The value of steam iron resistance is [tex]20ohm[/tex].
The relation between voltage, current and resistant shown below,
[tex]V=IR\\\\R=\frac{V}{I}[/tex]
Where V is voltage and I is current .
Given that, [tex]I=6A,V=1.20*10^{2}[/tex]
Substitute values in above relation.
[tex]R=\frac{1.2*10^{2} }{6}=20ohm[/tex]
Hence, the value of steam iron resistance is [tex]20ohm[/tex].
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The resistance of the steam iron can be calculated using Ohm's Law, expressed as the formula R = V/I. With a current of 6.00 A and a voltage of 1.20 ✕ 102 V, the steam iron has a resistance of 20.0 Ohms.
Explanation:The electrical characteristics of a device, including resistance, can usually be determined using Ohm's Law, which states that the current through a conductor between two points is directly proportional to the voltage across the two points. It can be expressed in the formula R = V/I, where R is resistance, V is voltage, and I is current.
In this case, our steam iron has a current (I) of 6.00 A and a voltage (V) of 1.20 ✕ 102 V. Substituting these values into the Ohm's Law formula gives us: R = (1.20 x 102 V) / 6.00 A.
This calculates to a resistance (R) of 20.0 Ohms for the steam iron.
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. A child has a toy tied to the end of a string and whirls the toy at constant speed in a horizontal circular path of radius R. The toy completes each revolution of its motion in a time period T. What is the magnitude of the acceleration of the toy? a. c. Zero d. 4T2R/T2 e. TR/T2
Explanation:
Formula for centripetal acceleration of an object is as follows.
a = [tex]\frac{v^{2}}{r}[/tex]
When an object is travelling in a circular path then it is difficult to measure its velocity.
Hence, for a circular object the formula for acceleration is as follows.
a = [tex]\frac{4 \pi^{2} r}{T^{2}}[/tex]
a = [tex]\frac{V^{2}}{r}[/tex], and V = [tex]\frac{d}{T} = \frac{2 \pi r}{T}[/tex]
a = [tex]\frac{(\frac{[2\pi r]}{T})^{2}}{r}[/tex]
= [tex]\frac{4 \pi^{2} r}{T^{2}}[/tex]
Thus, we can conclude that the magnitude of the acceleration of the toy is [tex]\frac{4 \pi^{2} r}{T^{2}}[/tex].
If you fire a bullet through a board, it will slow down inside and emerge at a speed that is less than the speed at which it entered. Does light, then, similarly slow down when it passes through glass and also emerge at a lower speed? Defend your answer.
Answer:
RUn he got gun
Explanation:
The net external force acting on an object is zero. Is it possible for the object to be traveling with a velocity that is not zero if the answer is yes state whether any conditions must be placed on the magnitude and direction of velocity?
Answer:
Yes, this is according to the Newton's first law of motion.
Neither its direction nor its velocity changes during this course of motion.
Explanation:
Yes, it is very well in accordance with Newton's first law of motion for a body with no force acting on it and it travels with a non-zero velocity.
During such a condition the object will have a constant velocity in a certain direction throughout its motion. Neither its direction nor its velocity changes during this course of motion.
An object can maintain a nonzero velocity in the absence of a net external force due to Newton's first law of motion. The object will continue to move with the velocity it has until a net force acts upon it. This state of motion with constant velocity is known as dynamic equilibrium.
Explanation:Yes, it is possible for an object to be traveling with a velocity that is not zero even if the net external force acting on it is zero. According to Newton's first law of motion, also known as the law of inertia, an object will maintain its state of motion unless acted upon by a net external force. This means that if there is no net external force on the object, its acceleration is zero, and it will continue moving at its current velocity, which can be nonzero. This constant velocity can be in any direction and of any magnitude, and it remains constant until acted upon by a net force.
For example, when your car is moving at a constant velocity down the street, even though it is moving, the net external force on it can be zero. The forces such as friction and air resistance are balancing out the driving force, leading to no net force on the car, which is a state of dynamic equilibrium.
A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force \vec{F} =(33 N)\hat{i} - (41 N)\hat{j} to the cart as it undergoes a displacement \vec{s} = (-9.4 m)\hat{i} - (3.1 m)\hat{j}.
Part A
How much work does the force you apply do on the grocery cart?
Express your answer using two significant figures.
W =
{\rm J}
Answer:
[tex]W=-183.1\ J[/tex]
Explanation:
Given:
force applied, [tex]\vec{F} =(33 N)\hat{i} - (41 N)\hat{j}[/tex]
displacement caused, [tex]\vec{s} = (-9.4 m)\hat{i} - (3.1 m)\hat{j}[/tex]
Work done by the force on the cart:
[tex]W=\vec F.\vec s[/tex]
[tex]W=[(33 N)\hat{i} - (41 N)\hat{j}].[(-9.4 m)\hat{i} - (3.1 m)\hat{j}][/tex]
[tex]W=-310.2+127.1[/tex]
[tex]W=-183.1\ J[/tex]
Negative work means that the force and displacement have an obtuse angle between them.
Answer:
-180 J
Explanation:
We are given that
Constant force=[tex]F=(33 N)\hat{i}-(41 N)\hat{j}[/tex]
Displacement=[tex]\vec{s}=(-9.4m)\hat{i}-(3.1m)\hat{j}[/tex]
We have to find the work done .
