The required height of the tin can is given by the height at the minimum value of the surface area function of the tin can
The height of the tin can must be 4 inchesReason:
The given information on the tin can are;
Volume of the tin can = 16·π in.³
The amount of tin to be used = Minimum amount
The height of the can in inches required
Solution;
Let h, represent the height of the can, we have;
The surface area of the can, S.A. = 2·π·r² + 2·π·r·h
The volume of the can, V = π·r²·h
Where;
r = The radius of the tin can
h = The height of the tin can
Which gives;
16·π = π·r²·h
16 = r²·h
[tex]h = \dfrac{16}{r^2}[/tex]
Which gives;
[tex]S.A. = 2 \cdot \pi \cdot r^2 + 2 \cdot \pi \cdot r \cdot \dfrac{16}{r^2} = 2 \cdot \pi \cdot r^2 + \pi \cdot \dfrac{32}{r}[/tex]
When the minimum amount of tin is used, we have;
[tex]\dfrac{d(S.A.)}{dx} =0 = \dfrac{d}{dx} \left( 2 \cdot \pi \cdot r^2 + \pi \cdot \dfrac{32}{r} \right) = \dfrac{4 \cdot \pi \cdot \left (r^3-8 \right)}{r^2}[/tex]
Therefore;
[tex]\dfrac{4 \cdot \pi \cdot \left (r^3-8 \right)}{r^2} = 0[/tex]
4·π·(r³ - 8) = r² × 0
r³ = 8
r = 2
The radius of the tin can, r = 2 inches
The
[tex]h = \dfrac{16}{2^2} = 4[/tex]
The height of the tin can, h = 4 inches
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The required height of can is 4 inches.
Given data:
The volume of cylindrical tin with top and bottom is, [tex]V = 16 \pi \;\rm in^{3}[/tex].
The volume of tin is,
[tex]V = \pi r^{2}h\\16 \pi = \pi r^{2}h\\[/tex]
here, r is the radius of can and h is the height of can.
[tex]16 =r^{2}h\\\\h=\dfrac{16}{r^{2}}[/tex]
The surface area of tin is,
[tex]SA = 2\pi r(r+h)\\SA = 2\pi r(r+\dfrac{16}{r^{2}})\\SA = 2\pi r^{2}+\dfrac{32 \pi}{r})[/tex]
For minimum amount of tin used, we have,
[tex]\dfrac{d(SA)}{dr} =0\\\dfrac{ d(2\pi r^{2}+\dfrac{32 \pi}{r})}{dr} = 0\\4\pi r-\dfrac{32 \pi}{r^{2}}=0\\4\pi r=\dfrac{32 \pi}{r^{2}}\\r^{3}=8\\r =2[/tex]
So, height of can is,
[tex]h=\dfrac{16}{2^{2}}\\h=4[/tex]
Thus, the required height of can is 4 inches.
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Region that contains the majority of molecules in the atmosphere
Answer:
Trophosphere
Explanation:
The troposphere is the atmospheric layer closest to the planet and is characterized because contains the largest percentage of the mass of the total atmosphere.
On special characteristic is that in this layer the temperature and water vapor content decrease a lot respect to the altitude. Also on this layer the Water vapor is important in order to regulate the air temperature since on this zone we have absorption of the solar energy.
The troposphere contains almost all the water vapor in the atmosphere. And specially on the tropics we have an accumulation of the water vapour.
All weather phenomena occur within the troposphere. Tropos means "change" and Troposphere means "region of mixing".
Above this layer, we have the tropopause, ranges in height from 5 miles near the poles up to 11 miles above the equator. And the height depends of the seasons, with an special characteristic: the is highest height occurs in the summer and lowest height occurs in the winter.
The troposphere contains almost 75% of the mass of the entire atmosphere. The air on this layer is composed by 78% nitrogen, 21% oxygen and 1% is made of argon, water vapor, and carbon dioxide.
So for this reason this is the Region that contains the majority of molecules in the atmosphere.
After a completely inelastic collision, two objects of the same mass and same initial speed move away together at half their initial speed. Find the angle between the initial velocities of the objects.
Answer:
120°
Explanation:
given,
mass of two objects are same , Let it be m
speed of both the objects = v
after inelastic collision the
mass are moving with velocity = v/2
let the angle between them be θ₁ and θ₂
writing the equation along y-axis
m v sinθ₁ - m v sinθ₂ = 0
m v sinθ₁ = m v sinθ₂
θ₁ = θ₂ = θ
writing the equation along x- axis
m v cos θ + m v cos θ = 2 m V
v cos θ = v/2
[tex]cos\theta = \dfrac{1}{2}[/tex]
[tex]\theta = cos^{-1}(\dfrac{1}{2})[/tex]
[tex]\theta = 60^0[/tex]
angle between the two objects = 2 θ
= 2 x 60° = 120°
A fluid flows through a pipe whose cross-sectional area changes from 2.00 m2 to 0.50 m2 . If the fluid’s speed in the wide part of the pipe is 3.5 m/s, what is its speed when it moves through the narrow part of the pipe?
Answer:
v₂ = 7/ (0.5)= 14 m/s
Explanation:
Flow rate of the fluid
Flow rate is the amount of fluid that circulates through a section of the pipeline (pipe, pipeline, river, canal, ...) per unit of time.
