Each of the images and properties represents or describes the a form.
A-DNA is thought to be one of the three biologically active double helix structures, along with B-DNA and Z-DNA.
This is a right-handed double helix that is very similar to the more common B-DNA form, but is shorter and has a base pair that is not perpendicular to the helix axis as in B-DNA. It has a more compact helical structure.
Form is the shape, an appearance, or composition of an object. In a broader sense, form is the way something happens. Forms also refer to forms (documents), documents (printed or electronic) that contain fields for writing or entering data. A form (educational), class, group, or group of students.
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3. A Tylenol has 80 mg of acetaminophen.
How many grams is that
\(\boxed{\sf 1g=1000mg}\)
\(\\ \sf\longmapsto 80g\)
\(\\ \sf\longmapsto \dfrac{80}{1000}\)
\(\\ \sf\longmapsto 0.08g\)
Three solid plastic cylinders all have radius 2.50cm and length 6.00cm. Find the charge of each cylinder given the following additional information about each one. Cylinder (b) carries charge with uniform density 15.0 nC/m² on its curved lateral surface only. Cylinder
By the charge density, the curved surface area of the cylinder charge is 0.141 nC.
We need to know about charge density to solve this problem. The charge density can be determined as
λ = Q / A
where λ is charge density, Q is charge and A is surface area.
The parameter given is the charge density and the solid cylinder shape which are :
λ = 15 nC/m²
r = 2.5 cm = 0.025 m
L = 6.0 cm = 0.06 m
Find the curved surface area of cylinder
A = 2πrL
A = 2π. 0.025. 0.06
A = 9.43 x 10¯³ m²
Find the charges
Q = λ x A
Q = 15 x 9.43 x 10¯³
Q = 0.141 nC
Hence, the curved surface area of the cylinder charge is 0.141 nC.
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Describe the three methods of thermal energy transfer and provide a sketch that depicts each:
Conduction:
Convection:
Radiation:
Answer:
a)the process by which heat or electricity is directly transmitted through the material of a substance when there is a difference of temperature
b)the movement caused within a fluid by the tendency of hotter and therefore less dense material to rise, and colder, denser material to sink under the influence of gravity, which consequently results in transfer of heat
c)radiation is the emission or transmission of energy in the form of waves or particles through space or through a material medium.
The specific heat of a solid Y is 11.5 cal/g°C. A sample of this
solid at 135 K is heated to 260 K. The solid absorbs 7.90 kcals.
What is the sample of solid in grams?
g
The sample of the solid in grams is 5.5g
HOW TO CALCULATE SPECIFIC HEAT CAPACITY:
The quantity of heat absorbed or released by a substance can be calculated using the following formula:Q = m × c × ∆T
Where;
Q = quantity of heat absorbed/released (cal)
m = mass of the substance (g)
c = specific heat (cal/g°C)
∆T = change in temperature (°C)
According to the information provided in this question:Q = 7.90Kcal = 7900cal
m = ?
c = 11.5 cal/g°C
T1 = 135K = 135K − 273.15 = -138.1°C
T2 = 260K = 260K − 273.15 = -13.15°C
∆T = -13.15 - (-138.1) = 124.95°C
Hence, the mass of the substance can be calculated as follows:m = Q ÷ c∆T
m = 7900 ÷ (11.5 × 124.95)
m = 7900 ÷ 1436.93
m = 5.5grams.
Therefore, the sample of the solid in grams is 5.5g.
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Alex rides a merry go round for 10 rotations. In each rotation she travels a distance of 15 meters. The ride lasts for 5 minutes. Alex then exits the ride at the same spot she boarded it. What is the magnitude of Alex's average velocity for the 5 minutes she spent on the ride
0 m/min
30 m/min
15 m/min
150 m/min
andy runs 40 meters east in 5 seconds and then 100 meters west in 15 seconds his average velocity is?
Answer:
\(\Large \boxed{3}\)
Explanation:
Info given;
∵ 40 meters east in 5 seconds
∵ 100 meters west in 15 seconds
=> \((100 -40)\)
=> Divided by;
=> \((5 + 15)\)
=> \(3 m/s\)
Answer; 3 m/s
(Average velocity is; 3 m/s )
A runner starts at point A, runs around a 1-mile track and finishes the run back at point A. Which of the following statements is true?