We know that
Work done=[tex]F\cdot s[/tex]
Using the formula
Work done=[tex](33i-41j)\cdot (-9.4i-3.1j)[/tex]
Work done =[tex]33i\cdot (-9.4)i+41j\cdot 3.1 j[/tex]
By using rule [tex]i\cdot i=j\cdot j=k\cdot k=1,i\cdot j=j\cdot k=k\cdot i=i\cdot k=k\cdot j=j\cdot i=0[/tex]
Work done=[tex]-310.2+127.1[/tex]
Work done=-183.1 J
We have to write answer in two significant figures.
When units digit 3 is less than 5 then digits on left side of 3 remains same and digits on right side of 3 and 3 will be replace by zero
Work done=-180 J
Hence, the work done =-180 J
Two large parallel conducting plates are separated by a distance d, placed in a vacuum, and connected to a source of potential difference V. An oxygen ion, with charge 2e, starts from rest on the surface of one plate and accelerates to the other. If e denotes the magnitude of the electron charge, the final kinetic energy of this ion is:
Answer:
2eVI
Explanation:
The final kinetic energy of an oxygen ion, with a charge of 2e, accelerated from one plate to another separated by a potential difference V, equals twice the product of the electron charge and the potential difference, or 2eV.
Explanation:The question involves an oxygen ion with a charge of 2e that is accelerated between two conducting plates separated by a distance d and connected to a potential difference V. This physical scenario falls under the domain of physics, specifically electromagnetism. The energy an ion gains when accelerated through a potential difference is called its kinetic energy. From conservation of energy, we understand that the kinetic energy of the accelerated ion should equal the work done on it by the electrical force, which is itself equal to the charge of the ion times the potential difference of the plates.
Therefore, the kinetic energy (KE) of the ion can be expressed as follows: KE = qV, where q is the charge of the particle and V is the potential difference. In this case, the charge is 2e (twice the electron charge), and the potential difference is V. Thus, the final kinetic energy of the ion is 2eV. It's important to note that this equation is derived from the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred or converted from one form to another.
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A lunar eclipse can only happen during a(1) new moon.(2) solstice.(3) first quarter moon.(4) full moon.(5) perihelion passage of the Sun.
Answer:
(4) full moon.
Explanation:
Lunar eclipse can only occur on a full moon night when the sun the earth and the moon are very much in a straight line.
During this period the the light of the sun that incidents on the moon is blocked by the earth and so we have the phases of the moon due to the relative motion of the three bodies which partially enables the light of the sun to reach the moon.
The moon appears orange-red during this time because the light that reaches the moon is after the refraction through the earth's atmosphere from which the other wavelengths have been absorbed by the earth's atmosphere.
Lunar eclipse can only occur at night and hence it can only be observed from about half of the earth.
A lunar eclipse occurs when the full moon moves into Earth's shadow, which can only happen when the Sun, Earth, and Moon are nearly aligned. This event is more common and widely visible than a solar eclipse, which requires a new moon and occurs when the Moon blocks the Sun.
A lunar eclipse occurs when the Moon enters the shadow of Earth. For a lunar eclipse to happen, the Sun, Earth, and Moon must be nearly in a straight line. The Moon must be in its full moon phase, as this is the only time when the alignment allows Earth's shadow to fall on the entire face of the Moon that is visible from Earth. This alignment does not occur at every full moon due to the inclination of the Moon's orbit. However, when it does, the shadow of Earth can cover about four full moons, given the length of Earth's shadow is about 1.4 million kilometers, and the distance to the Moon is roughly 384,000 kilometers.
It's important to differentiate between a lunar and a solar eclipse; the latter occurs when the Moon passes in front of the Sun, blocking it from view, and this can only happen during a new moon. The solar eclipse also requires the celestial bodies to be in the same plane called the ecliptic. Unlike a solar eclipse, a lunar eclipse is visible to all on the night side of Earth, making it an event observed more frequently from any given place on Earth.
The child then walks towards the center of the merry-go-round and stops at a distance 0.455 m from the center. Now what is the angular velocity of the merry-go-round? Answer in units of rad/s.
Final answer:
The new angular velocity of the merry-go-round can be calculated using the conservation of angular momentum. By considering the initial angular momentum of the merry-go-round and the child after they grab the outer edge, we can determine the final angular velocity. The new angular velocity is approximately 0.414 rad/s.
Explanation:
To calculate the new angular velocity of the merry-go-round, we can use the conservation of angular momentum. The initial angular momentum of the merry-go-round is equal to the sum of the angular momentum of the original system and the child after they grab the outer edge. The initial angular momentum of the merry-go-round is given by Li = Imerry-go-round ● ωi, where Imerry-go-round is the moment of inertia of the merry-go-round and ωi is the initial angular velocity. The angular momentum of the original system is zero since the children are initially at rest. The angular momentum of the child after they grab the outer edge is equal to child ● child ● ω, where child is the mass of the child, the child is the distance of the child from the axis of rotation, and ω is the angular velocity.
Applying the principle of conservation of angular momentum, we have:
Li = (Imerry-go-round + child ● child) ● ωf
Solving for ωf, we get:
ωf = Li / (Imerry-go-round + child ● child)
Substituting the given values, we have:
ωf = (1000.0 kg.m² ● 6.0 rev/min) / (1000.0 kg.m² + 22.0 kg ● 0.455 m)
Converting rev/min to rad/s, we get:
ωf = (1000.0 kg.m² ● (6.0 rev/min ● 2π rad/rev) / (60 s/min)) / (1000.0 kg.m² + 22.0 kg ● 0.455 m)
Simplifying the expression, we find that the new angular velocity of the merry-go-round is approximately 0.414 rad/s.
