The greatest possible distance from the mirror to the point where the reflected rays meet is infinite for a flat mirror, as the reflected rays never actually converge, creating a virtual image.
The greatest possible distance d from the mirror to the point where the reflected rays meet would be when the object distance do approaches the focal length f of the mirror from the right side, causing the image distance d to approach negative infinity, indicating that the reflected rays would never converge in real space. This situation describes the formation of a virtual image where the rays appear to come from. In a flat mirror, the focal length is technically at infinity since parallel rays remain parallel after reflection and never actually converge. Therefore, in a practical sense, when considering a flat mirror, the reflected rays appear to converge at a distance behind the mirror equal to the object's distance in front of the mirror, which is defined as d = -do.
The greatest possible distance (d) from the mirror to the point where the reflected rays meet is [tex]\( \frac{L}{2} \)[/tex], which is the focal length of the mirror when it is at the center of the spherical surface.
To understand why the greatest possible distance (d) from the mirror to the point where the reflected rays meet is [tex]\( \frac{L}{2} \)[/tex], let's consider the behavior of light rays reflecting off a mirror.
When a light ray reflects off a mirror, the angle of incidence is equal to the angle of reflection. This means that the path of the light ray before and after reflection are symmetric with respect to the normal to the mirror surface at the point of incidence.
For the reflected rays to meet at a point, they must converge. The best way to visualize this is to consider a light ray that travels parallel to the mirror's surface. After reflection, this ray will appear to come from a point behind the mirror, known as the focal point. The distance from the mirror to this focal point is the focal length of the mirror.
In the case of a flat mirror, the focal length is infinite because parallel rays remain parallel after reflection and do not converge to a single point. However, if we consider a spherical mirror, the focal length (f) is finite and is related to the radius of curvature (R) of the mirror by the equation [tex]\( f = \frac{R}{2} \)[/tex].
Now, let's consider the scenario where the mirror can be moved horizontally. The greatest possible distance (d) from the mirror to the point where the reflected rays meet would occur when the mirror is at the center of the spherical surface it is a part of. At this point, the focal point of the mirror would be at a distance equal to the focal length (f) from the mirror's surface.
Given that the length of the mirror is (L), and the radius of curvature (R) is equal to (L) (since the mirror is a segment of a sphere), we can substitute (R) with (L) in the focal length equation. Thus, the focal length (f) is [tex]\( \frac{L}{2} \)[/tex].
A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3 /s. determine the minimum power that must be supplied to the fan. take the density of air to be 1.18 kg/m3
Chinook salmon are able to move through water especially fast by jumping out of the water periodically. this behavior is called porpoising. suppose a salmon swimming in still water jumps out of the water with velocity 6.65 m/s at 48.1° above the horizontal, sails through the air a distance l before returning to the water, and then swims the same distance l underwater in a straight, horizontal line with velocity 3.79 m/s before jumping out again. (a) determine the average velocity of the fish for the entire process of jumping and swimming underwater. incorrect: your answer is incorrect. your response differs from the correct answer by more than 10%. double check your calculations. m/s (b) consider the time interval required to travel the entire distance of 2l. by what percentage is this time interval reduced by the jumping/swimming process compared with simply swimming underwater at 3.79 m/s?
Jeff’s father is installing a do-it-yourself security system at his house. He needs to get a device from his workshop that converts electric energy to sound energy. Which device is Jeff’s father looking for? switch motor buzzer bulb battery
Answer:
buzzer
Explanation:
The auditory cortex is located in which lobe of the brain?
The auditory cortex is located in temporal lobe of the brain.
What is auditory cortex?The auditory cortex is the part of the temporal lobe that processes hear-able data in people and numerous different vertebrates.
It is a piece of the auditory framework, carrying out fundamental and higher roles in hearing. For example, potential relations to language switching.
The auditory cortex is situated on the predominant transient gyrus in the fleeting curve and gets highlight point input from the ventral division of the average geniculate complex. Consequently, it contains an exact tonotopic map.
It is found generally at the upper sides of the fleeting curves in people, bending down and onto the average surface, on the prevalent transient plane, inside the sidelong sulcus and containing portions of the cross over worldly gyri, and the unrivaled worldly gyrus, including the planum polare and planum temporale.
Thus, auditory cortex is located in temporal lobe of the brain.
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Select all that apply.
