Answer:
a) The rocket reaches a maximum height of 737.577 meters.
b) The rocket will come crashing down approximately 17.655 seconds after engine failure.
Explanation:
a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.
1st Stage - Engine
Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ([tex]v[/tex]), measured in meters per second, by using this kinematic equation:
[tex]v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}[/tex] (1)
Where:
[tex]a[/tex] - Acceleration, measured in meters per square second.
[tex]\Delta s[/tex] - Travelled distance, measured in meters.
[tex]v_{o}[/tex] - Initial velocity, measured in meters per second.
If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]a = 2.35\,\frac{m}{s^{2}}[/tex] and [tex]\Delta s = 595\,m[/tex], the final velocity of the rocket is:
[tex]v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}[/tex]
[tex]v\approx 52.882\,\frac{m}{s}[/tex]
The time associated with this launch ([tex]t[/tex]), measured in seconds, is:
[tex]t = \frac{v-v_{o}}{a}[/tex]
[tex]t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }[/tex]
[tex]t = 22.503\,s[/tex]
2nd Stage - Gravity
The rocket reaches its maximum height when final velocity is zero:
[tex]v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})[/tex] (2)
Where:
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]v[/tex] - Final speed, measured in meters per second.
[tex]a[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]s_{o}[/tex] - Initial height, measured in meters.
[tex]s[/tex] - Final height, measured in meters.
If we know that [tex]v_{o} = 52.882\,\frac{m}{s}[/tex], [tex]v = 0\,\frac{m}{s}[/tex], [tex]a = -9.807\,\frac{m}{s^{2}}[/tex] and [tex]s_{o} = 595\,m[/tex], then the maximum height reached by the rocket is:
[tex]v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})[/tex]
[tex]s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}[/tex]
[tex]s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}[/tex]
[tex]s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]s = 737.577\,m[/tex]
The rocket reaches a maximum height of 737.577 meters.
b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:
[tex]s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}[/tex] (2)
Where:
[tex]s_{o}[/tex] - Initial height, measured in meters.
[tex]s[/tex] - Final height, measured in meters.
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]a[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]t[/tex] - Time, measured in seconds.
If we know that [tex]s_{o} = 595\,m[/tex], [tex]v_{o} = 52.882\,\frac{m}{s}[/tex], [tex]s = 0\,m[/tex] and [tex]a = -9.807\,\frac{m}{s^{2}}[/tex], then the time needed by the rocket is:
[tex]0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}[/tex]
[tex]-4.904\cdot t^{2}+52.882\cdot t +595 = 0[/tex]
Then, we solve this polynomial by Quadratic Formula:
[tex]t_{1}\approx 17.655\,s[/tex], [tex]t_{2} \approx -6.872\,s[/tex]
Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.
why would the bulb not light?
Im not sure this is correct but, i think thats because the wire connected at the negative side. The current go from positive to negative and thats why the bulb not light, also the bulb location is not at the center of the battery which explain why there's no current at the upper side.
Hope you understand. Sorry if my english sucks YEET
The light bulb isn't connected to the positive thing sticking out of the battery, both ends must connect back to the battery
A centrifuge is a device that rotates an object to produce an acceleration many times that of gravity Pilots are trained in such devices because they can simulate the same Accelerations that are experienced in certain types of flight. At which angular velocit would a space station of 30 m radius have to rotate to simulate Earth graivty?
Answer:
ω = 0.571 rad/s
Explanation:
given data
radius = 30 m
solution
we take here g = 9.8 m/s²
and g is express as
g = r × ω² ....................1
put here value and we get
9.8 = 30 × ω²
solve it we get
ω = 0.571 rad/s
The x-component and y-component of two vectors A & B are Ax = 9, Ay = 12,Bx =
15 & By = 20. Find:
/A+B/
Answer:
40Explanation:
Given the following
Ax = 9,
Ay = 12,
Bx = 15
By = 20
Get A and B
A = √9²+12²
A= √81+144
A = √225
A = 15
Get B;
B = √15²+20²
B = √225+400
B = √625
B = 25
get /A+B/
A+B = 15+25
/A+B/ = /40/
Hence the value of /A+B/ is 40
How old do you need to be in order to qualify to be a U.S. Senator
Answer: 30 Years Old
Explanation: The constitution has around three qualifications for service in the U.S. Senate, Your age must be at least 30 years.
