Answer: Synthetic polymers are lightweight.
Explanation:
A benefit of using synthetic polymers is the fact that synthetic polymers are lightweight.
A polymer is a molecule composed of many repeating subunits.
Synthetic polymers are artificial polymers created by humans.
Most of the synthetic polymers are not biodegradable (unlike natural fibers such as cotton).
Synthetic polymers are classified according to their use into plastics, elastomers and synthetic fibers.
The advantages of synthetic polymers include: hard to break, being lightweight, and they last for a long time.
In conclusion, a benefit of using synthetic polymers is the fact that synthetic polymers are lightweight.
Learn more in:
https://brainly.com/question/4047007?referrer=searchResults
A 3.7-mm-diameter wire carries a 20 A current when the electric field is 8.2×10−2 V/m .
What is the wire's resistivity? (in Ωm)
Answer:
ρ = 4.4 10⁻⁸ Ω m
Explanation:
For this exercise let's start by finding the value of the resistance using Ohm's law
V = I R
R = V / I
The voltage is related to the electric field
V = E L
let's substitute
R = E L / I
R = 8.2 10⁻² / 20
R = 4.1 10⁻³ L
now we can use the resistance relation
R = ρ L / A
the area of a circular wire is
A = π r² = π d² / 4
ρ = R π d² / (4 L)
let's calculate
ρ = (4.1 10⁻³ L) π 0.0037² / (4 L) = (4.1 10⁻³) π 0.0037² / (4)
ρ = 4.4 10⁻⁸ Ω m
The resistivity of the wire is 4.4 10⁻⁸ Ω m.
We know that;
V = I R
R = V / I
The voltage the electric field can be connected using the formula;
V = E L
Hence;
R = E L /I
R = 8.2 10⁻² / 20
R = 4.1 10⁻³ L
For resistivity;
R = ρ L / A
the area can be obtained from;
A = π r² = π d² / 4
ρ = R π d² / (4 L)
ρ = (4.1 10⁻³ L) π 0.0037² / (4 L) = (4.1 10⁻³) π 0.0037² / (4)
ρ = 4.4 10⁻⁸ Ω m
Learn more about electric field:https://brainly.com/question/1195122
A magnetic field is created by ____.
A moving electric charges
b elctromagnetic pulses
c a strong current
d a change in the current of a wire
Personal behavior is the only factor that determines if a person becomes ill.
Please select the best answer from the choices provided.
ОТ
OF
Answer:
FALSE
Explanation:
what is the mass of an object that is experiencing a net force of 200N and an acceleration of 500m/s2
Answer:
F=200N
a=500m/s2
Mass=?
Explanation:
F=ma
200=m*500
200/500=m
Mass=0.4kg
Let's start by calculating what acceleration the rocket must produce to launch into earth orbit. In order to attain orbit around earth, the ATLAS V rocket must accelerate up to a speed of about 7700 meters per second in about 4.2 minutes. What average acceleration is required to accomplish this
Answer:
30.56 m/s^2
Explanation:
Given that In order to attain orbit around earth, the ATLAS V rocket must accelerate up to a speed of about 7700 meters per second in about 4.2 minutes.
The average acceleration that is required to accomplish this will be
Average acceleration = change in velocity / time
Average acceleration = 7700/ 4.2 × 60
Average acceleration = 7700/252
Average acceleration = 30.56 m/s^2
Two cylindrical resistors are made from the same material. The shorter one has a length LL and diameter DD . The longer one has a length 16L16L and diameter 4D4D . How do their resistances compare? The resistance of the longer resistor is four times the resistance of the shorter resistor. The resistance of the longer resistor is twice the resistance of the shorter resistor. The resistance of the longer resistor is the same as the resistance of the shorter resistor. The resistance of the longer resistor is half the resistance of the shorter resistor. The resistance of the longer resistor is a quarter of the resistance of the shorter resistor.
