Answer:small barrel gun
Explanation:
Given
Muzzle velocity of bullet is greater in short barrel gun as compared to larger barrel gun
acceleration is given by change in velocity with respect to time
[tex]a=\dfrac{\Delta v}{\Delta t}[/tex]
In case of short barrel bullet time taken by bullet to reach its muzzle velocity is less therefore acceleration of small barrel bullet is more compared to long barrel bullet.
Final answer:
The muzzle velocity of a bullet can be calculated using the formula v = u + at. Using the given acceleration of 6.20 x 10⁵ m/s² and time of 8.10 x 10⁻⁴ s, the muzzle velocity is 502 m/s.
Explanation:
To calculate the muzzle velocity of a bullet, which is the velocity of the bullet as it leaves the barrel of a gun, we can use the formula for uniform acceleration: v = u + at, where v is the final velocity, u is the initial velocity (which is 0 m/s in this case, as the bullet starts from rest), a is the acceleration, and t is the time. Here, we are given an acceleration of 6.20 × 10⁵ m/s² and a time of 8.10 × 10⁻⁴ s. Plugging these values into the formula gives us the muzzle velocity.
Using the formula, we have: v = 0 m/s + (6.20 × 10⁵ m/s²)(8.10 × 10⁻⁴ s), which simplifies to v = 502 m/s. Therefore, the muzzle velocity of the bullet is 502 meters per second. This value represents how fast the bullet is traveling when it exits the gun barrel.
The neurotransmitter dopamine is most closely associated with
Since there's no answer choices available, I would say "reward mechanisms in the brain", "pleasure", "Schizophrenia" or "appetite".
Answer:
A neurotransmitter used in the parts of the brain involved in regulating movement and experiencing pleasure.
Explanation:
Dopamine is one of the chemicals in our brain which controls mood Interacting with the pleasure and reward center of our brain, dopamine, along with other chemicals like oxytocin and endorphins plays a very crucial role in how we feel
Find the magnitude of the torque produced by a 4.5 N force applied to a door at a perpendicular distance of 0.26 m from the hinge. Answer in units of N · m.
Answer:
The required torque is 1.17 N-m.
Explanation:
The given data :-
The magnitude of force ( F ) = 4.5 N.
The length of arm ( r ) = 0.26 m.
Here given that force is applied at perpendicular means ( ∅ ) = 90°.
The torque ( T ) is given by
T = F * r * sin∅
T = 4.5 * 0.26 * sin 90°
T = 4.5 * 0.26 * 1
T = 1.17 N-m.
The magnitude of the torque produced by the given force perpendicular to the lever is 1.17N.m.
Given the data in the question;
Force; [tex]F = 4.5N[/tex]Perpendicular distance or radius; [tex]r = 0.26m[/tex]Since the force is perpendicular to the lever, Angle; [tex]\theta = 90^o[/tex]Torque; [tex]T = \ ?[/tex]
Torque simply the measure of the force that can cause an object to rotate about an axis. It is expressed as:
[tex]T = rFsin\theta[/tex]
Where r is radius, F is force applied and θ is the angle between the force and the lever arm.
We substitute our given values into the equation;
[tex]T = 0.26m * 4.5N * sin90^o\\\\T = 0.26m * 4.5N * 1\\\\T = 1.17N.m[/tex]
Therefore, the magnitude of the torque produced by the given force perpendicular to the lever is 1.17N.m.
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slader A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s and it bounces back with this same speed. The ball is in contact with the wall 0.050 s. What is the magnitude of the average force exerted on the wall by the ball?
Answer: 800N
Explanation:
Given :
Mass of ball =0.8kg
Contact time = 0.05 sec
Final velocity = initial velocity = 25m/s
Magnitude of the average force exerted on the wall by the ball is can be calculated using the relation;
Force(F) = mass(m) * average acceleration(a)
a= (initial velocity(u) + final velocity(v))/t
m = 0.8kg
u = v = 25m/s
t = contact time of the ball = 0.05s
Therefore,
a = (25 + 25) ÷ 0.05 = 1000m/s^2
Therefore,
Magnitude of average force (F)
F=ma
m = mass of ball = 0.8
a = 1000m/s^2
F = 0.8 * 1000
F = 800N
What are the strengths and limitations of the doppler method?
Answer:
Strength: able to detect planets in a wide range of orbits, as long as orbits aren't face on
Limitations: yield only planet's mass and orbital properties
Explanation:
Strengths: yield planet's size
Limitations: few planets have the necessary orbital alignment to be detectable
The Doppler method is a valuable tool for detecting and characterizing exoplanets, particularly those with significant mass and short orbital periods. However, it has limitations in terms of its sensitivity to certain types of planets and the limited information it provides about exoplanet properties. So, option A is correct.
The Doppler method is particularly effective in detecting massive exoplanets that are relatively close to their host stars. It has been successful in identifying hundreds of exoplanets since its inception.The Doppler method is more sensitive to massive exoplanets that are closer to their host stars. This means that it may miss smaller, more distant planets or those with longer orbital periods. The method also has difficulty detecting exoplanets with orbits that are nearly aligned with the line of sight, resulting in underrepresentation of such planets in detection statistics.
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Complete question is below
What are the strengths and limitations of the doppler method?
A.it has limitations in terms of its sensitivity to certain types of planets and the limited information it provides about exoplanet properties.
B.it has NO limitations in terms of its sensitivity to certain types of planets and the limited information it provides about exoplanet properties
The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is constant. At 15 degrees Celsius the pressure is 101.3 kPa at sea level and 87.14 kPa at h = 1000m. What is the pressure at an altitude of 3000 m?
