To solve this problem, let us consider that the Earth is the origin, the initial reference point. Therefore the speed of rocket plus the missile would be 0.8 C
Now after the rocket had moved away from Earth, it fired a missile at a speed of 0.7 C. Now the reference made to this is relative to the rocket. We have established that our initial reference point is the Earth, therefore the real speed of the missile with reference to Earth is:
Speed of missile relative to Earth = 0.8 C + 0.7 C
Speed of missile relative to Earth = 1.5 C
Answer is:
A
Astronaut mark uri is space-traveling from planet x to planet y at a speed of relative to the planets, which are at rest relative to each other. when he is precisely halfway between the planets, a distance of 1.0 light-hour from each one as measured in the planet frame, nuclear devices are detonated on both planets. the explosions are simultaneous in mark's frame. what is the difference in the time of arrival of the flashes from the explosions as observed by mark?
The time of arrival of the flashes from the explosions will be different for Mark due to time dilation. Mark's velocity relative to the planets will cause a time difference between the observed arrival times. The Lorentz transformation formula can be used to calculate the time dilation factor.
Explanation:The difference in the time of arrival of the flashes from the explosions as observed by Mark can be calculated using the concept of time dilation. According to the theory of special relativity, time is relative and depends on the observer's frame of reference. As Mark is traveling at a speed relativistic to the planets, the time measured by Mark will be different from the time measured by an observer at rest on the planets.
In this scenario, Mark is traveling halfway between the planets, so the distance to each planet from Mark is 1.0 light-hour. The explosions on both planets are simultaneous according to Mark's frame of reference. However, due to time dilation, the time of arrival of the flashes from the explosions will be different as observed by Mark.
The time dilation factor can be calculated using the Lorentz transformation formula:
t' = t * sqrt(1 - (v^2 / c^2))
Where:
t' is the time measured by Mark
t is the time measured in the planet frame
v is Mark's velocity relative to the planets
c is the speed of light
Since Mark is traveling at a speed relativistic to the planets, his velocity v will be a significant fraction of the speed of light, resulting in a noticeable time dilation effect.
The common balance works on the principle of equality of
Answer:
HAve two arms suspended and one known weight
Explanation:
the common balance is a balance that we all have seen in examples of Justice, it was well known in the old times, but nowadays its more common seeing it as a jsutice symbol, it works in the principle of having two weights accross a balance and letting gravity work on them, while you have an object on one side of the balance of which you know the mass and you add mass to the object in the othar side of the balance to equalize the weights.
A grocery cart with a mass of 15 kg is pushed at constant speed along an aisle by a force fp = 12 n which acts at an angle of 17° below the horizontal. find the work done by each of the forces on the cart if the aisle is 14 m long.
Given:
Mass of cart: 15kg
Aisle length = 14m
Angle = 17° below the horizontal
Force fp = 12 N
So, the solution would be like this for this specific problem:
1) W(by applied force) = F(applied) x s x cosθ
The speed of sound at 0°C is 331.5 m/s. Calculate the speed of sound in the room at 20.0°C.
Answer:
v = 343.5 m/s
Explanation:
As we know that speed of sound at a given temperature "t" is given by the formula
[tex]v = 331.5 + 0.6 t[/tex]
now we know that
if t = 0 degree Celsius
then the speed of sound will be
v = 331.5 m/s
now at t = 20 degree Celsius
[tex]v = 331.5 + 0.6(20)[/tex]
[tex]v = 343.5 m/s[/tex]
so the speed will be 343.5 m/s
A bullet is fired horizontally with an initial velocity of 144.7 m/s from a tower 11 m high. if air resistance is negligible, what is the horizontal distance the bullet travels before hitting the ground?
Final answer:
To find the horizontal distance a bullet travels before hitting the ground when fired from a certain height, one must calculate the time of fall due to gravity and then multiply by the horizontal velocity. In the given scenario, the bullet would travel approximately 216.5 meters.
Explanation:
The problem of calculating the horizontal distance a bullet travels before hitting the ground involves understanding projectile motion, specifically when an object is fired horizontally from a certain height. The question states that a bullet is fired horizontally with an initial velocity of 144.7 m/s from a tower 11 m high, and we need to assume air resistance is negligible. To find the horizontal distance, we first need to determine the time it takes for the bullet to reach the ground.