Answer:
The resistance of the longer resistor is a quarter of the resistance of the shorter resistor.Explanation:
If Two cylindrical resistors are made from the same material, then their resistivity will be the same. Formula for calculating resistivity of a material is expressed as;
[tex]\rho = \frac{RA}{L} \ where \ A = \frac{\pi d^2}{4}[/tex] where;
R is the resistance
A is the cross sectional area of the material
L is the length of the material
For the shorter cylinder:
Length = L
diameter = D
[tex]\rho = \dfrac{R_s(\frac{\pi D^2}{4})}{L} \\\\\rho = \dfrac{R_s{\pi D^2}}{4L}[/tex]
For the longer cylinder:
Length = 16L
diameter = 4D
[tex]\rho = \dfrac{R_l(\frac{\pi (4D)^2}{4})}{16L} \\\\\\\rho = \dfrac{R_l(\frac{\pi (16D^2)}{4})}{16L} \\\\\rho = \dfrac{R_l{16\pi D^2}}{16L}\\\\\rho = \dfrac{R_l{\pi D^2}}{L}[/tex]
Since their resistivity are the same then;
[tex]\dfrac{R_s{\pi D^2}}{4L} = \dfrac{R_l{\pi D^2}}{L} \\\\ \dfrac{R_s}{4} = {R_l} \\\\R_s = 4R_l\\\\R_l = \frac{R_s}{4}[/tex]
Hence the resistance of the longer resistor is a quarter of the shorter resistor.
You drop a ball from a window located on an upper floor of a building. It strikes the ground with speed v. You now repeat the drop, but you ask a friend down on the ground to throw another ball upward at speed v. Your friend throws the ball upward at the same moment that you drop yours from the window. At some location, the balls pass each other. Is this location.
Answer:
y = y₀ (1 - ½ g y₀ / v²)
Explanation:
This is a free fall problem. Let's start with the ball that is released from the window, with initial velocity vo = 0 and a height of the window i
y = y₀ + v₀ t - ½ g t²
y = y₀ - ½ g t²
for the ball thrown from the ground with initial velocity v₀₂ = v
y₂ = y₀₂ + v₀₂ t - ½ g t²
in this case y₀ = 0
y₂2 = v t - ½ g t²
at the point where the two balls meet, they have the same height
y = y₂
y₀ - ½ g t² = vt - ½ g t²
y₀i = v t
t = y₀ / v
since we have the time it takes to reach the point, we can substitute in either of the two equations to find the height
y = y₀ - ½ g t²
y = y₀ - ½ g (y₀ / v)²
y = y₀ - ½ g y₀² / v²
y = y₀ (1 - ½ g y₀ / v²)
with this expression we can find the meeting point of the two balls
True or false: points that lie on the same plane are Collinear
A corvette starts from rest and travels 69.0 meters in 50 s. What is its acceleration?
Answer:
0.0552 m/s²
Explanation:
Given:
Δx = 69.0 m
v₀ = 0 m/s
t = 50 s
Find: a
Δx = v₀ t + ½ at²
69.0 m = (0 m/s) (50 s) + ½ a (50 s)²
a = 0.0552 m/s²
QUICK
a sound wave has a frequency of 85 Hz and a wavelength of 4.2 m. what is the wave speed of the sound wave?
A. 357 seconds
B. 89.2 m/s
C. 357 m/s
D. 20 m/s
Answer:C
Explanation:
I did the test
a sound wave has a frequency of 85 Hz and a wavelength of 4.2 m. Then the wave speed of the sound wave is 20.23 m/s. Hence option D is correct.
What is wave ?Wave is is a disturbance in a medium that carries energy as well as momentum . wave is characterized by amplitude, wavelength and phase. Amplitude is the greatest distance that the particles are vibrating. especially a sound or radio wave, moves up and down. Amplitude is a measure of loudness of a sound wave. More amplitude means more loud is the sound wave.
Wavelength is the distance between two points on the wave which are in same phase. Phase is the position of a wave at a point at time t on a waveform. There are two types of the wave longitudinal wave and transverse wave.
The wave speed is given by,
c = νλ
where λ is wavelength, ν is frequency.
Given,
frequency ν = 85 Hz
wavelength λ = 4.2 m
The speed of the wave,
c = 85 × 4.2 = 20.23 m/s
Hence option D is correct.
To know more about Speed :
https://brainly.com/question/28224010
#SPJ2.
What were the physical activities in your childhood that you still do today? Do you spend more time now in doing these activities as compared before?
If two tug boats are towing a ship with force of 5 tons each and the angle between the two ropes is 60 degrees, what is the resultant force on the ship? Explain how to use a force table to verify answer.
Answer:
8.6602 tons
Explanation:
We first draw the known vector forces.
