a) The amplitude of oscillation is 0.780 m.
b) The force constant of the spring can be determined by using the equation F=kx, where F is the force applied to the mass, k is the force constant, and x is the displacement of the mass. Since the displacement of the mass is 0.780 m, we can solve for k to get k = F/x = 0.780 N/m.
c) The position of the mass after one half a period is given by x(t) = (0.780 m)cos[(18.0 rad/s)t] = (0.780 m)cos[(9.0 rad/s)t] = 0.780 m.
d) The position of the mass one-third of a period after it has been released is given by x(t) = (0.780 m)cos[(18.0 rad/s)t] = (0.780 m)cos[(6.0 rad/s)t] = -0.500 m.
e) To determine the time it takes the mass to get to the position x=-0.500 m after it has been released, we can use the equation x(t) = (0.780 m)cos[(18.0 rad/s)t]. Since we know that x=-0.500 m when t=t0, we can solve for t0 to get t0=6.00 s. Therefore, it takes the mass 6.00 s to get to the position x=-0.500 m after it has been released.
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What is the term that describes waves that require a medium through which to travel?
Answer:
Mechanical waves are waves that require a medium. This means that they have to have some sort of matter to travel through. These waves travel when molecules in the medium collide with each other passing on energy. One example of a mechanical wave is sound.
Which of the following is an example of a scientific law?1.Experimental results that show that v = λf2.An explanation for why some electrons are affected when they are hit by photons and some electrons are not.3.A prediction that increasing the volume of an object will increase its ability to float in water4.None of the above
In general, a scientific law describes a phenomenon (or what phenomena will happen given specific circumstances) without explanations of the reasons. On the other hand, scientific theory aims to explain the reasons as to why phenomena occur.
Given this information, the most appropriate answer choice is:
3. A prediction that increasing the volume of an object will increase its ability to float in water
A car is travelling in a straight line and has its velocity uniformly reduced from 20 m * s ^ - 1 to 12m * s ^ - 1 in a distance of 80 m. The car the travels at the lower velocity for 1 minute, and then decelerates uniformly to rest in a further 12 sec. show the whole journey on a velocity-time graph and calculate
(i) the initial deceleration and the time taken to travel 80 m.
(ii) the final deceleration
(iii) the total displacement for the whole journey
i) The initial deceleration of the car is -1.6 m/s² and the time taken is 5 seconds
ii) The final deceleration is -1 m/s²
iii) The total dispalcement = 1016 m
What is the initial deceleration of the car?The initial deceleration of the car is given by the formula below:
v² = u² + 2as
where;
v is the initial velocityu is the final velocitya is acceleration/decelerations is the displacementSolving for a;
12² = 20² + 2 * a * 80
a = -1.6 m/s²
Time taken, t = v - u / a
t = 12 - 20 / (-1.6)
t = 5 seconds
Final deceleration:
a = v - u / t
a = 0 - 12 / 12
a = -1 m/s²
iii) Displacement at constant velocity = 12 * 1 * 60
Displacement at constant velocity = 720 m
Final displacement, s = ut + 0.5at²
s = 12 * 12 + 0.5 * 1 * 12²
s = 216 m
Total dispalcement = 80 + 720 + 216
Total dispalcement = 1016 m
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Rocks in the asteroid belt are traveling around the Sun at a distance of 3.2 Astronomical Units (AU). According to Kepler's Third Law, what is the orbital period of these rocks?
Recall Kepler's Third Law can be written as
P 2 = a 3
if the period is measured in (Earth) years and the distance is measured in AU. Follow the steps in Example 3.1 (just with a different distance value).
Pick the answer that's closest to your result:
The orbital period of the rocks in the asteroid belt is approximately 5.73 years.
What is the average distance, a, between the sun and the asteroid, expressed in astronomical units?
The distance from the Sun to the asteroid belt is between 2-4 AU, or between 300 and 600 million kilometers (186-272 million mi).
