Answer: 1 Newton
Explanation:
"Every action has an equal and opposite reaction."
Please mark as Brainliest if it is correct.
Force is an action-reaction principle. It stated that the force always exists in a pair. The reaction force of the wall on the swimmer will be 1N.
What is Newton's third law of motion?Newton's third law of motion state that every action has an equal and opposite reaction. It is an action-reaction principle. It stated that the force always exists in a pair.
Both occur in an action-reaction form. Hence for every action, there is an equal and opposite reaction.
[tex]\rm F_{action}=F_{reaction} \\\\ \rm F_{action= 1N[/tex]
[tex]\rm F_{reaction} = 1N[/tex]
Hence the reaction force of the wall on the swimmer will be 1N.
To learn more about Newton's third law refer to the link;
https://brainly.com/question/1077877
Please Help! Will mark brainliest.
Answer:W = m*g*h
19*9.8*32.4 = 6,032.9 rounded
honestly, I do not know if this is correct so please don't come back at me
hopefully this helps
Explanation: [do the following, if you think I am wrong]
just pick a formula,
plug in the number to the mass, gravity, and height
than multiply
get your answer, but don't forget to round to the nearest tenth
A star can give off white light. Why is this evidence that a star is a blackbody
radiator?
A. White light is made up of many different wavelengths of light.
B. The star reflects the white light.
C. The star absorbs the white light.
D. White light is only one wavelength of light.
O
Answer:
It's A. White light is made up of many different wavelengths of light.
write short note on fulcrum
Answer:
The definition of a fulcrum is a pivot point around which a lever turns, or something that plays a central role in or is in the center of a situation or activity.
The nucleus of an atom can be modeled as several protons and neutrons closely packed together.Each particle has a mass of 1.67 3 10227 kg and radius on the order of 10215 m.
(a) Use this model and the data provided to estimate the density of the nucleus of an atom.
(b) Compare your result with the density of a material such as iron. What do your result and comparison suggest about the structure of matter?
Answer:
Explanation:
Let n be number of total number of nucleons ( protons + neutrons )
Total mass inside nucleus = n x 1.67 x 10⁻²⁷ Kg
volume of nucleus = 4/3 π r³
= 1.33 x 3.14 x (10⁻¹⁵)³ m
= 4.17 x 10⁻⁴⁵ m³
Density = mass / volume
= n x 1.67 x 10⁻²⁷ / 4.17 x 10⁻⁴⁵
= .4 n x 10¹⁸ kg / m³
or of the order of 10¹⁸ kg/m³
b )
Density of iron = 7900 kg / m³
or of the order of 10⁴ kg / m³
So nucleus of a matter is about 10¹⁴ times denser than iron .
Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 80.5 N, Jill pulls with 81.7 N in the northeast direction, and Jane pulls to the southeast with 131 N. (Since the donkey is involved with such uncoordinated people, who can blame it for being stubborn
Answer:
F = 233.52 N, θ' = 351.41º
Explanation:
In this exercise we must find the net force applied on the donkey.