Answer:
64.59kpa
Explanation:
See attachment
________ is the characteristic color bands that represent the range of radiations emitted by an element when it is heated
Answer:
Emission spectrum
Explanation:
The emmison spectrum is unique to every elements
Light consists of electromagnetic radiation of different wavelengths. Therefore, when the elements or their compounds are heated either on a flame or by an electric arc they emit energy in the form of light.
The color band indicate the amount of energy emitted in form of electromagnetic radiation.
When electrons in an atom gain energy, the become excited, on falling and losing their energy, they return to their ground state releasing the measure of energy they absorbed in transitioning.
The characteristic color bands emitted by an element when heated are called line spectra, which act as a unique fingerprint for the element and are important for identifying various physical properties.
The characteristic color bands that represent the range of radiations emitted by an element when it is heated are known as the line spectra. When elements are heated to incandescence, like in Bunsen and Kirchhoff's experiments, they emit light with a series of sharp wavelengths characteristic of that element's atomic composition.
This light, when analyzed using a spectrometer or a simple prism, results in visible narrow bands of colors, each corresponding to a specific wavelength. For instance, hydrogen emits a red light with a strong line at 656 nm, and sodium is known to emit in the yellow portion of the spectrum at about 589 nm.
The line spectra serve as a unique fingerprint for each element and are crucial for understanding not only the chemical composition but also various physical properties such as temperature and density of the radiant gas. In terms of non-visible bands, color palettes are often chosen to represent different levels of brightness, energies of photons, and inferred physical properties from data modeling, which are essential in the study of astronomy and spectroscopy.
Two hypothetical discoveries in Part A deal with moons that, like Earth's moon, are relatively large compared to their planets. Which of the following best explains why finding 1 planet with such a moon is consistent with the nebular theory, while finding 6 planets with such moons is not consistent?
Answer:
Unusually large moons form in giant impacts, which are relatively rare events
Explanation:
Solution:
- Finding large moons comparable in size to their planets result from impacts of two astro-bodies. The probability of such an event occurring is very rare.
- Even at the best luck, one moon can be made from the result of giant impact. While the probability of 6 planets having moons of comparable sizes is close to impossible. The transition from an undifferentiated cloud to a star system complete with planets and moons takes about 100 million years.
A ship sets sail from Rotterdam, The Netherlands, intending to head due north at 6.5 m/s relative to the water. However, the local ocean current is 1.50 m/s in a direction 40.0º north of east and changes the ship's intended motion. What is the velocity of the ship relative to the Earth?
Answer:
Explanation:
velocity of ship with respect to water = 6.5 m/s due north
[tex]\overrightarrow{v}_{s,w}=6.5 \widehat{j}[/tex]
velocity of water with respect to earth = 1.5 m/s at 40° north of east
[tex]\overrightarrow{v}_{w,e}=1.5\left ( Cos40\widehat{i} +Sin40\widehat{j}\right)[/tex]
velocity of ship with respect to water = velocity of ship with respect to earth - velocity of water with respect to earth
[tex]\overrightarrow{v}_{s,w} = \overrightarrow{v}_{s,e} - \overrightarrow{v}_{w,e}[/tex]
[tex]\overrightarrow{v}_{s,e} = 6.5 \widehat{j}- 1.5\left (Cos40\widehat{i} +Sin40\widehat{j} \right )[/tex]
[tex]\overrightarrow{v}_{s,e} = - 1.15 \widehat{i}+5.54\widehat{j}[/tex]
The magnitude of the velocity of ship relative to earth is [tex]\sqrt{1.15^{2}+5.54^{2}}[/tex] = 5.66 m/s
An aminoacyl-trna that enters the a site of the ribosome will next occupy which site?
Answer:
The P site
Explanation:
After the formation of peptide linkage, the trna of P site becomes free, more to the E site and then slip away. The A site trna has dipeptide instead of a single aminoacid.
In the prescence of translocase and energy from GTP, the ribosome move in such a way the peptidyl bearing trna of A site come to lie with P sit. This exposes next codon to A site. A new aminoacid trna complex reaches the fresh codon exposed at A site.
An aminoacyl-tRNA molecule that enters the A site of the ribosome next moves to the P site during protein synthesis. This is part of the process known as translocation. After the P site, it moves to the E site, then exits the ribosome.
Explanation:In protein synthesis, an aminoacyl-tRNA that enters the A site of the ribosome will next occupy the P site. This transition between the A site (aminoacyl site) and the P site (peptidyl site) happens during the process known as translocation, which is driven by elongation factors. Once in the P site, the tRNA molecule will be linked with the growing polypeptide chain. After this the tRNA will move to the E site (exit site) and leave the ribosome.
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A block pushed along the floor with velocity v0xv0x slides a distance dd after the pushing force is removed. If the mass of the block is doubled but its initial velocity is not changed, what distance does the block slide before stopping?
Answer:
There is no change in distance
Explanation:
Given that block slides a distance dd with initial velocity [tex]v_{ox}[/tex][tex]v_{ox}[/tex]
Work energy theorem states that Work done W is the difference between final kinetic energy [tex]KE_{f}[/tex] and initial kinetic energy [tex]KE_{i}[/tex]
W = [tex]KE_{f}[/tex] - [tex]KE_{i}[/tex] ...........(equation 1)
In the given problem, the work is done by frictional force
Hence the work done is given by
[tex]W_{Friction}[/tex] = -[tex]F_{Friction}[/tex] x dd
[tex]F_{Friction}[/tex] is negative since it is resistive force and dd is the distance
[tex]W_{Friction}[/tex] = - μ mg X dd
where μ is coefficient of friction.