Using Kepler's Third Law, we have:
P² = a³
where P is the orbital period of the rocks in years, and a is the semi-major axis of their orbit in astronomical units (AU).
Substituting a = 3.2 AU, we get:
P² = (3.2 AU)³
P² = 32.768 AU³
P = \(\sqrt{32.768}\) years
P ≈ 5.73 years
Therefore, the orbital period of the rocks in the asteroid belt is approximately 5.73 years.
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A 75.0 kg man pushes on a 500,000 kg wall for 250 s but it does not move.
a. How much work does he do on the wall? ____________
b. How much energy is used?__________
c. How much power is exerted?____________
Since no work is done, the power exerted is zero. Therefore, the man exerts no power on the wall.
What is force?In physics, force is defined as any action that can change the motion of an object or cause an object to accelerate. Force is a vector quantity, meaning that it has both magnitude (size or strength) and direction. The unit of force in the International System of Units (SI) is the Newton (N), which is defined as the amount of force required to accelerate a mass of one kilogram at a rate of one meter per second squared (1 N = 1 kg × 1 m/s^2). Force can be measured using a variety of instruments, such as spring scales, strain gauges, or force plates. Some common types of forces include gravitational force, electromagnetic force, frictional force, and normal force. The study of forces and their effects on the motion of objects is known as mechanics and is a fundamental concept in physics.
Here,
a. The man does not do any work on the wall because the wall does not move. Work is only done when there is a displacement in the direction of the force applied.
b. Since no work is done, no energy is used or transferred.
c. The power exerted by the man can be calculated using the formula:
Power = Work / Time
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A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius = 2.0 m). When the mass is at the lowest point on the circle, the speed of the mass is 12 m/s. What is the magnitude of the force of the string on the mass at this position?
Answer:
F= 36 N
Explanation:
F= mv^2/r
F= .5*12^2/2
F= ( .5 * 144 )/2
F= 72 / 2
F= 36
A student is investigating the transfer of thermal energy. The student boils a pot water, removes the pot from the burner, and adds ice cubes to the water Which
direction is the thermal energy flowing in this system?
A from the solid ice to the air
OB from the liquid water to the pot
C from the liquid water to the solid ice
D
from the solid ice to the liquid water
Answer:
C from the liquid water to the solid ice
Explanation:
The direction of heat flow is from the liquid water to the solid ice. Thermal energy generally flows from a body at a high temperature to one at lower temperature values.
Heat is measure of the average kinetic energy of the particles of a medium. The ice is at a lower temperature The boil water has higher temperatureHeat will flow from the boiled water to the solid ice mass.
The thermal energy moves from the liquid water to the solid ice because heat energy moves from hotter body to colder body.
Direction of thermal energyThermal energy moves from hotter body to colder body so we know that in this experiment water is a hotter body whereas ice cubes are colder body so the heat energy will definitely moves from hot body to the cold body.
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An object's moment of inertia is 2. 0 kg. M2. Its angular velocity increases from 20 rad/s to 60 rad/s in 10 seconds. What is the net torque on the object?.
The net torque on the object is 8 N·m
We can use the equation:
Net torque = (change in angular momentum) / (time interval)
An object's angular momentum is determined by:
L = I * ω
where is the angular velocity and I is the moment of inertia.
So, the change in angular momentum is:
ΔL = I * Δω
where Δω is the change in angular velocity.
Substituting the given values, we get:
ΔL = 2.0 kg·m² * (60 rad/s - 20 rad/s) = 80 N·m·s
The time interval is given as 10 seconds.
So, the net torque is:
Net torque = ΔL / Δt = 80 N·m·s / 10 s = 8 N·m
Therefore, the net torque on the object is 8 N·m.
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Imagine you are riding on a yacht in the ocean and traveling at 20 mph. You then hit a golf ball at 100 mph from the deck of the yacht. You see the ball move away from you at 100mph, while a person standing on a nearby beach would observe your golf ball traveling at 120 mph (20 mph + 100 mph).