In the absence of air resistance, the horizontal motion of the bullet is at a constant velocity. Meanwhile, the vertical motion is influenced by gravity and can be calculated using the following kinematic equation for free fall:
t =√(2h/g)
Where:
t is the time in seconds
h is the height in meters
g is the acceleration due to gravity (approximately 9.81 m/s²)
Once we have the time, we can calculate the horizontal distance using the equation:
x = v * t
Where:
x is the horizontal distance
v is the horizontal velocity
t is the time
Applying these calculations:
t = √(2 * 11 m / 9.81 m/s²) = √(2.24 s²) = 1.497 s (approximately)
Then:
x = 144.7 m/s * 1.497 s = 216.5329 m (approximately)
Therefore, the bullet will travel approximately 216.5 meters horizontally before hitting the ground.
An ideal gas is at a pressure 1.00 Ã 105 n/m2 and occupies a volume 2.00 m3. if the gas is compressed to a volume 1.00 m3 while the temperature remains constant, what will be the new pressure in the gas?
Cl is highly reactive you would expect NaCl to
Have similar properties
Be highly reactive
To have completely different properties
Take on some of the properties of Cl
Although it might be true that the chemical identity of Cl (Chlorines) is known to be highly reactive, but when it combines with another element, it loses its former chemical identity. This means that Cl and NaCl would definitely have different chemical identities and therefore different reactivities. Another great example would be O (Oxygen). It is a gas in its natural state when alone but then it becomes a liquid when combined with H (Hydrogen) to form what we called water (H2O). Therefore the correct answer among the choices would be:
Cl is highly reactive you would expect NaCl “To have completely different properties”.
A test rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 49 m and acquired a velocity of 30m/s. How long did the burn phase last?
You have developed a method in which a paint shaker is used to measure the coefficient of static friction between various objects and a known surface. the shaker oscillates with a fixed amplitude of 40 mm , but you can adjust the frequency of the motion. you have affixed a horizontal tabletop (the known surface) to the shaker so that the tabletop oscillates with it. then you put an object on the tabletop and increase the frequency until the object begins to slip on the surface. part a if a frequency f = 1.75 hz is required before a penny positioned on the tabletop starts to slide, what is the coefficient of static friction between penny and tabletop?
Two children standing on opposite sides of a merry-go-round are trying to rotate it. they each push in opposite directions with forces of magnitude 10.2 n. (a) if the merry-go-round has a mass of 180 kg and a radius of 1.8 m, what is the angular acceleration of the merry-go-round? (assume the merry-go-round is a uniform disk.)
The angular acceleration of the merry-go-round can be found using Newton's second law for rotation and the moment of inertia. To calculate the total moment of inertia, we can approximate the child as a point mass and add it to the moment of inertia of the merry-go-round which is 1.8
Explanation:The angular acceleration of the merry-go-round can be found using Newton's second law for rotation which states that the torque is equal to the moment of inertia times the angular acceleration. In this case, the torque is equal to the product of the force applied by each child and the radius of the merry-go-round. The moment of inertia of the merry-go-round can be found using the equation [tex]I = 1/2 * MR^2,[/tex] where M is the mass and R is the radius of the merry-go-round.
Thus,
F = m a
a = F / m
a = 10.2 N / 180 kg
[tex]a = 0.057 m / s^2[/tex]
The relationship between angular velocity (a) and angular velocity (ω) is:
w = a / r
[tex]w = (0.057 m / s^2) / 1.8 m[/tex]
[tex]w = 0.031 rad / s^2[/tex]
To get an answer in terms of degrees per s^2, we multiply it with the conversion factor = 180˚ / π
[tex]w = (0.031 rad / s^2) (180 / \pi rad)[/tex]
[tex]w = 1.8 / s^2[/tex]
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The position of a 55 g oscillating mass is given by x(t)=(2.0cm)cos(10t), where t is in seconds. determine the velocity at t=0.40s
The velocity of the oscillating mass at time [tex]t = 0.40\,{\text{s}}[/tex] is [tex]\boxed{15\times{10^{-2}}\,{\text{m/s}}}[/tex] or [tex]\boxed{15\,{\text{cm/s}}}[/tex] or [tex]\boxed{0.15\,{\text{m/s}}}[/tex].