For this we use Newton's second law, where we create a reference frame with the horizontal x axis
let's decompose the forces
Jack
= 80.5 N
Jill
cos 45 = F_{2x} / F₂2
sin 45 = F_{2y} / F₂2
F_{2x} = F₂ cos 45
F_{2y} = F₂ sin 45
F_{2x} = 81.7 cos 45 = 57.77 N
F_{2y} = 81.7 sin 45 = 57.77 N
Jane
cos (270 + 45) = F_{3x} / F₃3
sin 315 = F_{3y} / F₃
F_{3x} = 131 cos 315 = 92.63 N
F_{3y} = 131 sin 315 = -92.63 N
the force can be found in each axis
X axis
F_{x} = F_{1x} + F_{2x} + F_{3x}
F_{x} = 80.5 +57.77 + 92.63
F_{x} = 230.9 N
Axis y
F_{y} = F_{1y} + F_{2y} + F_{3y}
F_{y} = 0 + 57.77 -92.63
F_{y} = -34.86 N
we can give the result in two ways
a) F = (230.9 i ^ - 34.86 j ^) N
b) in the form of module and angle
we use the Pythagorean theorem
F = √(Fₓ² + F_{y}²
F = √(230.9² + 34.86²)
F = 233.52 N
let's use trigonometry for the angle
tan θ = [tex]\frac{F_y}{F_x} }[/tex]
θ = tan⁻¹ (\frac{F_y}{F_x} })
θ = tan⁻¹ (-34.86 / 230.9)
θ = -8.59º
if we measure this angle from the positive side of the x-axis counterclockwise
θ' = 360 -θ
θ‘= 360- 8.59
θ' = 351.41º
You are working as a letter sorter in a U.S. Post Office. Postal regulations require that employees' footwear must have a minimum coefficient of static friction of 0.5 on a specified tile surface. You are wearing athletic shoes for which you do not know the coefficient of static friction. In order to determine the coefficient, you imagine that there is an emergency and start running across the room. You have a coworker time you, and find that you can begin at rest and move 4.38 m in 1.21 s. If you try to move faster than this, your feet slip. Assuming your acceleration is constant, does your footwear qualify for the postal regulation?
Answer:
μ = 0.66, therefore if it compiesy with the regulations
Explanation:
Let's solve this exercise in part, let's start by finding with kinematics the acceleration of man
y = v₀ t + ½ a t²
as it starts from rest the initial velocity is zero
y = ½ a t²
a = [tex]\frac{2y}{t^2}[/tex]
a =\frac{2 \ 4.38}{1.21^2}
a = 6.46 m / s²
Now let's use Newton's second law,
Axis y
N- W = 0
N = W
N = m g
X axis
on this axis the man exerts a backward force and by the law of action and reaction the floor exerts a forward force of the same magnitude, this forward force is the friction force.
fr = m a
the friction force has an expression
fr = my N
let's substitute
μ mg = m a
μ = a / g
let's calculate
μ = 6.46 / 9.8
μ = 0.66
therefore if you comply with the regulations
The deepest part of the ocean is the Challenger Deep, at 10,900 m. The depth was first measured in 1875 by the HMS Challenger by depth sounding (which does not involve sound waves). If you were to measure the depth by echo sounding (which does involve sound), what would you expect the time for a sound pulse at the surface to return in s, naively assuming a constant sound velocity throughout the ocean
Answer:
t = 14.53 s
Explanation:
The speed of a wave is constant and is given by
v = [tex]\sqrt{ \frac{B}{ \rho} }[/tex]
in this exercise they indicate that we assume the constant velocity, therefore we can use the uniform motion relations
v = x / t
t = x / v
in this case the sound pulse leaves the ship and must return so the distance is
x = 2d
where d is the ocean depth d = 10900m and the speed of sound in seawater is v = 1500 m / s
let's calculate
t = 2 10900/1500
t = 14.53 s
Two loudspeakers are about 10 mm apart in the front of a large classroom. If either speaker plays a pure tone at a single frequency of 400 HzHz, the loudness seems pretty even as you wander around the room, and gradually decreases in volume as you move farther from the speaker. If both speakers then play the same tone together, what do you hear as you wander around the room
Answer:
I hear points of low volume sound and points of high volume of sound.
Explanation:
This is because, since the two sources of sound have the same frequency and are separated by a distance, d = 10 mm, there would be successive points of constructive and destructive interference.
Since their frequencies are similar, we should have beats of high and low frequency.
So, at points of low frequency, the amplitude of the wave is smallest and there is destructive interference. The frequency at this point is the difference between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, f - f' = 400 Hz - 400 Hz = 0 Hz. So, the volume of the sound is low(zero) at these points.
Also, at points of high frequency, the amplitude of the wave is highest and there is constructive interference. The frequency at this point is the sum between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, (f + f') = 400 Hz + 400 Hz = 800 Hz. So, the volume of the sound is high at these points.