Also we know that Kinetic energy KE = [tex]\frac{1}{2}[/tex] [tex]mv^{2}[/tex]
[tex]KE_{f}[/tex] = 0 since final velocity is zero
[tex]KE_{i}[/tex] = [tex]\frac{1}{2}[/tex] [tex]m(v_{ox} v_{ox}) ^{2}[/tex]
Substituting the corresponding values equation 1 becomes
-μ mg x dd = 0 - [tex]\frac{1}{2}[/tex] [tex]m(v_{ox} v_{ox}) ^{2}[/tex]
dd= [tex]\frac{(v_{ox}v_{ox} )^{2} }{2g}[/tex]x (1/μ) ........... (equation 2)
To find distance when mass of block is doubled and initial velocity is not changed:
Equation 2 shows that the distance dd is independent of mass, therefore there is no change in distance.
The block will slide only half the distance before stopping when its mass is doubled while the initial velocity remains the same.
When a block is pushed along the floor with an initial velocity [tex]\(v_{0x}\)[/tex] and slides a distance [tex]\(d\)[/tex] before stopping, its stopping distance is determined by the initial velocity and the frictional force opposing its motion.
The force of friction is given by [tex]\(F_{\text{friction}} = \mu \cdot N\)[/tex], where [tex]\(\mu\)[/tex] is the coefficient of friction and [tex]\(N\)[/tex] is the normal force. In this case, we are assuming a constant coefficient of friction.
The deceleration of the block can be calculated using Newton's second law: [tex]\(F = m \cdot a\)[/tex]. When the mass of the block is doubled, the force required to decelerate it to a stop is also doubled.
Let's assume the initial velocity [tex]\(v_{0x}\)[/tex] remains the same. The deceleration [tex]\(a\)[/tex] is proportional to the force, so [tex]\(a\)[/tex] is also doubled when the mass is doubled. This is because [tex]\(F\)[/tex] is directly proportional to [tex]\(m\)[/tex].
The stopping distance [tex]\(d'\)[/tex] can be calculated using the equation of motion:
[tex]\[v_f^2 = v_i^2 + 2as\][/tex]
Where:
[tex]\(v_f\)[/tex] is the final velocity (0 m/s, as the block stops),
[tex]\(v_i\)[/tex] is the initial velocity [tex](\(v_{0x}\)),[/tex]
[tex]\(a\)[/tex] is the deceleration (doubled),
[tex]\(s\)[/tex] is the stopping distance [tex](\(d'\)).[/tex]
Plugging in the values:
[tex]\[0 = v_{0x}^2 + 2(2a)d'\][/tex]
Now, we can solve for [tex]\(d'\):[/tex]
[tex]\[2ad' = -v_{0x}^2\][/tex]
[tex]\[d' = \frac{-v_{0x}^2}{2a}\][/tex]
Since [tex]\(a\)[/tex] is doubled, the stopping distance [tex]\(d'\)[/tex] will be halved compared to the initial distance [tex]\(d\).[/tex] So, the block will slide only half the distance before stopping when its mass is doubled while the initial velocity remains the same.
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Which prediction of weather is the most accurate for the next two days if there is an occluded front over the area?
Question 2 options:
constant rain
dry warm weather
severe thunderstorms
cold but clear skies
The most accurate weather for the next two days would be dry warm weather or severe thunderstorms if there is an occluded front over the area.
Explanation:
The weather front that is created during the cyclogenesis process is an occluded front in meteorology. Cyclogenesis process is the development of extra-tropical cyclone and its intensification.
During the occurrence of this, the warm air is occluded (separated) from the center of cyclone at the surface of the earth.
The cold front rotates the storm as it intensifies and holds up the warm front. This develops an occluded front, that is the boundary which separates the new cold air mass and the older cool air mass that is already in warm front's north.
The ΔG of a reaction would be at the minimum when... Group of answer choices the equilibrium constant is equal to 1 (i.e., the reactant and product concentrations are always equal). the reactant and product concentrations don't change over time (the system is at equilibrium). the entropy has reached its maximum positive value. the reaction goes to completion. the reaction is very slow
Answer:
the equilibrium constant is equal to 1 (i.e., the reactant and product concentrations are always equal).
Explanation:
ΔG is a symbol related to Gibbs free energy, which is a physical quantity related to thermodynamics. ΔG refers to the difference between the change in enthalpy (and sometimes entropy) and the temperature of a chemical reaction.
Gibbs free energy is very useful for measuring the work done between the reactants in a reaction. It is calculated using the formula: ΔG = change in enthalpy - (temperature x change in entropy).
The ΔG of a reaction would have a minimum value (zero), if the equilibrium constant is equal to 1 (that is, the concentrations of the reagent and the product are always equal).
the equilibrium constant should be equal to 1 (i.e., the reactant and product concentrations are always equal).
Gibbs free energy:ΔG refers to a symbol that should be related to Gibbs free energy, i.e. physical quantity related to thermodynamics. ΔG means the difference between the change in enthalpy and the temperature of a chemical reaction. Now it is determined using the formula:
ΔG = change in enthalpy - (temperature x change in entropy).
In the case when the ΔG of a reaction would have a minimum value (zero), so if the equilibrium constant is equivalent to 1 i.e. the concentrations of the reagent and the product should always equal.
