The size of the image is determined by the distance between the object and the mirror (object distance) and the distance between the mirror and the screen (image distance).
Using the formula for image distance (1/f = 1/object distance + 1/image distance), we can find the focal length of the mirror:
1/f = 1/object distance + 1/image distance (twice the object size)
1/f = 1/object distance + 2
1/f = 1/object distance + 1/image distance (three times the object size)
1/f = 1/object distance + 3
We know that the image distance (the distance from the mirror to the screen) is 75 cm in both scenarios, so we can set up the equations using that information:
1/f = 1/object distance + 2 (twice the object size)
75 = 2 * object distance + f
1/f = 1/object distance + 3 (three times the object size)
75 = 3 * object distance + f
Solving for object distance in the first equation:
object distance = (75 - f) / 2
Substituting that into the second equation:
75 = 3 * ((75 - f) / 2) + f
Simplifying and solving for f:
75 = (225 - 3f) / 2 + f
150 = 225 - 3f
3f = 75
f = 25 cm
So the focal length of the mirror is 25 cm.
To find the distance the object is moved, we can use the equation for object distance:
object distance = (75 - f) / 2
object distance = (75 - 25) / 2
object distance = 25 cm
So, the object is moved 25 cm in the process.
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20N , 5m
2. If the crate does not move, does Ignacio perform any work? Does he use any
power? Explain.
th
probl
Study the illustration above, and then answer the following questions on a separate
sheet of paper.
1. Ignacio is trying to move this crate. What are the two equations you can use to
calculate the power Ignacio must use to move the crate?
3. What else do you need to know to find the power Ignacio uses when he moves
the crate?
4. Ignacio moves the crate in 10 s. How much power, in watts, does he use?
5. Carolina moves the crate in 5 s. How much power, in joules per second, does
she use?
6. Felix moves the crate twice as far as Ignacio and Carolina in 5 s. How much
power, in watts, does he use?
The quantity of internal and mechanical energy that things contain fluctuates as a result of work.
How can you tell if someone is working on something or not?Example of a Work, An item must be subjected to a force and move in the direction of the force in order for work to be done on it.The quantity of internal and mechanical energy that things contain fluctuates as a result of work. Energy is contributed to a system or an item when work is done on it. A system or item transfers part of its energy to something else when it does work. When a ball is thrown, a hand exerts force while an arm swings forward.To learn more about mechanical energy refer to:
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if half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration (in m/s2) it can achieve on wet concrete where the coefficient of kinetic friction is 0.5 and the coefficient of static friction is 0.7.
The maximum acceleration that the flatbed truck can achive on wet concrete is 5[tex]m/s^{2}[/tex]
The maximum acceleration a truck can achieve on wet concrete is determined by the coefficient of friction between the two drive wheels and the surface.
Taking into account the coefficients of static and kinetic friction between the drive wheels and a wet concrete surface—0.7 and 0.5, respectively—the maximum acceleration of a truck with half its weight supported by its two drive wheels can be calculated using Newton's Second Law of Motion.
According to Newton's Second Law, F = ma, where F is force, m is mass and a is acceleration.
Since the truck has half its weight supported by its two drive wheels, we know that F = 0.5 mg (where m is mass and g is gravitational acceleration). Thus, when solving for a (acceleration), we have:
a = F/m
= 0.5mg/m
= 0.5g
= 4.9 [tex]m/s^{2}[/tex] ≈ 5 [tex]m/s^{2}[/tex]
This means that on wet concrete with coefficients of static friction and kinetic friction at 0.7 and 0.5 respectively, the maximum acceleration a truck with half its weight supported by its two driving wheels can achieve is 5 [tex]m/s^{2}[/tex].
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Two Small Spheres Spaced 20.0 Cm Apart Have Equal Charge. &Nbsp; How Many Excess Electrons Must Be Present On Each Sphere If The Magnitude Of The Force Of Repulsion Between Them Is 4.57 X 10-21 N? &Nbsp; Number Of Electrons = ?This problem has been solved!You'll get a detailed solution from a subject matter expert that helps you learn core concepts.See AnswerTwo small spheres spaced 20.0 cm apart have equal charge.How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57 x 10-21 N?Number of Electrons = ?
