<span>When the question says the ball lands a distance of 235 meters from the release point, we can assume this means the horizontal distance is 235 meters.
Let's calculate the time for the ball to fall 235 meters to the ground.
y = (1/2)gt^2
t^2 = 2y / g
t = sqrt{ 2y / g }
t = sqrt{ (2) (235 m) / (9.81 m/s^2) }
t = 6.9217 s
We can use the time t to find the horizontal speed.
v = d / t
v = 235 m / 6.9217 s
v = 33.95 m/s
Since the horizontal speed is the speed of the plane, the speed of the plane is 33.95 m/s</span>
Answer:
72.54 degree west of south
Explanation:
flow = 3.9 m/s north
speed = 11 m/s
to find out
point due west from the current position
solution
we know here water is flowing north and ship must go south at an equal rate so that the velocities cancel and the ship just goes west
so it become like triangle with 3.3 point down and the hypotenuse is 11
so by triangle
hypotenuse ×cos(angle) = adjacent side
11 ×cos(angle) = 3.3
cos(angle) = 0.3
angle = 72.54 degree west of south
Complete question is;
An experiment is carried out to measure the extension of a rubber band for different loads.
The results are shown in the image attached.
What figure is missing from the table?
Answer:
17.3 cm
Explanation:
The image attached showed values for load, extension and initial length.
Now, the first length there is 15.2 cm and as such it's corresponding extension is 0 because it has no preceding measured length.
The second measured length is 16.2 cm. Since it's initial measured length is 15.2 cm, then the extension has a formula; final length - initial length.
This gives: 16.2 - 15.2 = 1 cm
This corresponds to what is given in the table.
For the next measured length, it is blank but we are given the extension to be 2.1 cm. Now, since the initial measured length is 15.2 cm.
Thus;
2.1 cm = Final length - 15.2 cm
Final length = 15.2 + 2.1
Final length = 17.3 cm
Explanation:
speed of light= c
wave length= L
frequency= f
c=Lf → L= c/f → L= 3 × 10⁸/ 27 × 10⁹ → L = 1/90 ≈ 0.011 m
<span>3933 watts
At 100 C (boiling point of water), it's density is 0.9584 g/cm^3. The volume of water lost is pi * 12.5^2 * 10 = 4908.738521 cm^3
The mass of water boiled off is 4908.738521 * 0.9584 = 4704.534999 grams.
Rounding to 4 significant figures gives me 4705 grams of water.
The heat of vaporization for water is 2257 J/g. So the total energy applied is
2257 J/g * 4705 g = 10619185 J
Now we need to divide that by how many seconds we've spent boiling water. That would be 45 * 60 = 2700 seconds.
Finally, the rate of heat transfer in Joules per second will be the total number of joules divided by the total number of seconds. So
10619185 J / 2700 s = 3933 J/s = 3933 (kg m^2/s^2)/s = 3933 (kg m^2/s^3)
= 3933 watts</span>
