It'll have a higher frequency.
The product of (wavelength) times (frequency) for a wave
is always the same number ... it's the wave speed.
So if one of them is small, the other one has to be big.
A great, helpful, useful definition of acceleration is
<em>A = (change in speed) / (time for the change)</em> . <== you should memorize this
This simple tool will directly solve all 3 problems.
The REASON for assigning these problems for homework is NOT to find the answers. It's to help YOU find out whether you know this definition, to let you go back and review it if you don't, and to give you a chance to practice using it if you do. Noticed that if you get the answers from somebody else, you lose all of these benefits.
The only wrinkle anywhere here is in #3, because when you use this definition, the unit of time has to be the same in both the numerator and the denominator.
So for #3, you have to EITHER change the km/hr to km/sec, OR change the 4sec to a fraction of an hour, before you plug anything into the definition.
Question: A loader sack of total mass
is l000 grams falls down from
the floor of a lorry 200 cm high
Calculate the workdone by the
gravity of the load.
Answer:
19.6 Joules
Explanation:
Applying
W = mgh........................ Equation 1
Where W = Workdone by gravity on the load, m = mass of the loader sack, h = height, g = acceleration due to gravity
From the question,
Given: m = 1000 grams = (1000/1000) kilogram = 1 kg, h = 200 cm = 2 m
Constant: g = 9.8 m/s²
Substitute these values into equation 1
W = (1×2×9.8)
W = 19.6 Joules
Hence the work done by gravity on the load is 19.6 Joules
Answer:
B
Explanation:
The net force is the force between action and reaction and when this forces are not the same an acceleration is spurred.
<h3><u>Answer</u>;</h3>
= F0 L ( 1 - 1/e )
<h3><u>Explanation;</u></h3>
Work done is given as the product of force and distance.
In this case;
Work done = ∫︎ F(x) dx
= F0 ∫︎ e^(-x/L) dx
= F0 [ -L e^(-x/L) ] between 0 and L
= F0 L ( 1 - 1/e )