The number of excess electrons in each sphere is equal to 1.24 x 10^-19 C / 1.6 x 10^-19 C/electron = 0.78 x 10^19 electrons.
How to calculate the number of excess electrons?The number of extra electrons in each sphere can be calculated using Coulomb's law. Coulomb's law states that the repulsive force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
F = k * (q1 * q2) / d^2
where F is the repulsive force, q1 and q2 are the charges on the two spheres, d is the distance between the spheres, and k is the Coulomb constant (8.99 x 10^9 Nm^2/C^2).
Since the repulsive force is 4.57 x 10-21 N and the distance between the spheres is 20 cm = 0.2 m, we can rearrange the equations to find the charge on each sphere.
q1 = sqrt(F * d^2 / k)
q1 = square root (4.57 x 10-21 N * (0.2 m)^2 / (8.99 x 10^9 Nm^2/C^2)) = 1.24 x 10^-19 C
So the number of extra electrons in each sphere is equal to 1.24 x 10^-19 C / 1.6 x 10^-19 C/electron = 0.78 x 10^19 electrons.
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4 A block of mass 500g is pulled from rest on a horizontal
frictionless bench by a steady force Fand travels 8m in 2 s.
Find
I
a the acceleration,
b the value of F.
The acceleration and force of the block is 4 m/s² and 2 N respectively.
What do you mean by acceleration?Acceleration is a measure of the change in an object's velocity over time. It is defined as the rate at which an object changes its velocity, either in speed or direction. Mathematically, acceleration is calculated as the change in velocity (delta v) divided by the change in time (delta t): a = Δv / Δt. In other words, it is the rate of change of velocity. Acceleration has units of meters per second squared (m/s^2) in the International System of Units (SI).
The acceleration of the block can be calculated using the formula:
a = Δv / t, where Δv is the change in velocity and t is the time interval.
Since the block starts from rest, its initial velocity is 0 m/s, and its final velocity can be calculated as v = a × t = 8 m/s.
So, a = [tex]\frac{v}{t}[/tex] = [tex]\frac{8m/s}{2s}[/tex] = 4 m/s².
The value of F can be calculated using Newton's second law, which states that the net force on an object is equal to its mass times acceleration:
F = ma = 500 g × 4 m/s² = 2 N.
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The part of the reader that consists of a laser, optical filters, light-collecting optics, and beam-shaping devices that is designed to project and guide a precisely controlled laser beam back and forth across the plate as the plate moves through the scan area is the Group of answer choices
The correct option is (b) i.e. optical system. It is made up of a laser, optical filters, light-collecting optics, and beam-shaping components.
A scanner assembly, as it relates to digital imaging and document scanning, typically consists of a laser, optical filters, light-collecting optics, and beam-shaping devices. The laser is used to project a precise, concentrated beam of light onto the surface of the plate or document being scanned. Optical filters are used to control the wavelength and intensity of the light being used. Light-collecting optics, such as lenses, are used to focus the beam of light onto the plate or document. Finally, beam-shaping devices, such as mirrors or prisms, are used to guide the beam of light as it scans across the plate or document. The scanner assembly is designed to move the beam of light back and forth across the plate in a controlled manner, capturing an image of the plate or document as it does so.
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Question - The part of the reader that consists of a laser, optical filters, light-collecting optics, and beam-shaping devices that is designed to project and guide a precisely controlled laser beam back and forth across the plate as the plate moves through the scan area is the
a) photodetector
b) optical system
c) drive mechanism
d) photostimulable luminescence
A source produced 15 waves in 3 seconds , The distance between a crust is 15cm . Find frequency and wavelength
Frequency and wavelength is 100m/s.
Exactly how is frequency calculated?Pictures for Calculate the frequency and wavelength. The formula for frequency expressed in terms of time is given as follows: f = 1/T, where T is the duration of a cycle expressed in seconds and f is the frequency expressed in hertz. The frequency formula is stated as f = / where denotes the wave speed and the wavelength of the wave.
Explanation:
In 3 seconds, the source is generating 15 waves.
In a second, there are 15/3 waves, or 5 waves, produced.
