The acceleration of an object in circular motion is calculated by the formula ac = v²/r. Given that in case a, both the speed and radius are twice that of case b, the acceleration in case a is twice that of case b. This shows the squared relationship between speed and acceleration in circular motion.
Explanation:In physics, the acceleration of an object moving in a circular path, also known as centripetal acceleration, is calculated by the formula ac = v²/r where v is the velocity or speed of the object and r is the radius of the circle.
In case a, the speed is twice that of case b and the circle's radius is also two times of case b. If we plug these values into our formula, we can see that the acceleration in case a (let's call it acA) could be represented as acA = (2v)² / 2r = 4v²/2r = 2v²/r. In case b, the acceleration (acB) would be represented as acB = v² / r.
Hence, comparing both, you can see the acceleration in case a is twice that in case b. Therefore, even if the speed and the radius are increased by the same factor, the acceleration increases by a factor of that increase due to the squared relationship between velocity and acceleration in circular motion.
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The acceleration in case A is twice that of case B.Given the conditions that the speed and radius of circle in case A are twice those in case B.
Let's analyze the relationship between acceleration, speed, and radius in circular motion. We know the acceleration of an object in uniform circular motion can be given by the equation:
|a| = |v|²/r
From the problem, we have two cases:
[tex]Case A: Speed (|v_A|) is \twice \ that \ of \ case \ B (|v_B|).\\\\Case A: Radius (|r_A|) \ is \ also \ two\ times \ that\ of \ case\ B (|r_B|).[/tex]
Let's denote the acceleration in case A as [tex]|a_A|[/tex] and in case B as [tex]|a_B|[/tex]
For case A:
[tex]|a_A| = |v_A|^2 / |r_A|[/tex]
For case B:
[tex]|a_B| = |v_B|^2 / |r_B|[/tex]
Given that [tex]|v_A| = 2|v_B|[/tex] and [tex]|r_A| = 2|r_B|[/tex], substituting these into the equations gives:
[tex]|a_A| = (2|v_B|)^2 / (2|r_B|)[/tex]
Simplifying this, we get:
[tex]|a_A| = 4|v_B|^2 / 2|r_B| = 2(|v_B|^2 / |r_B|) = 2|a_B|[/tex]
Therefore, the acceleration in case A is twice that of case B.
A roller coaster engineer is trying to design a new coaster to build at the theme park. The cars, when loaded with people, is expected to have a mass of 8000. He wants his roller coaster cars to be going 10 m/s when it reaches the bottom. Assuming that friction is negligible, how high up should the roller coaster start?
_________Number ___M___Units
Then Answer
Well, it turns out that friction is not negligible, the Engineer figured out that the car is going to lose 32000 Joules energy as it goes down the track. this means that he has to start up even higher. what is the actual starting height after taking into account the energy loss from friction?
__________Number _____M____Units
Thank you SOO much :D
1) The initial height of the roller coaster must be 5.1 m
2) The height of the roller coeaster must be 5.5 m
Explanation:
1)
We can solve this problem by using the law of conservation of energy.
In fact, in absence of frictional force, the total mechanical energy must be conserved, so we can write:
[tex]KE+PE = KE'+PE'[/tex]
where
KE = 0 is the initial kinetic energy, at the top (the car starts from rest, so it has zero speed and so zero kinetic energy)
PE = mgh is the initial potential energy, where
m = 8000 kg is the mass of the car
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
h is the initial height
[tex]KE'=\frac{1}{2}mv'^2[/tex] is the kinetic energy of the car at the bottom, where
v' = 10 m/s is the speed of the car at the bottom
PE' = 0 is the potential energy at the bottom (where h = 0)
Re-writing the equation with all the values and solving for h, we find the initial height of the track:
[tex]h=\frac{v^2}{2g}=\frac{10^2}{2(9.8)}=5.1 m[/tex]
2)
In this case, we also have to take into account the energy lost due to friction. So the equation of conservation of energy becomes
[tex]KE+PE = KE'+PE'+W[/tex]
where
W = 32,000 J is the work done by friction
Re-arranging the equation and solving again for h, we find:
[tex]mgh = \frac{1}{2}mv'^2 + W\\h=\frac{v^2}{2g}+\frac{W}{mg}=5.1+\frac{32,000}{(8000)(9.8)}=5.5 m[/tex]
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if you are traveling at 75 km/h how long will it take to travel 32 km?
