Explanation:
On the increase in the density of the water, the buoyant force acting on the body weight is less than the body weight. In the dead sea, the water is saline due to the presence of salt in the water which increases the density of the water. Thus, the force acting on the body is less than the bodyweight which results in the sinking of the body.Hence, the correct option is (b) that is "less than your weight".A basketball with a mass of 0.61 kg falls vertically to the floor where it hits with a velocity of −7 m/s. (We take the positive direction to be upward here.) The ball rebounds, leaving the floor with a velocity of 4.6 m/s. (Indicate the direction with the signs of your answers.)
Answer:
a. What impulse act on the ball during its collision with the floor?
b. If the ball is in contact with the floor for 0.04 s, what is the average force of the ball on he floor?
The answers to the question are
a. The impulse acting on the ball during its collision with the floor is
7.076 kg·m/s
b. The average force of the ball on the floor with contact of 0.04 s is
176.9 N
Explanation:
Mass of the ball = 0.61 kg
Initial velocity = -7 m/s
Final velocity = 4.6 m/s
Impulse = Δp = FΔt = mΔv
Where Δt =change in time
m = mass of Average force of he the ball
Δv = velocity change
Therefore impulse = m×(Final velocity - initial velocity)
= 0.61 × (4.6-(-7)) = 0.61×11.6
= 7.076 kg·m/s
b. Average force of the ball on the floor is given by
[tex]F_{Average}[/tex]×Δt = mΔv = 7.076 kg·m/s
where Δt = change in time = 4.6 m/s
Therefore [tex]F_{Average}[/tex] =mΔv /Δt = 7.076/0.04 = 176.9 kg·m/s²
= 176.9 N
Assume that a uniform magnetic field is directed into the monitor you are reading this on. If an electron is released with an initial velocity directed from the bottom edge to the top edge of the screen, which of the following describes the direction of the resultant force acting on the electron?a. out of the page
b. to the right
c. to the left
d. into the page
Answer:
b. to the right
Explanation:
In the given condition we have the magnetic field lines perpendicular to the direction of moving charges then here in this case we can apply the Fleming's left-hand rule to determine the force acting on the charges (here electrons).As the direction of magnetic field lines is into the screen and the electrons move form bottom of the screen towards up then we apply the Fleming's left-hand rule and find that the deviation will be towards right.We spread the index finger, thumb and the middle finger of our left hand mutually perpendicular to each other then we align the pointing of the index finger in the direction of the magnetic field and the middle finger in the direction of the current or the direction of the motion of the positive charge and if we have a negative charge then we take the direction to be opposite. Now the direction of the thumb pointing tells us the direction of the force.Final answer:
Using the left-hand rule for a negatively charged particle like an electron, the force on an electron moving upward with the magnetic field directed into the page is to the left. The correct answer is option c, to the left.
Explanation:
To determine the direction of the resultant force acting on an electron as it moves within a magnetic field, one can use the right-hand rule or, in the case of negatively charged particles like an electron, the left-hand rule. Considering that an electron is negatively charged, we apply the left-hand rule, with the thumb in the direction of the electron's velocity (from the bottom to the top edge of the screen), and the index finger in the direction of the magnetic field (into the page). The middle finger then points in the direction of the force experienced by the electron.
By this rule, for a magnetic field directed into the page and an electron moving upward, the resultant force will be to the left of the electron's initial direction of travel. The correct answer to the question is option c, to the left. This force will cause the electron to move in a circular path due to the magnetic Lorentz force.
Two cylindrical solenoids, A and B, each have lengths that are much greater than their diameters. The two solenoids have the same density of turns in their windings. Solenoid A has a diameter that is twice the diameter of solenoid B, but solenoid B has a length that is twice the length of solenoid A. How do their inductances compare with one another
Answer:
The inductance of solenoid A is twice that of solenoid B
Explanation:
The inductance of a solenoid L is given by
L = μ₀n²Al where n = turns density, A = cross-sectional area of solenoid and l = length of solenoid.
Given that d₁ = 2d₂ and l₂ = 2l₁ and d₁ and d₂ are diameters of solenoids A and B respectively. Also, l₁ and l₂ are lengths of solenoids A and B respectively.
Since we have a cylindrical solenoid, the cross-section is a circle. So, A = πd²/4.
Let L₁ and L₂ be the inductances of solenoids A and B respectively.
So L₁ = μ₀n²A₁l₁ = μ₀n²πd₁²l₁/4
L₂ = μ₀n²A₂l₂ = μ₀n²πd₂²l₂/4
Since d₁ = 2d₂ and l₂ = 2l₁, sub
L₁/L₂ = μ₀n²πd₁²l₁/4 ÷ μ₀n²πd₂²l₂/4 = d₁²/d₂² × l₁/l₂ = (2d₂)²/d₂² × l₁/2l₁ = 4d₂²/d₂² × l₁/2l₁ = 4 × 1/2 = 2
L₁/L₂ = 2
L₁ = 2L₂
So, the inductance of solenoid A is twice that of solenoid B
When Dr. Hewitt immerses an object in water the second time and catches the water that is displaced by the object, how does the weight lost by the object compare to the weight of the water displaced?