In the vertical jump, an Kobe Bryant starts from a crouch and jumps upward to reach as high as possible. Even the best athletes spend little more than 1.00 s in the air (their "hang time"). Treat Kobe as a particle and let ymax be his maximum height above the floor. Note: this isn't the entire story since Kobe can twist and curl up in the air, but then we can no longer treat him as a particle.
To explain why he seems to hang in the air, calculate the ratio of the time he is above ymax/2 moving up to the time it takes him to go from the floor to that height. You may ignore air resistance.
Answer:
the athlete spends 2.4 times more time at the upper part of his way than in the lower one.
Explanation:
Let’s find the velocity V1 of an athlete to reach half of the maximum height equation
V1 = v20 -2gh = v20 -2g(ymax)/2
Here, Vo is the initial velocity of athlete, v1 is the velocity of athlete at half the maximum height, g is the acceleration due to gravity, h=ymax /2 is half of the maximum height.
We can fund the maximum height that athlete can reach from the law of conservation of energy
KE = PE
1/2M v20 = mg ymax
ymax = v20 /2g
Then, substituting ymax into the first equation we get
V21 = v20 – v20/2 = v20/2
V1 = V0/∫2, we can find the time that the athlete needs to reach the maximum height (ymax) from the kinematic equation
V = V0 – gt
Here, V is the final velocity of an athlete at the maximum height; V0 is the initial velocity of an athlete
Since, V=0ms-1, we get t=V0/g
Similarly, we can find the time t1 that an athlete needs to reach maximum height from the Ymax/2:
T1 = V1/g =V0/g∫2
So, it is obvious that the time to reach Ymax from Ymax/2 is nothing more than the difference between t and t1:
t-t1 =V0/g(1-1/∫2)
finally, we can calculate the ratio of the time he is above Ymax/2 to the time it takes him to go from floor to that height.
T1/t-t1 =V0/g∫2V0 ×g∫2/∫2-1 =2.4
Answer; the athlete spends 2.4 times more time at the upper part of his way than in the lower one.
550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to compress the gas by a factor of 11.0, starting from its initial volume?
Final answer:
To calculate the work done to compress the gas by a factor of 11.0, we can use the relationship between work and volume changes. The work required to compress the gas by a factor of 11.0 is -11,000 J.
Explanation:
To calculate the work done to compress the gas by a factor of 11.0, we can use the relationship between work and volume changes. The work done on a gas during compression is given by the equation:
Work = Pressure ×Change in Volume
In this case, we know that the work required to compress the gas to half its initial volume is 550 J. Let's assume the initial volume of the gas is V. So the work done to compress the gas by a factor of 11.0 can be calculated as:
Work = Pressure × Change in Volume
550 J = Pressure ×(V/2 - V)
550 J = Pressure ×(-V/2)
Solving for Pressure, we get:
Pressure = -1100 J/V
Therefore, the work required to compress the gas by a factor of 11.0 is -1100 J/V× (11V - V) = -11,000 J (negative sign indicates work is done on the gas).
Merry-go-rounds are a common ride in park play-grounds. The ride is a horizontal disk that rotates about a vertical axis at their center. A rider sits at the outer edge of the disk and holds onto a metal bar while someone pushes on the ride to make it rotate. Suppose that a typical time for one rotation is 6.0 s and the diameter of the ride is 16 ft.
A) For this typical time, what is the speed of the rider in m/s?
B) What is the rider's radial acceleration, in m/s?
C) What is the rider's radial acceleration if the time for one rotation is halved?
The speed of the rider is 0.81 m/s, the radial acceleration is 0.338 m/s², and if the time for one rotation is halved, the speed becomes 0.405 m/s and the radial acceleration becomes 0.085 m/s².
Explanation:A) To calculate the speed of the rider in m/s, we can use the formula:
Speed = Distance / Time
The distance traveled by the rider in one rotation is equal to the circumference of the ride, which is given as the diameter multiplied by π (pi).
Therefore, the distance = 16 ft × π
To convert this distance to meters, we multiply by the conversion factor 0.3048 m = 1 ft.
So, the distance in meters = 16 ft × 0.3048 m/ft × π
Given that the time for one rotation is 6.0 s, we can now calculate the speed:
Speed = (16 ft × 0.3048 m/ft × π) / 6.0 s
Simplifying this equation gives us:
Speed ≈ 0.81 m/s
B) The radial acceleration of the rider can be calculated using the formula:
Radial Acceleration = (Speed)² / Radius
Given that the radius of the ride is half the diameter, which is 8 ft, we can substitute the values into the formula:
Radial Acceleration = (0.81 m/s)² / (8 ft × 0.3048 m/ft)
Simplifying this equation gives us:
Radial Acceleration ≈ 0.338 m/s²
C) If the time for one rotation is halved, the speed of the rider will also be halved because speed is distance divided by time. Therefore, the new speed would be 0.81 m/s / 2 = 0.405 m/s.
The radial acceleration can then be calculated using this new speed and the same formula as in part B:
Radial Acceleration = (0.405 m/s)² / (8 ft × 0.3048 m/ft)
Simplifying this equation gives us:
Radial Acceleration ≈ 0.085 m/s²
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Air in human lungs has a temperature of 37.0°C and a saturation vapor density of 44.0 g/m³.
(a) If 2.00 L of air is exhaled and very dry air inhaled, what is the maximum loss of water vapor by the person?
(b) Calculate the partial pressure of water vapor having this density, and compare it with the vapor pressure of 6.31 × 10³ N/m².
The maximum loss of water vapor per breath is 0.088 g. The partial pressure of water vapor is 6286.41 N/m², which closely matches the given vapor pressure of 6.31 × 10³ N/m², confirming the saturation condition.