A scientific theory _____.
is true all the time
is supported by evidence
can evolve over time
is a guess
A sailboat is heading directly north at a speed of 20 knots (1 knot 50.514 m/s). the wind is blowing toward the east with a speed of 17 knots. (a) determine the magnitude and direction of the wind velocity as measured on the boat. (b) what is the component of the wind velocity in the direction parallel to the motion of the boat?
A falling stone is at a certain instant 90 feet above the ground. two seconds later it is only 10 feet above the ground. if it was thrown down with an initial speed of 4 feet per second, from what height was it thrown?
The stone was originally thrown from a height of 234 feet. This solution was obtained using the physics kinematic equation for vertical motion.
Explanation:This problem can be solved using the kinematic equation:
Δy = V₀t + 1/2gt²
, where:
Δy is the displacement (final position - initial position)V₀ is the initial velocityt is the timeg is the acceleration due to gravity, which is -32 ft/s² (negative because it's acting downwards)From the statement we know: Δy = 10 ft - 90 ft = -80 ft after 2 seconds. We substitute these values and solve for the initial position, y₀. The equation becomes:
-80 = 4*2 + 1/2*(-32)*2²
. Solving gives us -80 = 8 - 64, so
y₀ = 80 + 64 + 90 = 234 ft
. Therefore, the stone was thrown from a height of 234 feet.
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Using the equations of motion under gravity, it's determined that the stone was thrown from a height of 162 feet considering it was thrown down with an initial velocity and passed specific heights in its journey.
Explanation:The question involves finding the original height from which a stone was thrown, given that it was thrown down with an initial speed and passes specific points in its descent. To solve this problem, we can use the equations of motion under the influence of gravity. The formula that relates the initial velocity (u), the acceleration due to gravity (g = 32 feet/second2 downward), the time taken (t), and the displacement (s) is s = ut + (1/2)gt2.
In this instance, the stone is observed to move from 90 feet above the ground to 10 feet above the ground in 2 seconds, with an initial downward speed of 4 feet/second. The drop in height (displacement) is 80 feet (90 - 10). We can insert these values into the formula to find the initial height: Let H be the initial height, the equation becomes H - 90 = 4(2) + (1/2)(32)(22). Simplifying, we find that H - 90 = 8 + 64, which resolves to H = 162 feet.
Hence, the stone was thrown from a height of 162 feet.
When a simple machine multiplies force, it decreases
a.potential energy.
b.distance moved.
c.both
d.neither?
Which item is made from an alloy?
steel tray
glass plate
credit card
copper wire
How many miles can you get on one tank of gas which holds 18 gallons and you get 22 miles per gallon
The wonders and mysteries of plants provide evidence for:
:creation
:accidental chance
:intelligent design
:evolution
:natural chemical origins
Answer:
Creation and intelligent design
Explanation:
Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant (and quite modest) acceleration. a train travels through a congested part of town at 7.0 m/s . once free of this area, it speeds up to 14 m/s in 8.0 s. at the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed. what is the final speed?
First let us calculate the acceleration.
v1 = v0 + a t1
where v1 is final velocity, v0 is initial velocity, a is acceleration and t is time
Calculating for a:
14 m/s = 7 m/s + a * 8 s
a = 0.875 m/s^2
Therefore the final speed is calculated similarly:
v2 = v1 + a t2
v2 = 14 m/s + (0.875 m/s^2) * 16 s
v2 = 28 m/s
A car traveling at a speed of v can brake to an emergency stop in a distance x. assuming all other driving conditions are all similar, if the traveling speed of the car doubles, the stopping distance will be
which matches mandeleevs prediction for the properties of eka-aluminum?
What part causes the disc brake caliper piston to retract when the brakes are released?
What units do chemists normally use for density of liquids and solids? for gas density? explain the differences?
We can define density as per unit volume.
The SI unit to measure the density of solid, liquid and gas is kilogram per cubic metre (kg/m3) and in the centimetre–gram–second system of units (cgs unit) is gram per cubic centimetre (g/cm3). Gas density is very dependent of pressure and temperature whereas the density of solids and liquids is not so dependent, that is why the gas density is given at a standard temperature and pressure.
Chemists typically use g/mL or g/cm³ for liquids and solids, and g/L for gases. Differences arise due to the varying densities and volumes of these states of matter.
Chemists use different units for density based on the state of matter being measured. For liquids and solids, the density is usually expressed in grams per milliliter (g/mL) or grams per cubic centimeter (g/cm³). This is because liquids and solids generally have higher densities and smaller volumes compared to gases, making these units convenient for laboratory measurements and calculations.
