The answer is D. 6 . This is because it is a positive slope so you can cancel out B and C and then, you count your boxes
Choice A is correct.======Kinetic energy equation: KE = (1/2)(m)(v²)This tells us that KE is directly proportional to mass and the square of velocity. In other words, the more mass and more velocity an object has, the more kinetic energy.If an object is sitting at the top of a ramp, there is no velocity and therefore no kinetic energy. Choices B and D are wrong.A golf ball has more mass than a ping-pong ball, so a ping-pong ball would have less kinetic energy than a golf ball rolling off the end of a ramp. Choice C is wrong.Choice A is correct.
Answer:
Option A is correct.
The wires will be arranged in order of increasing resistance.
Explanation:
The resistance of a wire is given by
r = (ρl)/A
where r = resistance of the wire
ρ = resistivity of the wire
L = length of the wire
A = cross sectional area of the wire
Provided all the other parameters are constant, resistance is inversely proportional to cross sectional area
r ∝ (1/A)
And the the cross sectional Area of the wire increases with increase in thickness & decreases with thickness
So, decreasing thickness ----> Decreasing Cross sectional Area ----> Increasing resistance.
Answer:
a) 4.49Hz
b) 0.536kg
c) 2.57s
Explanation:
This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:
for some time t you have:
x=0.134m
v=-12.1m/s
a=-107m/s^2
If you divide the first equation and the third equation, you can calculate w:
with this value you can compute the frequency:
a)
b)
the mass of the block is given by the formula:
c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:
Finally, the amplitude is:
Answer: option B: conduction.
Conduction is the heat transfer that happens between two bodies in direct contact, due to the collision of the molecules, atoms and electrons within the body (microscopical level).