This question is incomplete because the options are missing; here is the complete question:
A runner starts at point A, runs around a 1-mile track, and finishes the run back at point A. Which of the following statements is true?
A. The runner's displacement is 1 mile.
B. The runner's displacement is zero.
C. The distance the runner covered is zero.
D. The runner's speed was zero.
The answer to this question is B. The runner's displacement is zero
Explanation:
Displacement always implies a change of position; this means an object or individual moves from point A to point B, and therefore the original position is different from the final position. Additionally, in displacement, other related factors such as the total distance the body moved and the direction of movement. In the case presented, it can be concluded there was no displacement or the displacement is zero because even when the runner moved and ran two miles, he returned to the initial position, and without a change in the position, there is no displacement.
g say you tie a thick, dense rope to a thin, less dense string. if you send a single wave pulse through the string, will it be transmitted through to the rope/and or reflected back into the string? also, which orientation will any transmitted or reflected pulses have relative to the original pulse?
If you send a single wave pulse through the string, it will be transmitted through to the rope.
Let's consider a thin rope attached to a thick rope, with every rope held at opposite ends via people. And suppose that a pulse is introduced through the person conserving the top of the skinny rope. If that is the case, there will be an incident pulse visiting within the less dense medium toward the boundary with a more dense medium.
A part of the strength carried through the incident pulse is transmitted into the thick rope. The disturbance that keeps shifting to the right is called the transmitted pulse. A part of the energy carried via the incident pulse is reflected and returns toward the left giving up the thin rope. The disturbance that returns to the left after bouncing off the boundary is called the reflected pulse.
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A construction worker is carrying 40kg if he travels a distance of 50 meters how much work is being done
Answer:
Workdone = 19600 Nm
Explanation:
Given the following data;
Mass = 40 kg
Distance = 50 meters
We know that acceleration due to gravity is equal to 9.8 m/s².
To find the work done;
First of all, we would determine the force being exerted by the construction worker.
Force = mass * acceleration due to gravity
Force = 40 * 9.8
Force = 392 Newton
Next, we would determine the work done;
Workdone is given by the formula;
\( Workdone = force * distance\)
\( Workdone = 392 * 50 \)
Workdone = 19600 Nm
clarisee asks a couple of questions of montag that unsettle him what are they
Clarisse asked a couple of questions of Montag that unsettle him, some of these questions are:
Are you happyWhat do you think happens after we die?Have you ever stolen a book instead of burning it?Have you heard the rumor that firemen once put out fires instead of starting them?Who is Clarisse?In Ray Bradburys renowned dystopian tale Fahrenheit 451 lies the character of Clarisse McClellan - an intriguing young woman who lives adjacent to protagonist Guy Montag. A free spirited thinker at just seventeen years old she embodies an independence and curiosity that sets her apart from those around her.
Through dialogue with Montag - including cleverly crafted questions - Clarisse subtly plants seeds of doubt in his mind about their controlled society ultimately leading him down a path towards rebellion against oppressive government forces.
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a particle moving along the x axis has a position given by where 22 m/s, 3.8 m/s3 and is measured in seconds. what is the magnitude of the acceleration of the particle at the instant when its velocity is zero? please give your answer in units of m/s2.
Answer:
Therefore, the magnitude of the acceleration of the particle at the instant when its velocity is zero is approximately 18.258 m/s².
Explanation:
To find the magnitude of acceleration at the instant when velocity is zero, we need to differentiate the given velocity equation with respect to time (t) to obtain the acceleration equation.
Given:
Velocity equation: v(t) = 22 - 3.8t^2
Differentiating the velocity equation with respect to time, we get:
a(t) = d(v(t))/dt = -2 * 3.8t
To find the magnitude of acceleration at the instant when velocity is zero, we need to solve for t when v(t) = 0.
0 = 22 - 3.8t^2
3.8t^2 = 22
t^2 = 22/3.8
t^2 ≈ 5.789
Taking the square root of both sides, we find:
t ≈ √(5.789)
t ≈ 2.403
Now we can substitute this value of t into the acceleration equation to find the magnitude of acceleration at that instant:
a(t) = -2 * 3.8 * 2.403
a(t) ≈ -18.258
Therefore, the magnitude of the acceleration of the particle at the instant when its velocity is zero is approximately 18.258 m/s².
if an object's rotational kinetic energy is 64 j and it rotates with an angular speed of 9 radians/s, what is its moment of inertia?