So, as you wander around the room, I should hear points of high and low sound across the room.
how do positive and negative acceleration differ?
1. positive acceleration represents an object speeding up; negative acceleration represents an object slowing down
2. positive acceleration moves North or east; negative acceleration moves south or west
3. positive acceleration occurs when there is more velocity than speed; negative acceleration occurs when there is less velocity than speed.
4. positive acceleration occurs when an object changes its speed but not its direction; negative acceleration occurs when an object changes both its speed and direction
Answer:
1. positive acceleration represents an object speeding up; negative acceleration represents an object slowing down
Explanation:
Acceleration is clearly defined as the rate of change of velocity with time. When are body is speeding up as we say, it is accelerating. When a body is coming to rest, it is decelerating.
Positive acceleration occurs when the speed of a moving continues to increase.
Negative acceleration is when the speed of a moving body reduces drastically.
3. Do you think Lynn’s (the protagonist)actions were justifiable by her motives? Why or why not? Please help me Bad Genius the movie
Answer:
I do believe her actions were justified.
Explanation:
Due to the school charging extra fee from her father who makes a modest amount as a teacher. There was sum of money involved that could change how he lived and her.
I do not believe her actions where justified
She had a lot going for her. She could have skipped the hardship of helping grace and pass. She could have easily have gotten a good job with a degree and paid back all the debts owed. Alot of troubles could have been avoided just by doing her own thing.
15 points!!:-) Need help ASAP!
When does a compass NOT point towards magnetic north?
A.during a solar eclipse, which changes
Earth's magnetic field.
B. When there is another magnet close by.
C.When there is an unused battery close by.
D. When there is a coil of copper wire close by.
Answer:
b i think so because it makes senes
Answer: When there is another magnet close by.
Explanation: The needle of a compass is itself a magnet, and thus the north pole of the magnet always points north, except when it is near a strong magnet. ... When you take the compass away from the bar magnet, it again points north. So, we can conclude that the north end of a compass is attracted to the south end of a magnet.
.1 An 8-ft 3 tank contains air at an initial temperature of 808F and initial pressure of 100 lbf/in. 2 The tank develops a small hole, and air leaks from the tank at a constant rate of 0.03 lb/s for 90 s until the pressure of the air remaining in the tank is 30 lbf/in. 2 Employing the ideal gas model, determine the final temperature, in 8F, of the air remaining in the tank
Correct temperature is 80°F
Answer:
T_f = 38.83°F
Explanation:
We are given;
Volume; V = 8 ft³
Initial Pressure; P_i = 100 lbf/in² = 100 × 12² lbf/ft²
Initial temperature; T_i = 80°F = 539.67 °R
Time for outlet flow; t_o = 90 s
Mass flow rate at outlet; m'_o = 0.03 lb/s
Final pressure; P_f = 30 lbf/in² = 30 × 12² lbf/ft²
Now, from ideal gas equation,
Pv = RT
Where v is initial specific volume
R is ideal gas constant = 53.33 ft.lbf/°R
Thus;
v = RT/P
v_i = 53.33 × 539.67/(100 × 12²)
v_i = 2 ft³/lb
Formula for initial mass is;
m_i = V/v_i
m_i = 8/2
m_i = 4 lb
Now change in mass is given as;
Δm = m'_o × t_o
Δm = 0.03 × 90
Δm = 2.7 lb
Now,
m_f = m_i - Δm
Thus; m_f = 4 - 2.7
m_f = 1.3 lb
Similarly in above;
v_f = V/m_f