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The playing time of 16 popular songs is found to have a standard deviation of 54.5 seconds. Use a 0.05 significance level to test the claim that the songs are from a population with a standard deviation less than one minute (60 seconds). State the initial and final conclusion.
Answer:
fail to reject the null hypothesis; there is not sufficient evidence to support the claim that the songs are from a population with a standard deviation less than one minute
Explanation:
Before we understood that objects have a tendency to maintain their velocity in a straight line unless acted upon by a net force, people thought that objects have a tendency to stop on their own. This happened because a specific force was not yet understood. What was that force?
Answer:Friction force
Explanation:
The frictional force is responsible for stopping the moving object.
The friction force is provided by nature in the form of air resistance, fluid resistance, Surface resistance.
For surface resistance, the friction force is of two types namely kinetic friction force and static friction force.
Static friction acts when the object is stationary and a force is applied to cause the motion while kinetic friction acts when the body starts moving.
kinetic friction is lesser is in magnitude as compared to static friction.
The misunderstood force that causes objects to seem to stop on their own is known as friction. This force opposes the motion of objects. The misconception was fixed with Newton's first law of motion.
Explanation:The force that people did not understand before they realized that objects have a tendency to maintain their velocity in a straight line unless acted upon by a net force is friction. Friction is a force that opposes movement. It's the reason why objects seem to stop on their own when in contact with other objects or surfaces. The perception that an object moves and then eventually stops due to its own 'nature' occurs because friction was at play, causing it to slow down and finally stop. It was Sir Isaac Newton who, in his first law of motion, clarified this misperception by stating that an object in motion stays in motion unless acted upon by an external force.
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What pollutant forms when automobile emissions react with oxygen gas and ultraviolet rays?
Ozone is a molecule that consists of three oxygen atoms. When high-energy ultraviolet rays strike ordinary oxygen molecules (O2), they split the molecule into two single oxygen atoms, known as atomic oxygen. A freed oxygen atom then combines with another oxygen molecule to form a molecule of ozone. Hydrocarbons and nitrogen oxides come from great variety of industrial and combustion processes. Motor vehicle exhaust and industrial emissions, gasoline vapors, and chemical solvents are some of the major sources of NOx and VOC that acts as a precursor of ozone. In urban areas, the number of automobiles are more and therefore, more production of such harmful gases. These gases in presence of sunlight leads to the formation of bad ozone.
What is its maximum altitude above the ground? The answer is the maximum height above the ground
Answer:
Maximum altitude above the ground = 1,540,224 m = 1540.2 km
Explanation:
Using the equations of motion
u = initial velocity of the projectile = 5.5 km/s = 5500 m/s
v = final velocity of the projectile at maximum height reached = 0 m/s
g = acceleration due to gravity = (GM/R²) (from the gravitational law)
g = (6.674 × 10⁻¹¹ × 5.97 × 10²⁴)/(6370000²)
g = -9.82 m/s² (minus because of the direction in which it is directed)
y = vertical distance covered by the projectile = ?
v² = u² + 2gy
0² = 5500² + 2(-9.82)(y)
19.64y = 5500²
y = 1,540,224 m = 1540.2 km
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A solid cylinder and a hollow cylinder of equal mass and radius are at rest at the top of an inclined plane. They are released simultaneously and roll down the plane without slipping. Which object reaches the bottom of the incline first?
Answer:
Solid cylinder
Explanation:
Solid cylinder will reach first
If radius of each cylinder is r
mass is m
then,
Moment of inertia I= dm[tex]r^{2}[/tex]
Here d = measure as how close the mass is to the edge
Velocity of rolling cylinder is given by
v= [tex]\sqrt} \frac{2gh}{1+d}[/tex]
where,
g= 9.8 m/s2
h= height from ground
So from formula of velocity we can say that velocity will be maximum if denominator is minimum i-e if k is of small value -- or in other word if mass is away from edge i-e if mass is closer to the center !
As all the mass in hollow cylinder is near the edge so k value will be higher for it and hence it will have low velocity value. So it will reach later as compared to the solid cylinder in which mass is closer to the center and hence k is greater for solid cylinder.
A hollow aluminum cylinder 19.0 cm deep has an internal capacity of 2.000 L at 23.0°C. It is completely filled with turpentine at 23.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 91.0°C. (The average linear expansion coefficient for aluminum is 2.4 x10^-5/°C, and the average volume expansion coefficient for turpentine is 9.0 x10^-4/°C.)
(a) How much turpentine overflows?
in cm^3?
(b) What is the volume of turpentine remaining in the cylinder at 91.0°C? (Give you answer to four significant figures.)
in cm^3?
(c) If the combination with this amount of turpentine is then cooled back to 23.0°C, how far below the cylinder's rim does the turpentine's surface recede?
in cm?
Answer:
Explanation:
The approach of Linear expansivity was used in solving the problem and the detailed steps and appropriate calculation is carefully shown in the attached file.
Answer:
a. 112.608 cm³ b. 2010 cm³ c. 0.032 cm
Explanation:
Given
Volume of aluminium cylinder = 2.0 L= 2.0 × 10³ cm³
Length of aluminium cylinder = 19.0 cm
Volume of turpentine = 2.0 L= 2.0 10³ cm
Initial temperature of aluminium cylinder = 23.0 °C
Initial temperature of turpentine = 23.0 °C
Average linear expansion coefficient of aluminium α = 2.4 × 10⁻⁵ °C
Average volume expansion coefficient of turpentine γ₁ = 9.0 × 10⁻⁴ °C
(a) How much turpentine overflows in cm³?