\(\\ \sf\longmapsto K.E=\dfrac{1}{2}I\omega^2\)
\(\\ \sf\longmapsto 64=\dfrac{1}{2}I(9)^2\)
\(\\ \sf\longmapsto 81I=128\)
\(\\ \sf\longmapsto I=\dfrac{128}{81}\)
\(\\ \sf\longmapsto I=1.5kgm^2\)
For an investigation, a student measures the speed of a cart as it rolls down a ramp. The student then records data in the table shown above. Which of these best explains the student’s data?
A.
The speed of the cart decreases as the cart rolls down the ramp because of friction between the cart and the ramp.
B.
The speed of the cart increases as the cart rolls down the ramp because the force acting on the cart is greater than the force of gravity.
C.
The speed of the cart increases as the cart rolls down the ramp because the forces acting on the cart are unbalanced.
D.
The speed of the cart decreases as the cart rolls down the ramp because the forces acting on the cart are balanced.
Answer: it’s c
Explanation:I put that as my answer for an assignment and got it right
The correct option is the speed of the cart increases as the cart rolls down the ramp because the forces acting on the cart are unbalanced.
What is force?The push or pull on a mass-containing item changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
Given that in the question there is a child who is doing experiment for an investigation, a student measures the speed of a cart as it rolls down a ramp. The student then records data in the table given in question we see that speed increases as cart goes down.
The speed of the cart increases as the cart rolls down the ramp because the forces acting on the cart are unbalanced.
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An average 1.50 V Alkaline AA battery has 9,360 J of stored energy. If the battery is placed in a toy that requires 2.00 amperes of current to operate, how many minutes will the battery last before it runs out of energy?
Answer:
52 minutes
Explanation:
Given the following data;
Voltage = 1.5 V
Current = 2 Amperes
Energy = 9360 J
To find how many minutes will the battery last before it runs out of energy;
First of all, we would determine the power rating of the toy
Power = current * time
Power = 2 * 1.5
Power = 3 Watts
Next, we find the time;
Energy = power * time
9360 = 3 * time
Time = 9360/3
Time = 3120 seconds
To convert to minutes;
60 seconds = 1 minute
3120 seconds = 3120/60 = 52 minutes
A 0.230-kg baseball is thrown with a speed of 41 m/s.What is the ball's momentum?
The momentum of the 0.230-kg baseball thrown with a speed of 41 m/s is 9.53 kg m/s. This means that if the ball were to collide with an object with equal and opposite momentum, the combined momentum of the two objects would be zero.
Momentum is a measure of an object's motion and is calculated by multiplying its mass and velocity. In physics, it is an important concept as it is conserved in many cases and helps to predict how objects will interact with each other.
The momentum of an object can be calculated using the formula: p = mv, where p is momentum, m is the mass of the object, and v is its velocity. In this case, the mass of the baseball is 0.230 kg, and its velocity is 41 m/s. Therefore, the momentum of the baseball can be calculated as follows:
p = mv = 0.230 kg * 41 m/s = 9.53 kg m/s.
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What are the coordinates of M' after triangle MNO is reflected over the y-axis, and rotated 90 degrees clockwise with the origin as the center of rotation?(0,-6)(0,6)(6,0)(-6,0)
what is the standard unit of velocity
Answer:
Explanation:
Meters per second
Answer:
m/s (meters per second)
Explanation:
A body of mass 3 kg and volume 4 x 10^m' is hung from a balance
graduated in Newton. What is the reading of the balance when the
body is
a. In air;
b. Fully immersed in water
immersed in water.
Answer:
it's b I guess not sure
Explanation:
The reading of the balance for a body with mass 3kg and volume 4 × 10⁻⁴ in air and when fully immersed in water are 29.4N and 25.5N, respectively.
What is weight?
The weight of an object is the force which is acting on the object due to the influence of gravity. Weight is a vector quantity because it has both the magnitude and direction. The gravitational force acts on the object.
The reading of the balance when the body is in air, W = mg = (3) × (9.8) = 29.4 N
The reading of the balance when the body is fully immersed in water is:
F = mg − ρgV = (3) × (9.8) − (1000) × (9.8) × (0.0004) = 25.5 N
Therefore, the reading of the balance in the given conditions are 29.4N and 25.5N.