Further explanation:
Velocity of a particle or a mass at any instant is defined as the rate of change of position of particle with respect to time.
Mathematically,
[tex]V\left( t \right) = \dfrac{d}{{dt}}\left( {X\left( t \right)} \right)[/tex]
If position of a particle or mass is a function of time then velocity of mass at any instant will change with respect to time.
Given:
The position of an oscillating mass varies according to the function [tex]X(t)=({2.0\text{ cm}}})\cos({10t})[/tex].
Mass of an oscillating object is [tex]55\text{ g}[/tex].
Concept:
The velocity of mass at any instant is calculated by using the following relation
[tex]\begin{aligned}V(t)&=\frac{{dX\left( t \right)}}{{dt}}\\&=\frac{d}{{dt}}\left[{\left( {2.0{\kern 1pt} {\text{cm}}} \right)\cos \left( {10t} \right)} \right]\\&=-\left( {20\,{\text{cm/s}}}\right)\sin\left( {10t}\right)\\\end{aligned}[/tex]
Therefore the velocity of the mass at any instant is given by
[tex]V\left( t \right)=-\left( {20{\kern 1pt} {\text{cm/s}}} \right)\sin \left( {10t} \right)[/tex]
From the above expression of velocity it can be observed that velocity is changing with time according to the sin function.
Substitute [tex]0.40\,{\text{s}}[/tex] for t in the above expression
[tex]\begin{aligned}V\left( {0.4\,{\text{s}}} \right)&=-\left( {20\,{\text{cm/s}}} \right)\sin \left( 4 \right)\\&=-\left( {20\,{\text{cm/s}}} \right)\left( { - 0.757} \right)\\&=15.14\,{\text{cm/s}}\\\end{aligned}[/tex]
Thus, the velocity of the oscillating mass at time [tex]t = 0.40\,{\text{s}}[/tex] is [tex]\boxed{15\times{10^{-2}}\,{\text{m/s}}}[/tex] or [tex]\boxed{15\,{\text{cm/s}}}[/tex] or [tex]\boxed{0.15\,{\text{m/s}}}[/tex].
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Answer Details:
Grade: College
Subject: Physics
Chapter: Force
Keywords:
position, oscillating, 55 g, mass, time, x(t)=(2.0cm)cos(10t), t=4 s, determine, velocity, 15 cm/s, 0.15 m/s, rate, change in position.
Which of these should you not do when merging onto a freeway? A. Select a gap large enough to fit your vehicle. B. Change lanes smoothly. C. Change two lanes at once and speed up. D. Adjust your speed to create a safe following interval.
a piano has a mass of 185 kg, and the coefficient of friction between it and the floor is 0.39. What is the maximum force of friction between the piano and the floor?
707 N
523 N
1813 N
1208 N
Volcanoes are formed ________. question 3 options:
a.when the earth's core erupts onto the surface
b.where the earth's crust is especially thick
c.in subduction zones or in rift valleys
d.after earthquakes have damaged the earth's crust
e.as a result of water seeping into the core from the surface
The human body can survive a negative acceleration trauma incident (sudden stop) if the magnitude of the acceleration is less than 250 m/s2 (approximately 25g). if you are in an automobile accident with an initial speed of 76 km/h (47 mi/h) and are stopped by an airbag that inflates from the dashboard, over what distance must the airbag stop you if you are to survive the crash?
To survive an auto accident at a speed of 76 km/h, considering a max survivable acceleration of 250 m/s², the airbag must stop you over a minimum distance of approximately 0.89 meters.