Thus, the frequency is 5 Hz.
assuming that the wave length /2 = 10 cm
so that =20 cm
speed = frequency *wavelength v=5*20
v=100m/s.
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A physics demo launches a ball horizontally while dropping a second ball vertically at exactly the same time. Which ball hits the ground first?
Launched (horizontally) ball Dropped ball Both hit at the same time
In this physics demonstration, the ball that is dropped vertically and the one that is launched horizontally will both make contact with the ground at distinct times. The ball that falls vertically will first hit the ground.
This is because the force of gravity acts downward on both balls, causing them to accelerate toward the ground. However, the horizontally launched ball experiences the launch force as well, giving it an initial horizontal velocity. The ball's rate of descent to the ground is slowed by this horizontal velocity, resulting in a longer time spent on the ground.
On the other hand, the vertically dropped ball only experiences the force of gravity because it has no initial horizontal velocity. This indicates that it descends straight down at a steady rate and reaches the ground earlier than the horizontally launched ball.
What is speed?The rate at which an object's position changes over time is measured by its velocity. It has magnitude and direction because it is a vector quantity. Meters per second (m/s) is the International System of Units (SI) standard for measuring velocity.
Mathematically, velocity can be represented as:
v = x/t, where x is the object's change in position (or displacement), t is the change in time, and v is the velocity.
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a gardener mows a lawn with a push mower. the handle of the mower makes an angle of 32 degrees. if a 209 n force is applied along the handle of the 18 kg mower, what is the normal force exerted on the mower?
The normal force exerted on the mower is 290.8 N. The result is obtained by using the concept of Newton's first law.
What is Newton's first law?Newton's first law states that every object will remain at rest or keep constantly moving in a straight line unless an external force acts upon it.
The equation of Newton's first law can be expressed as
∑F = 0
A gardener mows a lawn with a push mower with
Angle, θ = 32 degreesForce, F = 209 NMass = 18 kgFind the normal force exerted on the mower?
We use g = 9.8 m/s².
See the illustration picture of forces directions in the attachment!
The direction of normal force is up. Using the Newton's first law, we will add parallel forces.
∑F = 0
Fn - F sin θ - W = 0
Fn = F sin θ + W
Fn = 209 sin 32° + 18 × 10
Fn = 209 (0.53) + 180
Fn ≈ 290.8 N
Hence, the mower will exert the normal force of 290.8 N.
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if a vehicle is of length 4.7 m on the average, what is the average spacing between vehicles at the above traffic density?
Average vehicle spacing at higher traffic density is 35.4 m for a vehicle with a length of 4.7 m.
Let's assume that Traffic flows at 96 km/h as per international data.
We need to find average vehicle spacing.
It indicates that a 96 km stretch of traffic gets cleared in one hour. In addition, 2400 cars may be supported in one lane for an hour. It implies that at least 96km must be taken up by 2400 automobiles. Congestion results from more vehicles moving at the same speed but occupying a smaller space as the number of automobiles increases.
Here, the key to the issue is that you must just consider length. 96 kilometres equal 96,000 metres at this time. If 2400 automobiles take up this much space [just lengthwise], 1 car will take up 96000/2400 m. it is 40 metres. Therefore, each automobile may use up 40m. But the automobile is 4.2 metres long.
It can thus also have a free space of 40 - 4.6 = 35.4 m.
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How do microorganisms work as decomposers in an ecosystem
Answer:
they eat what is decomposing speeding up the decomposition
A baseball pitcher can throw a 132 km/h (82 mph) curve ball that
rotates about 6. 0×102 rpm. What is the angular velocity of the
thrown ball? The pitcher's throwing motion lasts about 0. 15 s, and
the moment of inertia of the ball is 8. 0x10-5 kg-m?. What average
torque did the pitcher exert on the ball?
If someone can please solve this right now it’ll be a big help!!
According to its definition, the net force is the total of all the forces operating on an object.So, 108.46 N force is exerted on the ball.
What is the ball's angular velocity when it is thrown?According to its definition, the net force is the total of all the forces operating on an object.
Mass can be propelled forward by net force.In motion or at rest, another force exerts its influence on the body.When a system contains a sizable number of forces, the term "net force" is employed.