The formula for calculated the flow rate is:
Q= v*A Formula (1)
Where :
Q is the Flow rate (m³/s)
A is the cross sectional area of a section of the pipe (m²)
v is the speed of the fluid in that section (m/s)
Equation of continuity
The volume flow rate Q for an incompressible fluid at any point along a pipe is the same as the volume flow rate at any other point along a pipe:
Q₁= Q₂
Data
A₁ = 2m² : cross sectional area 1
v₁ = 3.5 m/s : fluid speed through A₁
A₂ = 0.5 m² : cross sectional area 2
Calculation of the fluid speed through A₂
We aply the equation of continuity:
Q₁= Q₂
We aply the equation of Formula (1):
v₁*A₁= v₂*A₂
We replace data
(3.5)*(2)= v₂*(0.5)
7 = v₂*(0.5)
v₂ = 7/ (0.5)
v₂ = 14 m/s
Using the Continuity Equation that states the mass flow rate must be constant in fluid flow, the speed of the fluid when it moves through the narrow part of the pipe is 14 m/s.
Explanation:The question pertains to the physics principle of fluid dynamics, specifically the concept of continuity in fluid flow. According to the Continuity Equation, the mass flow rate must remain constant across different sections of a pipe. In this particular scenario, the fluid is changing speed due to a reduction in pipe cross-sectional area. This principle is represented by the equation A₁V₁ = A₂V₂, where A₁ and A₂ represent the cross-sectional areas of the wide and narrow sections of the pipe respectively, and V₁ and V₂ represent the fluid's velocity in these sections.
Applying the equation to this scenario, we have A₁ (2.00 m²) multiplied by V₁ (3.5 m/s) equal to A₂ (0.50 m²) multiplied by V₂, the value we want to find. Solving for V₂, we get V₂ = A₁V₁ / A₂ = (2.00 m²)(3.5 m/s) / 0.50 m² = 14 m/s.
Hence, the speed of the fluid when it moves through the narrow part of the pipe is 14 m/s.
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A car and driver weighing 7130 N passes a sign stating...?
"Bridge Out 32 m Ahead." She slams on the brakes, and the car decelerates at a constant rate of 13.8 m/s^2. The acceleration of gravity is 9.8 m/s^2.
What is the magnitude of the work done stopping the car if the car just stops in time to avoid diving into the water?
Answer:
321 280 J
Explanation:
Work done = force * distance
The distance is 32 m
The force can be calculated using the second law of motion
F = ma = (7130 N ÷ 9.8 m/s²) * 13.8 m/s² = 10 040 N
Work done = force * distance
= 10 040 N * 32 m
= 321 280 J
Answer:
321 280
Explanation:
Work done = force * distanceThe distance is 32 mThe force can be calculated using the second law of motionF = ma = (7130 N ÷ 9.8 m/s²) * 13.8 m/s² = 10 040 NWork done = force * distance = 10 040 N * 32 m = 321 280 J
While driving down the road, a firefly strikes the windshield of a bus and makes a quite obvious mess in front of the face of the driver. This is a clear case of Newton's third law of motion. The firefly hit the bus and the bus hits the firefly. Which of the two forces is greater: the force on the firefly or the force on the bus?
Answer:
Both forces have the same magnitude
Explanation:
1) Notation
[tex]F_{bus}[/tex]= force exterted by the bus to the firefly
[tex]F_{firefly}[/tex]= force exterded by the firefly to the bus
2) Analysis for the situation
The reason why is from the Third law of Newton that states: "For every action, there is an equal reaction force on the opposite direction to the original action force".
For this special case on math terms we have:
[tex]F_{bus}=F_{firefly}[/tex]
The fact that the firefly splatter on this case is because since have a smaller mass, it is less able to stand up to the larger acceleration from the interaction.
3) Conclusion
Based on this analysis, each force would have the same magnitude, so none force is greater than the other.
The force on the firefly and the force on the bus are equal in magnitude but opposite in direction, demonstrating Newton's third law of motion.
Explanation:When a firefly collides with the windshield of a bus, we have a real-life illustration of Newton's third law of motion. This law states that for every action there's an equal and opposite reaction.
In the case of the firefly and the bus, both objects exert forces on each other that are equal in magnitude but opposite in direction.
The mess that is made when the firefly hits the bus might suggest to us that the force on the firefly is greater, but according to Newton's third law, the force on the firefly and the force on the bus are actually the same.
This symmetry in nature is fundamental to our understanding of physics and can be observed in many situations, like a professor walking across a room or a car accelerating down a road.
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A balloon is filled to a volume of 7.00*10^2 mL at a temperature of 20.0°C. The balloon is then cooled at constant pressure to a temperature of 1.00*10^2 K. What is the final volume of the balloon?
The final volume of the gas is 238.9 mL
Explanation:
We can solve this problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas (V) is proportional to its absolute temperature (T):
[tex]\frac{V}{T}=const.[/tex]
Which can be also re-written as
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
where
[tex]V_1, V_2[/tex] are the initial and final volumes of the gas
[tex]T_1, T_2[/tex] are the initial and final temperature of the gas
For the gas in the balloon in this problem, we have:
[tex]V_1 = 7.00\cdot 10^2 mL = 700 mL[/tex] is the initial volume
[tex]T_1=20.0^{\circ}C+273=293 K[/tex] is the initial absolute temperature
[tex]V_2[/tex] is the final volume
[tex]T_2 = 1.00\cdot 10^2 K = 100 K[/tex] is the final temperature
Solving for [tex]V_2[/tex],
[tex]V_2 = \frac{V_1 T_2}{T_1}=\frac{(700)(100)}{293}=238.9 mL[/tex]
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holiday lights are often connected in series and use special lamps that short out when the potential difference across a lamp increases to the line voltage. generate an explanation why and explain why these light sets might blow their fuses after many bulb
Explanation:
If there isn't the shorting mechanism, the whole set will be blown if anyhow one lamp burns out. Since having blown out several lamps and then shorted, the overall resistance of the remaining operating lamps will be decreased resulting in an increased working current that is adequate to blast the fuse.