Now imagine you are aboard the Hermes spacecraft traveling at 0.1c (1/10 the speed of light) past Mars and shine a laser from the front of the ship. You would see the light traveling at c (the speed of light) away from your ship.
An observer on the spacecraft would see the light beam moving away from the ship at the speed of light, while an observer on Mars would also see the light beam moving away from the ship at the speed of light. This is because the speed of light is always constant, regardless of the motion of the observer or the light source.
When the ball is hit from the deck of the yacht at 100 mph, it is moving at a speed relative to the yacht.
Since the yacht is moving at 20 mph, the ball would appear to be moving away from the yacht at 100 mph.
This is because the speed of the ball relative to the yacht is 100 mph, while the speed of the yacht relative to the ground is 20 mph.
Therefore, the total speed of the ball relative to the ground would be the sum of the speed of the yacht and the speed of the ball relative to the yacht, which is 120 mph (20 mph + 100 mph).
Now let's consider the Hermes spacecraft traveling at 0.1c past Mars and shining a laser from the front of the ship.
According to the theory of relativity, the speed of light is always the same for all observers, regardless of their motion or the motion of the light source.
So, regardless of the speed of the spacecraft, the light beam would travel away from the ship at the speed of light, c.
An observer on the spacecraft would see the light beam moving away from the ship at the speed of light, while an observer on Mars would also see the light beam moving away from the ship at the speed of light.
This is because the speed of light is always constant, regardless of the motion of the observer or the light source.
In summary, the theory of relativity tells us that the speed of light is always the same for all observers, regardless of their motion or the motion of the light source.
This means that the speed of light is a fundamental constant of the universe, and it plays a crucial role in our understanding of the laws of physics.
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A cyndrical flask of cross sectional area A is fitted with an airtight piston that is free to slide up and down .contained within the flask is an ideal gas. Initially the pressure applied by the piston is 140kpa and the base of the flask is 22cm. When additional mass is added to the piston the pressure increases to 380kpa. Assuming the system is always at the temperature of 290k find the new height of the piston
Assuming the system is always at the temperature of 290k, the new height of the piston be 16.21 cm.
What is pressure?The physical force applied to an object is referred to as pressure. Per unit area, a perpendicular force is delivered to the surface of the objects. F/A is the fundamental formula for pressure (Force per unit area). Pascals are a unit of pressure (Pa).
Absolute, atmospheric, differential, and gauge pressures are different types of pressure.
As height of the piston is proportional to volume of ideal gas. From Boyle's law of gases at constant temperature we can write:
PV = pv
⇒PH = ph
⇒ 140 kpa × 22 cm = 290 kpa × h
⇒ h = 22 × 140/190 cm = 16.21 cm.
Hence, the new height of the piston be 16.21 cm.
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If an elevator accelerates upward at 10 ms−2 , what is the average blood pressure in the brain? What is the average blood pressure in the feet? If the elevator accelerates downward with the same acceleration, what is the average blood pressure in the brain and feet? take g = 10
The average blood pressure in the brain will be higher than the average blood pressure in the feet. the average blood pressure in the brain will be lower than the average blood pressure in the feet.
When the elevator accelerates upward at \(10 ms^{-2}\), the blood pressure in the brain and feet of a person changes.
Similarly, when the elevator accelerates downward with the same acceleration, the blood pressure in the brain and feet of a person changes.
Let's discuss them one by one:Blood Pressure When Elevator Accelerates Upward at \(10 ms^{-2}\)
When the elevator accelerates upward at \(10 ms^{-2}\), the blood pressure in the brain of a person increases, while the blood pressure in the feet of a person decreases.
This happens due to the gravitational force acting on the body.
Since the gravitational force on the head is greater than the gravitational force on the feet, the blood pressure in the brain increases while the blood pressure in the feet decreases.