To solve the given problem, we need to perform the following calculations:
(a) Maximum Loss of Water Vapor
The saturation vapor density of water vapor at 37.0°C is 44.0 g/m³. Given that 2.00 L of air is exhaled, we first convert the volume to cubic meters (since the density is in g/m³):
[tex]2.00 L = 2.00 * 10^{-3} m^3[/tex]
Now, using the density to find the mass of water vapor exhaled:
[tex]mass = density * volume = 44.0 g/m^3 *2.00 * 10^{-3} m^3 = 0.088 g[/tex]
Thus, the maximum loss of water vapor per breath is 0.088 g.
(b) Partial Pressure of Water Vapor
To find the partial pressure of water vapor, we use the Ideal Gas Law: PV = nRT. First, we convert the given water vapor density into moles per unit volume:
Density (44.0 g/m³) divided by the molar mass of water (18.015 g/mol) gives us:
44.0 g/m³ ÷ 18.015 g/mol = 2.44 mol/m³
So the number of moles, n, per unit volume is 2.44 mol/m³. Now using the Ideal Gas Law:
P = nRT, where R is the universal gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin (310.15 K)
[tex]P = 2.44 mol/m^3 * 8.314 J/(mol.K) * 310.15 K = 6286.41 N/m^2[/tex]
Thus, the partial pressure of water vapor is 6286.41 N/m². Comparing this with the vapor pressure of 6.31 × 10³ N/m², we see they are very close, confirming the saturation condition.
A car accelerates uniformly from rest to 20 m/sec in 5.6 sec along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is 9,000 N, and (b) the weight of the car is 14,000 N.
Answer:
(a) [tex]P=33000W[/tex]
(b) [tex]P=51000W[/tex]
Explanation:
The average power is defined as the amount of work done during a time interval:
[tex]P=\frac{W}{t}(1)[/tex]
According to work-energy theorem, the work done is equal to the change in kinetic energy. So, we have:
[tex]W=\Delta K\\W=K_f-K_0\\W=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}\\(2)[/tex]
Recall that the weight is given by:
[tex]w=mg\\m=\frac{w}{g}(3)[/tex]
The car accelerates uniformly from rest ([tex]v_0=0[/tex]). Replacing (3) in (2), we have:
[tex]W=\frac{wv_f^2}{2g}[/tex]
(a) Finally, we replace this in (1):
[tex]P=\frac{wv_f^2}{2gt}\\P=\frac{9000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=33000W[/tex]
(b)
[tex]P=\frac{14000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=51000W[/tex]
(a) The average power required to accelerate the car of 9000 N is 32798.57 W.
(b) The average power required to accelerate the car of 14,000 N is 51020.40 W.
Given data:
The initial velocity of car is, u = 0 m/s. (Since car was initially at rest)
The final velocity of car is, v = 20 m/s.
The time interval is, t = 5.6 s.
The given problem is based on the concept of average power. The average power is defined as the amount of work done during a time interval. Then,
P = W/t
Here, W is the work done and its value is obtained from the work - energy theorem as,
[tex]W = \Delta KE\\\\W = \dfrac{1}{2}m(v^{2}-u^{2})[/tex]
Here, m is the mass.
(a)
For the weight of 9000 N, the mass of car is,
[tex]w = mg\\\\9000 = m \times 9.8\\\\m =918.36 \;\rm kg[/tex]
So, the Work is obtained as,
[tex]W =\dfrac{1}{2} \times 918.36 \times (20^{2}-0^{2})\\\\W =183672\;\rm J[/tex]
Then, the average power required to accelerate the car is,
P = W/t
P = 183672 / 5.6
P = 32798.57 W
Thus, we can conclude that the average power required to accelerate the car of 9000 N is 32798.57 W.
(b)
For the weight of 14,000 N, the mass of car is,
[tex]w = mg\\\\14,000 = m \times 9.8\\\\m =1428.57 \;\rm kg[/tex]
So, the Work is obtained as,
[tex]W =\dfrac{1}{2} \times 1428.57.36 \times (20^{2}-0^{2})\\\\W =285714.28\;\rm J[/tex]
Then, the average power required to accelerate the car is,
P = W/t
P = 285714.28 / 5.6
P = 51020.40 W
Thus, we can conclude that the average power required to accelerate the car of 14,000 N is 51020.40 W.
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A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is 1.60 105 Pa. Assuming that the top and bottom surfaces of the cap each have an area of 3.70 10-4 m2, obtain the magnitude of the force that the screw thread exerts on the cap in order to keep it on the bottle. The air pressure outside the bottle is one atmosphere.
Answer:
F tread = 21.8N
Explanation:
In order to find the force that the screw thread exert on the cap, use equation 11.3 taking into consideration that the cap is in equilibrium
Making the vertical net force equal zero .
Sum Fy= - F tread+ Inside -F outside=0
F tread = F inside- F out side = P inside A- P out side A =
(P inside- P outside) A.=
((160000pa)-(101000pa))* 0.00037
21.8N
A stretched rubber band has ___________ energy. a. elastic kinetic energy b. gravitational potential energy c. elastic potential energy d. gravitational potential energy
Answer: Option (c) is the correct answer.
Explanation:
An elastic object is defined as the object that is able to retain its shape when a force is applied on it.
For example, when we pull a rubber band then it stretches and when we withdraw the force applied on it then it retain its shape.
As we know that potential energy is the energy obtained by an object due to its position.
So, when we stretch a rubber band then it will have elastic potential energy as position of the rubber band is changing and since, it will retain it shape hence it has elastic potential energy.