A long straight rod experiences several forces, each acting at a different location on the rod. All forces are perpendicular to the rod. The rod might be in translational equilibrium, rotational equilibrium, both, or neither.
(1) If a calculation reveals that the net torque about the left end is zero, then one can conclude that the rod
(A) is definitely in rotational equilibrium.
(B) is in rotational equilibrium only if the net force on the rod is also zero.
(C) might not be in rotational equilibrium even if the net force on the rod is also zero.
(D) might be in rotational equilibrium even if the net force is not zero.
(2) If a calculation reveals that the net force on the rod is zero, then one can conclude that the rod
(A) is definitely in rotational equilibrium.
(B) is in rotational equilibrium only if the net torque about every axis through anyone point is found to be zero.
(C) might be in rotational equilibrium if the net torque about every axis through anyone point is found to be zero.
(D) might be in rotational equilibrium even if the net
torque about any axis through anyone point is not zero.
Answer:
1. C
2.C
Explanation:
1. The rod is perpendicular to every axis and forces are acting on every location. If the torque on the left side is zero, this indicates that forces with respect to their distance on the left side is zero and doesn't account for the net force at a point.
2. If the net torque about every point on every axis is zero, the rod will be rotational because each axis will yield a magnitude of zero which obeys the principle of rotation at a point.
When the net torque on a rod is zero about the left end, the rod is in rotational equilibrium. If the net force is zero, the rod is in translational equilibrium, and rotational equilibrium depends on the net torque being zero about every axis.
Equilibrium Conditions in Physics
To determine whether a rod is in equilibrium under the influence of several forces, we need to examine the conditions for both translational and rotational equilibrium. The conditions necessary for equilibrium are quite straightforward:
The net external force on the system must be zero (net F = 0), ensuring there is no linear acceleration.
The net external torque must also be zero (net T = 0), preventing any angular acceleration.
Addressing the questions:
If the net force on the rod is zero, it indicates that the rod is in translational equilibrium. However, for the rod to be in rotational equilibrium, we must also confirm that the net torque about every axis through anyone point is zero (Option B).
Each static equilibrium condition is critical in itself. While the zero net force ensures no linear acceleration, the zero net torque ensures no rotational acceleration. Both conditions are required for complete static equilibrium of an object.
The index of refraction for flint glass is 1.5. This means that
Answer:
Explanation:
This means that the ratio of the speed of the incident beam of light in a vacuum to the speed of light in the glass is 1.5
This means that the speed of light in vacuum is 1.5 times the speed of light in flint glass.
A horse runs a race on a oval track and starts and ends at the same point. He runs one mile track in 7 minutes.
What is the horse's velocity
A. 0 mph
B. 7 mph
C. 14 mph
D. 20 mph
The horse's velocity after completing a mile on an oval track and ending at the starting point is 0 mph. However, if considering speed, the horse runs at approximately 8.57 mph. The given options do not accurately represent this calculation.
Explanation:The question pertains to determining the velocity of a horse after running a complete race on an oval track, where the start and end points are the same. In physics, velocity is a vector quantity, meaning it has both magnitude and direction. Since the horse starts and finishes at the same point, the displacement over the course of the race is zero. Consequently, the average velocity is also zero because velocity is displacement divided by time.
In this scenario however, we are looking for speed, which is a scalar quantity and only considers magnitude, not direction. Speed is total distance divided by time. The horse runs a one mile track in 7 minutes. To find the speed, first convert 7 minutes to hours:
7 minutes = 7/60 hours = 0.1167 hours
Then divide the distance (1 mile) by the time in hours to find the speed:
Speed = 1 mile / 0.1167 hours = approximately 8.57 mph
Since none of the options matches this calculation, it seems there might be a mistake in the provided answer choices. If we had to choose from the given options and ignore the direction of the horse's travel, B. 7 mph would be the closest to the calculated value, albeit still not accurate.
Magma forms when solid rock in the crust and upper mantle. True or False
Answer:
True
Explanation:
Magma is known as melted rock deep within the Earth, normally coming from the melting of the upper mantle or crust. Magma is formed by the partial melting of the mantle and crust and this can occur in different ways. One way can be called heat-transfer melting. Rising magma or rock will bring heat with it, and so can melt the surrounding mantle or crustal rock. For example, magmas generated in the mantle tend to be around 1200 degrees Celsius, whereas the more silicate minerals such as quartz and orthoclase feldspar (common in continental crustal rocks) begin to partially melt at around 650-850 degrees Celsius. Therefore, the crustal rock will begin to partially melt due to the introduction of heat from rising magma. A Another way of melting rock is known as decompression melting. During decompression melting, rock from within the mantle is brought to the surface adiabatically (no exchange of heat or energy with its surroundings) and so the lithostatic pressure decreases. This means that the parcel of rising rock crosses the solidus, and so at this point the thermal vibration of the molecules is no longer counteracted by the lithostatic pressure and the rock begins to partially melt.