It is common to see birds of prey rising upward on thermals. The paths they take may be spiral-like. You can model the spiral motion as uniform circular motion combined with a constant upward velocity. Assume a bird completes a circle of radius 6.00 m every 5.00 s and rises vertically at a rate of 3.00 m/s. Part APart complete Find the speed of the bird relative to the ground. Express your answer using three significant figures. v
Answer:
8.11 m/s
Explanation:
The bird has two velocity, one velocity is in the upward direction (+y direction) and the other velocity is in the x direction (that is along the radius of circle of the bird).
the vertical speed [tex]v_y=3\ m/s[/tex]
The horizontal speed ([tex]v_x[/tex]):
[tex]v_x=\frac{2\pi R}{T} =\frac{2\pi *6\ m}{5\ s} =7.54\ m/s[/tex]
The bird's speed relative to the ground (v) is given by:
v = [tex]\sqrt{v_x^2+v_y^2}[/tex]
[tex]v=\sqrt{7.54^2+3^2} \\\\v=8.11\ m/s[/tex]
Two motorcycles are traveling due east with different velocities. However, 5.68 seconds later, they have the same velocity. During this 5.68-second interval, motorcycle A has an average acceleration of 3.87 m/s2 due east, while motorcycle B has an average acceleration of 18.2 m/s2 due east. (a) By how much did the speeds differ at the beginning of the 5.68-second interval, and (b) which motorcycle was moving faster?
Answer:
The answer is below
Explanation:
Let a be the initial velocity of motorcycle A and b be the initial velocity of motorcycle B.
After 5.68 seconds, both motorcycle had the same velocity (v), therefore for motorcycle A:
(a - v) / 5.68 = 3.87
a - v = 21.9816
v = a - 21.9816
For motorcycle B:
(b - v) / 5.68 = 18.2
b - v = 103.376
v = b - 103.376
Therefore:
a - 21.9816 = b - 103.376
b - a = -21.9816 + 103.376
b - a = 81.3944
a) The difference between their speeds at the beginning was 81.3944 m/s
b) Since b - a = 81.3944. This means that the initial velocity of motorcycle B is greater than that of motorcycle A by 81.3944 m/s.
Therefore motorcycle B was moving faster
Energy cannot be created nor destroyed, only changed from one form to another. How does listening to music on a radio obey the law of conservation of energy?
Group of answer choices
Sound energy is changed into potential energy.
Electrical energy is converted into other forms of energy, such as sound.
Electrical energy remains unchanged.
Electrical energy is gradually destroyed as the radio plays.
Answer: Electrical energy is converted into other forms of energy, such as sound.
Explanation: Using the law of conservation of energy we know that energy can never be destryped it can be transferred or be transformed into from one form to another.
Electrical energy is converted into other forms of energy, such as sound.
The law of conservation of energy or matter, states that energy can neither be created nor destroyed but can be converted from one form to another.
When listening to music on radio, the electric energy supplied to the radio will be converted to mechanical energy of the moving parts of the radio which is then converted to sound energy.
Thus, we can conclude that electrical energy is converted into other forms of energy, such as sound.
Learn more here:https://brainly.com/question/2828402
Temperature and salinity difference in ocean water cause...
Answer:
high and low tied maybe? not sure
SERIOUSLY HELP NOWWW LIKE NOWW I REALLY NEED THIS ANSWERED
Which of the following is the best way to measure the age of the Earth?
Examine rocks collected in my backyard.
Analyze maps of all known fossil sites in the United States.
Test the rate of decay of specific elements in rock samples.
Explore leaves embedded in ash from recent volcanic eruptions.
Answer:
the second one Test the rate of decay of specific elements in rock samples.
Which of the following is not a true statement?
A
B
C
D
Answer:
I think A
Explanation:
because I don't think a unknowned number can be a division problem
A dumped harmonic oscillator of a mass of 500 g has a period of 0.5 second. The amplitude of the oscillation is decreasing 2.0 % during each oscillation. a./ The initial amplitude is 10 cm. What will be the amplitude after 25 oscillation
Answer:
The answer is below
Explanation:
The amplitude decreases by 2% during each oscillation. Hence the decrease in amplitude can be represented by an exponential decay in the form:
y = abˣ; where x ad y are variables, a is the initial value and b is the factor.
Let y represent the amplitude after x oscillations. Since the initial amplitude is 10 cm, hence:
a = 10 cm, b = 2% = 0.02.
Therefore:
y = 10(0.02)ˣ
The amplitude after 25 oscillations is gotten by substituting x = 25 into the equation. Hence:
y = 10(0.02)²⁵
y= 3.355 * 10⁻⁴² cm
The amplitude after 25 oscillations is 3.355 * 10⁻⁴² cm
First to answer gets brainliest
77. A drag racing vehicle travels from 0 to 100 mph in 5 seconds north. What is the acceleration?
a).004
s2
b).0056 m/s2 c).0079"
d).01 m/s2
M
m
Answer:
a
Explanation:
i just took the test
When you jump, you push down on the earth and it pushes back up against you. The earth pushing up against you is what causes you to go into the air. Why doesn’t your push cause the earth to go down if your push on the earth is equal and opposite of the earth's push on you?
