
Explanation:
It is known that electric field is responsible for creating electric potential. As a result, it depends only on the electric field and not on the magnitude of charge.
So, when a charge is increased by a factor of 2 then electric potential will remain the same. Since, expression to calculate the electric potential is as follows.
U = qV
Since, the electric potential is directly proportional to the charge. Hence, when 0.2
tends to replaced by 0.4
then charge is increased by a factor of 2. Hence, the electric potential energy is doubled.
Thus, we can conclude that if that charge is replaced by a +0.4 µC charge then electric potential stays the same, but the electric potential energy doubles.