Answer:
The weight loss of object is equal to the weight of water displaced.
Explanation:
- When the object is immersed the force of Buoyancy acts against the weight and reducing the scale weight.
- The amount of Buoyancy Force is proportional to the fraction of Volume of object immersed in water; hence, the same amount is spilled/lost.
- The amount of water is proportional to the volume of object of fraction of object immersed in water will lead to the same fraction of water displaced and caught by Dr. Hewitt.
MIT’s robot cheetah can jump over obstacles 46. cm high and has speed of 12.0 km/h. a) If the robot launches itself at an angle of 60.° at this speed, what is its maximum height? b) What would the launch angle have to be to reach a height of 46. cm?
Answer:
(a) [tex]y_{max}=0.423m[/tex]
(b) [tex]\alpha =64.3^{o}[/tex]
Explanation:
Given data
[tex]v_{i}=12km/h=3.33m/s\\\alpha =60^{o}\\g=9.8m/s^{2}\\Required\\(a)y_{max}\\(b)Angle[/tex]
Solution
For Part (a)
As the velocity component in direction of y is given by:
[tex]v_{yi}=v_{i}Sin\alpha \\v_{yi}=3.33Sin60\\v_{yi}=2.88m/s[/tex]
The maximum displacement is given by:
[tex]v_{yf}^{2}=v_{yi}^{2}-2gy_{max}\\ y_{max}=\frac{(2.88)^{2}}{2(9.8)}\\ y_{max}=0.423m[/tex]
For Part (b)
To reach y=46cm =0.46m apply:
[tex]0=v_{yi}^{2}-2(9.8)(0.46)\\v_{yi}=3m/s\\As\\Sin\alpha =\frac{v_{yi}}{v_{i}}\\\alpha =Sin^{-1}(\frac{v_{yi}}{v_{i}})\\\alpha =Sin^{-1}(\frac{3}{3.33} )\\\alpha =64.3^{o}[/tex]
a) When launched at a 60° angle with a speed of 3.33 m/s, the robot cheetah can reach a maximum height of about 0.715 meters.
b) To reach a height of 0.46 meters, the launch angle needs to be approximately 31.74 degrees.
a) To find the maximum height (h_max) when the robot cheetah launches at a 60-degree angle with a speed of 3.33 m/s, we can use the following formula:
h_max = (u^2 * sin^2(θ)) / (2 * g)
Where:
u is the initial speed (3.33 m/s)
θ is the launch angle (60 degrees)
g is the acceleration due to gravity (9.8 m/s²)
First, calculate sin(60 degrees):
sin(60 degrees) = √3 / 2 ≈ 0.866
Now, plug in the values and calculate h_max:
h_max = (3.33^2 * (0.866)^2) / (2 * 9.8) ≈ 0.715 meters
b) To determine the launch angle (θ) required to reach a height of 0.46 meters (46 cm), we'll rearrange the same formula:
sin^2(θ) = (2 * g * h_desired) / u^2
Plug in the values:
sin^2(θ) = (2 * 9.8 * 0.46) / (3.33^2)
Now, find sin(θ):
sin(θ) ≈ √(0.2794) ≈ 0.5289
Finally, calculate θ:
θ ≈ arcsin(0.5289) ≈ 31.74 degrees
So, the launch angle required to reach a height of 0.46 meters is approximately 31.74 degrees.
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What kind of friction occurs when moving parts have ball bearings
Answer: Rolling friction
Explanation:
Rolling friction is said to act when an object rolls over a surface. Machines often have moving parts. These moving parts roll over each other as the machine works. Ball bearings are commonly used to reduce friction between the two surfaces in contact as the machine parts roll over each other. This is a typical example of rolling friction as applied to machine parts.
Answer:
Rolling friction.
Explanation:
Rolling friction is the force resisting the motion when a body ( such as a ball, tires, wheel) rolls on a surface. It is applicable for bodies whose point of contact keeps changing. It is the force that opposes the motion of a body which is rolling over the surface of another.
Rolling resistance, sometimes called rolling friction or rolling drag, is the force resisting the motion when a body rolls on a surface. It is mainly caused by non-elastic effects; that is, not all the energy needed for deformation of the wheel, roadbed, etc. is recovered when the pressure is removed.
A new class of objects can be created conveniently by ________; the new class (called the ________) starts with the characteristics of an existing class (called the ________), possibly customizing them and adding unique characteristicsof its own. Group of answer choices
Answer:
Option A is correct.
Inheritance, superclass, subclass
A new class of objects can be created conveniently by **inheritance**; the new class (called the **superclass**) starts with the characteristics of an existing class (called the **subclass**), possibly customizing them and adding unique characteristics of its own.