For gases, the density is typically expressed in grams per liter (g/L). Gases have much lower densities and occupy larger volumes compared to liquids and solids, so g/L is more appropriate. The larger unit (liter) helps to manage the lower mass of gases, ensuring the numerical values are easy to work with.
The choice of units helps chemists maintain consistency and accuracy when measuring and comparing densities across different substances and states of matter. Using units that reflect the typical scales of measurement for each state ensures that the values are neither too large nor too small, facilitating easier interpretation and communication of data.
As a train accelerates away from a station, it reaches a speed of 4.9 m/s in 5.1 s. If the train's acceleration remains constant, what is its speed after an additional 7.0s has elapsed?
The speed of the rain after accelerating for addition [tex]7\text{ s}[/tex] will be [tex]\boxed{11.62\text{ m/s}}[/tex].
Explanation:
Given:
The speed of the train after [tex]5.1\text{ s}[/tex] is [tex]4.9\text{ m/s}[/tex].
The initial speed of the train is [tex]0\text{ m/s}[/tex].
Concept:
The acceleration of a body is defined as the rate at which the velocity of the body in motion changes every second. If the acceleration of the body is in the direction of motion of the body, the body will be accelerated.
As the train stars from rest and accelerates away from the station, the speed of the train will increase according to the first equation of motion.
The expression for the first equation of motion for the motion of the train is:
[tex]\boxed{v_f=v_i+at}[/tex]
Here, [tex]v_f[/tex] is the final speed of the train, [tex]v_i[/tex] is the initial speed of the train, [tex]a[/tex] is the acceleration of the train and [tex]t[/tex] is the time taken by the train.
Substitute the values of velocity for first [tex]5.1\text{ s}[/tex] of motion of the train.
[tex]\begin{aligned}4.9&=0+a.(5.1)\\a&=\dfrac{4.9}{5.1}\text{ m/s}^2\\&=0.96\text{ m/s}^2\end{aligned}[/tex]
Now, the for the speed of the train after it travels for addition [tex]7.0\text{ s}[/tex] or a total of [tex]12.1\text{ s}[/tex].
[tex]\begin{aligned}v_f&=0+(0.96)(12.1)\text{ m/s}\\v_f&=11.62\text{ m/s}\end{aligned}[/tex]
Thus, The speed of the rain after accelerating for addition [tex]7\text{ s}[/tex] will be [tex]\boxed{11.62\text{ m/s}}[/tex].
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Answer Details:
Grade: Senior School
Subject: Physics
Chapter: Laws of motion
Keywords:
train, accelerates, constant, rest, equation of motion, initial, final, velocity, time taken, addition 7 seconds, acceleration.
Activation energy can be provided by the kinetic energy of moving molecules. true or false?
Answer:
False
Explanation:
Activation energy is simply the initial energy input which is needed to proceed a chemical reaction. The source of this energy is heat, which is obtained when reactant molecules absorb thermal energy from their surroundings. This thermal energy provides the kinetic energy of moving molecules, by speeding up the motion of the reactant molecules.
Therefore, the correct option is false.
A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00 m/s , releases a sandbag at an instant when the balloon is a height h = 40.0 m above the ground (Figure 1) . After it is released, the sandbag is in free fall. For the questions that follow, take the origin of the coordinate system used for measuring displacements to be at the ground, and upward displacements to be positive.
A) Compute the position of the sandbag at a time 1.05 s after its release.
B)Compute the velocity of the sandbag at a time 1.05 s after its release.
c) How many seconds after its release will the bag strike the ground?
(a). Position of sandbag at time [tex]1.05\text{ s}[/tex] after its release is [tex]\boxed{39.84\text{ m}}[/tex] above the ground.
(b). Velocity of the sandbag after time [tex]1.05\text{ s}[/tex] is [tex]\boxed{5.3\text{ m/s}}[/tex].
(c). The time taken after release the bag to strike the ground is [tex]\boxed{3.41\text{ s}}[/tex].
Further explanation:
Here, all the actions performed is under free fall. So, we will use the kinematic equations of motion for free falling body.
Given:
The velocity of rising of hot air balloon is [tex]5\text{ m/s}[/tex].
Height of hot air balloon when sandbag released is [tex]40\text{ m}[/tex].
Calculation:
Part (a)
Position of sandbag at time [tex]1.05\text{ s}[/tex] after its release.
When sandbag released the hot air balloon was rising up with the velocity of [tex]5\text{ m/s}[/tex].