2fcos30⁰
We have f to be equal to 5tons
Inserting into formula
Σfx = 2(5)cos30⁰
= 8.6602 tons
Σfy is equal to 0, this is because in the y direction, the forces cancel themselves out.
Therefore the resultant force on the ship is equal to 8.6602 tons
I hope this helps!
Please check attachment for diagram.
What is the velocity of a plane to travel 3000 miles from New York to California in 5.0 hours
Answer:
10 miles per minute.
When the sun provides energy for photosynthesis, an interaction with the __________ takes place.
A bucket is being lowered by a very light rope with a constant downward velocity. The tension in the rope must be
Answer:
The tension in the light rope must be equal to the weight of the bucket
Explanation:
Given that,
Constant velocity of bucket and direction of bucket in downward
We need to find the tension in the rope
Using given data,
When a bucket moves downward with a constant velocity then the net force does not applied on the bucket.
So, The weight of the bucket will be equal to the tension in the light rope
In mathematically,
[tex]T=mg[/tex]
Where, T = tension
m = mass of bucket
g = acceleration due to gravity
Hence, The tension in the light rope must be equal to the weight of the bucket.
Mathew has a filtration kit, which consists of a funnel, a flask, and filter papers. Which of these mixtures can he separate using filtration?
Answer:
C. Muddy Water
An airplane with a hot-wire anemometer mounted on its wing tip is to fly through the turbulent boundary layer of the atmosphere at a speed of 50 m/sec. The velocity fluctuations in the atmosphere are of order 0.5 m/sec, the length scale of the large eddies is about 100 m. The hot-wire anemometer is to be designed so that it will register the motion of the smallest eddies.What is the highest frequency the anemometer will encounter
Answer:
0.55 hz
Explanation:
Given that the plane fly through the turbulent boundary layer of the atmosphere at a speed of 50 m/sec. And the velocity fluctuations in the atmosphere are of order 0.5 m/sec, the length scale of the large eddies is about 100 m.
The maximum speed attained will be
Maximum speed = 50 + 0.5 = 5.5 m/s
The Length = 100m
Speed = FL
Where F = frequency
Substitute speed and distance length into the formula
55 = 100F
F = 55/100
F = 0.55 Hz
Therefore, the highest frequency the anemometer will encounter will be 0.55 Hz
Answer:
40079 Hz
Explanation:
1. The first step is to calculate energy dissipation ∈=u^3/l and here u is fluctuating velocity
∈=(0.5^3)/100 = 0.00125 m^2/s^3
2. Find out the length scale of the small eddies
η=(viscosity/∈)^1/4
η=(1.470e-5/0.00125)^1/4 = 0.00126 m
3. The frequency associated with these small-scale eddies will be the greatest frequency the anemometer will encounter, thus:
u_max=f_max * η
u_max = u + u' = 50+0.5=50.5 m/s
f_max = u_max/η = 50.5/0.00126 = 40079 Hz
This is the heighest frequency the anemometer will encounter.
A basketball leaves a player's hands at a height of 2.20 m above the floor. The basket is 2.70 m above the floor. The player likes to shoot the ball at a 36.0 ∘ angle. Of the shot is made from a horizontal distance of 9.10 m and must be accurate to ±0.23m (horizontally), what is the range of initial speeds allowed to make the basket
Answer:
The range of initial speeds allowed to make the basket is: [tex]9.954\,\frac{m}{s}\leq v \leq 10.185\,\frac{m}{s}[/tex].
Explanation:
We must notice that basketball depicts a parabolic motion, which consists of combining a constant speed motion in x-direction and free fall motion in the y-direction. The motion is described by the following kinematic formulas:
x-Direction
[tex]x = x_{o} + v_{o}\cdot t \cdot \cos \alpha[/tex]
y-Direction
[tex]y = y_{o} + v_{o}\cdot t\cdot \sin \alpha +\frac{1}{2}\cdot g\cdot t^{2}[/tex]
Where:
[tex]x_{o}[/tex], [tex]y_{o}[/tex] - Initial position of the basketball, measured in meters.
[tex]x[/tex], [tex]y[/tex] - Final position of the basketball, measured in meters.