We calculate the volume expansion of the aluminium cylinder and that of the turpentine and then subtract the difference.
The volume expansion of material V = V₀(1 + γΔθ)
where V = final volume, V₀ = initial volume, γ = volume expansion and Δθ = temperature change.
For aluminium V₀ = 2000 cm³, γ = 3α = 3 × 2.4 × 10⁻⁵ /°C = 7.2 × 10⁻⁵ /°C
Δθ = θ₂ - θ₁ = 91 °C - 23 °C = 68 °C
V₁ = 2000(1 + 7.2 × 10⁻⁵ /°C × 68°C) = 2000(1 + 0.004896) = 2000(1.004896) = 2009.792 cm³
For turpentine V₀ = 2000 cm³, γ = 9.0 × 10⁻⁴ /°C
Δθ = θ₂ - θ₁ = 91 °C - 23 °C = 68 °C
V₂= 2000(1 + 9.0 × 10⁻⁴ /°C × 68°C) = 2000(1 + 0.0612) = 2000(1.0612) = 2122.4 cm³.
The volume of turpentine that overflows = V₂ - V₁ = 2122.4 cm³ - 2009.792 cm³ = 112.608 cm³
(b) What is the volume of turpentine remaining in the cylinder at 91.0°C? (Give you answer to four significant figures.) in cm³?
The volume of turpentine remaining is the final volume of the turpentine minus overflow = 2122.4 cm³ - 112.608 cm³ = 2009.792 cm³ = 2010 cm³ to four significant figures.
(c) If the combination with this amount of turpentine is then cooled back to 23.0°C, how far below the cylinder's rim does the turpentine's surface recede in cm?
The linear expansion for the aluminium to 91 C from 23 C is L = L₀(1 + Δθ)
For aluminium L₀ = 19 cm, α = 2.4 × 10⁻⁵ /°C =
Δθ = θ₂ - θ₁ = 91 °C - 23 °C = 68 °C
L = 19(1 + 2.4 × 10⁻⁵ /°C × 68°C) = 19(1 + 0.001632) = 19(1.001632) = 19.031 cm
The linear expansion for the aluminium from 91 C to 23 C is L = L₀(1 + Δθ)
For aluminium L₀ = 19 cm, α = 2.4 × 10⁻⁵ /°C =
Δθ = θ₂ - θ₁ = 23 °C - 91 °C = -68 °C
L₁ = 19.031(1 + 2.4 × 10⁻⁵ /°C × -68°C) = 19.031(1 - 0.001632) = 19.031(0.998368) = 18.999 cm
How far below the rim it recedes is L - L₁ = 19.031 cm - 18.999 cm = 0.032 cm.
Which one of the following crystal structures has the fewest slip directions and therefore the metals with this structure are generally more difficult to deform at room temperature?
a. BCC
b. FCC
c. HCP
d. BCT
The HCP structure has the fewest slip directions, making HCP metals like Mg and Zn more difficult to deform at room temperature compared to those with BCC or FCC structures.
Explanation:The crystal structure that has the fewest slip directions and is thus generally more difficult to deform at room temperature is the hexagonal close-packed (HCP) structure. Metals with the HCP structure include Cd, Co, Li, Mg, Na, and Zn. The nature of HCP's stacking arrangement, with alternating type A and type B close-packed layers (ABABAB...), results in fewer slip systems compared to body-centered cubic (BCC) and face-centered cubic (FCC) structures. BCC structures, found in metals like K, Ba, Cr, Mo, W, and Fe, have a coordination number of 8, whereas FCC structures have a coordination number of 12 and include metals like Ag, Al, Ca, Cu, Ni, Pb, and Pt. The closer atomic packing found in FCC, identified as CCP or cubic close-packed, is reflected in the ABCABCABC... stacking sequence, which affords a higher number of slip directions.
A proton moves 10.0 cm on a path parallel to the direction of a uniform electric field of strength 3.0 N/C. What is the change in electrical potential energy?
Answer:
ΔPE= -4.8×10⁻²⁰J
Explanation:
Given data
Electric field of strength E=3.0 N/C
Charge of proton q=1.60×10⁻¹⁹C
Proton moves distance d=10 cm=0.10 m
To find
Change in electrical potential energy ΔPE
Solution
As we know that:
ΔPE= -qEd
[tex]=-(1.60*10^{-19}C )(3.0N/C)(0.10m)\\=-4.8*10^{-20}J[/tex]
ΔPE= -4.8×10⁻²⁰J
Answer:
Explanation:
Given:
D = 10 cm
= 0.1 m
E = 3.0 N/C
Qp = 1.602 × 10^-19 C
U = Q × V
But,
V = E × D
= 3 × 0.1
= 0.3 V
U = 1.602 × 10-19 × 0.3
= 4.806 × 10^-20 J.
During a certain comet’s orbit around the Sun, its closest distance to the Sun is 0.6 AU, and its farthest distance from the Sun is 35 AU. At what distance will the comet’s orbital velocity be the largest?
Answer:
At the closest point
Explanation:
We can simply answer this question by applying Kepler's 2nd law of planetary motion.
It states that:
"A line connecting the center of the Sun to any other object orbiting around it (e.g. a comet) sweeps out equal areas in equal time intervals"
In this problem, we have a comet orbiting around the Sun:
- Its closest distance from the Sun is 0.6 AU
- Its farthest distance from the Sun is 35 AU
In order for Kepler's 2nd law to be valid, the line connecting the center of the Sun to the comet must move slower when the comet is farther away (because the area swept out is proportional to the product of the distance and of the velocity: [tex]A\propto vr[/tex], therefore if r is larger, then v (velocity) must be lower).