Answer:
The answer is true
Explanation:
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Which type of indirect evidence can substitute for direct measurements of past climate change?
Answer:
Paleoclimate proxies
Explanation:
Paleoclimate proxies are materials which are preserved and which can be analysed and correlated with parameters that relates to the environment and climate change. These materials are physical, biological or chemical and are preserved in paleoclimate archives
Scientists always seek clues from paleoclimate archives to understand how climate varied over the recent past few millions of years
Space probes may be separated from their launchers by exploding bolts. (They bolt away from one another.) Suppose a 4800-kg satellite uses this method to separate from the 1500-kg remains of its launcher, and that 5000 J of kinetic energy is supplied to the two parts. What are their subsequent velocities using the frame of reference in which they were at rest before separation?
The subsequent velocities of the satellite and the remains of the launcher can be calculated using the principle of conservation of momentum. The resultant velocities are approximately 8.70×10−² m/s and 81.5 m/s.
Explanation:The subsequent velocities of the satellite and the remains of the launcher can be calculated using the principle of conservation of momentum. Since there are no external forces acting on the system, the initial momentum of the system is equal to the final momentum. Therefore, the final velocities of the satellite and the launcher can be calculated based on their masses and the kinetic energy supplied to them.
The mass of the satellite, m1 = 4800 kg, and the mass of the launcher, m2 = 1500 kg. The total initial kinetic energy supplied to the system is given as 5000 J. To calculate the velocities, we need to find the ratio of the kinetic energies of the satellite and the launcher.
Let v1 be the velocity of the satellite and v2 be the velocity of the launcher. According to the conservation of momentum, m1 * v1 + m2 * v2 = 0. Also, the total initial kinetic energy supplied to the system is given as 5000 J, where the kinetic energy of the satellite is (1/2) * m1 * v1^2 and the kinetic energy of the launcher is (1/2) * m2 * v2^2.
Using these equations, we can solve for v1 and v2. The subsequent velocities are approximately 8.70×10−² m/s in the direction of motion of the less massive satellite and 81.5 m/s, respectively.
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The satellite moves at approximately 0.704 m/s and the remains of the launcher move at approximately -2.253 m/s in the opposite direction.
To solve for the velocities of the satellite and the launcher remains after separation, we will use the principles of conservation of momentum and the given kinetic energy.
1. Conservation of Momentum
[tex]\[m_s v_s + m_l v_l = 0\][/tex]
Since the total momentum is zero:
[tex]\[4800 \, v_s + 1500 \, v_l = 0\][/tex]
Solving for one velocity in terms of the other:
[tex]\[v_l = - \frac{4800}{1500} v_s = -3.2 v_s\][/tex]
2. Conservation of Energy
The total kinetic energy provided to the system is 5000 J. The kinetic energy of the system is the sum of the kinetic energies of both parts:
[tex]\[\frac{1}{2} m_s v_s^2 + \frac{1}{2} m_l v_l^2 = 5000\][/tex]
Substituting [tex]\( v_l = -3.2 v_s \)[/tex]:
[tex]\[\frac{1}{2} (4800) v_s^2 + \frac{1}{2} (1500) (-3.2 v_s)^2 = 5000\][/tex]
[tex]\[2400 v_s^2 + 750 (10.24) v_s^2 = 5000\][/tex]
[tex]\[2400 v_s^2 + 7680 v_s^2 = 5000\][/tex]
[tex]\[10080 v_s^2 = 5000\][/tex]
Solving for [tex]\( v_s^2 \)[/tex]:
[tex]\[v_s^2 = \frac{5000}{10080} = 0.496 \, \text{m}^2/\text{s}^2\][/tex]
[tex]\[v_s = \sqrt{0.496} \approx 0.704 \, \text{m/s}\][/tex]
3. Determine [tex]\( v_l \)[/tex]
Using [tex]\( v_l = -3.2 v_s \)[/tex]:
[tex]\[v_l = -3.2 \times 0.704 \approx -2.253 \, \text{m/s}\][/tex]
The subsequent velocities of the satellite and the launcher remains after separation are:
- Velocity of the satellite: [tex]\( v_s \approx 0.704 \, \text{m/s} \)[/tex]
- Velocity of the launcher remains: [tex]\( v_l \approx -2.253 \, \text{m/s} \)[/tex]
Objects 1 and 2 attract each other with a gravitational force of 72.0 units. If the mass of Object 1 is one-fourth the original value AND the mass of object 2 is tripled AND the distance separating Objects 1 and 2 is halved, then the new gravitational force will be _____ units.
When the mass of Object 1 is reduced to one-fourth, the mass of Object 2 is tripled, and the separating distance is halved, the new gravitational force between these objects will be three times the original force. Thus, the new gravitational force will be 216.0 units.
Explanation:
The question asks us to determine the new gravitational force between two objects given that the mass of Object 1 is reduced to one-fourth of its original mass (so M1 becomes 0.25M1), the mass of Object 2 is tripled (so M2 becomes 3M2), and the distance separating the objects is halved (so R becomes 0.5R). Since gravitational force is calculated using the equation F = G*(M1*M2)/(R^2), where G is the universal gravitational constant, we can plug our new values into this equation.