[tex]v_{o}[/tex] - Initial speed of the basketball, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]\alpha[/tex] - Tilt angle, measured in sexagesimal degrees.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If we know that [tex]x_{o} = 0\,m[/tex], [tex]y_{o} = 2.20\,m[/tex], [tex]\alpha = 36^{\circ}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]x = (9.10\pm0.23)\,m[/tex] and [tex]y = 2.70\,m[/tex], the system of equation is reduce to this:
[tex](9.10\pm 0.23)\,m = 0\,m + v_{o}\cdot t \cdot \cos 36^{\circ}[/tex]
[tex]9.10\pm 0.23 = 0.809\cdot v_{o}\cdot t[/tex] (Ec. 1)
[tex]2.70\,m = 2.20\,m + v_{o}\cdot t \cdot \sin 36^{\circ} + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]
[tex]0.50 = 0.588\cdot v_{o}\cdot t-4.904\cdot t^{2}[/tex] (Ec. 2)
At first we clear [tex]v_{o}\cdot t[/tex] in (Ec. 1):
[tex]v_{o}\cdot t = \frac{9.10\pm 0.23}{0.809}[/tex]
[tex]v_{o}\cdot t = 11.248\pm 0.284[/tex]
(Ec. 1) in (Ec. 2):
[tex]0.5 = 0.588\cdot (11.248\pm 0.284)-4.904\cdot t^{2}[/tex]
Now we clear the time in the resulting expression:
[tex]4.904\cdot t^{2} = 0.588\cdot (11.248\pm 0.284)-0.5[/tex]
[tex]t = \sqrt{\frac{0.588\cdot (11.248\pm 0.284)-0.5}{4.904} }[/tex]
There are two solutions:
[tex]t_{1} = \sqrt{\frac{0.588\cdot (11.248- 0.284)-0.5}{4.904} }[/tex]
[tex]t_{1} \approx 1.101\,s[/tex]
[tex]t_{2} = \sqrt{\frac{0.588\cdot (11.248+ 0.284)-0.5}{4.904} }[/tex]
[tex]t_{2}\approx 1.131\,s[/tex]
The initial velocity is cleared within (Ec. 2):
[tex]v_{o}=\frac{0.50+4.904\cdot t^{2}}{0.588\cdot t}[/tex]
The bounds of the range of initial speed is determined hereafter:
[tex]t_{1} \approx 1.101\,s[/tex]
[tex]v_{o} = \frac{0.50+4.904\cdot (1.101)^{2}}{0.588\cdot (1.101)}[/tex]
[tex]v_{o} = 9.954\,\frac{m}{s}[/tex]
[tex]t_{2}\approx 1.131\,s[/tex]
[tex]v_{o} = \frac{0.50+4.904\cdot (1.131)^{2}}{0.588\cdot (1.131)}[/tex]
[tex]v_{o} = 10.185\,\frac{m}{s}[/tex]
The range of initial speeds allowed to make the basket is: [tex]9.954\,\frac{m}{s}\leq v \leq 10.185\,\frac{m}{s}[/tex].
A hockey puck initially travelling to the right at 34 m/s. It moves for 7 before
coming to a stop. How far did it move in 7 seconds?
You can use kinematic equations
Answer:
[tex]x=119m[/tex]
Explanation:
Hello,
In this case, since the hockey puck was moving at 34 m/s and suddenly stopped (final velocity is zero) in 7 seconds, we can first compute the acceleration via:
[tex]a=\frac{v_f-v_o}{t}=\frac{0m/s-34m/s}{7s}\\ \\a=-4.86m/s^2[/tex]
In such a way, we can compute the displacement via:
[tex]x=\frac{v_f^2-v_o^2}{2a}\\ \\x=\frac{0^2-(34m/s)^2}{2*-4.86m/s^2}\\ \\x=119m[/tex]
Best regards.
What is Bill's average running speed?
Answer:
Hello!
Sorry you haven't put up an image of your question! Without it we can't answer your question!
Explanation:
Maybe put up another one and it'll be answered!
:D
A trumpet player on a moving railroad flatcar moves toward a second trumpet player standing alongside the track both play a 490 Hz note. The sound waves heard by a stationary observer between the two players have a beat frequency of 2.0 beats/s. What is the flatcar's speed? Take the speed of sound to be 343 m/s.