So, initial velocity of sandbag will be [tex]5\text{ m/s}[/tex] in upward direction.
So, the time taken by the sand bag to reach at its top position is given by,
[tex]\boxed{v = u - gt}[/tex] …… (1)
Here, [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity, [tex]g[/tex] is the acceleration due to gravity and negative sign is due upward motion of sandbag, [tex]t[/tex] is the time required to reach at top position.
Substitute values for v and u in equation (1).
[tex]\begin{aligned}0&=5-9.8t\\9.8t&=5\\t&=0.51\text{ s}\\\end{aligned}[/tex]
So, the distance travel by sandbag to top position can be calculated as,
[tex]\boxed{{v^2}={u^2}-2g{s_1}}[/tex]
Substitute values for [tex]v[/tex] and u in above equation.
[tex]\begin{aligned}{0^2}&={5^2}-2\times9.8\times{s_1}\\19.6{s_1}&=25\\{s_1}&=1.27\text{ m}\\\end{aligned}[/tex]
After that sandbag will start falling.
The time remain from the given time is,
[tex]\begin{aligned}{t_1}&=1.05-0.51\\{t_1}&=0.54\text{ s}\\\end{aligned}[/tex]
The distance travel by sandbag in [tex]0.54\text{ s}[/tex] in downward direction can be calculated as,
Substitute [tex]0[/tex] for [tex]u[/tex] and [tex]0.54\text{ s}[/tex] for [tex]t[/tex] in above equation.
[tex]\begin{aligned}{s_2}&=0\times0.54+\frac{1}{2}\times9.8{\left({0.54}\right)^2}\\&=1.43\text{ m}\\\end{aligned}[/tex]
So, the position of the sandbag after [tex]1.05\text{ s}[/tex] from the ground can be calculated as,
[tex]\begin{aligned}h&=40+{s_1}-{s_2}\\&=40+1.27-1.43\\&=39.84\text{ m}\\\end{aligned}[/tex]
Part (b)
Velocity of the sandbag after time [tex]1.05\text{ s}[/tex].
The velocity of the sandbag after time [tex]t[/tex] can be calculated as,
[tex]\boxed{v=u+gt}[/tex]
Substitute the values for [tex]u[/tex] and t in above equation.
[tex]\begin{aligned}v&=0+9.8\times0.54\\&=5.3\text{ m/s}\\\end{aligned}[/tex]
Thus, the velocity of the sandbag after time [tex]1.05\text{ s}[/tex] is [tex]\boxed{5.3\text{ m/s}}[/tex].
Part (c)
The time taken after release the bag to strike the ground.
The total distance the top most position of the bag and the ground is,
[tex]\begin{aligned}S&=40+{s_1}\\&=40+1.27\\&=41.27\text{ m}\\\end{aligned}[/tex]
Now, time taken by the bag to strike the ground from its top most position,
[tex]\boxed{S=ut+\dfrac{1}{2}g{t_2}^2}[/tex]
Substitute [tex]41.27{\text{ m}}[/tex] for [tex]S[/tex] and [tex]0[/tex] for [tex]u[/tex] in above equation.
[tex]\begin{aligned}41.27&=0\timest+\dfrac{1}{2}\times9.8{t_2}^2\\41.27&=4.9{t_2}^2\\{t_2}^2&=\dfrac{{41.27}}{{4.9}}\\&=2.9{\text{ s}}\\\end{aligned}[/tex]
Now, the total time taken by bag to strike the ground from the instant of release is,
[tex]\begin{aligned}T&={t_2}+t\\&=2.9+0.51\\&=3.41\text{ s}\\\end{aligned}[/tex]
Thus, the time taken after release the bag to strike the ground is [tex]\boxed{3.41\text{ s}}[/tex].
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Answer detail:
Grade: Senior School
Subject: Physics
Chapter: Kinematics
Keywords:
Hot air balloon, constant velocity, height of, position of sandbag, velocity of sandbag, total time, rising up, 5m/s, 40m, 1.05 s, displacement, balloonist, strike the ground.
Which of the following survey questions would be an example of question-wording bias? A. Do you think background checks before buying a gun is a good idea? B. Do you think there should be a ban on assault rifles? C. Do you think hand guns should be registered? D. Do you think we should ignore our constitutional rights and let the government take citizens' guns away?