v_f = 8/1.3
v_f = 6.154 ft³/lb
Again;
Pv = RT
Thus;
T_f = P_f•v_f/R
T_f = (30 × 12² × 6.154)/53.33
T_f = 498.5°R
Converting to °F gives;
T_f = 38.83°F
The final temperature, in °F, of the air remaining in the tank is 38.83°F
It is given that volume V = 8 ft³
Initial Pressure Pi = 100 lbf/in² = 100 × 12² lbf/ft²
Initial temperature Ti = 80°F = 539.67 °R
Time for outlet flow; to = 90 s
Mass flow rate at outlet; m'o = 0.03 lb/s
Final pressure; Pf = 30 lbf/in² = 30 × 12² lbf/ft²
Now, from ideal gas equation,
Pv = RT
where v is initial volume, R is ideal gas constant = 53.33 ft.lbf/°R
[tex]v = RT/P\\ \\ v_i = 53.33 *539.67/(100*12^2)\\ \\ v_i = 2 ft^3/lb [/tex]
The initial mass is;
[tex]m_i = V/v_i\\ \\ m_i = 8/2\\ \\ m_i = 4 lb [/tex]
Now change in mass is given as;
Δm = [tex]m'_o*t_o[/tex]
Δm = 0.03 × 90
Δm = 2.7 lb
[tex]m_f[/tex] = [tex]m_i[/tex] - Δm
[tex]m_f[/tex] = 4 - 2.7
[tex]m_f[/tex] = 1.3 lb
now,
[tex]v_f = V/m_f\\ \\ v_f = 8/1.3\\ \\ v_f = 6.154 f^3/lb [/tex]
From the gas equation
Pv = RT
Final state:
[tex]T_f = P_fv_f/R\\\\ T_f = (30*12^2*6.154)/53.33\\\\ T_f = 498.5^oR [/tex]
Converting to °F:
[tex]T_f[/tex] = 38.83°F is the final temperature.
Learn More:
https://brainly.com/question/18518493
To understand the behavior of the electric field at the surface of a conductor, and its relationship to surface charge on the conductor. A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge.
Answer:
Explanation:
The electric field inside of a conductor is 0 because the conduction electrons are pushed to the outer edges of the conductor. The surface of the conductor still has charge.
Based on your average reaction time, how much time would it take to react to a traffic situation and stop a car traveling at 60 mph (1 mph equals 0.45 m/s) if you could decelerate the car at a rate of -3.4m/s2?
What distance would you travel (in meters) as the car came to a stop in the above situation?
Avg Reaction time: 0.218 ms
Answer:
d = 106.41 m
Explanation:
Given that,
Initial speed of the car, u = 60 mph = 26.9 m/s
The deceleration in the car, a = -3.4 m/s²
The average reaction time, t = 0.218 m/s
It finally stops, final velocity, v = 0
We need to find the distance covered by the car as it come to a stop.
Using third equation of motion to find.
[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-26.9^2}{2\times (-3.4)}\\\\d=106.41 m[/tex]
So, the car will cover 106.41 m as it comes to a stop.
Which cell line is pointing to the body?
Answer:
The answer is B .........number 2
Explanation:
a sensor light installed on the edge of a home can detect motion for a distance of 50 feet in front and with a range of motion of 200 degrees. what is the arc length of the area covered
Answer:
4363.3231 feets²
Explanation:
Given that :
Distance, r = 50 ft
θ = 200°
The arc length of area covered :
Arc length = θ/360° * πr²
Arc length = (200/360) * 50 ft ^2 * π
Arc length = 0.5555555 * 2500 * π
Arc length = 4363.3231 feets²
What quantity measures the amount of space an object occupies?
A. Volume B.Temperature C. Mass D. Density
Answer:
mas
Explanation:
mass is the amount of space something occupies.
What is the frequency of a wave of a light is with a wavelength of 4 x 10-7 m?
Answer:
7.5 × 10^14 Hz
Velocity of light = 3×10^8m/s
Frequency = (3×10^8)/(4 x 10^-7)
= 7.5 × 10^14 Hz
16. An object has a gravitational potential energy 41,772.5 Jof and has a mass of 1550 kg. How high is it
above the ground?
Plz help
Answer:
2.75 m.
Explanation:
From the question given above, the following data were obtained:
Potential energy (PE) = 41772.5 J
Mass (m) of object = 1550 kg
Height (h) =?