The time taken is approximately 26 minutes
Explanation:
The motion of the body in this problem is a uniform motion (= at constant velocity), therefore we can use the following equation:
[tex]v=\frac{d}{t}[/tex]
where
v is the speed
d is the distance covered
t is the time taken
In this problem, we have
v = 75 km/h is the speed
d = 32 km is the distance to be covered
Solving for t, we find the time needed:
[tex]t=\frac{d}{v}=\frac{32}{75}=0.43 h[/tex]
Converting into minutes,
[tex]t=0.43 h \cdot 60 = 25.8 min \sim 26 min[/tex]
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A 5.5Kg block is hanging from a rope that is wrapped around the outside of a 13Kg flywheel disk witha radius of 33cm that is hagning form the ceilin. Friction in the flywheel provides a constant torque of 2.5Nm. When the block is released what is the magnitude of its acceleration as it falls
Answer:
[tex]3.9m/s^{2}[/tex]
Explanation:
Using second law of motion
[tex]a =\frac {m1 * g - \frac {T}{r}}{m1 + 0.5 * m2}[/tex] where m1 is mass of block, m2 is mass of flywheel, g is acceleration due to gravity whose value is taken as [tex]9.81 m/s^{2}[/tex], T is torque and r is radius
Substituting 5.5 Kg for m1, 13 Kg for m2, 0.33 m for r, 2.5 Nm for T we obtain
[tex]a = \frac {5.5 \times 9.81 - \frac {2.5}{0.33}}{(5.5 + 0.5 \times13)}=3.9m/s^{2}[/tex]
To find the acceleration of the 5.5 kg block, we consider both gravitational force and frictional torque acting on a 13 kg flywheel. The calculated acceleration of the block is approximately 2.73 m/s².
To determine the magnitude of the acceleration of the 5.5 kg block as it falls, we need to consider several forces and torques involved in the system. These include the gravitational force on the block, the tension in the rope, the rotational inertia of the flywheel, and the frictional torque opposing the motion.
First, let’s identify the key forces:
Gravitational force on the block (Fg): Fg = mblock * g = 5.5 kg * 9.8 m/s² = 53.9 NFrictional torque (τfriction): τfriction = 2.5 NmThe rotational inertia (I) of the flywheel (a disk) is given by:
I = 0.5 * mflywheel * r² = 0.5 * 13 kg * (0.33 m)² = 0.70785 kg·m²
Applying Newton’s second law for rotation, we have:
∑τ = I * α → τtension - τfriction = I * α
Here, τtension is the torque due to the tension in the rope, which is equal to T * r. Hence, we have:
T * r - τfriction = I * α
Also, the linear acceleration (a) of the block is related to the angular acceleration (α) of the flywheel by the equation:
a = α * r
Combining these equations, we get:
T * r - τfriction = I * (a / r)
Solving for T in terms of a:
T = (I * a / r²) + (τfriction / r)
Newton’s second law for the falling block gives us:
mblock * g - T = mblock * a
Substitute T from the earlier equation into this one and solve for a:
mblock * g - [(I * a / r²) + (τfriction / r)] = mblock * a
Rearranging terms to isolate a:
a = [mblock * g - (τfriction / r)] / [mblock + (I / r²)]
Substitute the known values:
mblock = 5.5 kgg = 9.8 m/s²τfriction = 2.5 NmI = 0.70785 kg·m²r = 0.33 ma = [5.5 kg * 9.8 m/s² - (2.5 Nm / 0.33 m)] / [5.5 kg + (0.70785 kg·m² / (0.33 m)²)]
a ≈ 2.73 m/s²
Thus, the magnitude of the acceleration of the block as it falls is approximately 2.73 m/s².