Explanation:To determine the distance an airbag must stop you to survive a crash at an initial speed of 76 km/h we must first convert this speed into meters per second:
76 km/h = (76*1000 meters) / (3600 seconds) = 21.11 m/s.Using the formula for deceleration (a = Δv / Δt) where Δv is the change in velocity and Δt is the change in time, and the fact that the maximum survivable acceleration is 250 m/s2, we can calculate the minimum time (Δt) needed to decelerate:
Δv = 21.11 m/s (since the final velocity is 0 m/s).250 m/s2 = 21.11 m/s / Δt, solving for Δt gives Δt = 21.11 / 250 seconds.Then, we can use the formula for distance covered under constant acceleration (d = 0.5 * a * Δt2) to find the distance:
d = 0.5 * 250 m/s2 * (Δt2).d = 0.5 * 250 m/s2 * (21.11 / 250)2 meters.After calculation, d ≈ 0.89 meters.To survive the crash, the airbag must stop you over a distance of at least 0.89 meters.
Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 163 v. (b) calculate the speed of an electron that is accelerated through the same potential difference.
Battery life is what distinguishes one type of mobile computer from another.
The starship enterprise returns from warp drive to ordinary space with a forward speed of 50 km/s. to the crew’s great sur- prise, a klingon ship is 100 km directly ahead, traveling in the same direction at a mere 20 km/s. without evasive action, the enterprise will overtake and collide with the klingons in just slightly over 3.0 s. the enterprise’s computers react instantly to brake the ship. what magnitude acceleration does the enterprise need to just barely avoid a collision with the klingon ship? assume the acceleration is constant.
The magnitude of the acceleration the Enterprise needs to just barely avoid a collision with the Klingon ship is 3.27 x 10^10 m/s^2.
Explanation:To find the magnitude acceleration the Enterprise needs to just barely avoid a collision with the Klingon ship, we can use the impulse-momentum relation. The change in momentum can be determined by subtracting the initial momentum from the final momentum. The mass of the Enterprise is given as 2 x 10^9 kg. The initial momentum can be calculated by multiplying the mass of the Enterprise by its initial speed, and the final momentum can be calculated by multiplying the mass of the Klingon ship by its final speed. Using the formula for impulse, which is the change in momentum divided by the time interval, we can solve for the acceleration.
The initial momentum of the Enterprise is (2 x 10^9 kg) x (50 km/s) = 1 x 10^11 kg m/s. The final momentum of the Klingon ship is (100 km) x (20 km/s) = 2 x 10^8 kg m/s. The change in momentum is (2 x 10^8 kg m/s) - (1 x 10^11 kg m/s) = -9.80 x 10^10 kg m/s. The time interval is given as 3.0 s.
Using the formula for impulse, we have:
Force x Time Interval = Change in Momentum
Force = Change in Momentum / Time Interval
Force = (-9.80 x 10^10 kg m/s) / (3.0 s) = -3.27 x 10^10 N
The magnitude of the acceleration the Enterprise needs to just barely avoid a collision with the Klingon ship is 3.27 x 10^10 m/s^2. The negative sign indicates that the acceleration is in the opposite direction of the initial velocity of the Enterprise.
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To avoid the collision, the Enterprise needs an acceleration of approximately 2.22 km/s². This calculation involves initial speeds, distances, and applying the constant acceleration formula. The relative speed and distance are critical factors in determining the necessary braking acceleration.
To prevent a collision with the Klingon ship, we need to determine the magnitude of acceleration required by the starship Enterprise. Here’s a step-by-step breakdown:
The initial relative speed between the Enterprise and the Klingon ship: [tex]\Delta v = 50 km/s - 20 km/s = 30 km/s.[/tex]The distance to the Klingon ship:[tex]d = 100 km[/tex].The time to collision without evasive action:[tex]t = 3 s[/tex].Using the equation for constant acceleration: [tex]d = v_i * t + 0.5 * a * t^2[/tex]. Here, d is the distance, v_i is the initial speed, a is acceleration, and t is time.Since the Enterprise needs to stop before reaching 100 km, [tex]v_i * t[/tex] must be subtracted from d:[tex]100 km = (30 km/s) * 3 s + 0.5 * a * (3 s)^2[/tex]
Simplifying the equation:
[tex]100 km = 90 km + 4.5a[/tex]
Solve for a:
[tex]10 km = 4.5a[/tex]
[tex]a = - (10 km) / (4.5 s^2) = -2.22 km/s^2[/tex]
Thus, the magnitude of the required acceleration is approximately 2.22 km/s².