Thus, we can conclude that a stretched rubber band has elastic potential energy.
A stretched rubber band has elastic potential energy. Option C
What is elastic potential energy?A rubber band that has been stretched holds elastic potential energy. The energy held in an object when it is stretched or distorted is known as elastic potential energy. The deformation of a rubber band's molecular structure causes it to gain potential energy when it is stretched.
Rubber bands are comprised of materials that are elastic and can be stretched before snapping back into place. Stretching the rubber band causes its structure to distort and its molecular structure to store potential energy. This potential energy is turned into kinetic energy when the rubber band is released, causing it to rebound and return to its original shape.
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Consider an electrical transformer that has 10 loops on its primary coil and 20 loops on its secondary coil. What is the current in the secondary coil if the current in the primary coil is 5.0 A?A. 5.0 AB. 10.0 AC. 2.5 AD. 20.0 A
Answer:
C. 2.5 A
Explanation:
Transformer: A transformer is an electromechanical device that is used to change the voltage of an alternating current.
The current and the number of loops in a transformer is related as shown below
Ns/Np = Ip/Is........................... Equation 1
Where Ns = Secondary loop, Np = primary loop, Ip = primary current, Is = secondary current.
Making Is the subject of the equation
Is = NpIp/Ns........................ Equation 2
Given: Np = 10 loops, Ns = 20 loops, Ip = 5.0 A.
Substitute into equation 2
Is = (10×5.0)/20
Is = 50/20
Is = 2.5 A.
Hence the current in the primary coil = 2.5 A.
The right option is C. 2.5 A
Final answer:
The current in the secondary coil of an electrical transformer with 10 loops in the primary coil and 20 loops in the secondary coil, given a primary current of 5.0 A, is 2.5 A.
Explanation:
The question asks for the current in the secondary coil of an electrical transformer, given the current in the primary coil and the number of loops in both the primary and secondary coils. To find the current in the secondary coil, we use the principle of conservation of energy in a transformer, which states that the power input to the primary coil (P1) equals the power output from the secondary coil (P2), assuming an ideal scenario without any losses. The formula for power is P = IV, where I is current and V is voltage. Therefore, the ratio of the currents in the primary and secondary coils is inversely proportional to the ratio of the number of turns in the primary and secondary coils: I1/I2 = N2/N1. Given that N1 = 10 loops and N2 = 20 loops with a primary current (I1) of 5.0 A, we find that I2 = 2.5 A. Thus, the correct answer is C. 2.5 A.
A car starts from rest and uniformly accelerated to a speed of 40 km/h in 5 s . The car moves south the entire time. Which option correctly lists a vector quantity from the scenario?
Explanation:
Speed= distance / time
Making distance the subject of the formula.
Distance = speed × time
Convert km/hr to m/ s
(40× 1000)/3600= 11.1m/s
Distance = 11.1m/s × 5s
Distance= 55.6m
So it is vector ,although the question is not complete.
Answer:
speed: 40 km/h
distance: 40 km
acceleration 8 km/h/s south
velocity: 5 km/h north
Two cellists, one seated directly behind the other in an orchestra, play the same 220-Hz note for the conductor who is directly in front of them. What is the smallest non-zero separation that produces constructive interference?
Answer:
d= 1.56 m
Explanation:
In order to have a constructive interference, the path difference between the sources of the sound, must be equal to an even multiple of the semi-wavelength, as follows:
⇒ d = d₂ - d₁ = 2n*(λ/2)
The minimum possible value for this distance, is when n=1, as it can be seen here:
dmin = λ
In any wave, there exists a fixed relationship between the wave speed, the frequency and the wavelength:
v = λ*f
If v = vsound = 343 m/s, and f = 220 1/s, we can solve for λ:
λ =[tex]\frac{v}{f} = \frac{343 m/s}{220(1/s)} = 1.56 m[/tex]
⇒ dmin =λ = 1.56 m
A gamma ray burst produces radiation that has a period of 3.6x10-21 s. What wavelength does this radiation have?
Answer:
The radiation wavelength is 1.08 X 10⁻¹² m
Explanation:
Frequency is the ratio of speed of photon to its wavelength
F = c/λ
where;
c is the speed of the photon = 3 x 10⁸ m/s
λ is the wavelength of gamma ray = ?
F is the frequency of the gamma ray = 1/T
T is the period of radiation = 3.6x10⁻²¹ s
[tex]\frac{1}{T} = \frac{c}{\lambda}[/tex]
λ = T*C
λ = 3.6x10⁻²¹ * 3 x 10⁸
λ = 1.08 X 10⁻¹² m
Therefore, the radiation wavelength is 1.08 X 10⁻¹² m
Vector A has a magnitude of 5.0 m and points east, vector B has a magnitude of 2.0 m and points north, and vector C has a magnitude of 7.0 m and points west. The resultant vector A + B + C is given by
Answer:
The answer to your question is Vr = 2.83 m, to the Northwest
Explanation:
Data
Vector A = 5 m
Vector B = 2.0 m
Vector C = 7.0 m
Process
1.- Calculate the ∑Vx and ∑Vy
∑Vx = 5m - 7m = -2m Vectors substract because they are in opposite directions
∑Vy = 2m
2.- Calculate the resultant vector with the Pythagorean theorem
Vr² = Vx² + Vy²
Vr² = (-2)² + (2)²
Vr² = 4 + 4
Vr² = 8
Vr = 2.83 m
3.- Calculate the direction
tan Ф = 2/-2 = 1
tan⁻¹Ф = 45° to the Northwest
Final answer:
The resultant vector A + B + C has a magnitude of approximately 2.83 m and it points north-west, after subtracting the east-west components and adding the north-south component with no opposition.