A 1500kg car, moving at a speed of 20m/s comes to a halt. How much work was done by the brakes?
The work done by the brakes is [tex]-3.0\cdot 10^5 J[/tex]
Explanation:
According to the work-energy theorem, the work done by the brakes on the car is equal to the change in kinetic energy of the car. Therefore:
[tex]W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]
where
W is the work done
m is the mass of the car
v is the final speed
u is the initial speed
For the car in this problem, we have:
m = 1500 kg
u = 20 m/s
v = 0 (the car comes to a halt)
Substituting, we find the work done:
[tex]W=0-\frac{1}{2}(1500)(20)^2=-3.0\cdot 10^5 J[/tex]
And the work is negative because the brakes apply a force in the opposite direction to the motion of the car.
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Which of the following properties did Rutherford use in his experiment?
A. the negative charge of the alpha particles and the random distribution of protons
B. the negative charge of the alpha particles and the positive charge of the gold foil
C. the positive charge of the alpha particles and the negative charge of the electrons
D. the positive charge of the electrons in a uniform negative charge
Answer:
(C) The positive charge of the alpha particles and the negative charge of the electrons are the properties Rutherford used in his experiment
Explanation:
In his scattering experiment, scientist Ernest Rutherford used the property of positively charged alpha particles and negatively charge electrons. He performed this experiment by passing some of the alpha particles through a gold foil. The result was that the some alpha particles scattered while some passed through the gold foil without collision.
He concluded that the alpha particles are centrally positively charged and needed a large amount repelling force. This experiment of Rutherford is also known as Rutherford model of atom. This experiment helped him in doing so many other discoveries.
If the airman had a mass of 80 kg, find the magnitude of the air drag acting on him when he reached terminal velocity of 54 m/s
The magnitude of the air drag is 784 N
Explanation:
An object falling down reaches the terminal velocity when the magnitude of the air drag acting on it becomes equal to the weight of the object. Mathematically, this condition can be written as:
[tex]F_D = mg[/tex]
where
[tex]F_D[/tex] is the magnitude of the air drag
m is the mass of the object
g is the acceleration of gravity
In this problem, we have
m = 80 kg is the mass of the airman
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Substituting into the formula, we find:
[tex]F_D = (80)(9.8)=784 N[/tex]
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A toy car has an initial acceleration of 2 m/s" across a horizontal surface after it is released from rest. After the car travels for a timet
consisting of only the car an open system or a closed system, and why?
Open system because the acceleration of the car is not constant.
Open system, because an external force is applied to the car that causes it to accelerate.
Closed system, because the speed of the car is as expected in the case where an object has uniform acceleration for a timet
Closed system because mechanical energy was not removed from the system as a result of a net force
Rie
B
7:53 PM
12/11/2019
The car is considered as an open system, because an external force is applied to the car that causes constant acceleration of car. Hence, option (B) is correct.
The system isolated from its surrounding is known as closed-system. While, in an open system, the flow of information between the system and surrounding takes place, to cause any change in the system.
As per the concept of open and closed system:
The given problem indicates that car has initial acceleration of [tex]2 \;\rm m/s^{2}[/tex], travels for time t . Which means the car is accelerating constantly.And for a constant acceleration, there must be some external force acting on the car.Hence, the car must be considered an an open system, because the acceleration of car is related with the force, which is being applied from the outside of frame of reference.
Thus, we can conclude that considering only car is an open system because external force is applied to the car that causes it to accelerate. Hence, option (B) is correct.
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what is the magnitude of an electric field which will balance the weight of an electron on the surface of earth ?
The magnitude of the electric field must be [tex]5.59\cdot 10^{-11} N/C[/tex]
Explanation:
In order for the electron to be in equilibrium, the force of gravity acting on the electron must be equal to the force due to the electric field.