Therefore, the average blood pressure in the brain will be higher than the average blood pressure in the feet.
Blood Pressure When Elevator Accelerates Downward at \(10 ms^{-2}\) When the elevator accelerates downward at \(10 ms^{-2}\), the blood pressure in the brain of a person decreases, while the blood pressure in the feet of a person increases.
This also happens due to the gravitational force acting on the body. Since the gravitational force on the head is less than the gravitational force on the feet, the blood pressure in the brain decreases while the blood pressure in the feet increases.
Therefore, the average blood pressure in the brain will be lower than the average blood pressure in the feet.
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1. Two small, identical conducting spheres A and B are a distance R apart; each carries the same charge Q. What is the force sphere B is exerting on sphere A? (2 marks) b. An identical sphere with zero charge, sphere C, makes contact with sphere A and is then moved very far away. What is the net force now acting on sphere A? (3 marks) c. Sphere C next makes contact with sphere A and is then moved far away. What is the force on sphere A in this third case? (2 marks)
Answer:
the same question I want to know
a. Spheres A and B carry the same charge \(\rm \( Q \)\), the force can be written as:
\(\rm \[ F = \frac{k Q^2}{R^2} \]\), b. The net force on sphere A now is the force due to sphere B, which is: \(\rm \[ F = \frac{k \frac{Q}{2} \cdot Q}{R^2} = \frac{k Q^2}{2R^2} \]\), c. The magnitude of the force on sphere A in this third case is \(\rm \( \frac{3}{8} \)\) of the original force.
a) The magnitude of the force (F) sphere B exerts on sphere A can be calculated using Coulomb's law:
\(\rm \[ F = \frac{k Q_1 Q_2}{R^2} \]\)
where:
k is Coulomb's constant \(\rm \( k = \frac{1}{4\pi\epsilon_0} \)\) where \(\rm \( \epsilon_0 \)\) is the vacuum permittivity constant),
\(\rm \( Q_1 \)\) and \(\rm \( Q_2 \)\) are the charges on spheres A and B respectively, and
\(\rm \( R \)\) is the distance between the two spheres.
Since both spheres A and B carry the same charge \(\rm \( Q \)\), the force can be written as:
\(\rm \[ F = \frac{k Q^2}{R^2} \]\)
b) When an identical sphere C makes contact with sphere B, they share the charge equally. Sphere B now carries \(\rm \( \frac{Q}{2} \)\) charge, and sphere C carries \(\rm \( \frac{Q}{2} \)\) charge.
When sphere C is moved far away, it exerts no force on sphere A. So, the net force on sphere A now is the force due to sphere B, which is:
\(\rm \[ F = \frac{k \frac{Q}{2} \cdot Q}{R^2} = \frac{k Q^2}{2R^2} \]\)
c) When sphere C makes contact with sphere A, they both share the charge equally. Each sphere now carries \(\rm \( \frac{Q}{2} \)\) charge.
When sphere C is moved far away, the net force on sphere A now is the force due to the charge on sphere A itself, which is:
\(\rm \[ F = \frac{k \frac{Q}{2} \cdot \frac{Q}{2}}{R^2} \\\\\rm = \frac{3}{8} \frac{k Q^2}{R^2} \]\)
So, the magnitude of the force on sphere A in this third case is \(\rm \( \frac{3}{8} \)\) of the original force.
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An object is moving with an initial velocity of 5.5m/s.It is then subject to a constant acceleration of 2.5 m/s for 11s. How far will it have traveled during the time of its acceleration?
The distance traveled by the object during the time of acceleration is 211.75 m.
What is distance?Distance can be defined as the total lenght between two points.
T o calculate the distance traveled by the object during the time of acceleration, we use the formula below.