Explanation:
To find the resultant vector A + B + C, we need to consider the directions and magnitudes of each vector. Vector A has a magnitude of 5.0 m and points east, vector B has a magnitude of 2.0 m and points north, and vector C has a magnitude of 7.0 m and points west. We can calculate the overall resultant vector by adding up the components in the east-west direction and the north-south direction separately.
Since east and west are opposite directions, we subtract the magnitudes of vectors A and C, which point in these directions:
East-West component: 5.0 m (east) - 7.0 m (west) = -2.0 m (west)Vector B points north and has no opposing southward vector, so its component remains unchanged:
North-South component: 2.0 m (north)Now, to find the resultant vector's magnitude, we can use the Pythagorean theorem:
Resultant magnitude = \\(\sqrt{(-2.0)^2 + 2.0^2} m\\) = \sqrt{4 + 4} m = \sqrt{8} m = 2.83 m (to two decimal places)
The resultant vector has a magnitude of approximately 2.83 m and it points north-west, considering that the east-west component is in the west direction while the north-south component points directly north.
In which of these models is heat being added to the molecules? 2 points Molecules are moving fast. As they run into slower molecules, they slow down. Molecules are moving slow. As they run into faster molecules, they speed up. The molecules are moving at different speeds. As they run into each other, some molecules slow down while others speed up. The molecules bounce around. Every time they collide with another molecule, they slow down.
Molecules are moving slow, as they run into faster molecules, they speed up.
Explanation:
The model that best depicts heat being added to the molecules is that slow moving molecules run into faster ones and they speed up.
When heat is added to a body, the kinetic energy of the system increases.
Slow moving particles have low kinetic energy among them. Heat causes gain in kinetic energy. The slow moving particles first begins to vibrate and as time proceeds starts colliding with other ones. The overall entropy of the system increases as they run into faster molecules.learn more;
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An electroscope is charged by touching its top with positive glass rod. The electroscope leaves spread apart and the glass rod is removed. Then a negatively charged plastic rod is brought close to the top of the electroscope, but it does not touch. What happens to the leaves?
Answer: The leaves will spread.
Explanation: Since the top of the electroscope has been touched initially by a positive glass rod, the charge on it is positive ( this is charging by conduction).
By bringing a negative charged rod towards ( but not touching the top) the top of electroscope means we will be charging the electroscope by induction. Charging by induction implies that an opposite charge of the conductor (negative charged rod) we are using to charge will be formed on the conductor that we want to charge (already positive charged electroscope)
In this case, our plastic rod is negative and it is brought towards the top of the electroscope ( which is already positively charged), there will be an induced positive charge on the electroscope.
So we have positive charge on the electroscope by a positive rod ( charging by conduction) and a positive charge from a negative rod ( charging by induction) thus the leaves will spread meaning the charges are repelling based on the fact that opposite charges attract and like charges repel
In the presence of a negatively charged rod, the positively charged electroscope will have its positive charges attracted toward the negative charge. This re-distributes the charges within the electroscope causing the leaves to move closer together.
Explanation:An electroscope is a device used to detect and measure electric charge. After it is charged with a positive charge (from the glass rod), the leaves spread apart due to like charges repelling each other. When a negatively charged plastic rod is brought near (but not touching) the electroscope, the positive charges in the electroscope are attracted to the negative charge of the rod. This causes the leaves of the electroscope to move closer together as most of the positive charges are attempting to move toward the top of the electroscope to be closer to the negative charge.
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You watch distant Sally Homemaker driving nails into a front porch at a regular rate of 1 stroke per second. You hear the sound of the blows exactly synchronized with the blows you see. And then you hear one more blow after you see the hammering stop. Explain how you calculate that Sally is 340 m away from you.?
Answer:
The velocity of sound of an echo is given as:
v = 2d/t, where d is the distance the sound source and the reflecting surface.
The time take for the stroke to be heard is 2s, because the rate is one stoke per second(one stroke in 1s). It means it will be heard after 2s, after the reflection of the sound wave.
v is speed of sound in air(Value 240m/s)
Therefore, d = vt/2 = (340 x 2)/2
d = 340m.
Sally is 340m away.
Explanation:
The above question is an application of echo.
Echo is a sound heard after the reflection of sound wave.
The distance covered will be twice the distance between the source of the sound and the reflecting surface. This is because the sound will travel a certain distance to a reflecting surface and travels back equal distance to the source after getting reflected.
Sound is detected when a sound wave causes the tympanic membrane (the eardrum) to vibrate (see the figure ). Typically, the diameter of this membrane is about 8.40 {\rm mm} in humans.
a.)how much energy is delivered to the eardrum each second when someone whispers (20.0 {\rm dB}) a secret in your ear?
b.)To comprehend how sensitive the ear is to very small amounts of energy, calculate how fast a typical 2.00 {\rm mg} mosquito would have to fly (in {\rm mm/s}) to have this amount of kinetic energy.
Answer:
a) Energy delivered per second to the tympanic membrane = 5.54 × 10⁻¹⁵ J/s
b) velocity of mosquito that will generate that amount of energy, v = 0.0000744 m/s = 0.0744 mm/s.
Explanation:
a) [D] = 10 log (I/I₀)
I₀ = 10⁻¹² W/m²
Given the sound intensity level in decibels, we need to obtain the corresponding sound intensity.