It is given that,
Initial velocity of the baseball, u = 0
The baseball's final speed was 97 mph, or 43.36 m/s.
Mass of baseball, m = 0.15 kg
Distance, s = 1.3 m
(a) Let a be the acceleration of the baseball. Use third equation of motion as :
a = 43.36 /2*1.3
a = 723.11 m/s
Force is given by :
F = m a
F = 0.15 * 723.11= 108.46 N
So, 108.46 N force is exerted on the ball.
(b) The pitcher would need to increase the force so as to alter the required force to reach the same speed if the mass of the ball is decreased.
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In the Soapbox Derby in, young participants build non-motorized cars with very low-friction wheels. Cars race by rolling down a hill. Assume that the track begins with a 51-ft-long (1 m= 3.28 ft) section tilted 14 ∘ below horizontal.
What is the maximum possible acceleration of a car moving down this stretch of track?
If a car starts from rest and undergoes this acceleration for the full l, what is its final speed in m/s ?
(a) The maximum possible acceleration of a car moving down this stretch of track is 2.38 m/s².
(b) The final speed of the car is 8.6 m/s.
What is the coefficient of friction on the incline?The coefficient of friction acting on the inclined surface is calculated by applying the following formula.
μ = tanθ
where;
θ is the angle of inclination of the inclineμ = tanθ
μ = tan (14)
μ = 0.25
The maximum possible acceleration of a car moving down this stretch of track is calculated as follows;
ma = μmg cosθ
a = μg cosθ
a = ( 0.25 x 9.8 m/s² ) x cos (14)
a = 2.38 m/s²
The final speed of the car when accelerating at maximum acceleration is calculated as follows;
s = distance travelled by the car = length of the incline
s = ( 1 m / 3.28 ft ) x ( 51 ft ) = 15.55 m
v² = u² + 2as
where;
u is the initial speed of the car = 0v is the final speedv² = 0 + 2as
v² = 2as
v = √ ( 2as )
v = √ ( 2 x 2.38 x 15.55 )
v = 8.6 m/s
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a cubical gaussian surface surrounds two charges q1 6.8 10 12 c and q2 2.0 10 12 c as the drawing shows what is the electric flux passing through the surface how do i get the answer as 0.54 n m 2 c a see answer
For the given statement the Electric flux through the surface is 0.54 N m 2 /C.
Note: we are considering the distance between two charges to be 0.1 m
Electric flux in electromagnetism is a measurement of the electric field passing through a certain surface. The SI unit for the scalar quantity known as electric flux is newton-meters squared per coulomb (N m 2 /C). N · m 2 /C )
The electric flux passing through a surface can be calculated using the equation:
Φ = E * A = (k * q1 + k * q2) / r^2 * A
k = Coulomb's constant = 8.99 x 10^9 N * m^2 / C^2
r = distance between the charges = 0.1 m
A = surface area = 0.1 m * 0.1 m = 0.01 m^2
Now if we calculate:
Φ = (k * q1 + k * q2) / r^2 * A = [(0.01 m^2)]* [(8.99 x 10^9 N * m^2 / C^2) *( 6.8 x 10^-12 C)] + [(8.99 x 10^9 N * m^2 / C^2) * (2 x 10^-12 C)] / (0.1 m)^2 = 0.54 N * m^2 / C.
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how many decibels are emitted from a jet plane with a sounf intensity of 8.5*10^2 watts per square meter
The jet plane emits 149 914 decibels at 8.2 .
What is the jet plane?
The Principal Differences. The key difference between a jet and an airplane is that to be a jet, a plane has to have a jet engine. Essentially, every jet is an airplane but not every airplane is a jet. A jet engine could either be a turbojet or turbofan, while a non-jet engine is usually a turboprop.A jet aircraft (or simply jet) is an aircraft (nearly always a fixed-wing aircraft) propelled by jet engines.While the typical VLJ and light jet carry a maximum of six passengers, a mid-size business jet offers comfortable seating for up to 10 passengers.A Jet aircraft flies much faster at higher altitudes, as high as 33,000–49,000 ft, than a propeller-powered aircraft. The Jet engine consists of an engine with a rotary air compressor powered by a turbine, with the leftover power providing thrust via a propelling nozzle.To learn more about jet refers to:
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the motor has a resistance of 9.7 ohms . when the motor switch is closed (meaning that branch of the circuit now has current flowing through it), what power does the bulb consume?