Potential energy is the energy possess by an object due to its location. Mathematically, potential energy is expressed as shown below:
PE = mgh
Where
PE => potential energy
m => mass of the object
g => acceleration due to gravity
h => height to which the object is located.
With the above formula, we can obtain the height to which the object is located as follow:
Potential energy (PE) = 41772.5 J
Mass (m) of object = 1550 kg
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) =?
PE = mgh
41772.5 = 1550 × 9.8 × h
41772.5 = 15190 × h
Divide both side by 15190
h = 41772.5 / 15190
h = 2.75 m
Thus, the object is located at 2.75 m above the ground.
What school did Ronald McNair go to and what kind of science did he work in
Answer:
McNair graduated as valedictorian of Carver High School in 1967. In 1971, he received a Bachelor of Science degree in engineering physics, magna cu.m laude, from the North Carolina Agricultural and Technical State University in Greensboro, North Carolina.
A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.4 cm. Assuming heat to be transferred uniformly from all surfaces, determine (a) the amount of heat this resistor dissipates during a 24-hour period, (b) the heat flux, and (c) the fraction of heat dissipated from the top and bottom surfaces.
Answer:
(a) The resistor disspates 103680 joules during a 24-hour period.
(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.
(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.
Explanation:
(a) The amount of heat dissipated ([tex]Q[/tex]), measured in joules, by the cylindrical resistor is the power multiplied by operation time ([tex]\Delta t[/tex]), measured in hours. That is:
[tex]Q = \dot Q \cdot \Delta t[/tex] (1)
If we know that [tex]\dot Q = 1.2\,W[/tex] and [tex]\Delta t = 86400\,s[/tex], then the amount of heat dissipated by the resistor is:
[tex]Q = (1.2\,W)\cdot (86400\,s)[/tex]
[tex]Q = 103680\,J[/tex]
The resistor disspates 103680 joules during a 24-hour period.
(b) The heat flux ([tex]Q'[/tex]), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ([tex]A[/tex]), measured in square meters:
[tex]Q' = \frac{\dot Q}{A}[/tex] (2)
[tex]Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }[/tex] (3)
Where:
[tex]D[/tex] - Diameter, measured in meters.
[tex]h[/tex] - Length, measured in meters.
If we know that [tex]\dot Q = 1.2\,W[/tex], [tex]D = 4\times 10^{-3}\,m[/tex] and [tex]h = 2\times 10^{-2}\,m[/tex], the heat flux of the resistor is:
[tex]Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }[/tex]
[tex]Q' \approx 4340.589\,\frac{W}{m^{2}}[/tex]
The heat flux of the resistor is approximately 4340.589 watts per square meter.
(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ([tex]r[/tex]), no unit, is the ratio of the top and bottom surfaces to total surface:
[tex]r = \frac{\frac{\pi}{2}\cdot D^{2}}{A}[/tex] (3)
If we know that [tex]A \approx 2.765\times 10^{-4}\,m^{2}[/tex] and [tex]D = 4\times 10^{-3}\,m[/tex], then the fraction is:
[tex]r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}[/tex]
[tex]r = 0.045[/tex]
The fraction of heat dissipated from the top and bottom surfaces is 0.045.
20 points!!!! A 2,00ON steel rod that is 5 meters long is placed in a corner between the floor and a wall, and balanced at an angle using a cord attached to the wall The rod is balanced such that its top end is 2.38 meters away from the wall, The cord is 40 cm long, and it is attached to the wall at a height of 75 cm above the floor. The diagram to the right shows the situation If the lower end of the rod does not slip from the corner, what is the tension in the cord?
Answer:
WE NEED TO ADD ALL 40+2.38 +75+5
Explanation:
PLSE GIVE SOME POINTS DUDE
Which of the following is NOT a characteristic of noble gases?
unreactive
odorless
solid at room temperature
colorless
What makes electromagnets useful for sorting metals in recycling centers?
O A. The current can be turned on to pick up items containing all
metals and turned off to drop them.
O B. The current can be turned off to pick up items containing all
metals and turned off to drop them.