That's a great question !
The answer is: It does !
A push on an object causes the object to accelerate in the direction of the force.
The less mass the object has, the more the force accelerates it.
Now, when you jump, the forces on you and on the Earth are equal forces.
The up force on you causes you to accelerate up by some amount.
The down force on the Earth causes the Earth to accelerate down by some amount.
The Earth's mass is something like 5,972,000,000,000,000,000,000,000 kg, while your mass is something like 50 kg.
The Earth has something like 119,400,000,000,000,000,000,000 times as much mass as you have.
So your acceleration is something like 119,400,000,000,000,000,000,000 times as great as the Earth's acceleration.
==> The Earth's downward acceleration, caused by your jump, is there. It's just too small to notice.
BUT . . . That's the reason why seismometers (instruments to detect and measure the vibrations from distant earthquakes) have to be located as far as possible from cities and busy roads.
In places that are too close to cities and roads, the Earth's surface is always vibrating, wiggling, jiggling, heaving and weaving, in reaction to the forces of people walking around, cars and trucks driving around, even rain falling down. And kids jumping up and down !
In such places, these people-motions are louder and stronger than the vibrations coming from distant earthquakes. Seismometers wouldn't work there.
Two people stand across from one another at the top edges of identical buildings, 50 meters above the ground. One person throws a rock straight upwards with an initial speed of 15 m/s, while the other person throws their rock straight downwards with an initial speed of 15 m/s. If both of the rocks miss the building and continue down to the street,
a. How long will it take for the rock thrown upwards to reach the ground?
b. How long will it take for the rock thrown downwards to reach the ground?
c. How fast will the upward thrown rock be travelling just before it hits the ground? The downward thrown rock?
Answer:
C
Explanation:
Hope this helps!!!!
A wire has a length of 0.50 m and measures about 0.50 mm in its cross-sectional radius. At normal temperature, what is its resistance in Ohms, if Aluminum has a resistivity of 2.82x10^-8 Ohms*meter
Answer:
Explanation:
For resistance , the expression is as follows .
R = ρ L / S where ρ is specific resistance , L is length of wire and S is cross sectional area .
cross sectional area = π x ( .5 x 10⁻³ )²
S = .785 x 10⁻⁶ m²
Putting the values
R = 2.82 x 10⁻⁸ x .50 / .785 x 10⁻⁶
= 1.796 x 10⁻² ohm .
Help me I don't know what I'm doing
Answer:
C the metal handle because it is a good conductor
Answer:
D.
Explanation:
Although the metal handle will last longer, if heated up enough it could burn her hand.
a body accelerates uniformly from rest at 2m/s^2 for 5 seconds. Calculate its averege velocity in this time
HERE IS YOUR ANSWER!
) Explain why the foil is attracted at first by the charged rod. Consider any charge that exists in the uncharged foil. (You will consider later interaction between the rod and foil in the last question. Just think about the first interaction when answering this question.) [
Answer:
The attraction is due to the induced charge.
Explanation:
When we approach a charged rod to a sheet, an induced load is produced in the sheet that is of the same magnitude as the rod of opposite sign, this is because the charges of different sign attract each other, this explains the initial attraction.
This induced load occurs if importing the plate load
The attraction is due to the induced charge.
A positively charged oil drop of mass is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates spaced 16 cm apart. If the mass of the drop is 8.0 10-15 kg and it remains stationary when the potential difference between the plates is 2.44 kV, what is the magnitude of the charge on the drop
Answer:
the magnitude of the charge on the drop is 5.163 × 10⁻¹⁸ C
Explanation:
Given that;
mass of the drop m = 8.0×10⁻¹⁵ kg
distance d = 16 cm = 0.16 m
potential difference between the plates V = 2.44 kV = 2440 v
acceleration of gravity g = 9.81 m/s²
the magnitude of the charge on the drop = ?
weight is balanced by the electrostatic force
weight = mg = 8.0×10⁻¹⁵ kg × 9.81 m/s² = 7.848 × 10⁻¹⁴
we know that; V = Ed
E = V/d = 2440 / 0.16 = 15200 v/m
Electrostatic force = qE
so weight = qE
q = weight / E
q = 7.848 × 10⁻¹⁴ / 15200
q = 5.163 × 10⁻¹⁸ C
Therefore, the magnitude of the charge on the drop is 5.163 × 10⁻¹⁸ C
Nikki was walking around a department store shopping one day, and did not realize that the shirt she was wearing looked just like the shirts worn by employees. When a stranger asked, "do you work here," she thought it was funny. The other customers' assumption that Nikki was a store employee demonstrates the Gestalt principle of _______.