Formula:
s = ut+at²/2............ Equation 1Where:
s = Distanceu = Initial velocitya = Accelerationt = TimeFrom the question,
Given:
u = 5.5 m/st = 11 sa = 2.5 m/s²Substitute these values into equation 1
s = (5.5×11)+(2.5×11²)/2s = 60.5+151.25s = 211.75 mHence, the distance traveled by the object is 211.75 m.
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(1 of 5)
Drag the phase slider left and right to move the planet into and out of transit mode. The
normalized flux (relative amount of star light reaching an observer on Earth) when the
planet is off to the side (not transiting) is
while the normalized flux when the planet
is transiting is. (Hint: examine the plot at the top of the applet while you move the
phase slider. The number you are looking for is on the left of the plot.)
O 1.000, 0.990
O 0.990,1.000
O 1.00000, 0.99995
O 0.99995,1.00000
Drag the phase slider left and right to move the planet into and out of transit mode. The normalized flux (the relative amount of starlight reaching an observer on Earth) when the planet is off the side (not transiting) is 0.990 while the normalized flux when the planet is transiting is 1.000. Thus, the correct option for this question is B.
What is Normalized flux?Normalized flux may be characterized as a type of key methodology in spectral data mining. It is important for the efficiency and accuracy of the automatic processing of massive astronomical spectral data, information extraction, and sharing.
A transit occurs when a planet passes between a star and its observer. Transits within our solar system can be observed from Earth when Venus or Mercury travel. The frequency of the normalized flux when the planet is transiting is completely different from when the planet is off the side.
Therefore, the correct option for this question is B.
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Answer: 1.000,0.990 (correct)
Explanation:
PLZ EXPLAIN AND I WILL GIVE YOU BRANILEST
How do two interacting objects exert equal and opposite forces on each other when they collide, even though they have different masses?
Answer:
Did a little research.
Explanation:
The Law of Action-Reaction:
Newton's third law of motion is naturally applied to collisions between two objects. In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction. Such forces often cause one object to speed up (gain momentum) and the other object to slow down (lose momentum). According to Newton's third law, the forces on the two objects are equal in magnitude. While the forces are equal in magnitude and opposite in direction, the accelerations of the objects are not necessarily equal in magnitude. In accord with Newton's second law of motion, the acceleration of an object is dependent upon both force and mass. Thus, if the colliding objects have unequal mass, they will have unequal accelerations as a result of the contact force that results during the collision.Consider the collision between the club head and the golf ball in the sport of golf. When the club head of a moving golf club collides with a golf ball at rest upon a tee, the force experienced by the club head is equal to the force experienced by the golf ball. Most observers of this collision have difficulty with this concept because they perceive the high speed given to the ball as the result of the collision. They are not observing unequal forces upon the ball and club head, but rather unequal accelerations. Both club head and ball experience equal forces, yet the ball experiences a greater acceleration due to its smaller mass. In a collision, there is a force on both objects that causes an acceleration of both objects. The forces are equal in magnitude and opposite in direction, yet the least massive object receives the greatest acceleration.
Consider the collision between a moving seven ball and an eight ball that is at rest in the sport of table pool. When the seven ball collides with the eight ball, each ball experiences an equal force directed in opposite directions. The rightward moving seven ball experiences a leftward force that causes it to slow down; the eight ball experiences a rightward force that causes it to speed up. Since the two balls have equal masses, they will also experience equal accelerations. In a collision, there is a force on both objects that causes an acceleration of both objects; the forces are equal in magnitude and opposite in direction. For collisions between equal-mass objects, each object experiences the same acceleration.
Cs-124 has a half-life of 30.8 s
A) If we have 7.5 μg initially, how many Cs nuclei are present?
B) How many nuclei are present 2.6 min later?
A) There are 3.70 x 10¹⁶ Cs nuclei present initially and B) There are 2.27 x 10¹⁵ Cs nuclei present 2.6 min later.