So, the new gravitational force F' = G*(0.25M1 * 3M2)/(0.5R)^2. This simplifies further to F' = G*(0.75M1M2)/(0.25R^2) = 3 *(G*M1M2/R^2) = 3 * original force. Therefore, if the original force was 72.0 units, the new gravitational force is 3 * 72.0 = 216.0 units.
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To find the new gravitational force between the objects, we can use Newton's Law of Universal Gravitation.
Explanation:The new gravitational force can be calculated using Newton's Law of Universal Gravitation, which states that the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Let's denote the original mass of Object 1 as m1, the original mass of Object 2 as m2, and the original distance separating them as d. The original gravitational force can be represented as:
F1 = G * (m1 * m2) / d²
After the given changes, the new mass of Object 1 is 1/4 * m1, the new mass of Object 2 is 3 * m2, and the new distance is 1/2 * d. Plugging these values into the equation, we find the new gravitational force:
F2 = G * ((1/4 * m1) * (3 * m2)) / (1/2 * d)²
Simplifying this equation will give us the answer.
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Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 689 rad/s. One such flywheel is a solid, uniform cylinder with a mass of 507 kg and a radius of 1.2 m that rotates about its central axis. What is the kinetic energy of the flywheel after charging
Answer:
k = 86,646,076.92 J
Explanation:
We know that:
K = [tex]\frac{1}{2}IW^2[/tex]
where K is the kinetic energy, I is the moment of inertia and W is the angular velocity.
First we have to find the I with the equation:
I = [tex]\frac{1}{2}MR^2[/tex]
where M is the mass and R the radius of the cylinder.
so:
I = [tex]\frac{1}{2}(507 kg)(1.2m)^2[/tex]
I = 365.04 kg*m^2
Now we replace all the data in the first equation as:
K = [tex]\frac{1}{2}IW^2[/tex]
K = [tex]\frac{1}{2}(365.04)(689)^2[/tex]
k = 86,646,076.92 J
Calculate the force of friction that keeps an 80-kg person sitting on the edge of a horizontal rotating platform when the person sits 2 m from the center of the platform and has a tangential speed of 3 m/s.
Answer:
360N
Explanation:
The frictional force = centripetal force = mass * centripetal acceleration = ma
Since centripetal acceleration = V^2/ r
Frictional force = mv^2 / r
Substitute the values of m = 80kg, radius = 2m and tangential speed = 3m/s
Frictional force = 80 * (3)^2/ 2 = 360N
The force of friction that keeps the person sitting on the edge of the rotating platform is 360 N.
To calculate the force of friction that keeps an 80-kg person sitting on the edge of a horizontal rotating platform, we need to find the centripetal force required to keep the person moving in a circular path. The centripetal force is provided by the frictional force between the person and the platform.
The formula for centripetal force [tex](\(F_c\))[/tex] is given by:
[tex]\[ F_c = \frac{mv^2}{r} \][/tex]
Let's plug in the values:
[tex]\[ F_c = \frac{(80\text{ kg})(3\text{ m/s})^2}{2\text{ m}} \][/tex]
[tex]\[ F_c = \frac{(80\text{ kg})(9\text{ m}^2/\text{s}^2)}{2\text{ m}} \][/tex]
[tex]\[ F_c = \frac{720\text{ kg}\cdot\text{m}/\text{s}^2}{2} \][/tex]
[tex]\[ F_c = 360\text{ N} \][/tex]
Therefore, the force of friction that keeps the person sitting on the edge of the rotating platform is 360 N.
The answer is: [tex]360\text{ N}[/tex]
1. A 0.250 kg baseball sits on the ledge of a window in Treadwell hall. If the ball has 18.5
Joules of potential energy, how high is the ball?
2. How long would it take to hit the ground?
Answer:
h = 7.54 m
t = 1.24 s
Explanation:
1.Let g = 9.81 m/s2 is the gravitational acceleration. Since the formula for potential energy is:
[tex]E_p = mgh[/tex]
where m = 0.25 is the ball mass and h is the height. We can solve for h
[tex]h = \frac{E_p}{mg} = \frac{18.5}{0.25*9.81} = 7.54m[/tex]
2. The time it take for the ball to reach a distance of 7.54m with a gravitational acceleration of 9.81m/s2:
[tex]h = \frac{gt^2}{2}[/tex]
[tex]t^2 = \frac{2h}{g} = \frac{2*7.54}{9.81} = 1.54[/tex]
[tex]t = \sqrt{1.54} = 1.24 s[/tex]
Imagine that the apparent weight of the crown in water is Wapparent=4.50N, and the actual weight is Wactual=5.00N. Is the crown made of pure (100%) gold? The density of water is rhow=1.00 grams per cubic centimeter. The density of gold is rhog=19.32 grams per cubic centimeter.
Answer:
Explanation:
Actual weight, Wo = 5 N
Apparent weight, W = 4.5 N
density of water = 1 g/cm^3 = 1000 kg/m^3
density of gold, = 19.32 g/cm^3 = 19.32 x 1000 kg/m^3
Buoyant force = Actual weight - Apparent weight
Volume x density of water x g = 5 - 4.5
V x 1000 x 9.8 = 0.5
V = 5.1 x 10^-6 m^3
Weight of gold = Volume of gold x density of gold x gravity
W' = 5.1 x 10^-6 x 19.32 x 1000 x 9.8 = 0.966 N
As W' is less than W so, it is not pure gold.