The force of gravity on the electron located on the Earth's surface is:
[tex]F_G = mg[/tex]
where
[tex]m=9.11\cdot 10^{-31} kg[/tex] is the electron mass
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
The force due to the electric field is
[tex]F_E = qE[/tex]
where
[tex]q=1.6\cdot 10^{-16}C[/tex] is the electron charge
E is the magnitude of the electric field
Since the two forces must be balanced,
[tex]F_G = F_E[/tex]
So we find:
[tex]mg=qE\\E=\frac{mg}{q}=\frac{(9.11\cdot 10^{-31})(9.81)}{1.6\cdot 10^{-19}}=5.59\cdot 10^{-11} N/C[/tex]
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3. Consider a locomotive and the rest of a freight train to be a single object. Suppose the locomotive is pulling the train up a hill. Describe the action and reaction forces that cause the locomotive to move up the hill, such as the reaction force and gravity.
This question has been posted before put the answers were not correct.
Answer:
Action - Pulling up the train.
Reaction - Friction on the locomotive
Explanation:
Locomotive is pulling the train upwards ,
Which is the action force applied by the locomotive,
As a reaction locomotive will be pulled by the train which is the reaction of pulling
Now, considering it as a action on locomotive , friction force will act on it as a reaction upwards which will result to move it upwards.
For train action is pulling up by locomotive and reaction will be friction acting on it downwards.
A truck accelerates to a velocity of 38 m/s over 755 m of road
1) The acceleration is [tex]0.96 m/s^2[/tex]
2) The time taken is 39.6 s
Explanation:
1)
Since the motion of the truck is a uniformly accelerated motion (=constant acceleration), we can use the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance covered
For the truck in this problem,
u = 0 (it starts from rest)
v = 38 m/s
s = 755 m
Solving for a, we find the acceleration:
[tex]a=\frac{v^2-u^2}{2s}=\frac{38^2-0}{2(755)}=0.96 m/s^2[/tex]
2)
For this part we can use the following suvat equation
[tex]v=u+at[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time taken for the velocity to change from u to v
In this problem,
u = 0
v = 38 m/s
[tex]a=0.96 m/s^2[/tex]
Solving for t,
[tex]t=\frac{v-u}{a}=\frac{38-0}{0.96}=39.6 s[/tex]
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The truck's acceleration, starting from rest to a velocity of 38 m/s over 755 m, is approximately 0.956 m/s². It takes about 39.75 seconds to cover this distance.
a. To determine the acceleration of the truck, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (38 m/s),
u = initial velocity (0 m/s since the truck was initially at rest),
a = acceleration,
s = displacement (755 m).
Rearranging the equation to solve for acceleration (a), we get:
a = (v^2 - u^2) / (2s)
Substitute the given values:
a = (38^2 - 0^2) / (2 * 755)
a = 1444 / 1510
a ≈ 0.956 m/s^2
b. To find the time taken (t), we can use the kinematic equation:
v = u + at
Since the truck started from rest (u = 0), this simplifies to:
t = v / a
Substitute the values:
t = 38 / 0.956
t ≈ 39.75 s
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The question probable may be:
A Truck accelerates to a velocity of 38 m/s over 755 m of road.
a. If the truck was initially at rest, what was its acceleration?
b. How long willthe truck take to travel this distance?
Andre really likes his new car, and he knows it has a certain amount of mechanical energy. Which types of energy are included in the mechanical
energy of the car? Choose the two that apply.
A. electrical energy from the battery
B. kinetic energy from any movement the car has
C. potential energy based on its position
D. thermal energy from when fuel burns in the engine
Option - B and Option - C are correct.
We have Andre's new car.
We have to determine what types of Energies are included in the Mechanical energies of the car.
What is Mechanical Energy?The energy possessed by an object due to its motion or its position is called Mechanical Energy.
According to the question -
The Mechanical energy of the new car would include the following two types of energy -
Kinetic Energy of Car - The kinetic energy of the car is the energy possessed by the car by virtue of its motion. Mathematically -[tex]$E(K) = \frac{1}{2} mv^{2}[/tex]
Potential Energy of Car - The potential energy of the car is due to its position. Assume that the car of mass 'm' is parked over the mountain at height 'h'. Then its potential energy will be -U(h) = mgh
Hence, Option B and C are correct.