On the other hand, when the the comet is closer to the Sun the line must move faster ([tex]A\propto vr[/tex], if r is smaller, v must be higher). Therefore, the comet's orbital velocity will be the largest at the closest distance to the Sun, 0.6 A.
Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 5.50 m in front of one of the speakers perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible? Express your answer with the appropriate units.
Answer:
242.85 Hz
Explanation:
For maximum intensity of sound, the path difference,ΔL = (n + 1/2)λ/2 where n = 0,1,2...
Since Abby is standing perpendicular to one speaker, the path length for the sound from the other speaker to him is L₁ = √(2.00² + 5.50²) = √(4.00 + 30.25) = √34.25 = 5.85 m.
The path difference to him is thus ΔL = 5.85 m - 5.50 m = 0.35 m.
Since ΔL = (n + 1/2)λ/2 and for lowest frequency n = 0,
ΔL = (n + 1/2)λ/2 = (0 + 1/2)λ/2 = λ/4
ΔL = λ = v/f and f = v/4ΔL where f = frequency of wave and v = velocity of sound wave = 340 m/s.
f = 340/(4 × 0.35) = 242.85 Hz
The condition of constructive interference and the speed of a wave allows finding the frequency for the first constructive interference that Abby hears is:
f = 971.4 Hz
The interference of the coherent sound wave occurs when two wave fronts are added at a point and the intensity depends on the path difference of these waves, we have two extreme cases:
Destructive. In this case the sum of the intensity gives a resultant of zero. Constructive. The sum of the intensities gives a maximum and is described by the expression.Δr = [tex]2n \ \frac{\lambda}{2}[/tex]
Where Δr is the path difference, λ is the wavelength and n is an integer.
In the attachment we can see the distance from Abby is to the speakers for the closest y = 5.50 m, the distance to the furthest speaker.
Let's use the Pythagoras' theorem,
r =[tex]\sqrt{x^2 + y^2}[/tex]
Let's calculate.
r = [tex]\sqrt{2^2 + 5.5^2}[/tex]
r = 5.85 m
The difference in path is:
Δr = r-y
Δr = 5.85 - 5.5
Δr = 0.35 m
let's find the wavelength.
Δr = 2n [tex]\frac{\lambda}{2}[/tex]
λ = [tex]\frac{\Delta r}{n}[/tex]
The first constructive interference occurs for n = 1
λ= Δr
λ = 0.35 m
Wave speed is proportional to wavelength and frequency.
v = λ f
f = [tex]\frac{v}{\lambda }[/tex]
Let's calculate.
f = [tex]\frac{340}{0.35}[/tex]
f = 971.4 Hz
In conclusion using the condition of constructive interference and the speed of a wave we can find the frequency for the first constructive interference that Abby hears is:
f = 971.4 Hz
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Strategic planning is long-range, formulated by top management, and made as if the company operated in a vacuum Group of answer choices True False
Answer:
THE ANSWER IS: TRUE
Explanation:
Strategic planning is indeed long-range and formulated by top management but not under the assumption that the company operates in a vacuum. Effective strategic planning acknowledges and incorporates external factors.
Explanation:The statement is partially true and partially false. It is accurate that strategic planning is long-range and formulated by top management. These types of plans typically look several years into the future to set comprehensive goals for the company. However, the statement is false in suggesting that strategic planning is done as if a company operates in a vacuum. In truth, effective strategic planning must incorporate external influences like market trends, competitive activities, and regulatory changes.
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Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 9 m into 32 cm. of water. Assuming that he stops just as he reaches the bottom of the water and estimating his mass to be 62 kg, find the magnitudes of the impulse on him from the water.
Answer:
823.46 kgm/s
Explanation:
At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.
So, mgh = 1/2mv²
From here, his velocity just as he reaches the surface of the water is
v = √2gh
h = 9 m and g = 9.8 m/s²
v = √(2 × 9 × 9.8) m/s
v = √176.4 m/s
v₁ = 13.28 m/s
So his velocity just as he reaches the surface of the water is 13.28 m/s.
Now he dives into 32 cm = 0.32 m of water and stops so his final velocity v₂ = 0.
So, if we take the upward direction as positive, his initial momentum at the surface of the water is p₁ = -mv₁. His final momentum is p₂ = mv₂.
His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,
J = (62 × 0 + 62 × 13.28) kgm/s = 0 + 823.46 kgm/s = 823.46 kgm/s
So the magnitude of the impulse J of the water on him is 823.46 kgm/s
The circuit to the right consists of a battery ( V 0 = 64.5 V) (V0=64.5 V) and five resistors ( R 1 = 711 (R1=711 Ω, R 2 = 182 R2=182 Ω, R 3 = 663 R3=663 Ω, R 4 = 534 R4=534 Ω, and R 5 = 265 R5=265 Ω). Find the current passing through each of the specified points. -g
Answer:
The current in R₁ is 0.0816 A.
The current at H point is 0.0243 A.
Explanation:
Given that,
Voltage = 64.5
Resistance is
[tex]R_{1}=711\ \Omega[/tex]
[tex]R_{2}=182\ \Omega[/tex]
[tex]R_{3}=663\ \Omega[/tex]
[tex]R_{4}=534\ \Omega[/tex]
[tex]R_{5}=265\ \Omega[/tex]
Suppose, The specified points are R₁ and H.