Answer:
The value is [tex]v_s = 1.394 \ m/s[/tex]
Explanation:
From the question we are told that
The frequency of the second player is [tex]f_2 = 490 \ Hz[/tex]
The beat frequency is [tex]f_b = 2.0 \ Hz[/tex]
The speed of sound is [tex]v_s = 343 \ m/s [/tex]
Generally the frequency of the note played by the first player is mathematically represented as
[tex]f_1 = f_2 + f_b[/tex]
=> [tex]f_1 = 490 + 2.0 [/tex]
=> [tex]f_1 = 492 Hz[/tex]
From the relation of Doppler Shift we have that
[tex]f_1 = \frac{ f_2 (v+ v_o )}{v-v_s }[/tex]
Here [tex] v_o\ is\ the\ velocity\ of\ the\ observer\ with\ value\ 0 \ m/s [/tex]
So
[tex]492 = \frac{ 490 (343+0 )}{343 -v_s }[/tex]
=> [tex]v_s = 1.394 \ m/s[/tex]
ou are walking down a straight path in a park and notice there is another person walking some distance ahead of you. The distance between the two of you remains the same, so you deduce that you are walking at the same speed of 1.05 m/s. Suddenly, you notice a wallet on the ground. You pick it up and realize it belongs to the person in front of you. To catch up, you start running at a speed of 2.75 m/s. It takes you 18.5 s to catch up and deliver the lost wallet. How far ahead of you was this person when you started running
Answer:
The value is [tex]d = 31.45 \ m [/tex]
Explanation:
Generally the relative speed at which you are moving with respect to the person ahead of you is mathematically represented as
[tex]v_r = v_s - v_c[/tex]
substituting 1.05 m/s for [tex] v_c [/tex] and 2.75 m/s for [tex]v_s[/tex]
So
[tex]v_r = 2.75 - 1.05[/tex]
=> [tex]v_r = 1.7 \ m/s [/tex]
Generally the distance by which the person is ahead of you is mathematically represented as
[tex]d = v_r * t[/tex]
substituting 18.5 s for [tex] t [/tex]
[tex]d = 1.7 * 18.5[/tex]
=> [tex]d = 31.45 \ m [/tex]
If 13 is added to a number, the result is 43 less than twice the number. Find the number,
Explanation:
13+x=43>2x
13+x=43>2x+43>2x
13+x=86>4x
x-4x=86-13
3x=73
x=73/3
x=24.333
x=24.4
Using the equation for Impact, can you explain the following:
Why are car steering rods designed to collapse?
Why are highway guard rails designed to crumple up on impact?
Why are traffic saftey barrels filled with water or sand?
Explanation:
Equation for Impact
FΔt = ΔP,
F = force
Δt = Impact of time
ΔP = Change in momentum
Car steering is engineered to fail in order to maximize the time of contact and hence reduce the initial impact and mitigate the damage incurred.
Road guard railing crumple on contact to maximize impact time and hence reduce impact intensity and mitigate damage.
Road safety containers are loaded with liquid or sand as they improve the period of impact.
Air that initially occupies 0.22 m3 at a gauge pressure of 86 kPa is expanded isothermally to a pressure of 101.3 kPa and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.)
Answer:
total work done = -5960.8 J
Explanation:
given data
initial volume v1 = 0.22 m³
initial pressure p1 = 86 kPa
final pressue p2 = 101.3 kPa
solution
we apply here isothermal expansion that is express as
p1 × v1 = p2 × v2 ......................1
put here value
86 × 0.22 = 101.3 × v2
v2 = 0.1867 m³
and
work done will be here
w1 = p1 × v1 × ln([tex]\frac{p1}{p2}[/tex]) ....................2
w1 = 86 × 10³ × 0.22 × [tex]ln(\frac{86}{101.3})[/tex]
w1 = -3.097 × 10³ J
and
it is cooled to initial volume at constant pressure so here work done will be
w2 = p(v2 - v1) .................3
w2 = 86 × 10³ × ( 0.1867 - 0.22 )
w2 = -2863.8 J
so
total work done is
total work done = w1 + w2
total work done = -3097 + -2863.8
total work done = -5960.8 J
A 0.091-in-diameter electrical wire at 90°F is covered by 0.02-in-thick plastic insulation (k = 0.075 Btu/h·ft·°F). The wire is exposed to a medium at 50°F, with a combined convection and radiation heat transfer coefficient of 2.5 Btu/h·ft2·°F. Calculate the critical radius (rcr) of the plastic insulation (in inches).
Answer:
The critical radius of the plastic insulation is 0.72 inches.