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What distance will a car traveling 65km/hr travel in 3 hours
Answer:
it will cover 195km in 3 hours
a jet has a mass of 40000 kilograms the thrush pushing force of each of the four engines is 20000 Newtons what is the Jets acceleration when taking off
The combination of foods and beverages that constitute an individual’s complete dietary intake over time is known as
Eating Pattern
Mediterranean Diet
Dietary Guidelines
Healthy Intake
Answer:
The combination of foods and beverages that constitute an individual’s complete dietary intake over time is known as Eating pattern.
So, option A is the correct answer.
Explanation:
The Eating pattern is also called Dietary Pattern. Normal eating pattern may be three meals a day, may increased or changed according to the individual, and may refer to overeating.
Dietary guidelines are nutritional guidelines which advice an individual, greater than 2 years, consume a healthy and nutritional diet.
Mediterranean diet is the diet typical to Mediterranean countries, in which there is high consumption of vegetables, fruits, seeds, legumes, nuts and olive. It consists of moderate amounts of carbohydrates and proteins and a moderate to high amount of fats.
A healthy intake is a healthy diet in which the energy from sugars is less than 10% of the total energy intake.
Hence, Option A is the correct answer.
According to Newton's Third Law, If one object exerts a force on a second object, the second object exerts a force on the first object that is:
Question 2 options:
equal in magnitude and opposite in direction
equal in magnitutde and in the same direction
different in magnitude compared to the original force
the same direction as the original force
Answer: equal in magnitude but opposite in direction to the force that it exerts.
Explanation:
Answer:
The second object exert a force that is equal in magnitude and opposite in direction.
Explanation:
Newton's third law of motion states that for every action there is a reaction which is equal in magnitude to the action but acts in opposite directions as the action.
an object weighs 50 N on Jupiter and another object weighs 50 N on the earth. Which has greater mass?
The object on Earth has greater mass
Explanation:
The weight of an object is given by
[tex]W=mg[/tex]
where
W is the weight
m is the mass of the object
g is the strength of the gravitational field at the location of the object
The strength of the gravitational field on Jupiter ([tex]g_J[/tex]) is much larger than that on Earth ([tex]g_E[/tex]), so we can write
[tex]g_J > g_E[/tex]
The weight of the first object on Jupiter is
[tex]W_J = m_1 g_J = 50 N[/tex] (1)
where [tex]m_1[/tex] is the mass of the first object, while the weight of the second object on Earth is
[tex]W_E = m_2 g_E = 50 N[/tex] (1)
while [tex]m_2[/tex] is the mass of the second object.
By dividing eq.(1) by (2), we get
[tex]\frac{m_1 g_J}{m_2 g_E}=1\\\frac{m_1}{m_2}=\frac{g_E}{g_J}[/tex]
And since [tex]g_J > g_E[/tex], this means that [tex]m_2 > m_1[/tex], so the object on Earth has greater mass.
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A deputy sheriff rides a horse while directing traffic, what energy is being used?
Answer:
Kinetic Energy
Explanation:
In relation to the question, as the sheriff rides a horse, the sheriff is in motion and a body in motion is said top possess kinetic energy.
Kinetic energy is that energy of the body that it gains due to the motion or speed of the body which it gains and maintains.
In other words, we can say that the amount of work needed to accelerate an object from its position of rest and set it into motion at some certain velocity.
Imagine a place in the cosmos..far from all gravitational and frictional influences..Suppose that you visit that place (just suppose)...and throw a rock.
What will the rock do? Why?
After the initial push, the rock will keep moving forever at constant velocity (constant speed in a straight line)
Explanation:
We can answer this question by using Newton's first law of motion:
"An object at rest (or in motion at constant velocity) will stay at rest (or will keep moving at constant velocity) unless acted upon unbalanced forces" (Law of inertia)
In this problem, we have a rock in a place very far from any force that can act on it. This means that there are no unbalanced force acting on it, so the rock will keep its state of motion forever.