Using Archimedes' Principle, we determined the density of the crown to be 10.00 g/cm³, which is significantly lower than the density of pure gold (19.32 g/cm³). Therefore, the crown is not made of pure gold.
This problem involves determining whether a crown is made of pure gold using Archimedes' Principle.
According to the given information,
the actual weight of the crown is Wactual = 5.00N, and its apparent weight in water is Wapparent = 4.50N.Using Archimedes' Principle, we know that the buoyant force equals the weight of the water displaced:
Buoyant Force = Wactual - Wapparent = 5.00N - 4.50N Buoyant Force = 0.50NSince the density of water (ρw) is 1.00 g/cm³, and using the relation 1N = 0.10197 kg in gravitational terms, the weight of the water displaced in kilograms is
weight of the water displaced =0.50N / 9.81 m/s² weight of the water displaced = 0.05096 kg, which converts to 50.96 g.The volume of water displaced, which is also the volume of the crown (V), is
V = mass / density V = 50.96 g / 1.00 g/cm³ V = 50.96 cm³Density of the crown = mass / volume.
The mass of the crown (m) is calculated from its weight in air,
m = 5.00N / 9.81 m/s² m = 0.5096 kg m = 509.6 gThus, the density of the crown = 509.6 g / 50.96 cm³ = 10.00 g/cm³.
Since the density of pure gold is 19.32 g/cm³ and the calculated density is 10.00 g/cm³, the crown is not made of pure gold.
A student sees her physical science professor approaching on the sidewalk that runs by her dorm. She gets a water balloon and waits. When the professor is 2 s from being directly under her window 11 m above the idewalk, she drops the balloon. You finish the story. (Assume g # 10 m/s*)
Answer:Misses the Target
Explanation:
Given
Distance between window and Professor is h=11 m
she dropped the balloon 2 sec earlier
time taken by balloon to cover the 11 m
using
[tex]h=ut+\frac{at^2}{2}[/tex]
where h=height
u=initial velocity
t=time
a=acceleration
[tex]11=0+\frac{9.8t^2}{2}[/tex]
[tex]t=\sqrt{\frac{2\times 11}{10}}[/tex]
[tex]t=\sqrt{2.2}[/tex]
[tex]t=1.483 s[/tex]
Therefore she misses the Professor by 0.51 s
Final answer:
The water balloon will hit the ground before the professor is directly underneath the window since it takes approximately 1.48 seconds to fall 11 meters, which means the student is unlikely to hit the professor.
Explanation:
The student's question involves calculating the time it takes for a water balloon dropped from a height to hit a target. This is a problem in projectile motion and free fall within the context of physics, specifically under the topic of mechanics. As the professor is 2 seconds away from the point directly below the student's window, which is 11 meters above the sidewalk, we can determine if the balloon will hit the target by calculating how long it takes for the balloon to reach the ground.
Using the equation of motion for free fall s = 1/2gt^2, where g is the acceleration due to gravity (approximated to 10 m/s^2), and s is the distance (11 meters), we can solve for t (time): s = 1/2gt^2 implies t = sqrt(2s/g). Substituting the given values, we get t = sqrt(2*11/10), which gives t = sqrt(2.2), or approximately t = 1.48 seconds. Because the professor is 2 seconds away from being underneath the window, the water balloon will hit the ground before the professor reaches the point below the student's window, meaning the student's attempt to hit the professor with the balloon will likely fail unless the professor's speed changes.
Anissa slides down a playground slide sloped at 25o. The coefficient of kinetic friction between Anissa and the slide is 0.15. If Anissa’s mass is 55 kg, how fast is she going 2.0 s after starting from rest?
To find the speed of Anissa after 2.0 seconds, we need to consider the forces acting on her due to gravity and friction. By calculating the force parallel to the slide using the coefficient of kinetic friction and the normal force, we can determine the acceleration of Anissa. Finally, using the equations of motion, we can find her speed after 2.0 seconds.
Explanation:To find the speed of Anissa after 2.0 seconds, we need to consider the forces acting on her. The force of gravity pulling her down the slide can be divided into two components: a force parallel to the slide and a force perpendicular to the slide. The perpendicular force is canceled out by the normal force, leaving only the parallel force to consider. This force can be calculated using the coefficient of kinetic friction.
The force parallel to the slide is given by the equation F_parallel = μ_k * N, where F_parallel is the force parallel to the slide, μ_k is the coefficient of kinetic friction, and N is the normal force. The normal force can be calculated using the equation N = m * g * cos(θ), where m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of the slide.
Once we have the force parallel to the slide, we can use Newton's second law to find the acceleration of the object: F_parallel = m * a. Finally, we can use the equation v = u + a * t to find the speed of Anissa after 2.0 seconds, where v is the final velocity, u is the initial velocity (which is 0 since she starts from rest), a is the acceleration, and t is the time.
Plugging in the given values: μ_k = 0.15, m = 55 kg, g = 9.8 m/s^2, θ = 25 degrees, and t = 2.0 s, we can calculate the normal force, force parallel to the slide, acceleration, and final velocity.
The final velocity will depend on the length of the slide and other factors that are not mentioned in the question.
A time-dependent but otherwise uniform magnetic field of magnitude B0(t) is confined in a cylindrical region of radius 6.5 cm. Initially the magnetic field in the region is pointed out of the page and has a magnitude of 3.5 T, but it is decreasing at a rate of 21 G/s. Due to the changing magnetic field, an electric field will be induced in this space which causes the acceleration of charges in the region. What is the direction of acceleration of a proton placed in at the point P1, 1.5 cm from the center?