O C. The current can be turned on to pick up items containing iron and
turned off to drop them.
D. The current can be turned off to pick up items containing iron and
turned on to drop them.
C
It is right because I took this and I got this answer correct
4. Sally applies a horizontal force of 462 N with a rope to drag a wooden crate across a floor with a constant speed. The rope tied to the crate is pulled at an angle of 56.00 . c.What work is done by the floor through force of friction between the floor and the crate
Answer:
-6,329.5Joules
Explanation:
Complete question:
Sally applies a horizontal force of 462N with a rope to drag a wooden crate across a floor with a constant speed the rope tied to the crate is pulled at an angle of 56.0degree and sally moves the crate 24.5m. What work is done by the floor through the force of friction between the floor and crate.
Work done = Fd cos theta
F is the horizontal force
d is the distance covered
theta angle of inclination
Substituting into the formula
Workdone = 462(24.5)cos 56
Workdone = 11,319(0.5592)
Workdone = 6,329.5Joules
Hence the workdone by sally is 6,329.5Joules
The work done by friction will be opposite the work done by sally, hence work done by the floor through force of friction between the floor and the crate is -6,329.5Joules
how is friction involved in the movement of space
Answer:
Friction can stop or slow down the motion of an object.
Explanation:
The slowing force of friction always acts in the direction opposite to the force causing the motion.
Can someone please help me get this right pleaseee I’ll mark brainless .
Answer:i think it is c
Explanation:
Answer:
Explanation: i think its c to try it
You are on the Pirates of the Caribbean attraction in the Magic Kingdom at Disney World. Your boat rides through a pirate battle, in which cannons on a ship and in a fort are firing at each other. While you are aware that the splashes in the water do not represent actual cannonballs, you begin to wonder about such battles in the days of the pirates. Sup-pose the fort and the ship are separated by 75.0 m. You see that the cannons in the fort are aimed so that their cannon-balls would be fired horizontally from a height of 7.00 m above the water.
(a) You wonder at what speed they must be fired in order to hit the ship before falling in the water.
(b) Then, you think about the sludge that must build up inside the barrel of a cannon. This sludge should slow down the cannonballs. A question occurs in your mind: if the can-nonballs can be fired at only 50.0% of the speed found ear-lier, is it possible to fire them upward at some angle to the horizontal so that they would reach the ship?
Answer:
a) v₀ₓ = 62.76 m / s, b) θ₁ = 17.6º, θ₂ = 67.0º
Explanation:
We can solve this exercise using the projectile launch ratios
a) Let's find the time it takes for the bullet to reach the water level
y = y₀ + v_{oy} t - ½ g t²
when it reaches the water its height is zero y = 0, as the bullet is fired horizontally its initial vertical velocity is zero
0 = y₀ + 0 - ½ g t²
t =[tex]\sqrt{2y_o/g}[/tex]
t = [tex]\sqrt{2 \ 7 /9.8}[/tex]
t = 1,195 s
now we can calculate the speed with the horizontal movement
