The food package will strike the ground at 11 degrees below the horizontal.
<h3>Time for the food package to hit the ground</h3>
The time for the food package to hit the ground is calculated as follows;
h = vt + ¹/₂gt²
<em>let the initial velocity be horizontal</em>
4900 = 0(t) + (0.5 x 9.8)t²
4900 = 4.9t²
t² = 4900/4.9
t² = 1,000
t = √1,000
t = 31.62 s
<h3> Final speed of the food package when it hits ground</h3>
vf(y) = vo + gt
vf(y) = 0 + (31.62 x 9.8)
vf(y) = 309.88 m/s
<h3>Angle of projection</h3>
The horizontal component of the speed will be constant, while vertical component will change
Angle below the horizontal = 90 - 79 = 11⁰
Thus, the food package will strike the ground at 11 degrees below the horizontal.
Learn more about angle of projection here: brainly.com/question/10671136
Answer:
45.89m/s²
Explanation:
Given
Distance S = 305m
Time t = 3.64s
To get the acceleration during this run, we will apply the equation of motion:
S = ut+1/2at²
Substitute the given parameters into the formula and calculate the value of a
305 = 0+1/2 a(3.64)²
304 = 1/2(13.2496)a
304 = 6.6248a
a = 304/6.6248
a = 45.89m/s²
Hence the average acceleration during this run is 45.89m/s²
The dependent variable was the time and the independent variable was thecars
C . plate a is negatively charged and plate b is positively charged
Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