A) To solve this problem, we can use the following equation,
\(N = N'2^{\frac{-t}{T_{1/2}}}\)), number of Cs nuclei present is N, initial number of Cs nuclei present is N', elapsed time is t, half-life of Cs-124 is T½. First, we need to convert the initial mass of Cs-124 to the number of nuclei present,
7.5 μg Cs-124(1g/10⁶μg)(6.022x10²³nuclei/1g)
= 4.52 x 10¹⁵ Cs-124 nuclei.
Using the equation above, we can find the number of Cs-124 nuclei present after 0 s,
\(N = 4.52 * 10^{15} * 2^{\frac{-0}{30.8}}}\)
= 4.52 x 10¹⁵ Cs-124 nuclei
Therefore, initially, there are 4.52 x 10¹⁵ Cs-124 nuclei present.
B) After 2.6 min (156 s), we can again use the same radioactivity equation to find the number of Cs-124 nuclei present,
\(N = 4.52 * 10^{15} * 2^{\frac{-156}{30.8}}}\)
N = 1.60 x 10¹⁴ Cs-124 nuclei
Therefore, 2.6 min later, there are 1.60 x 10¹⁴ Cs-124 nuclei present.
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What does a producer need to make its own food
Answer:
A producer needs the sun to make its own food, because producers use photosynthesis. In photosynthesis you use the sun to turn the air people breathe out and water into glucose and oxygen.
Explanation:
Hope this helps ^-^
A spaceship hovering over the surface of Uranus drops an object from a height of 19 m. How much longer does it take to reach the surface than if dropped from the same height on Earth? Neglect air resistance in both cases. [The acceleration due to gravity on Uranus is 88.9% of that on Earth, gUranus = (0.889)g.] answer in : s
h= height = 19 m
g.e = gravity on earth = 9.8 m/s^2
g.u = gravity on Uranus = 9.8 x (88.9/100) = 8.71 m/s^2
Apply :
h = vot + 1/2 gt^2
h= 1/2 g t^2
t = √(2h/g)
Earth:
t= √(2(19)/9.8) = 1.97 s
Uranus
t= √(2(19)/8.71) = 2.09 s
Tu - Te = 2.09 - 1.97 = 0.12 s
Answer : 0.12 s
give bullying cases a
stratified random sampling
in thesis
The main goal of a student bullying survey is to identify and quantify many aspects, including rates of bullying, student and staff attitudes towards bullying, the types of bullying, and more in order to tackle them.
Over 70% of young people say they have witnessed bullying take place in their school, according to StopBullying.gov, a website run by the U.S. Department of Health & Human Services. Nearly 50% of students in grades 4 through 12 reported experiencing bullying in a given month.
The two most prevalent kinds of bullying are verbal and social, which can take the form of name-calling, mocking, rumours, property theft, sexual comments and gestures, or even physical assault.
Social bullying occurs more frequently than physical bullying, and cyberbullying, while it is on the rise, is still less common.
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The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m
Answer:
\(v=(6ti+6k)\ m/s\)
Explanation:
Given that,
The position of a particle is given by :
\(r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m\)
Let us assume we need to find its velocity.
We know that,
\(v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s\)
So, the velocity of the particle is \((6ti+6k)\ m/s\).
Using the formula: w = mg , calculate your mass. g is the acceleration due to gravity. On earth, g = 9.8 m/s^2Show your work.
Given,
W=mg
The acceleration due to gravity, g=9.8 m/s²
Let us assume that the weight is W Newtons.
The weight is the product of the mass and the acceleration due to gravity. Thus, in the above equation m is the mass.
Thus on rearranging the given equation,
\(m=\frac{W}{g}\)On substituting the known values,
\(m=\frac{W}{9.8}\)Thus the mass is given by W/9.8, where W is the value of the weight.
1. Two charges are separated by a distance of 1 cm. One charge has a value of 7 micro Coulombs. The other charge has a value of 10 micro Coulombs. What is the force between them, in pounds. Make sure to include the sign of the force which will be positive if the charges repel each other and negative if they attract each other.