Answer:
Explanation:
Due to changing magnetic field , there will be emf induced in the region . EMF induced will create electric field which will be circular in shape and will be uniform along its circular path
The magnitude of circular electric field can be calculated as follows
We should apply Faraday law of electro magnetic induction
e = - dФ / dt = - ∫ E dl
Here Ф = π r² B
π r² dB / dt = - ∫ E dl
π r² dB / dt = E x 2π r
E = - r / 2 x dB / dt
For a circular electric field having a particular radius , magnitude of field will be constant .
The direction of electric field will be known by lenz's law
In the given case , magnetic field is upwards and it is reducing , therefore electric field induced will be such as to prevent this change of flux.
So electric field will be anticlock-wise . Hence direction of acceleration will also be anticlock-wise on proton at 1.5 cm from the centre.
During a medieval siege of a castle, the attacking army uses a trebuchet to hurl heavy stones at the castle walls. If the trebuchet launches the stones with a velocity of +30.0 m/s at an angle of 50.0°, how long does it take the stone to hit the ground? What is the maximum distance that the trebuchet can be from the castle wall to be in range? How high will the stones go? Show all your work.
The projectile reaches a maximum height of 26.4 m.
The question is demanding us to find the;
i) Time of flight:
T = 2usinθ/g
u = initial velocity
θ = angle of projection
g = acceleration due to gravity
T = 2 × 30.0 × sin 50.0°/10
T = 4.6 s
ii) The range;
R = u^2sin2θ/g
R = (30.0)^2 × sin 2( sin 50.0°)/10
R = 88.6 m
ii) The maximum height;
H = u^2sin^2θ/2g
H = (30.0)^2 × sin^2(50.0°)/2 × 10
H = 26.4 m
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Aluminum Rod#1 has a length L and a diameter d. Aluminum Rod#2 has a length 2L and a diameter 2d. If Rod#1 is under tension T and Rod#2 is under tension 2T, how do the changes in length of the two rods compare?
A) Rod #2 has quadruple the change in length that Rod #1 has.
B) Rod #2 has double the change in length that Rod #1 has.
C) Rod #1 has double the change in length that Rod #2 has.
D) They are the same.
E) Rod #1 has quadruple the change in length that Rod #2 has.
Answer:D
Explanation:
For Rod 1
length,diameter and Tension is L, d and T
for Rod 2 length,diameter and Tension is 2L, 2d and 2T
Change in Length is given by
[tex]\Delta =\frac{PL}{AE}[/tex]
where P=load
L=length
A=area of cross-section
E=young's Modulus
[tex]\Delta _1=\frac{TL}{\frac{\pi d^2}{4}E}[/tex]
since both are aluminium rod therefore E is common
[tex]\Delta _2=\frac{2T\cdot 2L}{\frac{\pi (2d)^2}{4}E}[/tex]
[tex]\Delta _2=\frac{TL}{\frac{\pi d^2}{4}E}=\Delta _1[/tex]
Thus extension in both the rods are same
The change in length of a rod under tension or compression depends on several variables. The change in length is directly proportional to the force and the original length and inversely proportional to the cross-sectional area. Therefore, if Rod #1 is under tension T and Rod #2 is under tension 2T, the change in length of Rod #2 will be double the change in length of Rod #1.
Explanation:The change in length of a rod under tension or compression depends on several variables. These variables include the force applied (T or 2T), the original length of the rod (L or 2L), and the cross-sectional area of the rod (πd^2/4 or π(2d)^2/4). The change in length is directly proportional to the force and the original length and inversely proportional to the cross-sectional area. Therefore, if Rod #1 is under tension T and Rod #2 is under tension 2T, the change in length of Rod #2 will be double the change in length of Rod #1.
A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.80 s apart.How far away did the impact occur? The speed of sound in air is 343 m/s, and in concrete is 3000 m/s.
Answer:
309 m
Explanation:
time apart = 0.8 s
speed of sound in air = 343 m/s
speed of sound in concrete = 3000 m/s
lets assume it time it takes to travel trough concrete = T
the time it takes to travel through air = T + 0.8 since they are 0.8 s apart
the distance traveled by both waves is the same, so we can equate the distance for both waves
distance = speed x time
for concrete, distance = 3000 x Tfor air, distance = 343 x (T + 0.8)now equating the two distance together we have
3000 x T = 343 x (T + 0.8)
3000T = 343T + 274.4
3000T - 343T = 274.4
T = 0.103 s
recall that distance = 3000 x T = 3000 x 0.103 = 309 m
What formal regions are located in the western hemisphere
Answer: Part of Europe, part of Africa, part of Antarctic and the entire America.
Explanation: Western hemisphere or west hemisphere encompasses all regions located west of the longitude 0 °, or meridian of Greenwich.
The formal regions located in the western hemisphere is Europe, Africa, Antarctic and America.
Final answer:
Western Europe can be divided into several smaller regions, including northern Europe, southern Europe, Central Europe, and the British Isles. The western hemisphere also includes various formal regions such as Central America, the Caribbean islands, and parts of South America.
Explanation:
Western Europe can be divided into several smaller regions, including northern Europe, southern Europe, Central Europe, and the British Isles. These regions are based on geographical location or cultural similarities. Additionally, the western hemisphere includes various formal regions such as Central America, the Caribbean islands, and parts of South America.