20 = 10 log (I/(10⁻¹²))
2 = log (I/(10⁻¹²))
100 = (I/(10⁻¹²))
I = 10⁻¹⁰ W/m²
Power experienced by the tympanic membrane of the ear due to the sound intensity = Intensity × Area of the membrane
Area of the membrane = πD²/4 = π(8.4 × 10⁻³)²/4 = 5.54 × 10⁻⁵ m²
Power = 10⁻¹⁰ × 5.54 × 10⁻⁵ = 5.54 × 10⁻¹⁵ W
Energy delivered per second to the tympanic membrane = 5.54 × 10⁻¹⁵ J/s
b) Kinetic energy = mv²/2
5.54 × 10⁻¹⁵ = (2 × 10⁻⁶)v²/2
v² = (2 × 5.54 × 10⁻¹⁵)/(2 × 10⁻⁶)
v = 0.0000744 m/s = 0.0744 mm/s.
The definition of decibels and the relationship between work and kinetic energy allows us to find the results for the questions about the system:
a) The energy supplied to the eardrum is I = 5.67 10⁻¹⁵ W.
b) The speed of the mosquito with this energy is: v= 7.53 10⁻² m/s
Given parameters.
Eardrum diameter d= 8.40 mm = 8.40 10-3 m. Mosquito mass m= 2.00 mg = 2.00 10-6 kg Whisper sound intensity. Beta = 20dBTo find.
a) The energy per second.
b) The kinetic energy of a mass mosquito
Decibels definition.The intensity of sound is in a very wide range of magnitudes, to simplify its use, the decibel is defined as a logarithmic unit.
[tex]\beta = 10 \ log (\frac{I}{I_o})[/tex]
where β are the decibels, I the intensity and I₀ the reference intensity. In the case of humans, the sensitivity threshold is of the order of 10⁻¹² W/m²
The intensity of the expression is:
[tex]\frac{I}{I_o} = 10^{\beta/10 }[/tex]
[tex]I = I_o\ 10^{\beta/10}[/tex]
Let's calculate
I = [tex]10^{-12} \ 10^{20/10}[/tex]
I = 10⁻¹⁰ W/m²
The intensity is defined by the energy deposited per unit of time and area.
I = [tex]\frac{P}{A}[/tex]
P = I A
Let's calculate the area of the eardrum.
A = π r² = [tex]\pi \ \frac{d^2}{4}[/tex]
Let's calculate.
A = [tex]\frac{\pi}{4} (8.4 \ 10^{-3} )^2[/tex]
A = 5.67 10⁻⁵ m²
Let's calculate the power.
P = 10⁻¹⁰ 5.67 10⁻⁵
P = 5.67 10⁻¹⁵5W
b) Power is work per unit of time.
P = [tex]\frac{W}{t}[/tex]
W= P t
The work is equal to the change in kinetic energy, if we assume that the mosquito starts from rest.
W = ΔK = [tex]K_f - K_o[/tex]
W = ½ mv²
v² = [tex]\frac{2 K}{m}[/tex]
Let's calculate
v² = = 5.67 10⁻⁹
v= 7.53 10⁻⁵ m/s
v= 7.53 10⁻² mm/s
In conclusion using the definition of decibels and the relationship of work and kinetic energy we can find the results for the questions about the system are:
a) The energy supplied to the eardrum is I = 5.67 10⁻¹⁵ W.
b) The speed of the mosquito with this energy is: v= 7.53 10⁻² mm/s
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A 72.9 kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.21 m/s in 0.80 s. It travels with this constant speed for 4.97 s, undergoes a uniform downward acceleration for 1.52 s, and comes to rest. What does the spring scale register (a) before the elevator starts to move?
Answer:
(a) 72.9 kg
Explanation:
Before the elevator starts to move, only gravitational force exerts on the man, this force is generated by the man mass and the gravitational acceleration, which in turn register in the scale. So the scale would probably indicate the man mass, which is 72.9 kg.
A 4.0-kg mass is moving with speed 2.0 m/s. A 1.0-kg mass is moving with speed 4.0 m/s. Both objects encounter the same constant braking force, and are brought to rest. Which object travels the greater distance before stopping?
Answer
given,
mass of object 1 = 4 Kg
Speed of object 1 = 2 m/s
mass of object 2 = 1 Kg
speed of object 2 = 4 m/s
KE of the object 1 = [tex]\dfrac{1}{2}MV^2[/tex]
= [tex]\dfrac{1}{2}\times 4 \times 2^2[/tex]
= 8 J
KE of the object 2 = [tex]\dfrac{1}{2}MV^2[/tex]
= [tex]\dfrac{1}{2}\times 1 \times 4^2[/tex]
= 8 J
Kinetic energy of both the object is same hence,Work done by both the object will also be same.
It is given that braking force is same in both cases.
So, distance travel by Both the object will be same.
Both of them cover the same amount of distance.
Given that;
4 kg mass is moving with speed 2.0 m/s
1.0 kg mass is moving with speed 4.0 m/s.
Both objects are brought to a halt by the same steady braking force.
Their momentum will be proportionate to their weight in the opposite direction = 1:4 ratio
Distance before coming to rest
v² - u² =2as
So,
s1/s2 = 16/16
Both of them cover the same amount of distance.
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Using Newton's Version of Kepler's Third Law II The Sun orbits the center of the Milky Way Galaxy every 230 million years at a distance of 28,000 light-years. Use these facts to determine the mass of the galaxy. (As we'll discuss in Chapter Dark Matter, Dark Energy, and the Fate of the Universe, this calculation actually tells us only the mass of the galaxy within the Sun's orbit.) M= solar billion years
Final answer:
To find the mass of the Milky Way galaxy, we apply a version of Kepler's Third Law using the orbital period and radius of the Sun's orbit. We convert units to meters and seconds, calculate the Sun's orbital velocity, and use this alongside the gravitational constant to estimate the galaxy's mass within the Sun's orbit.