Lighting uses typically range from 2 to 100 watts (W), depending on size and kind.
A 60W bulb consumes how much electricity?A 60W light bulb was previously used as an example. By virtue of its 60W rating, it consumes 60 joules of energy every second it is turned on. This indicates that 216,600 joules of energy would be expended over the length of 3,600 seconds, or 60 minutes.Lighting uses typically range from 2 to 100 watts (W), depending on size and kind. LED bulbs only require 2 to 18 W, compared to the 25 to 100 W used by conventional incandescent lamps. 110 volts or less and typically less than 1 amp are drawn by light bulbs.Lighting uses typically range from 2 to 100 watts (W), depending on size and kind.To learn more about electricity refer to:
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Two point charges are placed on the x axis. (Figure 1)The first charge, q1 = 8. 00nC , is placed a distance 16. 0m from the origin along the positive x axis; the second charge, q2 = 6. 00nC , is placed a distance 9. 00m from the origin along the negative x axis. EAx,EAy = 0,0. 300 N/C An unknown additional charge q3 is now placed at point B, located at coordinates (0 m, 15. 0 m ). Find the magnitude and sign of q3 needed to make the total electric field at point A equal to zero
The magnitude of charge q3 is 3*10^4 N/C and it has a positive sign.
To find the magnitude and sign of the additional charge q3 needed to make the total electric field at point A equal to zero, we can use the principle of superposition of electric fields. This principle states that the total electric field at any point in space is the vector sum of the electric fields due to each individual charge.
We can start by calculating the electric field due to each of the two point charges q1 and q2 at point A, which is located at the origin. The electric field due to a point charge q at a distance r from it is given by the equation:
E = k*q/r^2
Where k is the Coulomb constant and q is the charge.
For q1, the electric field at point A is:
E1 = kq1/(16m)^2 = 910^9 * 810^-9 / 256 = 310^4 N/C
For q2, the electric field at point A is:
E2 = kq2/(9m)^2 = 910^9 * 610^-9 / 81 = 610^4 N/C
Since q2 is on the negative x-axis, the electric field is pointing in the negative x direction.
Now we can calculate the total electric field at point A by adding these two fields vectorially.
E = E1 + E2 = 310^4 N/C - 610^4 N/C = -3*10^4 N/C
The total electric field at point A is pointing in the negative x direction.
To make the total electric field equal to zero, we need to add an additional charge q3 at point B that has the same magnitude as the total electric field but pointing in the opposite direction. The magnitude of q3 is -3*10^4 N/C, and since it is pointing in the positive x-direction, the sign of q3 is positive.
So, the magnitude of q3 is 3*10^4 N/C and it has a positive sign.
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A 7. 00-kg box is attached to a 3. 00-kg box by rope 1. The 7. 00-kg box is pulled by rope 2 with a force of 25. 0 N. Determine the acceleration of the boxes and the tension in rope 1. The coefficient of friction between the ground and the boxes is 0. 120
Answer:
net force=applied force
Explanation:
...........................................
Two tiny conducting spheres are identical any carry charges of `-20.0 mu C` and `+50.0 muC`. They are separated by a distance of 2.50 cm. The spheres are brought into contact and then separated to a distance of 2.50 cm. Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive or repulsive.
A. 1.30 x 10^(2) N, attractive
B. 3.24 x 10^(3) N, attractive
C. 3.24 x 10^(3) N, repulsive
D. 1.44 x 10^(4) N, repulsive
The magnitude of the force that each sphere before contact and after contact are 1.438 × 10¹⁶ N and 3.5 × 10¹⁴ N respectively.
"Coulomb's law states that the force of attraction or repulsion between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them." It runs on the same track as the two charges, which are thought of as point charges.
Q₁ = -20 C
Q₂ = +50 C
Distance d = 2.5 cm equals 0.025 metres
F = (k Q₁ Q₂)/d² = (8.99 × 10⁹ × 20×50) /(0.025)² = 1.438×10¹⁶ N
Given that, the charges have opposing signs, the force is alluring.