An 8.7 g bullet is accelerated in a rifle barrel 105 cm long to a speed of 465 m/s. Use the work-energy theorem to find the average force exerted on the bullet while it is being accelerated. Answer in units of N.
Answer:896 N
Explanation:
Given
mass of bullet [tex]m=8.7 gm[/tex]
Length of barrel [tex]d=105 cm=1.05 m[/tex]
final velocity [tex]v=465 m/s[/tex]
initial velocity [tex]u=0 m/s[/tex]
Work done by all the forces is equal to change in kinetic Energy
[tex]F\cdot d=\frac{1}{2}\times mv^2-0[/tex]
[tex]F_{avg}\times 1.05=\frac{1}{2}\times 8.7\times 10^{-3}\times 465^2[/tex]
[tex]F_{avg}=\frac{0.5\times 8.7\times 10^{-3}\times 465^2}{1.05}[/tex]
[tex]F_{avg}=895.789 Na\pprox 896 N[/tex]
Keller is driving at a speed of 60 MPH. At noon he sees the following sign: Belmont 3 miles Arville 17 miles Blison 35 miles Longwood 52 miles Swan's Lake 120 miles If he continues to drive the same speed, how far away will Longwood be at 12:20pm?
Answer:
32 miles
Explanation:
Assuming that the speed of 60 mph is constant
Distance = Speed × Time
Time is 20 minutes
Converting to hours
[tex]\frac{20}{60}=\frac{1}{3}\ h[/tex]
[tex]Distance=60\times \frac{1}{3}\\\Rightarrow Distance=20\ mi[/tex]
The car covered 20 mi in 20 minutes.
Distance to Longwood would be [tex]52-20=32\ mi[/tex]
Distance to Longwood is 32 miles
A chunk of aluminum at 91.4°C was added to 200.0 g of water at 15.5°C. The specific heat of aluminum is 0.897 J/g°C, and the specific heat of water is 4.18 J/g°C. When the temperature stabilized, the temperature of the mixture was 18.9°C. Assuming no heat was lost to the surroundings, what was the mass of aluminum added?
The mass of the aluminum added is calculated through the principle of conservation of energy, specifically thermal energy. By considering the heat lost by the aluminum and gained by the water, we can rearrange the equation for heat transfer and find that the mass of the aluminum is approximately 37.9 grams.
Explanation:In this physics question, we're looking at a thermodynamic process involving a chunk of aluminum and water. Given the known values of their respective specific heats, the mass of water, and their final equilibrium temperature, we're aiming to find the mass of the aluminum.
We begin by understanding that in a closed system, the heat gained by one body is equal to the heat lost by another. In this case, the aluminum is losing heat, and the water is gaining it. The equation for heat transfer (Q = mcΔT), where m is mass, c is specific heat, and ΔT is change in temperature.
The heat gained by the water = mass of water * specific heat of water * change of temperature in water = 200g * 4.18J/g°C * (18.9°C - 15.5°C) = 2836.4J.
This is equal to the heat lost by the aluminum. Solving the analogous heat equation for the mass of the aluminum gives us the answer:
m = Q / (c * ΔT) = 2836.4J / (0.897J/g°C * (91.4°C - 18.9°C)) = 37.9g
So the mass of the aluminum is approximately 37.9 grams.
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What is the name of moving fields of angular blocks which are formed by gravity and the freezing-thawing cycle? A. Scree B. Talus C. Bedrock slumps D. Bedrock slides E. Rock glaciers
Answer:
Option (E)
Explanation:
The Rock glaciers are a type of landform that is formed due to the downward motion of angular blocks under the influence of gravity and the continuous cycle of freeze and thaw method. It mostly occurs in the steep slopes where the soil particles and the rock materials are prone to landslide. Here, the materials are mixed and covered with ice. This also occurs in the region where the permafrost layer undergoes creeping.
This downward motion of ice erodes the soil particles and carries with them further, from the site of talus cone or moraines and extends outward.
Thus, the correct answer is option (E).
Answer:
one of it is e
Explanation:
A mass of 0.40 kg, hanging from a spring with a spring constant of 80 N/m, is set into an up-and-down simple harmonic motion. What is the speed of the mass when moving through the equilibrium point? The starting displacement from equilibrium is 0.10 m.
The speed of the mass when moving through the equilibrium point is 1.22 m/s.
Calculation of the mass speed:
But before that the angular frequency is
[tex]= \sqrt{\frac{80}{0.4} }[/tex]
= 14.14 rad/s
Now
The mass speed should be
[tex]= 14.14 \sqrt{(0.10)^2 - (0.05)^2}[/tex]
= 1.22 m/s
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The speed of a 0.40-kg mass moving through the equilibrium point in simple harmonic motion with a spring constant of 80 N/m and an initial displacement of 0.10 m is 2 m/s.
Explanation:The speed of the mass when moving through the equilibrium point in a scenario of simple harmonic motion can be determined using the formula for the maximum speed of an object in simple harmonic motion. This formula is v = √(k/m) * A, where v is the speed, k is the spring constant, m is the mass, and A is the amplitude, which in this case is the initial displacement of the object from equilibrium.
So, to calculate the speed of this 0.40-kg mass with a spring constant of 80 N/m as it passes through equilibrium, we plug those values into the formula:
v = √[(80 N/m) / (0.40 kg)] * (0.10 m) = 2 m/s.
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