Answer:
=9.72 m/s
Explanation:
From the Newton's laws of motion;
x=2(v²cos∅sin∅)/g
Using geometry we see that 2 cos∅sin∅ = sin 2∅
Therefore, x= (v²sin 2∅)g, where v is the take off speed x the range and ∅ the launch angle.
Making v the subject of the formula we obtain the following equation.
v=√{xg /(sin 2∅)}
x=7.80
∅=27.0
v=√{7.8×9.8/sin(27×2)}
v=√94.485
v=9.72 m/s
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Answer:
<u>Conventions used in SI to indicate units are as follows:</u>
- Only singular form of units are used. for example: use kg and not kgs.
- Do not use full stop after the abbreviations of any unit. for example: do not use kg. or cm.
- Use one space between last numeric digit and SI unit. for example: 10 cm, 9 km.
- Symbols and words should not be mixed. for example: use Kilogram per cubic and not kilogram/m3.
- While writing numerals, only the symbols of the units should be written. for example: use 10 cm and not Ten cm.
- Units named after a scientist should be written in small letters. for example: newton, henry.
- Degree sign should not be used when the kelvin unit is used. for exmaple: use 37° and not 37°k
The kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s
Given parameters
To find
Kinematics is the part of physics that establishes the relationships between the position, velocity, and acceleration of bodies.
In this case we have a vertical launch
y = y₀ + v₀ t - ½ g t²
Where y and y₀ are the final and initial positions, respectively, v₀ the initial velocity, g the acceleration of gravity (g = 9.8 m / s²) and t the time
With the ball in hand, its position is zero
0 = 0 + v₀ t - ½ g t²
v₀ t - ½ g t² = 0
v₀ = ½ g t
Let's calculate
v₀ = ½ 9.8 2.4
v₀ = 11.76 m / s
In conclusion using kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s
Learn more about vertical launch kinematics here:
brainly.com/question/15068914
Answer:
E = 12640.78 N/C
Explanation:
In order to calculate the electric field you can use the Gaussian theorem.
Thus, you have:
ФE: electric flux trough the Gaussian surface
Q: net charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:
r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m
Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:
Finally, you obtain for E:
hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C
