Answer:
The horizontal distance of the target should be 2721,4 meters.
Explanation:
First of all we need to find the time that the emergency package hits the ground after the moment of release:
y=0 (because when it hits the ground it is on the level of 0m);
The emergency package hits the ground after 24,74 seconds from release.
Lets assume that package preserves his 110 m/s horizontal speed during the free fall. The targets horizontal distance is:
2721,4 meters
Answer:
We have two possible values for the second tuning fork
443Hz and 337Hz
Explanation:
When two waves of slightly different frequency interfere, they produce beats and the frequency of the beats (fb) is:
with f1 the frequency of the first tuning fork and f2 the frequency of the second tuning fork, so we should solve
note that it is the absolute vale of the rest between f1 and f2 so we have two equations and two possible values:
and
Work is force times distance
pressure is force over area
hence pressure times volume will give us force times distance aka work
1.1 atm is 111.458 kPa
si units
111.458 kPa times 2.3 L equals 256.35 J
They are made from pure elements , silicon or germanium, or compounds like gallium arsenide
Answer:
the magnitude of the electric force on the projectile is 0.0335N
Explanation:
time of flight t = 2·V·sinθ/g
= (2 * 6.0m/s * sin35º) / 9.8m/s²
= 0.702 s
The body travels for this much time and cover horizontal displacement x from the point of lunch
So, use kinematic equation for horizontal motion
horizontal displacement
x = Vcosθ*t + ½at²
2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²
a = -2.23 m/s²
This is the horizontal acceleration of the object.
Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only
Therefore,the magnitude of the electric force on the projectile will be
F = m*|a|
= 0.015kg * 2.23m/s²
= 0.0335 N
Thus, the magnitude of the electric force on the projectile is 0.0335N