In this situation, the rock is initially thrown by the astronaut. After the initial push, which accelerates the rock up to a certain velocity, there will be no more forces acting on the rock. This means that the rock will continue moving at a constant velocity forever, so at a constant speed in a straight line.
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Most of the Earth's volume is contained in the
A. mantle
B.
crust
C.
inner core
D.
outer core
Most of the Earth's volume is contained in the "mantle". The correct answer would be option (A).
What is a mantle?The mantle is a layer of semi-solid rock that lies between the Earth's thin crust and its liquid outer core. The mantle is approximately 2,900 kilometers (1,800 miles) thick and makes up about 84% of the Earth's total volume.
The crust, mantle, and core are the three coatings of the Earth. The crust is made up of minerals and solid rocks. The mantle lies under the crust and is largely made up of solid rocks and minerals, but it is punctured by pliable patches of semi-solid magma.
It is composed primarily of silicate rocks and is considered to be the source of the Earth's heat and most of its volcanic activity.
The mantle also plays a key role in plate tectonics, which is the theory that explains the movement of the Earth's continents and ocean floor.
Thus, the correct answer would be option (A) mantle.
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A rock is dropped from a cliff.
What will eventually happen to the rock's motion?
A. Gravity will cause the rock to slow down before it reaches Earth.
B. Air resistance will cause the rock to accelerate as it falls.
C. The rock will continue falling until Earth exerts a contact force on the rock.
D. Friction will cause the rock to continue in its state of motion.
Answer:
C
Explanation:
Using the equation for force (due to weight) and your mass , calculate your force
F=m x 9.8m/s2
F=69.3 x 9.8 m/s2 =???
The weight of the object is 679.1 N
Explanation:
The weight of an object is given by:
[tex]W=mg[/tex]
where
W is the weight
m is the mass of the object
g is the acceleration of gravity
For an object near the Earth's surface, the acceleration of gravity is
[tex]g=9.8 m/s^2[/tex]
The mass of the object in this problem is
m = 69.3 kg
Therefore, its weight is
[tex]W=(69.3)(9.8)=679.1 N[/tex]
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A Tesla Roadster car accelerates from rest at a rate of 7.1m/s for a time of 3.9s
Calculate the distance it travels in this time.
The distance covered is 54.0 m
Explanation:
Since the motion of the car is a uniformly accelerated motion, we can use the following suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
u is the initial velocity
t is the time
a is the acceleration
s is the distance covered
For the car in this problem, we have
u = 0 (it starts from rest)
[tex]a=7.1 m/s^2[/tex] is the acceleration
t = 3.9 s is the time
Substituting, we find s:
[tex]s=0+\frac{1}{2}(7.1)(3.9)^2=54.0 m[/tex]
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Can someone explain the gas laws?
* ex...what happens to pressure if volume and temp. increase*
Explanation:
Use the Ideal Gas Law:
PV = nRT
where P is absolute pressure,
V is volume,
n is number of moles,
R is the universal gas constant,
and T is absolute temperature.
First, identify which variables in the problem are constant (note that R is always a constant).
Next, solve for the constants and use that equation to write a proportion.
Finally, plug in the given values and solve for the unknown variable.
For example, if n and R are the constants:
nR = PV/T
Therefore:
P₁V₁/T₁ = P₂V₂/T₂
Need help with 3 and 4
3) The boulder travelled for a distance of 10 m
4) The work done on the car is zero
Explanation:
3)
The work done by a force on an object is given by:
[tex]W=Fd cos \theta[/tex]
where
F is the magnitude of the force
d is the displacement
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
In this problem we have:
F = 24 N
W = 240 J
[tex]\theta=0^{\circ}[/tex], assuming that the force is applied in the same direction as the displacement
Therefore, we can find d, the displacement of the boulder:
[tex]d=\frac{W}{F cos \theta}=\frac{240}{(24)(cos 0)}=10 m[/tex]
4)
As in the previous exercise, the work done is
The work done by a force on an object is given by:
[tex]W=Fd cos \theta[/tex]
In this problem, we have:
F = 9000 N is the force applied
d = 0 is the displacement, since the car has not moved
Therefore, the work done on the car is
[tex]W=(9000)(0)(cos 0)=0[/tex]
So, no work is done on the car.