Answer:
D. Do you think we should ignore our constitutional rights and let the government take citizens' guns away?
Explanation:
A question wording biased is what happens when the question states directly a point of biew and suggests the interviewed a certain answer that is clear once you´ve heard the question, in this case it is obvious that the question is against the ban on guns, because it is already judging any decision that the congress could make on it and suggesting a point of view to the interviewed.
A mass weighing 16 pounds is attached to a spring whose spring constant is 49 lb/ft. what is the period of simple harmonic motion?
Final answer:
To find the period of simple harmonic motion for the given mass and spring constant, first convert units, then apply the period formula to get approximately 0.6283 seconds.
Explanation:
The period of simple harmonic motion for a mass-spring system can be calculated using the formula for the period T of a simple harmonic oscillator:
T = 2π√(m/k)
where m is the mass in kilograms, k is the spring constant in newtons per meter (N/m), and π is approximately 3.1416. To use this formula, we must convert the mass from pounds to kilograms and the spring constant from pounds per foot to newtons per meter. The period T is the time it takes for one complete cycle of oscillation.
First, convert 16 pounds to kilograms (1 pound is approximately 0.453592 kilograms):
16 pounds × 0.453592 = 7.257 kilograms
Now, convert the spring constant from lb/ft to N/m (1 lb/ft is approximately 14.5939 N/m):
49 lb/ft × 14.5939 = 715.6011 N/m
Using the conversion values:
T = 2π√(7.257 kg / 715.6011 N/m) = 2π√(0.010139 kg/N·m)
Perform the calculation to determine the period:
T = 2π√(0.010139) ≈ 2π√(0.01) ≈ 2π(0.1) ≈ 0.6283 seconds
Thus, the period of simple harmonic motion for the given mass-spring system is approximately 0.6283 seconds.
A ferris wheel of radius 100 feet is rotating at a constant angular speed Ï rad/sec counterclockwise. using a stopwatch, the rider finds it takes 5 seconds to go from the lowest point on the ride to a point q, which is level with the top of a 44 ft pole. assume the lowest point of the ride is 3 feet above ground level.
Angular speed is 0.188 rad/s ,
Tangential speed is 18.8 ft/s ,
Time for one revolution is 33.4 s.
Given :
Ferris wheel of radius 100 ft.
The lowest point of the ride is 3 feet above ground level.
Solution :
Refer the attached diagram for better understanding.
From the diagram we know that,
[tex]\rm y = 100-(44-3)=59 \; ft[/tex]
y = 59 ft
Now applying pythagorean theorem,
[tex]x^2 + 59 ^2= 100^2[/tex]
[tex]x=\sqrt{100^2-59^2}[/tex]
[tex]\rm x = 80.7403\;ft[/tex]
Now to calculate angle [tex]\theta\\[/tex],
[tex]\rm cos\theta = \dfrac{59}{100}=0.59[/tex]
[tex]\rm \theta = 53.84^\circ=0.9397\;radians[/tex]
Now the arc length pq is given by,
[tex]\rm S =pq=0.9397\times100[/tex]
[tex]\rm S = 93.97\; ft[/tex]
Now the angular velocity is given by,
[tex]\omega = \dfrac {0.9397}{5}[/tex]
[tex]\rm \omega = 0.188\;rad/sec[/tex]
Now the tangential velocity is given by,
[tex]\rm v={100}\times{0.188}[/tex]
[tex]\rm v = 18.8\;ft/sec[/tex]
Now the time for a single revolution is given by,
[tex]\rm T = \dfrac{2\pi}{0.188}[/tex]
[tex]\rm T= 33.4\; sec[/tex].
Therefore, angular speed is 0.188 rad/sec , tangential speed is 18.8 ft/s ec and time for one revolution is 33.4 sec.
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Which statement about matter is correct? A) In matter, molecules never stop moving. B) In the solid state, molecules stop moving. C) Pressure and temperature do not affect matter. D) Liquids have a lower level of energy than solids.
Answer: A
Explanation:
The name of the group of science that deals with earth and its neighbors in space is called
Select the correct statement to describe when a sample of liquid water vaporizes into water vapor. Question 12 options: Temperature increases and molecular motion increases while shape becomes less defined. Temperature decreases and molecular motion increases while shape becomes less defined. Temperature decreases and molecular motion decreases while shape becomes more defined. Temperature increases and molecular motion decreases while shape becomes more defined.