The charge on each sphere after contact,
Q = Q₁ + Q₂/2 = -20+ 50/2 = -20 + 25 = 5 C
Magnitude of force = k Q²/d² = (8.99× 10⁹ × 5²)/(0.025)² = 224.75× 10⁹/(0.000625) = 359600× 10⁹ = 3.5 × 10¹⁴ N
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which diagram best represents the position of a ball , at equal time intervals as it falls freely
The diagram that represents the position of a ball, at equal time intervals, as it falls freely is diagram B.
What is free fall kinematics?The acceleration of an item in free fall is -9.8 m/s/s. (The minus symbol represents a downward acceleration.) For any freely falling item, the acceleration in the kinematic equations is -9.8 m/s/s, whether or not this is explicitly stated.
vf = g x t where g is the acceleration of gravity. The value for g on Earth is 9.8 m/s/s. This formula is used to determine the object's speed after being dropped from rest for any period of time.
Therefore, the correct option is B. diagram B.
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The question is incomplete. Your most probably complete question is given below: the options are attached below.
Regardless of which method of delivery a speaker might think is appropriate for a speaking event, the audience has expectations about how messages should be delivered as well. This is a situation that speakers need to carefully consider before selecting a delivery method for their speech.
-true
-false
Delivery method for their speech is True.
How does the context of the speaking event affect the audience's expectations for the delivery of a speech?The context of the speaking event can greatly affect the audience's expectations for the delivery of a speech. For example, in a formal setting such as a business conference, the audience may expect a more professional and formal delivery, with the use of slides or visual aids and a well-structured speech. In contrast, at a casual event such as a family gathering, the audience may expect a more relaxed and conversational delivery, with less emphasis on visual aids and a more informal structure. Additionally, the topic of the speech, the background of the audience and the purpose of the speech can also affect the audience's expectations. Therefore, a speaker should be aware of the context of the event, and adjust their delivery method accordingly.
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To determine the density of a piece of glass, a forensic technician places the glass in a liquid with a density of 2. 1 g/mL. The glass sinks. What should the technician do now?
a. Record that the density is greater than 2. 1 g/mL and stop the analysis.
b. Add a liquid with a density less than 2. 1 g/mL until the glass begins to float.
c. Add a liquid with a density greater than 2. 1 g/mL until the glass begins to float.
d. Record that the density is less than 2. 1 g/mL and stop the analysis
The technician can either note that the density is more than 2.1 g/mL and halt the analysis, or he can add more liquid until the glass starts to float.
The density of a material is an indicator of how heavy it is in relation to its size. An object will float when submerged in water if its density is lower than that of the water, while sinking if its density is higher. A differentiating characteristic that is unrelated to a substance's volume is its density. By suspending a glass sample in a liquid solution, one can calculate the density of an unknown glass sample. The density of the liquid is then determined, either directly or by comparison with a glass sample.
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a stream 1 km wide has a constant current of 4 km/h. at what angle to the shore should a person navigate a boat, which is maintaining a constant speed of 16 km/h, in order to reach a point directly opposite? (give your answer for acute angle in decimal degrees, rounded to one decimal place.)
angle to the shore should a person navigate a boat, which is maintaining a constant speed of 16 km/h, in order to reach a point directly opposite is 75.5 degrees.
Since the stream is flowing parallel alongside the shore, it takes the vector form of v {stream} = [0,4]
the boat velocity is perpendicular to the stream, it's takes the vector form of v{boat} = [-16,0]
The resultant velocity of the boat is the sum of the velocity of the boat and stream, thus it takes the form
v{total} = [-16,0] + [0,4]
= [ -16,4]
the cosine of angle to the shore,
cos {theta} = v{total} . v {stream} / ||v{total} || x ||v {stream}||
= [ -16,4] . [0,4] / (16x4)
= 16/64
= 0.25
angle = 75.7 degrees.
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A pipe discharges storm water into a creek. Water flows horizontally out of the pipe at 1.9 m/s, and the end of the pipe is 3.1 m above the creek.
How far out from the end of the pipe is the point where the stream of water meets the creek?