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Helppp please pleaseeee fr
how many movements are there in a classical sonata?
In a classical sonata, there are typically four movements. Each movement serves a distinct purpose and has its own musical structure, character, and tempo.
These movements are arranged in a specific order and contribute to the overall narrative and emotional journey of the piece.The first movement of a classical sonata is usually in sonata-allegro form.
characterized by contrasting themes, development, and a recapitulation. It sets the stage for the entire composition and establishes the main musical ideas.
The second movement is often a slower and more lyrical piece, such as an adagio or andante. It provides a contrast to the energetic first movement, allowing for introspection and emotional depth.
The third movement is usually a minuet and trio or scherzo and trio. It serves as a lively and rhythmic interlude, often with a dance-like quality, adding a touch of lightness and playfulness to the sonata.
The final movement, known as the fourth movement, is typically a fast and spirited piece, such as an allegro or presto. It showcases virtuosity, excitement, and brings the sonata to a thrilling conclusion.
Overall, these four movements in a classical sonata provide a diverse and well-balanced musical experience, showcasing the range of emotions, technical skill, and compositional craftsmanship of the era.
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a 0.350 kg lump of clay is dropped from a height of 1.15 m onto the floor. it sticks to the floor and does not bounce.
When a 0.350 kg lump of clay is dropped from a height of 1.15 m onto the floor and sticks to the floor without bouncing, we can analyze the situation using the principle of conservation of mechanical energy.
Initially, the clay has gravitational potential energy due to its height above the ground. As it falls, this potential energy is converted into kinetic energy. When the clay hits the floor and sticks to it, its kinetic energy is completely absorbed and converted into other forms, such as thermal energy and sound energy.
The clay's potential energy is given by the equation PE = mgh, where m is the mass of the clay (0.350 kg), g is the acceleration (9.8 m/s²), and h is the height (1.15 m).
Thus, the initial potential energy is PE = 0.350 kg × 9.8 m/s² × 1.15 m.
Since the clay sticks to the floor and does not bounce, it comes to a complete stop and does not have any kinetic energy remaining. Therefore, the total mechanical energy of the system is conserved. In this case, the mechanical energy is converted into other forms of energy upon impact with the floor.
When a 0.350 kg lump of clay is dropped from a height of 1.15 m onto the floor and sticks to the floor without bouncing, we can analyze the situation using the principle of conservation of mechanical energy. Initially, the clay has gravitational potential energy due to its height above the ground. As it falls, this potential energy is converted into kinetic energy.
When the clay hits the floor and sticks to it, its kinetic energy is completely absorbed and converted into other forms, such as thermal energy and sound energy. The clay's potential energy is given by the equation PE = mgh, where m is the mass of the clay (0.350 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height (1.15 m).
Thus, the initial potential energy is PE = 0.350 kg × 9.8 m/s² × 1.15 m
= 3.9435 J.
Since the clay sticks to the floor and does not bounce, it comes to a complete stop and does not have any kinetic energy remaining. Therefore, the total mechanical energy of the system is conserved. In this case, the mechanical energy is converted into other forms of energy upon impact with the floor. The clay's potential energy is converted into thermal energy and sound energy due to the deformation and sticking of the clay.
This conversion of energy is due to the work done by the forces involved in the collision. The clay's kinetic energy is completely absorbed, resulting in its conversion into these other forms of energy.
When a 0.350 kg lump of clay is dropped from a height of 1.15 m and sticks to the floor without bouncing, its gravitational potential energy is converted into kinetic energy as it falls. Upon impact with the floor, the clay's kinetic energy is completely absorbed and converted into thermal energy and sound energy.
This conversion of energy is a result of the work done by the forces involved in the collision. The principle of conservation of mechanical energy states that the total mechanical energy of a system remains constant as long as no external forces act on it.
In this case, the mechanical energy of the clay-floor system is conserved, and the initial potential energy of the clay is converted into other forms of energy upon impact.
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PLEASE HELP!! Points will be given for suitable answer! Thank you!
Answer:
its lost 91
Explanation:
if your face is 25.0 cm away from the ball's front surface, where is your image? follow the sign conventions
Your image is located 25.0 cm behind the ball's front surface, following the sign conventions.