Explanation:
To determine the mass of the Milky Way galaxy using the information given, we can refer to a version of Kepler's Third Law tailored for galactic scales, which allows us to estimate the mass of the galaxy based on the orbital period and radius of an orbiting object—in this case, our Sun. The version of the law used in galactic dynamics is:
M = (v^2 x R) / G
where M is the mass of the galaxy within the Sun's orbit, v is the orbital speed of the Sun, R is the radius of the Sun's orbit, and G is the gravitational constant.
Calculation Steps:
First, we need to convert the orbital period from million years to seconds, as follows: 230 million years x (365.25 days/year) x (24 hours/day) x (3600 seconds/hour).
Next, convert the radius of the Sun's orbit from light-years to meters using the fact that one light-year is approximately 9.461 x 10^15 meters.
Now, we can calculate the orbital velocity of the Sun using the circumference of its orbit (2 x π x R) and the orbital period found in step 1 to obtain v = (2 x π x R) / period.
Finally, apply Kepler's Third Law to find the mass M using the velocity v from step 3, the radius R from step 2, and the known value of the gravitational constant G.
Performing these calculations would result in an estimate for the Milky Way's mass within the Sun's orbit.
Important Note
It is critical to understand that these calculations only provide the mass within the Sun's orbit. There is additional mass outside the Sun's orbit, much of which is thought to be dark matter, that is not accounted for in this simple model.
A shoreline runs north-south, and a boat is due east of the shoreline. The bearings of the boat from two points on the shore are 110° and 100°. Assume the two points are 550 ft apart. How far is the boat from the shore?
Answer:
2930.90 ft
Explanation:
*Attached are two rough sketches I made to represent the problem.
In diagram 2, the bearings are represented relative to the boat's position.
To find x, the distance between the boat and point having bearing 110° to the boat, we can use sine rule:
(sin 10°) / 550 = (sin 100) / x
=> x = (550 * sin 100°) / sin 10°
x = 3119 ft
Having found this, we can now find the distance between the host and the shore, as represented in diagram 1.
Using trigonometric function of SOHCAHTOA, we have that:
cos 20° = y / 3119
=> y = 3119 * cos 20°
y = 2930.90 ft
Hence, the distance between the boat and the shore is 2930.90 ft
Match the following kinds of lights in order from the longest wavelength to the shortest wavelength on the EM spectrum:
Group of answer choices
1
2
3
4
5
6
7
Answers
radio
infrared
gamma ray
microwave
x-ray
ultraviolet
visible
Answer:
From longest to shortest wavelength:
1) Radio waves
2) Microwaves
3) Infrared
4) Visible light
5) Ultraviolet
6) X-rays
7) Gamma rays
Explanation:
Electromagnetic waves are periodic oscillations of the electric and the magnetic field in a plane perpendicular to the direction of motion the wave itself.
All electromagnetic waves travel in a vacuum with the the same speed, which is know as the speed of light; it is one of the fundamental constants of nature, and its value is
[tex]c=3.0\cdot 10^8 m/s[/tex]
Electromagnetic waves are classified into 7 different types, depending on their wavelength/frequency. From longest to shortest wavelength (and so, from lowest to highest frequency, since frequency is inversely proportional to wavelength), we have (with their correspondant wavelength):
Radio waves (>1 m)
Microwaves (1 mm - 1 m)
Infrared (750 nm - 1 mm)
Visible light (380 nm - 750 nm)
Ultraviolet (10 nm - 380 nm)
X-rays (0.01 nm - 10 nm)
Gamma rays (<0.01 nm)
The electromagnetic spectrum spans from radio waves with the longest wavelength to gamma rays with the shortest. The order is: radio waves, microwaves, infrared, visible light, ultraviolet, x-rays, and gamma rays.
Explanation:The question asks you to match the various types of light to their respective wavelengths on the EM spectrum. The Electromagnetic Spectrum (EM Spectrum) arranges different types of electromagnetic radiation in order of their wavelengths. Light types in order of longest to shortest wavelengths are as follows:
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Calculate the efficiency of an engine with an input temperature of 755 K and exhaust temperature of 453 K.
Answer:
40%
Explanation:
The other person got it right up until making it opposite because its a percentage. The equation is correct but you'd just need to take the 60% answer and subtract it from 100% because 60% is equal to how much effiency the exhaust is taking away, thus making your answer 40%
John performs an experiment on an electric circuit. He increases the voltage from 25 volts to 50 volts while keeping the resistance constant. What will be the effect of John's changes on the current?
The current will double
Explanation:
The relationship between voltage and current in an electric circuit is given by the following equation (Ohm's law):
[tex]V=IR[/tex]
where
V is the voltage
I is the current
R is the resistance
making R the subject,
[tex]R=\frac{V}{I}[/tex]
Since in this experiment the resistance is kept constant, we can write:
[tex]\frac{V_1}{I_1}=\frac{V_2}{I_2}[/tex]
where
[tex]V_1=25 V[/tex] is the voltage in the 1st experiment
[tex]V_2=50 V[/tex] is the voltage in the 2nd experiment
[tex]I_1,I_2[/tex] are the currents in the 1st and 2nd experiment
We can re-arrange the equation as
[tex]\frac{I_2}{I_1}=\frac{V_2}{V_1}=\frac{50}{25}=2[/tex]
This means that the current will double in the 2nd experiment.
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