2. 12 gauge copper wire is normally used in house wiring. When aluminum wire is used one needs to use a smaller gauge size to obtain the same resistance, 40 ft of 12 gauge copper wire was calculated. What would the resistance be if 10 gauge aluminum wire were used?
3. A 12 V automobile battery can supply 51 amps for one hour and cost $194. What is the cost of this electricity in cents per kWh?
4. Most of the body's resistance is in its skin. When wet, salts go into ion form, and the resistance is lowered. Thus, the resistance of the skin can go from 100,000 ohms when dry to 300 ohms when wet. What is the current that would be carried through the body, in milliAmperes, if you touched a 240 V power line while dry? Currents over 10 mA are almost always deadly.
1. The force between the two charges is 1.78 × 10⁻⁵ pounds, with opposite signs indicating attraction between the charges.
2. The resistance of 10 gauge aluminum wire over a 40 ft distance would be 0.506 ohms.
3. The cost of electricity from the automobile battery is 38.6 cents per kWh.
4. The current that would be carried through the body is 0.8 mA if dry.
1. The force between two point charges can be calculated using Coulomb's law, which states that the force is proportional to the product of the charges and inversely proportional to the square of the distance between them.
Using the values given, the force can be calculated as F = (k * q1 * q2) / r², where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them. Plugging in the values, the force can be calculated as 1.78 × 10⁻⁵ pounds, with opposite signs indicating attraction between the charges.
2. The resistance of a wire is determined by its length, cross-sectional area, and resistivity. The resistivity of aluminum is higher than that of copper, so a larger cross-sectional area is required to achieve the same resistance. Using the gauge size conversion chart, 10 gauge aluminum wire has a cross-sectional area of 5.26 mm², which is approximately 83% of the cross-sectional area of 12 gauge copper wire.
Thus, the resistance of 10 gauge aluminum wire over a 40 ft distance can be calculated as R = (rho * L) / A, where rho is the resistivity of aluminum, L is the length, and A is the cross-sectional area. Plugging in the values, the resistance can be calculated as 0.506 ohms.
3. To calculate the cost of electricity per kWh, the total cost and the total amount of energy supplied must be known. Since the battery supplies 12 V and 51 A for one hour, the total energy supplied can be calculated as E = V * I * t, where V is the voltage, I is the current, and t is the time.
Plugging in the values, the total energy supplied can be calculated as 612 watt-hours (Wh). Since one kWh is equal to 1000 Wh, the total energy supplied can be converted to 0.612 kWh. Dividing the total cost by the total energy supplied gives the cost per kWh, which is 38.6 cents.
4. The current through the body can be calculated using Ohm's law, which states that current is equal to voltage divided by resistance. Using the values given, the resistance can be either 100,000 ohms or 300 ohms depending on whether the skin is dry or wet.
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a bicycle wheel with a radius of 30 cm and 16 spokes that is mounted with its axle fixed in a horizontal position. A 50-g mass is hung from a string wrapped around the periphery of the tire. The wheel is held stationary with the weight hanging as shown and then released. The wheel starts to spin and spins faster and faster until the string slips off 5 seconds after release.
What is the acceleration of the hanging mass (in m/s2) while the wheel's speed is increasing?
The acceleration of the hanging mass while the wheel's speed is increasing is 5,625 m/s².
What is acceleration?Acceleration is a vector quantity that measures the rate of change of an object's velocity. It is defined as the rate at which an object's velocity changes over time. Acceleration can be positive, negative, or zero. Acceleration can be caused by forces such as gravity, friction, or thrust.
The acceleration of the hanging mass is determined by the torque (τ) and the moment of inTherefore, the acceleration of the hanging mass can be calculated as a = τ/I = (F×r)/(mr²) = F/(mr). In this case, F = 50 g, r = 30 cm and m = 16 spokes. We can convert the mass from grams to kilograms by dividing by 1000, giving us m = 0.016 kg.