According to figure,
R₂,R₃,R₄ and R₅ are connected in parallel
We need to calculate the resistance
Using parallel formula
[tex]\dfrac{1}{R}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}+\dfrac{1}{R_{4}}+\dfrac{1}{R_{5}}[/tex]
Put the value into the formula
[tex]\dfrac{1}{R}=\dfrac{1}{182}+\dfrac{1}{663}+\dfrac{1}{534}+\dfrac{1}{265}[/tex]
[tex]\dfrac{1}{R}=\dfrac{35501}{2806615}[/tex]
[tex]R=79.05\ \Omega[/tex]
R and R₁ are connected in series
We need to calculate the equilibrium resistance
Using series formula
[tex]R_{eq}=R_{1}+R[/tex]
[tex]R_{eq}=711+79.05[/tex]
[tex]R_{eq}=790.05\ \Omega[/tex]
We need to calculate the equivalent current
Using ohm's law
[tex]i_{eq}=\dfrac{V}{R_{eq}}[/tex]
Put the value into the formula
[tex]i_{eq}=\dfrac{64.5}{790.05}[/tex]
[tex]i_{eq}=0.0816\ A[/tex]
We know that,
In series combination current distribution in each resistor will be same.
So, Current in R and R₁ will be equal to [tex]i_{eq}[/tex].
The current at h point will be equal to current in R₅
We need to calculate the voltage in R
Using ohm's law
[tex]V=I_{eq}\timesR[/tex]
Put the value into the formula
[tex]V=0.0816\times79.05[/tex]
[tex]V=6.45\ Volt[/tex]
In resistors parallel combination voltage distribution in each part will be same.
So, [tex]V_{2}=V_{3}=V_{4}=V_{5}=6.45 V[/tex]
We need to calculate the current at H point
Using ohm's law
[tex]i_{h}=\dfrac{V_{5}}{R_{5}}[/tex]
Put the value into the formula
[tex]i_{h}=\dfrac{6.45}{265}[/tex]
[tex]i_{h}=0.0243\ A[/tex]
Hence, The current in R₁ is 0.0816 A.
The current at H point is 0.0243 A.
A bungee cord is 30m long and when stretched a distance x itexerts a restoring force of magnitude kx. Your father-in-law (mass95.0kg) stands on a platform 45.0m above the ground , and one endof the cord is tied securley to his ankle and the other end to theplatform. You have promised him that when he steps off he will falla maximum of 41m before the cord stops him. You had several bungeecords to select from and tested them by stretching them out, tyingone end to a tree and pulling the other end with a force of 380.0N.When you did this, what distance will the bungee cord you shouldselect have stretched?
Answer:
The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.
Explanation:
As from the given data
the length of the rope is given as l=30 m
the stretched length is given as l'=41m
the stretched length required is give as y=l'-l=41-30=11m
the mass is m=95 kg
the force is F=380 N
the gravitational acceleration is g=9.8 m/s2
The equation of k is given by equating the energy at the equilibrium point which is given as
[tex]U_{potential}=U_{elastic}\\mgh=\dfrac{1}{2} k y^2\\k=\dfrac{2mgh}{y^2}[/tex]
Here
m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so
[tex]k=\dfrac{2mgh}{y^2}\\k=\dfrac{2\times 95\times 9.8\times 41}{11^2}\\k=630.92 N/m[/tex]
Now the force is
[tex]F=kx\\[/tex] or
[tex]x=\dfrac{F}{k}\\[/tex]
So here F=380 N, k=630.92 N/m
[tex]x=\dfrac{F}{k}\\x=\dfrac{380}{630.92}\\x=0.602 m[/tex]
So the distance is 0.602 m
The student requires help to calculate the stretch distance of a bungee cord when a force is applied, which is determined using Hooke's Law, F = kx, where F is the applied force and k is the spring constant.
Explanation:The student is asking about the properties of a bungee cord that would be suitable for their father-in-law's jump so that he will fall safely without hitting the ground. Specifically, the question involves calculating the amount by which the bungee cord would stretch when a force of 380.0 N is applied to it. The restoring force that the bungee cord exerts is given as kx, where k is the spring constant and x is the stretch distance.
To find the correct bungee cord, one must solve for x in the equation F = kx given that F (the force applied) is 380.0 N. The spring constant k is the unknown in this problem and must be determined based on the requirement that the maximum fall before the cord stops the father-in-law is 41 m, factoring in his weight and the initial unstretched length of the bungee cord (30 m).
This problem requires knowledge in the area of Hooke's Law and the work-energy principle, as it deals with forces and the properties of elastic materials. The actual calculation is not provided because it would require additional information about the bungee cords the student has access to, such as their specific spring constants.
A voltage source of 10 V is connected to a series RC circuit where R = 2.0 ´ 106 W , and C = 3.0 µF. Find the amount of time required for the current in the circuit to decay to 5% of its original value. Hint: This is the same amount of time for the capacitor to reach 95% of its maximum charge.