x = v₀ₓ t
v₀ₓ = x / t
v₀ₓ = 75.0 / 1.195
v₀ₓ = 62.76 m / s
b) if the speed of the bullets is half of that found
v₀ = 62.76 / 2 = 31.38 m / s
let's write the expressions for the distance
x = v₀ cos θ t
y = y₀ + v_{oy} sin θ t - ½ g t²
t = [tex]\frac{x}{v_o \ cos \theta}[/tex]
we substitute
[tex]0 = y_o + v_o sin \theta \ \frac{x}{v_o \cos \thetay} - 1/2 g \ (\frac{x}{v_o \ cos \theta})^2[/tex]
[tex]0 = y_o + x tan \theta - \frac{1}{2} g \ \frac{x^2}{ v_o^2 \ cos^2 \theta}[/tex]
let's use the identified trigonometry
sec² θ = 1 + tan² θ
sec θ = 1 / cos θ
we substitute
[tex]0 = y_o + x tan \theta - \frac{g x^2}{2 v_o^2} ( 1 + tan^2 \theta)[/tex]
[tex]\frac{g x^2}{2v_o^2} tan^2 \theta - x tan \theta + \frac{gx^2}{2v_o^2} - y_o = 0[/tex]
we change variable
tan θ = H
[tex]\frac{gx^2}{2 v_o^2 } H^2 - x H + \frac{gx^2}{2v_o^2}-y_o =0[/tex]
we subtitle the values
[tex]\frac{9.8 \ 75^2}{2 \ 31.38^2} H^2 - 75 H + \frac{9.8 \ 75^2}{2 \ 31.38^2}-7 =0[/tex]
27.99 H² - 75 H + 20.99 = 0
H² - 2.679 H + 0.75 = 0
we solve the quadratic equation
H = [2.679 ± [tex]\sqrt{2.679^2 - 4 0.75}[/tex]] / 2
H = [2,679 ± 2,044] / 2
H₁ = 0.3175
H₂ = 2.3615
now we can find the angles
H₁ = tan θ₁
θ₁ = tan⁻¹ H₁
θ₁ = tan⁻¹ 0.3175
θ₁ = 17.6º
θ₂ = 67.0º
for these two angles the bullet hits the boat
In traveling a distance of 2.3 km between points A and D, a car is driven at 83 km/h from A to B for t seconds and 41 km/h from C to D also for t seconds. If the brakes are applied for 4.4 seconds between B and C to give the car a uniform deceleration, calculate t and the distance s between A and B.
Answer:
- time t taken for car to travel is 64.57 s
- distance travelled between A and B is 1.4887 km
Explanation:
Given the data in the question;
[tex]U_{BC}[/tex] = 83 km/h = ( 83×1000 / 60×60) = 23.0555 m/s
[tex]U_{CD}[/tex] = 41 km/h = ( 41×1000 / 60×60) = 11.3888 m/s
now, we calculate the acceleration;
a = ( [tex]U_{BC}[/tex] - [tex]U_{CD}[/tex] ) / t
we substitute
a = ( 23.0555 - 11.3888 ) / 4.4
a = 11.6667 / 4.4
a = 2.6515 m/s²
Now equation for displacement from BC
[tex]S_{BC}[/tex] = [tex]U_{BC}[/tex]t + 1/2.at²
we substitute
[tex]S_{BC}[/tex] = 23.0555×4.4 + 1/2×a×(4.4)²
we substitute -2.6515m/s² for a
[tex]S_{BC}[/tex] = 23.0555×4.4 + 1/2×(-2.6515)×(4.4)²
= 101.4442 - 25.6665
[tex]S_{BC}[/tex] = 75.7792 m
Now, for total distance covered = 2.3km = ( 2.3×1000) = 2300m
so
[tex]S_{AB}[/tex] + [tex]S_{BC}[/tex] + [tex]S_{CD}[/tex] = 2300 m
we substitute substitute
[tex]S_{AB}[/tex] + 75.7792 m + [tex]S_{CD}[/tex] = 2300 m
[tex]S_{AB}[/tex] + [tex]S_{CD}[/tex] = 2300 - 75.7792
[tex]S_{AB}[/tex] + [tex]S_{CD}[/tex] = 2224.2208 m
so we substitute 23.0555t for [tex]S_{AB}[/tex] and 11.3888t for [tex]S_{CD}[/tex]
23.0555t + 11.3888t = 2224.2208
34.4443t = 2224.2208
t = 2224.2208 / 34.4443
t = 64.57 s
Therefore, time t taken for car to travel is 64.57 s
Distance Between A to B
[tex]S_{AB}[/tex] = t × [tex]U_{AB}[/tex]
we substitute
[tex]S_{AB}[/tex] = 64.57 s × 23.0555
[tex]S_{AB}[/tex] = 1488.69 m
[tex]S_{AB}[/tex] = 1.4887 km
Therefore, distance travelled between A and B is 1.4887 km
kinetic energy portfolio in part 2 the independent changes to----?