The distance from the end of the pipe is the point where the stream of water meets the creek is 1.52 m.
What is the time taken for the water to flow?
The time taken for the water to fall from the given height is calculated as follows;
t = √ ( 2h / g )
where;
g is acceleration due to gravityh is the height of fallt = √ ( 2 x 3.1 / 9.8 )
t = 0.8 s
The horizontal distance travelled by the stream of water is calculated as;
x = vt
x = 1.9 m/s x 0.8 s
x = 1.52 m
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a 700 kg car moving at 20 m/s collides with a stationary truck with mass 1400 kg. the two vehicles interlock as a result of the collision. what is the speed of the car/truck combination? what type of collision is this (elastic, inelastic, or explosion)?
The speed of the car truck combination is calculated to be 0.667 m/s and it is a type of inelastic collision.
Mass of the car m₁ = 700 kg
Velocity of the car u₁ = 20 m/s
Mass of the stationary truck m₂ = 1400 kg
Velocity of the stationary truck u₂ = 0
The momentum before collision is, m₁ u₁ = 700× 20 = 1400 kg m/s
As, we know from the principle of conservation of linear momentum,
momentum before the collision of trucks is equal to momentum after the collision of trucks.
The momentum after the collision is (m₁ + m₂) vf = 1400 kg m/s
(700 + 1400) vf = 1400
2100 vf = 1400
vf = 0.667 m/s
Now, let us compare kinetic energies before and after collision.
Kinetic energy before collision = 1/2 m₁ u₁² = 1/2 × 700× 20² = 14×10⁴ kg m/s
Kinetic energy after collision =1/2 (m₁+m₂) vf₂ = 1/2(700 + 1400) 0.667² = 467.13 kg m/s
As kinetic energy is not conserved, the type of collision is said to be an inelastic collision.
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oak is 0.8 as dense are water and therefore floats in water. what weight of water will be displaced by a 50 kg floating oak beam?
613.25 N is the weight of water will be displaced by a 50 kg floating oak beam, if oak is 0.8 times dense of water.
Density of water, = 1000 kg/m³
Density of oak, = 0.8 × 1000 = 800 kg/m³
Floating of an object is possible if the upward force exerted by the water is equal to the weight of the object. And this upward force is equal to the weight of water displaced by the floating object.
Volume of the water displaced by the 50 kg oak, V = 50/800 = 0.0625 m³
Mass of the displaced water = 0.0625 × 1000 = 62.5 kg
Weight of the displaced water = 62.5 × 9.81 = 613.13 N.
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how much charge flows from each terminal of a 12.0 v battery when it is connected to a 7 microf capacitor?
The amount of the charge flowing through the terminals of the battery is found to be 0.084C.
The voltage of the battery is given to be 12 V and the capacitance of the capacitor is given to be 7 micro F.
As per the standard relations, the amount of the charge flowing through the terminal of the battery will be given by,
Q = CV
Where,
Q is the charge through the battery,
C is the capacitance of the capacitor,
V is the voltage of the battery.
Putting all the values,
Q = 12 x 7/1000
Q = 0.084 C.
So, the charge flowing is 0.084 C.
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A boat has a speed of 2 m/s. It accelerates for half a minute to reach a speed of 8 m/s. Work out the acceleration of the boat.
The accelaration of the boat, given the speed of the boat and the time taken to reach the speed, is 0.2 m / s².
How to find the accelaration ?Acceleration can be found by dividing the change in the speed of an object by the time it takes.
The change in speed is:
= 8 m / s - 2 m / s
= 6 m /s
The change in time is:
= 1 / 2 x 60 seconds per minute
= 30 seconds
The acceleration is:
= 6 / 30
= 0.2 m / s²
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Answer:
0.2
Explanation:
Acceleration = change in velocity / time = (8 m/s – 2 m/s) / 30s = 0.2 m/s2
a meter stick is supported at each end by a spring scale. a heavy mass is then hung on the meter stick so that the spring scale on the left hand side reads four times the value of the spring scale on the right hand side. if the mass of the meter stick is negligible compared to the hanging mass, how far from the right hand side is the large mass hanging?