At a certain temperature, a 29.5-l contains holds four gases in equilibrium. their masses are: 3.5 g so3, 4.6 g so2, 22.6 g n2, and 0.98 g n2o. what is the value of the equilibrium constant at this temperature for the reaction of so2 with n2o to form so3 and n2 (balanced with lowest whole-number coefficients)?
The equilibrium constant for the reaction of SO2 with N2O to form SO3 and N2 is approximately 22.2.
The given problem involves finding the equilibrium constant for the reaction of SO2 with N2O to form SO3 and N2. The balanced chemical equation is:
SO2 (g) + N2O (g) ↔ SO3 (g) + N2 (g)
First, we need to calculate the number of moles of each gas using their masses and molar masses:
[tex]\text{Molar mass of }SO_2: 64.07 g/mol\\\text{Molar mass of }SO3: 80.07 g/mol\\\text{Molar mass of }N_2: 28.02 g/mol\\\text{Molar mass of }N_2O: 44.01 g/mol[/tex]
Using the formula: moles = mass / molar mass, we find:
[tex]\text{Moles of }SO_2 = 4.6 g / 64.07 g/mol = 0.0718 mol\\\text{Moles of }SO_3 = 3.5 g / 80.07 g/mol = 0.0437 mol\\\text{Moles of }N_2 = 22.6 g / 28.02 g/mol = 0.8065 mol\\\text{Moles of }N_2O = 0.98 g / 44.01 g/mol = 0.0223 mol[/tex]
The equilibrium constant expression, [tex]K_c[/tex], for the reaction is:
[tex]K_c = [SO_3][N_2] / [SO_2][N_2O][/tex]
In a 29.5 L container, the concentrations of each gas are:
[tex]SO_2= 0.0718 mol / 29.5 L = 0.00243 M\\SO_3 = 0.0437 mol / 29.5 L = 0.00148 M\\N_2 = 0.8065 mol / 29.5 L = 0.0273 M\\N_2O= 0.0223 mol / 29.5 L = 0.000756 M[/tex]
Now substitute these concentrations into the equilibrium expression:
[tex]Kc = (0.00148 M) \times (0.0273 M) / (0.00243 M) \times (0.000756 M)\\Kc = 22.2[/tex]
Efforts to relate gamma-ray bursts with specific sources has had what results so far?
X-rays are high energy electrons that can cause damage when exposed under extreme conditions. The best technology that can block it is using a lightweight type of metal foam. It can take in high energy collisions which also exhibits high forces. it does not only block x-rays but also, neutron radiation and gamma rays. A gamma ray primarily consists of pure energy and no mass is true. In fact it consists of high energy photons and massless. They have no charge and when exposed to any kind of material, it just passes through as if nothing is blocking it.
The acceleration of a motorcycle is given by ax(t)=at−bt2, where a=1.50m/s3 and b=0.120m/s4. the motorcycle is at rest at the origin at time t=0. calculate the maximum velocity it attains.
The maximum velocity it attains is 39.1 m/s
Further explanationThe velocity is changing over the course of time. Velocity is the rate of motion in a specific direction. Whereas acceleration is the rate of change of velocity of an object with respect to time. Maximum velocity is reached when you stop accelerating, To calculate velocity using acceleration, we start by multiplying the acceleration by the change in time
The acceleration of a motorcycle:
where [tex]a = 1.50 \frac{m}{s^{3}}[/tex] and [tex]ax* (t) = at - b*t^{2}[/tex]
[tex]a = 0.120 \frac{m}{s^{4}}[/tex]
The motorcycle is at rest at the origin at time t=0.
The maximum velocity it attains = ?
Answer:
[tex]v(t) = (1/2)At^2 - (1/3)Bt^3 v(0)[/tex]
but v(0) = 0.
[tex]x(t) = (1/6)At^3 - (1/12)Bt^4 x(0)[/tex]
but x(0) = 0.
[tex]v(t) = (0.75 \frac{m}{s^{3}}) t^2 - (0.04 \frac{m}{s^{4}})t^3[/tex]
Next step is find its position as a function of time.