Answer: Similarity
By.
What is the value of the angle of inclination of the slide?
Answer:
63°
that's my answer
but then I am sorry if I'm wrong
Explanation:
90-27 = 63°
A 20 kg box has an initial velocity of 2 m/s starting at the bottom of a 30-degree inclined plane. A person pushes on the box directly up the frictionless inclined plane so that it travels up the inclined plane at a constant velocity of 2 m/s. Calculate the how much is done by the person after 5 seconds have past.
Answer:
Explanation:
The box is moving with constant velocity so acceleration of box is zero . That means net force on the box is zero .
The weight component acting on box parallel to incline plane
= mg sin 30⁰ = 20 x 9.8 x sin 30 = 98 N
This force is acting down the plane , hence to make the net force zero acting on box , force exerted by person will also be 98 N up the incline .
Force exerted by person = 98 N
distance travelled in 5 s
= velocity x time
= 2 x 5 = 10 m
Work done by person
= 98 x 10
= 980 J .
Most new jobs in the United States will be in the _____.
in the service producing sector
kenneth ran a marathon (26.2 miles) in 5.5 hours. What was Kenneth’s average speed? (Round your answer to the nearest tenth.)
0.2 mph
4.8 mph
5.5 mph
144.1 mph
it is actually science i couldn't find the word science
Answer:
4.8mph
Explanation:
Speed= Distance/time
Speed= 26.2/5.5
= 4.76mph
( To the nearest tenth ) = 4.8mph
Answer:
38.7 mph
Explanation:
I just add all the numbers together then divided them by 4, which is the amount of numbers you gave.
calculate the force needed to push the ball up a 4 m ramp if the work is equal to 16 joules.
a bus initially at rest accelerated of 4m/S2 at the end of 10 second find average velocity
Answer:
V = 20m/s
Explanation:
Given the following data;
Acceleration = 4m/s²
Time = 10 secs
Initial velocity = 0m/s (since it's at rest).
To find the average velocity, we would use the first equation of motion;
[tex] V = U + at[/tex]
Where;
V is the final velocity.
U is the initial velocity.
a is the acceleration.
t is the time measured in seconds.
Substituting into the equation, we have;
V = 0 + 4*10
V = 40m/s
To find the average velocity;
Average velocity = (U + V)/2
Average velocity = (0 + 40)/2
Average velocity = 40/2
Average velocity = 20m/s
WILL MARK BRAINLIEST IF CORRECT. Worth Lots Of Points. NEEDS TO BE WITH EXPLANATION PLEASE. PLEASE HELP!
Why do people eat bo oty
Answer: I don't know my dude
Explanation:
In a crash test, a car with a mass of 1600 kg is initially moving at a speed of 20 m/s just before it collides with a barrier. The final speed of the car after the collision is zero. The original length of the car is 4.50 m , but after the collision, the smashed car is only 3.60 m long.
Required:
a. What is the average speed Of the car during the period from first contact with the barrier to the moment the car comes to a stop? You may assume the force that the barrier exerts on the car is constant during this period.
b. How much time elapses between the moment the car makes first contact with the barrier and the moment it comes to a stop?
c. Making the very rough approximation that the large force that the barrier exerts on the car is approximately constant during contact, determine the approximate magnitude of this force?
Answer:
The answer to the given points can be defined as follows:
Explanation:
In point 1:
[tex]\bold{v_f^2= v_i^2+2as}\\\\\to v_f=0\\\\\to v_i=20 \frac{m}{s}\\\\\to s= 4.50\ m -3.60 \ m \\\\[/tex]
[tex]=0.9 \ m \\[/tex]
put the value in the above formula:
[tex]\to 0= 20^2+2 \times a \times 0.9\\\\\to -1.8\ a=400\\\\\to -a= \frac{400}{1.8} \\\\ \to a= -222.22\ \ \frac{m}{s^2}[/tex]
[tex]\bold{v_f=v_i+at}\\\\\to 0=20+ (-222.22)t\\\\\to 222.22t=20\\\\\to t=\frac{20}{222.22}\\\\\to t= 0.0900 \ s\\\\\to v_{avg}=\frac{s}{t}=\frac{0.9}{t}= 10\ \frac{m}{s}[/tex]
for point 2:
[tex]t= 0.0900 \ s -\text{found above}[/tex]
for point 3:
[tex]\to |a| = 222.22 \frac{m}{s^2} \text{found above}\\\\\to \bold{|F| = m \cdot |a|}\\\\[/tex]
[tex]=1600 \ kg \times 222.22 \ \frac{m}{s^2} \\\\= 3.55\times 10^{5} \ N[/tex]