The mass is dangling 4/5 of the length of the metre stick away from the right side if the mass of the metre stick is insignificant in comparison to the hanging mass.
The turning force on an item is measured by torque, which is computed by dividing the force by the separation from the pivot point.
Let's designate the forces on the left and right spring scales as F1 and F2, respectively.
F1 = 4F2, as we are aware.
The left side's torque is F1 * L, while the right side's torque is F2 * L (L is the distance from the right side).
Accordingly, F1 * L = F2 * L 4 F2 * L = F2 * L, the total torque on the left side must match the total torque on the right side.
It is evident that L = 4L/5.
In other words, the mass is dangling 4/5 of the way down the metre stick from the right side.
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a car can sustain a maximum acceleration of 1 m/s/s, a maximum deceleration of 2 m/s/s, and a top speed of 20 m/s. determine the minimum amount of time it will take for this car to travel one kilometer if it starts from rest and finishes at rest, i.e., stops at the one kilometer mark g
If it starts from rest can travel 1000m, in a straight path, and again it has to come to rest at the end of 1000m, then minimum 70s will be taken.
According to the question, the car starts off at a stop, travels up to a top speed of 20 m/s, and then comes to a stop. This implies that the vehicle must have first reached its maximum speed before braking to a stop.
Let, [tex]T_1[/tex] and [tex]T_2[/tex] be the time for acceleration (attaining maximum velocity) and deceleration (come to the rest) respectively.
Then,
[tex]T_1[/tex] = (Maximum velocity - initial velocity) / acceleration
⇒ [tex]T_1=\frac{20-0}{1}=20sec[/tex]
Then, distance travelled in time [tex]T_1[/tex],
⇒ [tex]S_1=UT_1+\frac{1}{2}aT_1^2[/tex]
⇒ [tex]S_1=0+\frac{1}{2}*1*20^{2}[/tex]
⇒ [tex]S_1= 200m[/tex]
Similarly,
⇒ [tex]T_2=\frac{0-20}{-1}=20sec[/tex]
⇒ [tex]S_2=UT_2+\frac{1}{2}aT_ 2^2[/tex]
⇒ [tex]S_2=(20*20)-(\frac{1}{2}*1*20^2)[/tex]
⇒ [tex]S_2=400-200=200m[/tex]
∴ Distance travelled at uniform velocity is
⇒ [tex]S = 1000-(S_1+S_2)[/tex]
⇒ [tex]S=1000-(200+200)=1000-400=600m[/tex]
This 600m is travelled at a speed of 20m/s.
and time of travel for constant velocity is
⇒ [tex]T_3=\frac{S}{V}[/tex]
⇒ [tex]T_3=\frac{600}{20}=30sec[/tex]
Hence, total time for travel
⇒ [tex]T=T_1+T_2+T_3=(20+20+30)sec=70sec[/tex]
On a V-T graph, it can be displayed here(See picture).
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A fox fleeing from a hunter encounters a 0.745 m tall fence and attempts to jump it. The fox jumps with an initial velocity of 7.50 m/s at an angle of 45.0°, beginning the jump 1.99 m from the fence. By how much does the fox clear the fence? Treat the fox as a particle.
The fox which is fleeing from a hunter encounters a 0.745 m tall fence and this fox can clear the fence with a height of up to 0.895 meters.
What is Velocity?Velocity can be defined as the rate of change in the displacement of an object. It is a vector quantity as it has both the magnitude and direction. The change in velocity of an object is known as acceleration.
Given,
Initial velocity (u) = 7.50 m/s
Launch angle, θ = 45°
Height of fence = 0.745 m
Trajectory of Projectile is given by:
y = x tanθ - gx²/ 2u²cos²θ
For x = 1.99 m
y = 1.99 × tan 45° - 9.8 × (1.99)²/ 2 × (7.5)² × (cos45°)²
y = 1.99 × 1 - 9.8 × 3.96/ 2 × 56.25 × (1)²
y = 1.99 - 38.80/112.5
y = 1.99 - 0.345
y = 1.64m
At x = 1.99 m, the value of y = 1.64 m
Therefore, fox clear fence by a margin = 1.64 - 0.745 = 0.895 meters.
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