[tex]x(t)= x(t) = (0.25 \frac{m}{s^{3}})t^3 - (0.01 \frac{m}{s^{4}})t^4[/tex]
Then, calculate the maximum velocity it attains.
Max velocity will be attained when
[tex]At = Bt^2[/tex]
T= 1.50/0.120 = 12.5 seconds
V(t) = 0.750(12.5)^2 – 0.04(12.5)^3 = 39.1 m/s
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Subject: physics
Chapter: the maximum velocity
Keywords: the maximum velocity
a car slows down uniformly from a speed of 21.0 m/s to rest in 6.00 s. how far did it travel in that time?
The keyword slows down “uniformly” means that the car had a constant deceleration. Therefore we can use the formulas for linear motion.
a = (vf – vi) / t ---> 1
s = vi t + 0.5 a t^2 ---> 2
where a is acceleration, v is velocity with notation f and i for final and initial, s is the distance travelled, and t stands for time
First we solve for acceleration using equation 1:
a = (0 – 21) / 6
a = - 3.5 m / s^2 (negative means deceleration)
Then we can now calculate for the value of s using equation 2:
s = 21 (6) + 0.5 (-3.5) (6)^2
s = 63 m
Therefore the car travelled 63 m before it came to a stop.
The car traveled a distance of 252.0 meters in that time.
Explanation:To determine the distance the car traveled, we can use the equation for uniformly accelerated motion: x = ut + (1/2)at^2. In this case, the initial velocity u is 21.0 m/s, the acceleration a is -21.0 m/s^2 (negative because the car is slowing down), and the time t is 6.00 s. Plugging these values into the equation, we get:
x = (21.0 m/s)(6.00 s) + (1/2)(-21.0 m/s^2)(6.00 s)^2
x = 126.0 m - 378.0 m = -252.0 m
The negative sign indicates that the distance is in the opposite direction of the initial motion. So, the car traveled a distance of 252.0 meters in that time.
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Bright white shadings on infrared images indicate cloud tops that have relatively ________ temperatures
). with the input voltage range set at +/- 500mv, what is the smallest difference in voltage that can be resolved? show your calculation.
The primary coil of an ideal transformer has 100 turns and its secondary coil has 400 turns. if the ac voltage applied to the primary coil is 120 v, what voltage is present in its secondary coil?
The formula used in calculations relating to transformers is:
Secondary voltage (Vs)/ Primary voItage (VP) = Secondary turns (nS)/ Primary turns (nP)
Substituting the given values to find for Vs,
Vs / 120 V = 400 turns / 100 turns
Vs = 480 V
The voltage in the secondary coil of the transformer is 480 volts in this scenario, which is obtained by using the transformer equation to adjust the primary voltage according to the ratio of the number of turns in the secondary and primary coils.
Explanation:The subject of this question is an ideal transformer, which is a device that changes the voltage of an alternating current (AC) in a process known as electromagnetic induction based on Faraday's law. The output voltage (Vs) changes according to the ratio of the number of turns in the secondary coil (Ns) to the number of turns in the primary coil (Np). This relationship is given by the transformer equation: Vs/Vp = Ns/Np.
In your case, the number of turns in the primary coil (Np) is 100 and in the secondary coil (Ns) is 400. The given primary voltage (Vp) is 120 V.
By rearranging the transformer equation for Vs, we get: Vs = Vp * (Ns/Np).
Therefore, substituting the given values in this equation, we find: Vs = 120 V * (400 / 100) = 480 V. This implies that the voltage in the secondary coil of your transformer is 480 volts.
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A certain automobile manufacturer claims that its super-deluxe sports car will accelerate from rest to a speed of 42.3 m/s in 8.02 s. find the acceleration of the car. assume that the acceleration of the car is constant. answer in units of m/s 2 .
Answer:
5.27 m/s²
Explanation:
Given data
Initial velocity (v₀): 0 m/s (rest)Final velocity (vf): 42.3 m/sElapsed time (t): 8.02 sAcceleration (a): ?We can determine the acceleration of the car using the following kinematic expression.
a = Δv / t
a = vf - v₀ / t
a = (42.3 m/s - 0 m/s) / 8.02 s
a = 5.27 m/s²
The acceleration of the car is 5.27 m/s².