When dealing with sign conventions in optics, positive distances are measured in the direction of the light propagation, and negative distances are measured opposite to it. In this case, your face is 25.0 cm away from the ball's front surface, which is considered a positive distance.
Since the ball acts like a mirror, your image will appear at the same distance but in the opposite direction, making it a negative distance. Therefore, your image is located 25.0 cm behind the ball's front surface, following the sign conventions. This ensures that your image and face are equidistant from the ball's front surface, maintaining a symmetrical relationship in the optical setup.
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A 19.7 kg sled is pulled with a 42.0 N force at a 43.0° angle, across ground where μ₁ = 0.130.
What is the normal force on the sled?
The following information is provided in the problem: A sled with a weight of 19.7 kg is pulled with a force of 42.0 N at an angle of 43.0° across ground where μ₁ = 0.130. We need to find out the normal force that is exerted on the sled.
Let us examine each of the forces acting on the sled.The weight of the sled is equal to its mass multiplied by the acceleration due to gravity. Therefore, the weight of the sled is:mg = 19.7 kg x 9.8 m/s² = 193.06 N.The force exerted on the sled can be divided into two components: one that is parallel to the ground and one that is perpendicular to the ground.The force parallel to the ground is:F₁ = 42.0 N x cos(43.0°) = 30.56 N.The force perpendicular to the ground is:F₂ = 42.0 N x sin(43.0°) = 28.30 N.The frictional force is equal to the coefficient of friction multiplied by the normal force. Therefore, we need to find the normal force on the sled in order to calculate the frictional force. Since the sled is not accelerating vertically, the normal force is equal to the weight of the sled plus the force perpendicular to the ground. Therefore, N = mg + F₂N = 193.06 N + 28.30 N = 221.36 N.The frictional force is:Fr = μ₁ x NFr = 0.130 x 221.36 N = 28.77 N.Thus, the normal force exerted on the sled is 221.36 N.For such more question on perpendicular
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Describe how a change in resistance would affect the current in a circuit.
Explanation:
The relation between potential difference, current and resistance flowing in a circuit is given by using Ohm's law. It can be given by :
V = IR
\(I=\dfrac{V}{R}\)
Resistance opposes the flow of electric current in the circuit. It means that, if resistance is more, less current will flow through the circuit.
if the current density in a wire is given by j=alpha*r,where alpha is a constant and r is the distance from the center of the wire, 0
Given:
The current density is,
\(J=ar\)a is a constant, and the radius of the wire is R.
To find:
The current in the wire
Explanation:
The current in the wire is,
\(\begin{gathered} I=\int JdA \\ =\int ardA \end{gathered}\)We know,
\(\begin{gathered} A=\pi r^2 \\ dA=2\pi rdr \end{gathered}\)So,
\(\begin{gathered} I=\int_0^Rar\times2\pi rdr \\ =2\pi a\int_0^Rr^2dr \\ =2\pi a\times\frac{R^3}{3} \end{gathered}\)Hence, the required current is,
\(\frac{2\pi aR^3}{3}\)the drawing shows an exaggerated view of a rifle that has been"sighted in" for a 91.4-meter target. If the muzzle speed of thebullet is v0 = 427 m/s, what are the two possible anglesθ1 and θ2 between the rifle barreland the horizontal such that the bullet will hit the target? One ofthese angles is so large that it is never used in target shooting.(HInt: the following trigonometric identity may be useful: 2 sinθ cos θ = sin 2 θ.)
The two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.
In the given case, the figure shows an exaggerated view of a rifle that has been sighted in for a 91.4-meter target. Let the muzzle speed of the bullet be v0 = 427 m/s.
Now, we are required to find the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target.
It is known that the horizontal displacement of the bullet from the gun can be given by the equation: x = v0 t cosθ ..........(i)and the vertical displacement of the bullet from the gun can be given by the equation: y = v0 t sinθ - (1/2) g t^2..........(ii).
Here, t is the time of flight of the bullet and g is the acceleration due to gravity.
As the bullet hits the target, its final vertical displacement from the gun is equal to the height of the target, i.e.,y = 91.4m.Now, we can substitute equations (i) and (ii) in place of t and y in equation (ii) to get:x tanθ - (g/2v0^2) x^2 sec^2θ = 91.4 ..........(iii)This is a quadratic equation in tanθ.