Plugging these values into the equation, we get a = 50/(0.03×0.016) = 5,625 m/s².
Therefore, the acceleration of the hanging mass while the wheel's speed is increasing is 5,625 m/s².
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If the battery is 4.0V the voltmeter reading across R is 2.0V and the resistance per unit length of wire AX is 2 ohms per metre calculate the current in the circuit when AP is 40.0cm(neglect the internal resistance of the battery)
The current in the circuit is 2.5 A.
Voltage across the wire AX, V = 4 - 2 = 2V
Resistance per unit length of wire AX, R/l = 2 Ω/m
Length of the wire, l = 0.4 m
Resistance of the wire,
R = R/l x R
R = 2 x 0.4 = 0.8 Ω
According to Ohm's law, the current in the wire AX,
I = V/R
I = 2/0.8
I = 2.5 A
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An acorn falls from rest from the top of a 19m tall oak tree. How long does it take for the acorn to fall to the ground? How fast is the acorn going before it hits the ground?
Answer:
We can solve this problem using the kinematic equation:
y = 1/2 * g * t^2
where y is the height of the tree, g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken to fall to the ground.
We can solve for t using:
t = sqrt(2y/g)
Plugging in the values, we get:
t = sqrt(2(19)/9.8)
t = 2.19 seconds
So, it takes 2.19 seconds for the acorn to fall from the tree to the ground.
To find the velocity of the acorn just before it hits the ground, we can use:
v = g * t
Plugging in the values, we get:
v = 9.8 * 2.19
v = 21.46 m/s
So, the acorn is going approximately 21.46 m/s just before it hits the ground.
Explanation:
The slope of a speed vs.time graph is equal to the object’s
Answer:
acceleration
Explanation:
speed=v,
In speed vs time graph, slope = dv/dt = acceleration
2 part question
2. Write out the final answer for the coefficient of friction as the average of the values in the table (average all of your coefficients of frictions, and write that average below).
Answer for question #2
The average is = 0.2559975 or 0.256
3. Using similar methods as in questions #1 and #2, estimate the mass of the object in the present, the mystery mass. Include a description of the process and the mathematics of estimating the mass. Use the average coefficient of friction written in #2 to solve for the mystery mass.
4. Set an object (anyone you choose) to be moving at a constant speed. Describe the procedure for keeping the object moving at a constant speed. Write down the speed at which the object is moving, as well.
A uniform rod of length 0.7 m and mass 10 kg rotates freely about a horizontal axis passing through one end of the rod a bullet of mass 20 g traveling horizontally with velocity v=200 m/s penetrates the end of the rod. and hooked on the rod find the angular velocity of the rod when the bullet hits the bar
Answer:
ω = 1.7 rad/s
Explanation:
Conservation of angular momentum
Assuming the rod is initially hanging vertically at rest.
Initial angular momentum is carried by the bullet only
L = Iω = (mR²)(v/R) = mvR = 0.020(200)(0.7) = 2.8 kg•m²/s
the same angular momentum exists after impact, only the moment of inertia has increased by that of the rod. I = ⅓mR²
2.8 = (⅓(10)(0.7²) + 0.020(0.7²))ω
2.8 = (1.64313333...)ω
ω = 1.70406134...
PLEASE HELP!! Newton's second law of motion states that force equals mass times acceleration.
How would this law apply to a runner in a track and field event?
The runner needs to take a few steps to slow down after crossing the finish line.
To win the race, the runner must slow down on the turns and speed up on the straight sections.
The shoes of the runner help increase his mass so he can run faster.
Newton's second law of motion does not apply to track and field.
who will it benefit if they do butterfly groin stretch
Answer:
ummmmm mmmmm butterflies :)
Explanation:
Answer:
It is a great stretch for athletes who play field or court sports, runners, and anyone who has tight hips or a history of groin injuries.
Its a great flexibility exercise
Explanation:
Hope that helps