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Leah is making a chart to compare and contrast renewable and non-renewable resources. Which of the following would best complete the chart? Renewable Resources Non-renewable Resources __ Used faster than they can be replaced Most sources do not emit greenhouse gases Availability decreases over time Includes solar power, wind power, and geothermal power Includes fossil fuels, natural gas, and coal
Renewable resources can be replaced as quickly as they are used, while non-renewable resources are being used up faster than they can be made by nature.
Explanation:Renewable resources are those that can be replaced by natural processes as quickly as humans use them. Examples include solar power, wind power, and geothermal power. On the other hand, non-renewable resources are consumed or used up faster than they can be made by nature. Examples include fossil fuels such as coal, natural gas, and oil. These resources take millions of years to form and are being used up at a much faster rate than they can be replenished.
Amazon has hired you to help design a new fleet of robots to work in their warehouse. You are trying to decide how powerful the motor that allows the robots to climb vertically up the selves should be. assume that the robot itself will have a mass of 15 kg and needs to be able to carry an object with a mass of 5 kg up to the top of a 20 m shelf in 8 seconds. What minimum wattage should you use?
The minimum power is 490 W
Explanation:
The work that needs to be done by the robot in order to climb the shelves is equal to the gain in gravitational potential energy of the robot + object, therefore:
[tex]W=mg \Delta h[/tex]
where
m = 15 kg + 5 kg = 20 kg is the total mass of the system
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]\Delta h = 20 m[/tex] is the change in height
Substituting,
[tex]W=(20)(9.8)(20)=3,920 J[/tex]
Now we can calcualte the power (wattage) that the robot must have, using the equation
[tex]P=\frac{W}{t}[/tex]
where
W = 3,920 J is the work done
t = 8 s is the time interval
Substituting,
[tex]P=\frac{3920}{8}=490 W[/tex]
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A force does work on a 50 g particle as the particle moves along the following straight paths in the xy-plane: 25 J from (0 m, 0 m) to (5 m, 0 m); 35 J from (0 m, 0 m) to (0 m, 5 m); -5 J from (5 m, 0 m) to (5 m, 5 m); -15 J from (0 m, 5 m) to (5 m, 5 m); and 20 J from (0 m, 0 m) to (5 m, 5 m).
Is this a conservative force?
If the zero of potential energy is at the origin, what is the potential energy at (5m, 5m)?
1) Yes, it is a conservative force
2) The potential energy at (5m, 5m) is 20 J
Explanation:
1)
A force is defined to be conservative if the work done by the force when moving an object does not depend on the path taken, but only on the initial and final position of the object.
Let's verify if this condition is met for the force in this problem:
- For going from (0 m, 0 m) to (5 m, 5 m), the work done is 20 J (A)
Then we can go from (0 m, 0 m) to (5 m, 5 m) through a different path:
- from (0 m, 0 m) to (5 m, 0 m) (work done: 25 J)
- from (5 m, 0 m) to (5 m, 5 m) (work done: -5 J)
Total work done in the second path: 25 J + (-5 J) = 20 J --> same as (A)
We can also go from (0 m, 0 m) to (5 m, 5 m) through a different path:
- from (0 m, 0 m) to (0 m, 5 m) (work done: 35 J)
- from (0 m, 5 m) to (5 m, 5 m) (work done: -15 J)
Total work done in the third path: 35 J + (-15 J) = 20 J --> same as (A)
So, the force is conservative, since the work done does not depend on the path taken.
2)
For a conservative force, the change in potential energy of the object moved by the force is equal to the work done by the force in moving the object from A to B.