Answer:
0.5 s
Explanation:
The discharge current in a series RC circuit is given by
[tex]i = \frac{V_{0} }{R} e^{-\frac{t}{RC} }[/tex] where i₀ = V₀/R
For the current i to decay to 5%i₀, i/i₀ = 0.05. Where R = resistance = 2.0 × 10⁶ Ω and C = capacitance = 3.0 μF = 3.0 10⁻⁶ F
So,
[tex]i = {i_{0}e^{-\frac{t}{RC} }[/tex]
[tex]{\frac{i}{i_{0} } = e^{-\frac{t}{RC}[/tex]
[tex]0.05 = e^{-\frac{t}{2 X 10^{6} X 3 X 10^{-6}}\\[/tex]
[tex]0.05 = e^{-\frac{t}{6}[/tex]
taking natural logarithm of both sides
-t /6 = ln(0.05)
t = -ln(0.05)/6 = 0.5 s
So it takes the current 0.5 s to reach 5% of its original value
In an RC circuit, the time required for the current to decay to 5% of its original value, or for the capacitor to reach 95% of its maximum charge, is approximately three times the circuit's time constant. Here, the time constant is 6 seconds, so the time required is approximately 3 * 6 = 18 seconds.
Explanation:In an RC circuit, the time required for the current to decay to a certain percentage of its original value is connected to the circuit's time constant. The time constant is calculated by the formula τ = R*C. In your case, R equals 2.0 * 106 ohms and C equals 3.0 * 10-6 farads, so τ = 2.0 * 106 * 3 * 10-6 = 6 seconds.
The time required for the current to decay to 5% of its original value or for the capacitor to reach 95% of its maximum charge in an RC circuit is roughly three time constants, or 3τ.From the calculation above, τ equals 6 seconds, so the time required is approximately 3 * 6 = 18 seconds.
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A bob of mass m is suspended from a fixed point with a massless string of length L (i.e., it is a pendulum). You are to investigate the motion in which the string moves in a cone with half-angle θ.
What tangential speed, v, must the bob have so that it moves in a horizontal circle with the string always making an angle from the vertical?
How long does it take the bob to make one full revolution (one complete trip around the circle)?
Express your answer in terms of m, L, θ and acceleration due to gravity g.
The tangential speed v that the bob must have is: v = [tex]\sqrt{gLsin(\theta)tan(\theta)[/tex]
And the time for one full revolution is: T = [tex]\frac{2\pi \sqrt{Lsin(\theta)}}{\sqrt{g tan(\theta)}}[/tex]
To solve this problem, we need to find the tangential speed, v, that the bob must have to move in a horizontal circle and the time, T, it takes to make one complete revolution. Here’s a step-by-step approach to the solution:
Determine the radius of the circular motion (R):
Given that the string length is L and the angle from the vertical is θ, the radius of the horizontal circle can be expressed as:
R = L sin(θ)
Analyze the forces acting on the bob:
The forces acting on the bob are:
The tension in the string (T), which has components both in the radial (horizontal) and vertical directionsThe gravitational force (mg), acting vertically downwardEquations of motion:
In the vertical direction, since there is no vertical acceleration, the vertical component of the tension must balance the weight of the bob:
Tcos(θ) = mg
In the horizontal direction, the horizontal component of the tension provides the centripetal force necessary for the circular motion:
Tsin(θ) = mv²/R
Solve for the tension (T):
From the vertical force balance equation:
T = mg /cos(θ)
Substitute T into the horizontal force equation and solve for v:
mgsin(θ) /cos(θ) = mv²/R
Simplifying, we get:
gtan(θ) = v² /R
Substitute R = L sin(θ):
gtan(θ) = Lsin(θ)v²
Solving for v, we get:
v = [tex]\sqrt{gLsin(\theta)tan(\theta)}[/tex]
Determine the period of the motion (T):
The time for one complete revolution (T) is the circumference of the circle divided by the tangential speed v:
T = 2πR /v
Substitute R and v:
T = [tex]\frac{2\pi Lsin(\theta)}{\sqrt{gLsin(\theta)tan(\theta)}}[/tex]
Simplifying further:
T = [tex]\frac{2\pi \sqrt{Lsin(\theta)}}{\sqrt{g tan(\theta)}}[/tex]
Therefore, the tangential speed v: v = [tex]\sqrt{gLsin(\theta)tan(\theta)[/tex] ,and the period of revolution is: T = [tex]\frac{2\pi \sqrt{Lsin(\theta)}}{\sqrt{g tan(\theta)}}[/tex].
Awning windows can be 100% openable and are best used where extreme weather conditions require a tight seal when the window is closed, although these windows are in common use everywhere.1. True2. False
Answer:
The correct option is false
Explanation:
Awning windows are easy to open windows (made of glass) that are hinged at the top and are opened from the bottom. These windows form a slant (of about 45 degree) after opening hence do not open 100%; this slant opening is advantageous in preventing rain from entering the building.
The windows can also be tightly sealed (from the inside) during extreme/cold weather conditions but are not commonly used everywhere in the world because of there limitations (slight openings) which can prevent proper ventilation in hot regions.
A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 15-m-long cable. You are (unwisely!) standing directly beneath the point from which the wrecking ball is hung when you notice that the ball has just been released and is swinging directly toward you.
Answer:
1.9 s
Explanation:
We are given that
Length of cable=l=15 m
We have to find the time you have to move out of the way.
We know that
Time period,T=[tex]2\pi\sqrt{\frac{l}{g}}[/tex]
Where g=[tex]9.8m/s^2[/tex]
By using the formula
[tex]T=2\pi\sqrt{\frac{15}{9.8}}[/tex]
[tex]T=2\times 3.14\times \sqrt{\frac{15}{9.8}}=7.77 s[/tex]
Time you have to move out
[tex]t=\frac{T}{4}=\frac{7.77}{4}=1.9 s[/tex]
Hence,time you have to move out of the way=7.77 s