A bird is flying due east. Its distance from a tall building is given by x(t)=30.0m+(11.7m/s)t−(0.0450m/s3)t3. A) What is the instantaneous velocity of the bird when t = 8.00 s
To find the bird's instantaneous velocity at t = 8.00 s, the derivative of the position function x(t) is calculated to get the velocity function v(t), and then t = 8.00 s is substituted into this function to obtain the velocity at that instant.
Explanation:The question asks for the instantaneous velocity of a bird at a specific time given its position as a function of time, x(t). To find the instantaneous velocity, we need to take the derivative of the position function with respect to time. Given the position function x(t) = 30.0 m + (11.7 m/s)t - (0.0450 m/s3)t3, the derivative of this function will give us the velocity function v(t).
Therefore, v(t) = dx(t)/dt = 11.7 m/s - 3*(0.0450 m/s3)*t2. Plugging in t = 8.00 s into this velocity function, we get v(8.00 s) = 11.7 m/s - 3*(0.0450 m/s3)*(8.00 s)2, which we can calculate to find the instantaneous velocity at t = 8.00 seconds.
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A student walks 1.0 mi west and then 1.0 mi north. afterward, how far is she from her starting point?
A ball thrown horizontally at 22.2 m/s from the roof of a building lands 36.0 m from the base of the building. how high is the building?
The height of the building is 12.898 meters.
First, we calculate the time of flight using the horizontal distance and the horizontal velocity:
[tex]\[ t = \frac{d}{v_x} \][/tex]
where [tex]\( t \)[/tex] is the time of flight, [tex]\( d \)[/tex] is the horizontal distance (36.0 m) and is the horizontal velocity (22.2 m/s). Plugging in the values:
[tex]\[ t = \frac{36.0 \text{ m}}{22.2 \text{ m/s}} \][/tex]
[tex]\[ t = 1.6216 \text{ s} \][/tex]
Now, we use the time of flight to find the height of the building using the vertical motion equation:
[tex]\[ h = \frac{1}{2} g t^2 \][/tex]
where [tex]\( h \)[/tex] is the height of the building, [tex]\( g \)[/tex] is the acceleration due to gravity (approximately [tex]\( 9.81 \text{ m/s}^2 \)[/tex]), and [tex]\( t \)[/tex] is the time of flight we just calculated. Plugging in the values:
[tex]\[ h = \frac{1}{2} \times 9.81 \text{ m/s}^2 \times (1.6216 \text{ s})^2 \][/tex]
[tex]\[ h = \frac{1}{2} \times 9.81 \text{ m/s}^2 \times 2.6297 \text{ s}^2 \][/tex]
[tex]\[ h = 4.905 \text{ m/s}^2 \times 2.6297 \text{ s}^2 \][/tex]
[tex]\[ h = 12.898 \text{ m} \][/tex]
How long does it take for the ball to hit the wall?
At what height does the ball hit the wall?
The cannonball takes approximately 4.40 seconds to hit the wall and strikes it at a height of around 157.15 meters.
Let's solve this step by step:
Step 1: Calculate the time of flight
To find how long it takes for the cannonball to hit the wall, we need to consider the horizontal motion.
Initial horizontal velocity ([tex]u_x[/tex]): 89 m/s * cos(40°)[tex]u_x[/tex] = 89 * 0.766[tex]u_x[/tex] ≈ 68.174 m/sDistance to the wall (d): 300 mTime of flight (t) = d / [tex]u_x[/tex]t = 300 / 68.174t ≈ 4.40 secondsSo, it takes approximately 4.40 seconds for the ball to hit the wall.
Step 2: Calculate the height at which the ball hits the wall
For vertical motion:
Initial vertical velocity ([tex]u_y[/tex]): 89 m/s * sin(40°)[tex]u_y = 89 \times 0.643[/tex][tex]u_y \approx 57.227\ m/s[/tex]Vertical displacement (y) after time t: [tex]y = u_y \times t - 0.5 \times g \times t^2[/tex] (where g is the acceleration due to gravity, approximately 9.8 m/s²)
[tex]y = 57.227 \times 4.40 - 0.5 \times 9.8 \times (4.40)^2[/tex][tex]y \approx 251.79 - 94.64[/tex][tex]y \approx 157.15 meters[/tex]Therefore, the ball hits the wall at a height of approximately 157.15 meters.
The complete question is as follows:
A cannon elevated at 40 degrees is fired at a wall 300 m away on level ground. The initial speed of the cannonball is 89 m/sec.
1) how long does it take for the ball to hit the wall?
2) at what hight does the ball hit the wall?
What is a type of science that studies earth and and space