He electric potential at a certain distance from a point charge can be represented by v. what is the value of the electric potential at twice the distance from the point charge?
What is the mass, in kilograms, of an avogadro's number of people, if the average mass of a person is 150 lb ?
Strontium has density of 2.64 g/cm3 and crystallizes with the face-centered cubic unit cell. part a calculate the radius of a strontium atom.
The radius of a strontium atom can be calculated using its given density and the properties of its face-centered cubic unit cell. The mass and volume involved in computing for density pertain to the unit cell, with the atomic mass of strontium and Avogadro's number used to determine atomic mass in grams. The radius is indirectly determined through the side length of the cubic unit cell.
Explanation:The question is asking for the radius of a strontium atom, given that the strontium atom crystallizes in a face-centered cubic unit cell and its density is provided. For the face-centered cubic unit cell, we can approximate that there are four atoms in the unit cell: one-eighth of an atom at each of the eight corners (8 × 1/8 = 1 atom) and one-half of an atom on each of the six faces (6 × 1/2 = 3 atoms).
The atomic mass of strontium (Sr) is approximately 87.62 g/mol. To calculate the radius, we know that density = mass/volume. The mass of strontium is given by the number of atoms per unit cell times the atomic mass of strontium (converted to grams using Avogadro's number), divided by the volume of one unit cell. The side length of the unit cell, 'a', is related to the radius of the strontium atom, 'r', by the equation a = √2 * 4r. By substituting the given density and calculating for 'r', we can determine the radius of the strontium atom.
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The crew of an enemy spacecraft attempts to escape from your spacecraft by moving away from you at 0.259 of the speed of light. but all is not lost! you launch a space torpedo toward the foe at 0.349 of the speed of light with respect to you. at what speed in kilometers per second does the enemy crew observe the torpedo approaching its spacecraft?
The enemy crew observes the torpedo approaching its spacecraft at a speed of 0.55c or 165,000 kilometers per second.
Explanation:To determine the speed at which the enemy crew observes the torpedo approaching its spacecraft, we need to use relativistic velocity addition. In this case, the velocity of the torpedo as observed by the enemy crew can be calculated by adding the velocities of the torpedo with respect to you and the velocity of the enemy crew's spacecraft with respect to you. Using the formula for relativistic velocity addition, the velocity of the torpedo as observed by the enemy crew is:
v_t &= (v_{torpedo} + v_{enemy}) / (1 + v_{torpedo} * v_{enemy} / c^2)
where v_t is the velocity of the torpedo as observed by the enemy crew, v_{torpedo} is the velocity of the torpedo with respect to you (0.349c), v_{enemy} is the velocity of the enemy crew's spacecraft with respect to you (0.259c), and c is the speed of light. Plugging in the values, we have:
v_t &= (0.349 * c + 0.259 * c) / (1 + 0.349 * 0.259)
Simplifying the expression, we find that the velocity of the torpedo as observed by the enemy crew is approximately 0.55c or 165,000 kilometers per second.
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The enemy crew observes the torpedo approaching at approximately 29,658 km/s. This calculation uses the relativistic velocity addition formula and accounts for the relative velocities of the spacecraft and the torpedo.
This problem involves the concept of relative velocity in special relativity. We will use the relativistic velocity addition formula to find the speed of the torpedo as observed by the enemy spacecraft:
Relativistic velocity addition formula:
u' = (u + v) / (1 + uv/c²)
Here:
u = 0.349c (speed of the torpedo relative to your spacecraft)
v = -0.259c (speed of the enemy spacecraft relative to your spacecraft; negative because it's moving away)
c = speed of light
Substituting the values into the formula:
u' = (0.349c - 0.259c) / (1 - (0.349 × 0.259))
u' = (0.090c) / (1 - 0.090491)
u' ≈ 0.09886c
Therefore, the enemy crew observes the torpedo approaching at approximately 0.09886 times the speed of light. To convert this to kilometers per second (km/s):
c ≈ 300,000 km/s
u' ≈ 0.09886 × 300,000 km/s ≈ 29,658 km/s
The enemy crew observes the torpedo approaching at approximately 29,658 km/s.