On solving this equation using the quadratic formula, we get:tanθ = [-b ± √(b^2 - 4ac)]/2aWhere,a = -gx^2/(2v0^2) = -4.9x^2/v0^2, b = x, and c = -91.4.
Rearranging the terms, we get:2a tanθ^2 + b tanθ - 91.4 = 0On substituting the given values, we get:2(-4.9x^2/v0^2) tanθ^2 + x tanθ - 91.4 = 0θ1 and θ2 are the two possible angles which can be found by solving the above quadratic equation.
Using the trigonometric identity given in the hint, we can write: sin 2θ = 2 sinθ cos θ = 2 tanθ/ (1 + tan^2θ)Now, we can substitute tanθ = (-b ± √(b^2 - 4ac))/2a in the above equation to get: sin 2θ = (-4bx ± 2x√(b^2 - 4ac))/(b^2 + 4a^2)Now, we can substitute the given values to get: sin 2θ1 = -0.999sin 2θ2 = 0.998.
Thus, we get two values of sin 2θ, one is close to -1 and the other is close to 1. As sin 2θ = -1 when 2θ = -π/2 + nπ and sin 2θ = 1 when 2θ = π/2 + nπ, where n is an integer, we get two possible values of θ for each of these two cases.
Hence, the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.
As one of these angles is so large that it is never used in target shooting, we only need to consider the other angle.
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The two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.
In the given case, the figure shows an exaggerated view of a rifle that has been sighted in for a 91.4-meter target. Let the muzzle speed of the bullet be v0 = 427 m/s.
Now, we are required to find the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target.
It is known that the horizontal displacement of the bullet from the gun can be given by the equation: x = v0 t cosθ ..........(i)and the vertical displacement of the bullet from the gun can be given by the equation: y = v0 t sinθ - (1/2) g t^2..........(ii).
Here, t is the time of flight of the bullet and g is the acceleration due to gravity.
As the bullet hits the target, its final vertical displacement from the gun is equal to the height of the target, i.e.,y = 91.4m.Now, we can substitute equations (i) and (ii) in place of t and y in equation (ii) to get:x tanθ - (g/2v0^2) x^2 sec^2θ = 91.4 ..........(iii)This is a quadratic equation in tanθ.
On solving this equation using the quadratic formula, we get:tanθ = [-b ± √(b^2 - 4ac)]/2aWhere,a = -gx^2/(2v0^2) = -4.9x^2/v0^2, b = x, and c = -91.4.
Rearranging the terms, we get:2a tanθ^2 + b tanθ - 91.4 = 0On substituting the given values, we get:2(-4.9x^2/v0^2) tanθ^2 + x tanθ - 91.4 = 0θ1 and θ2 are the two possible angles which can be found by solving the above quadratic equation.
Using the trigonometric identity given in the hint, we can write: sin 2θ = 2 sinθ cos θ = 2 tanθ/ (1 + tan^2θ)Now, we can substitute tanθ = (-b ± √(b^2 - 4ac))/2a in the above equation to get: sin 2θ = (-4bx ± 2x√(b^2 - 4ac))/(b^2 + 4a^2)Now, we can substitute the given values to get: sin 2θ1 = -0.999sin 2θ2 = 0.998.
Thus, we get two values of sin 2θ, one is close to -1 and the other is close to 1. As sin 2θ = -1 when 2θ = -π/2 + nπ and sin 2θ = 1 when 2θ = π/2 + nπ, where n is an integer, we get two possible values of θ for each of these two cases.
Hence, the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.
As one of these angles is so large that it is never used in target shooting, we only need to consider the other angle.
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Three charges are on a line. A positive charge is on the far left labeled q Subscript 1 baseline positive 6 Coulombs. The second charge is 2 m away and is labeled q Subscript 2 baseline negative 4 Coulombs. The third charge is 1 m away and is labeled q Subscript 3 baseline = positive 3 Coulombs.
What is the electrical force between q2 and q3?
Recall that k = 8.99 × 109 N•meters squared over Coulombs squared..