Explanation:
Given that,
Diameter = 0.091 in
Thickness = 0.02 in
Initial temperature = 90°F
Final temperature = 50°F
Heat transfer coefficient = 2.5 Btu/h.ft²°F
Material conductivity = 0.075 Btu/h.ft °F
We need to calculate the critical radius of the plastic insulation
Using formula of critical radius
[tex]r_{cr}=\dfrac{2K}{h}[/tex]
Where, k = Material conductivity
h = Heat transfer coefficient
Put the value into the formula
[tex]r_{cr}=\dfrac{2\times0.075}{2.5}[/tex]
[tex]r_{cr}=0.06\ ft[/tex]
[tex]r_{cr}=0.72\ inches[/tex]
Hence, The critical radius of the plastic insulation is 0.72 inches.
A column of soldiers, marching at 100 steps per minute, keep in step with the beat of a drummer at the head of the column. It is observed that the soldiers in the rear end of the column are striding forward with the left foot when the drummer is advancing with the right. What is the approximate length of the column? (Take the speed of sound to be 343 m/s.)
Answer:
The value is [tex]D = 205.8 \ m [/tex]
Explanation:
The time taken for the column to take a step mathematically represented as
100 steps => 1 minutes => 60 seconds
1 step => t
=> [tex]t = 0.6 \ s [/tex]
Generally the length of the column is mathematically represented as
[tex]D = v * t[/tex]
substituting 343 m/s for v we have
[tex]D = 343 * 0.6 [/tex]
=> [tex]D = 205.8 \ m [/tex]
half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value after
Answer:
44 years
Explanation:
Use half life equation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is time,
and T is the half life.
0.25 A₀ = A₀ (½)^(t / 22)
0.25 = (½)^(t / 22)
t / 22 = 2
t = 44
44 sec is half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value.
What is half-life?The period of time it takes for one-half of a radioactive isotope to decay is known as the half-life. A given radioactive isotope's half-life is constant; it is unaffected by external factors and independent of the isotope's starting concentration.
Use half life equation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is time,
and T is the half life.
0.25 A₀ = A₀ (½)^(t / 22)
0.25 = (½)^(t / 22)
t / 22 = 2
t = 44 sec
44 sec is half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value.
To learn more about half life refer to the link:
brainly.com/question/24710827
#SPJ2
(7%) Problem 14: A robot cheetah can jump over obstacles. Suppose the launch speed is vo = 4.74 m/s, and the launch angle is 0 = 25.5
degrees above horizontal.
What is the maximum height in meters?
A watermelon is dropped from the edge of the roof of a build- ing and falls to the ground. You are standing on the sidewalk and see the watermelon falling when it is 30.0 m above the ground. Then 1.50 s after you first spot it, the watermelon lands at your feet. What is the height of the building
Answer:
The hight of the building is 38.16 m
Explanation:
These two pieces of information given, first, the watermelon is 30 m above the ground and after 1.50 s the watermelon has been spotted. Now we are required to find the height of the building.
Use the below formula to find the height of buildings.
S = ut + ½ gt^2
30 =1.5u + (1/2) × 9.8 (1.5)^2
u = 12.65 m/sec
v^2 – u^2 = 2gs
(12.65)^2 = 2×9.8 s’
S’ = 8.16 m
h = s + s’
h = 30 + 8.16 = 38.16 m
The hight of the building is 38.16 m.
The height of the building is 38.16 m.
Given data:
The height above the ground is, h = 30.0 m.
The time interval after observation of first spot is, t = 1.50 s.
We need to find the height of building. And since two pieces of information given, first, the watermelon is 30 m above the ground and after 1.50 s the watermelon has been spotted. So, using the second kinematic equation of motion as,
[tex]h = ut + \dfrac{1}{2}gt^{2}[/tex]
Here, u is the initial speed. Solving as,
[tex]30 = (u \times 1.50) + \dfrac{1}{2} \times 9.8 \times (1.50)^{2}\\\\u =12.65 \;\rm m/s[/tex]
Now landing distance (s') is calculated using the third kinematic equation of motion as,
[tex]v^{2} =u^{2}+2(-g)s\\\\0^{2} =12.65^{2}+2(-9.8)s\\\\s = 8.16 \;\rm m[/tex]
Then the height of building is given as,
H = h + s
H = 30 m + 8.16 m
H = 38.16 m
Thus, we can conclude that the height of the building is 38.16 m.
Learn more about the kinematic equations of motion here:
https://brainly.com/question/16556016