An investigation that has been done by more than one scientist with similar findings has been replicated.
<h3>What does replication in science mean?</h3>
The expression 'replication in science' makes reference to the same experimental result through different hypothesis testing procedures, which is fundamental to reinforcing a scientific idea.
In conclusion, an investigation that has been done by more than one scientist with similar findings has been replicated.
Learn more about replication in science here:
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Answer:
4.80 m
Explanation:
We are given the mass of the high jumper, its initial velocity, and the acceleration of gravity. We are trying to find the vertical displacement of the high jumper.
Let's set the upwards direction to be positive and the downwards direction to be negative.
List out the relevant known variables.
- v₀ = 9.7 m/s
- a = -9.8 m/s²
- Δx = ?
We still need one more variable in order to use the constant acceleration equations. Since we are trying to find the max height of the jumper, we can use the fact that at the top of its trajectory, its final velocity will be 0 m/s.
4. v = 0 m/s
Using these four variables, let's find the constant acceleration equation that contains these variables:
Substitute the known values into the equation and solve for Δx.
- (0)² = (9.7)² + 2(-9.8)Δx
- 0 = 94.09 + (-19.6)Δx
- -94.09 = -19.6Δx
- Δx = 4.80
The high jumper can jump to a max height of 4.80 m.
Answer:
B meet A 0.01 km east of flagpole
Explanation:
given data
distance A = 5.7 km west
velocity V1 = 8.9 km/h
distance B = 4.5 km east
velocity V2 = 7 km/h
to find out
How far runners from the flagpole, when paths cross
solution
we know A and B are 5.7 + 4.5 = 10.2 km apart
and we consider here B will run distance x km for meet
so time will be for B is
time B = distance / velocity
time B = x / 7 ...................1
and
for A distance for meet = ( 10.2 - x ) km
so time A = distance / velocity
time A = ( 10.2 - x ) / 8.9 .............2
now equating equation 1 and 2
time A = time B
x / 7 = ( 10.2 - x ) / 8.9
x = 4.490
so distance of B run for meet is 4.490 km
so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km
so B meet A 0.01 km east of flagpole
Construct a vector diagram. It will be a right-angled triangle. One vector (the hypotenuse) represents the heading of the boat, one represents the current and one represents the resultant speed of the boat, which I'll call x. Their magnitudes are 20, 3 and x. Let the required angle = theta. We have:
<span>theta = arcsin(3/20) = approx. 8.63° </span>
<span>The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line. </span>
<span>x = sqrt(20^2 - 3^2) </span>
<span>= sqrt(400 - 9) </span>
<span>= sqrt 391 </span>
<span>The boat's crossing time = </span>
<span>0.5 km/(sqrt 391 km/hr) </span>
<span>= (0.5/sqrt 391) hr </span>
<span>= approx. 0.025 hr </span>
<span>= approx. 91 seconds</span>