Here have:
Point A: (0m, 0m)
Point B: (5m, 5m)
So, the change in potential energy is equal to the work done from A to B:
[tex]W=\Delta U = U_B-U_A[/tex]
where
[tex]U_B[/tex] is the potential energy in point B
[tex]U_A[/tex] is the potential energy in point A
In this problem, the potential energy at the origin (point A) is zero, so
[tex]U_A = 0[/tex]
Also we know that the work done is
W = 20 J
So, we find
[tex]U_B = W+U_A = 20 + 0 = 20 J[/tex]
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The force exerted is not conservative as different amounts of work are being done to move the particle along different paths between the same points. Potential energy isn't defined in the usual sense because the force isn't conservative, but if taken as the average of the works done, the potential energy at (5,5) could be taken as 20J.
Explanation:A conservative force is defined as one in which the work done on a particle by the force as the particle moves from one point to another is independent of the path taken. Specifically, in the case of a conservative force, the work done is only dependent on the initial and final positions. In the given scenario, different amounts of work are being done to move the particle along different paths between the same points. This indicates that the force is not a conservative one.
The potential energy at a point in a field due to a conservative force is equal to the work done by the force to move the particle from a reference point (in this case, the origin point) to that point. Since the force in this scenario isn't conservative, potential energy isn't defined in the usual sense. Nevertheless, if we insist on measuring it, we could take the average of the works done: for instance, between (0,0) and (5,5) the works done are 35+(-15)=20 and -5+25=20, so we could say the potential energy at (5,5) is 20J considering the origin as the reference point with zero potential energy.
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A 0.0625 tank contains 0.0925kg nitrogen at a gauge pressure of 5.17atm.Find the temperature of the gas in degree Celsius
-271.96 °C
Explanation:We are given;
Volume of the tank as 0.0625 L Mass of nitrogen gas as 0.0925 kg or 92.5 g Pressure of the gas as 5.17 atmwe are required to calculate the temperature of the gas.
Step 1: Calculate the number of moles of nitrogen gasMoles = Mass ÷ Molar mass
Molar mass of nitrogen gas = 28.0 g/mol
Therefore;
Moles of N₂ = 92.5 g ÷ 28.0 g/mol
= 3.304 moles
Step 2: Calculate the temperature of the gas;According to the ideal gas equation;
PV = nRT , where n is the number of moles and R is the ideal gas constant, 0.082057 L.atm/mol.K
Rearranging the formula;
T = PV ÷ nR
= ( 5.17 atm × 0.0625 L) ÷ (3.304 moles × 0.082057)
= 1.19 K
But, °C = K - 273.15
Therefore;
T = 1.19 K - 273.15
= -271.96 °C
Thus, the temperature of the gas will be -271.96 °C
As the skydiver falls to Earth, she experiences positive acceleration due to
What is Snell law of refraction
Answer:
Snell's law tells us the degree of refraction and relation between the angle of incidence, the angle of refraction and refractive indices of given pair of media. We know that light experiences the refraction or bending when it travels from one medium to another medium. Snell's law predicts the degree of the bend.
Explanation:
Final answer:
Snell's law expresses the relationship between the incidence and refraction angles and the indices of refraction of two media, represented by the equation n₁ sin θ₁ = n₂ sin θ₂.
Explanation:
Snell's law describes the relationship between angles and indices of refraction when light passes from one transparent medium to another.
It is an essential concept in the study of optics and can be mathematically expressed as n₁ sin θ₁ = n₂ sin θ₂, where n₁ and n₂ are the refractive indices of the first and second medium, respectively, and θ₁ and θ₂ are the angles of incidence and refraction.
This law is named after the Dutch mathematician Willebrord Snell who described it in 1621, but it was also discovered by the Arabian physicist Ibn Sahl in the 10th century.
Through thorough experiments, it became evident that light rays change direction due to the differing speeds of light in various media, a concept Snell himself was not aware of but was able to confirm through the refractive indices.