1.0 × 1011 N
–1.1 × 1011 N
–1.6 × 1011 N
1.8 × 1011 N
Answer:
B) –1.1 × 1011 N
Explanation:
Since the second charge is 2 m away and is labeled q₂ = - 4 Coulombs. The third charge is 1 m away and is labeled q₃ = + 3 Coulombs, the electrical force between q₂ and q₃ is -1.1 × 10¹¹ N
To answer the question, we need to find the electrical force of attraction between two charges and this is given by Coulomb's law
Coulomb's lawThis states that the electrical force of attraction between two charges, F is directly proportional to the product of the charges q and Q and inversely proportional to the square of their distance apart, d.
So, mathematically F = kqQ/d²
Now, given that there are three charges are on a line. A positive charge is on the far left labeled q₁ = + 6 Coulombs. The second charge is 2 m away and is labeled q₂ = - 4 Coulombs. The third charge is 1 m away and is labeled q₃ = + 3 Coulombs.
The electrical force of attraction between q₂ and q₃So, the electrical force of attraction between q₂ and q₃ is F = kq₂q₃/d² where
k = 8.99 × 10⁹ Nm²/C², q₂ = - 4 C, q₃ = + 3 C and d = 1 m (since q₂ and q₃ are 1 m apart)Substituting the values of the variables into the equation, we have
F = kq₂q₃/d²
F = 8.99 × 10⁹ Nm²/C² × (-4 C) × (+ 3 C)/(1 m)²
F = 8.99 × 10⁹ Nm²/C² × (-12 C²)/(1 m)²
F = -107.88 × 10⁹ Nm²/1 m²
F = -1.0788 × 10¹¹ N
F ≅ -1.1 × 10¹¹ N
Since the second charge is 2 m away and is labeled q₂ = - 4 Coulombs. The third charge is 1 m away and is labeled q₃ = + 3 Coulombs, the electrical force between q₂ and q₃ is -1.1 × 10¹¹ N
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A ball is thrown straight up from the top of a 224 foot tall building with an initial speed of 80 feet per second. The height of the ball as a function of time can be modeled by the function h(t)=-16t2+80t+224. When will the ball reach a height of 308 ft?'
The ball will reach a height of 308 ft at approximately 2.7 seconds.
To find when the ball reaches a height of 308 ft, we need to solve the equation h(t) = 308 ft. The equation for the height of the ball as a function of time is given by h(t) = -16t^2 + 80t + 224.
Setting h(t) equal to 308 ft:
-16t^2 + 80t + 224 = 308
Rearranging the equation:
-16t^2 + 80t - 84 = 0
Dividing through by -4 to simplify the equation:
4t^2 - 20t + 21 = 0
We can solve this quadratic equation using factoring or the quadratic formula. Factoring is not possible, so we'll use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 4, b = -20, and c = 21.
Plugging in the values into the quadratic formula:
t = (-(-20) ± √((-20)^2 - 4(4)(21))) / (2(4))
t = (20 ± √(400 - 336)) / 8
t = (20 ± √64) / 8
t = (20 ± 8) / 8
There are two possible solutions:
t1 = (20 + 8) / 8 = 28 / 8 = 3.5
t2 = (20 - 8) / 8 = 12 / 8 = 1.5
However, we are interested in the time when the ball reaches a height of 308 ft, which is a positive value. Therefore, the ball will reach a height of 308 ft at approximately t ≈ 2.7 seconds.
The ball will reach a height of 308 ft at approximately 2.7 seconds.
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The parallax angle for the star Hadar is 0.010 arcseconds. How far away is Hadar?
The parallax angle for the star Hadar is 0.010 arcseconds. 100 PC far away is Hadar.
Parallax angle is a displacement or distinction in the apparent function of an item considered alongside two one-of-a-kind lines of sight and is measured by the attitude or semi-perspective of inclination among those two lines. because of foreshortening, close-by objects display a bigger parallax than farther items when found from distinct positions, so parallax may be used to determine distances.
To a degree of big distances, including the gap between a planet or a star from Earth, astronomers use the precept of parallax. right here, the term parallax is the semi-perspective of inclination between sightlines to the megastar, as observed while Earth is on contrary sides of the solar in its orbit. those distances shape the lowest rung of what's called "the cosmic distance ladder", the first in a succession of methods by which astronomers determine the distances to celestial items, serving as a basis for other distance measurements in astronomy forming the higher rungs of the ladder.
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