We will be using this equation for this problem
d = ut + ½.at²
<span>Given:</span>
<span>initial velocity, u = 0 (falling from rest) </span>
<span>acceleration, a = +9.80 m/s²(taking down as the convenient positive direction) </span>
<span>Time = 1.0s, 2.0s, 3.0s, 4.0s, 5.0s </span>
<span>Using .. d = ½.at² each time (each calculation is the distance from the top) </span>
<span>For 1.0s .. d = ½ x 9.80 x (1²) = 4.90 m </span>
<span>For 2.0s .. d = ½ x 9.80 x (2²) = 19.60 m </span>
<span>3.0s .. d = 44.10m (you show the working for the rest) </span>
<span>4.0s .. d = 78.40 m </span>
<span>5.0s .. d = 122.50m </span>
<span>Plot distance (displacement from the top) on the y-axis against time on the x-axis (label axes and give units for each).The line of best fit will be a smoothly upward curving line getting progressively steeper. Do not join graph points with straight lines.</span>
Answer:
is this a yes or no question
Answer:
She is incorrect
Step-by-step explanation:
Step 1: substitute
135 into x in both equations
Step 2: Solve the equations. Once you do this you will get y = 5375 for the first one and y = 4720 for the second one, therefore she is incorrect
Answer:

Step-by-step explanation:
The given expression is

If we evaluate the expression as it is we will get;

But we want to insert parenthesis so that we will use PEDMAS to obtain 37 as the result.
The only way this can be achieved is when we insert the parenthesis as shown below;

Speed of current ------ y km/h
<span>distance with the current = 4(x+y) </span>
<span>distance against the current = 5(x-y) </span>
<span>we know that 3.5(x+y) = 70 </span>
<span>x+y = 20 </span>
<span>but 4(x+y) = 4(20) = 80 </span>
<span>then 5(x-y) = 80 </span>
<span>x-y = 16 </span>
<span>x+y=20 </span>
<span>x-y=16 </span>
<span>add them </span>
<span>2x = 36 </span>
<span>x = 18 , then y = 2 </span>