Explanation:
A class is a term in object oriented programming that is extensible program code used to create, name and implement what you want the objects to do.
Inheritance is used to describe the making of new classes from already existing classes. The general properties of the old classes remain in the new classes created from them.
Hence, the new classes (because they're usually an improvement on the old classes as they hold on to the wanted properties of the old classes and additional great ones are added or modified) are called superclasses and the old, already existing ones are called subclasses.
The missing words from the given paragraph are inheritance, superclass and subclass respectively.
The given problem is based on the concept of Object Oriented Programming (OOPs). A class is a term in object oriented programming that is extensible program code used to create, name and implement what you want the objects to do.
Inheritance is used to describe the making of new classes from already existing classes. The general properties of the old classes remain in the new classes created from them.
Hence, the new classes (because they're usually an improvement on the old classes as they hold on to the wanted properties of the old classes and additional great ones are added or modified) are called super classes and the old, already existing ones are called subclasses.
So, the complete paragraph will be,
A new class of objects can be created conveniently by inheritance; the new class (called the superclass) starts with the characteristics of an existing class (called the subclass), possibly customizing them and adding unique characteristics of its own.
Thus, we can conclude that the missing words from the given paragraph are inheritance, superclass and subclass respectively.
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geography Which nation in the region has seen its growth rate accelerate, while the majority of the region has seen a decline in growth rates?
Answer:
For further investigation see also the most recent World Population Data ... and Latin America and the Caribbean, and the regions of Melanesia, ... While Germany's death rate exceeds its birth rate, its population ... Population growth accelerated.
To answer this question, suppose that each vehicle is moving at 7.69 m/s and that they undergo a perfectly inelastic head-on collision. Each driver has mass 82.6 kg. The total vehicle masses are 810 kg for the car and 4280 kg for the truck. Note that these values include the masses of the drivers. If the collision time is 0.129 s, (a) what force does the seat belt exert on the truck driver?
Answer:
Force(F) = -80,955.01 N
Explanation:
We need to first determine the impulse that the truck driver received from the car during the collision
So; m₁v₁ - m₂v₂ = (m₁m₂)v₀
where;
m₁ = mass of the truck = 4280 kg
v₁ = v₂ = speed of the each vehicle = 7.69 m/s
m₂ = mass of the car = 810 kg
Substituting our data; we have:
(4280×7.69) - (810×7.69) = (4280+810)v₀
32913.2 - 6228.9 = (5090)v₀
26684.1 = (5090)v₀
v₀ = [tex]\frac{26684.1}{5090}[/tex]
v₀ = 5.25 m/s
NOW, Impulse on the truck = m (v₀ - v)
= 4280 × (5.25 - 7.69)
= 4280 × (-2.44)
= -10,443.2 kg. m/s
Force that the seat belt exert on the truck driver can be calculated as:
Impulse = Force × Time
-10,443.2 kg. m/s = F (0.129)
F = [tex]\frac{-10,443.2}{0.129}[/tex]
Force(F) = -80,955.01 N
Thus, the Force that the seat belt exert on the truck driver = -80,955.01 N
Nearly all conventional piston engines have flywheels on them to smooth out engine vibrations caused by the thrust of individual piston firings. Why does the flywheel have this effect
Answer:
This is because, the flywheel has a very large moment of Inertia and hence sudden piston torques have negligible effect on the flywheel, but every piston combined has a significant torque. This smoothens out the vibrations.
Explanation:
Final answer:
A flywheel reduces engine vibrations by utilizing its angular momentum to smooth out the intermittent power delivery from piston firings, thus ensuring steady rotation of the crankshaft and maintaining engine speed during periods when power is not supplied.
Explanation:
The primary reason why a flywheel reduces engine vibrations in conventional piston engines is due to its ability to store rotational energy. When the pistons fire, they impart energy to the crankshaft, causing the engine to turn. This energy, however, is delivered in pulses rather than smoothly, which can lead to uneven rotation and vibrations. The flywheel, being a heavy rotating disk, has significant angular momentum. This momentum ensures that the flywheel continues to spin between these pulses of power, effectively smoothing out the rotation of the engine's crankshaft. This results in a reduction of the vibrations that occur due to the intermittent nature of power delivery from individual piston firings.
Flywheels also help maintain engine speed during the periods when no power is supplied, such as between combustion events in a four-stroke engine. This steady rotation is crucial for the engine's performance and longevity. Additionally, flywheels play a role in energy conservation, as they can store energy and release it when needed, which is particularly important in stopping and starting scenarios.
You heat a 5.2 gram lead ball (specific heat capacity = 0.128 Joules/g-deg) to 183. ∘ C. You then drop the ball into 34.5 milliliters of water (density 1 g/ml; specific heat capacity = 4.184 J/g-deg ) at 22.4 ∘ C. What is the final temperature of the water when the lead and water reach thermal equilibrium?
Answer:
The final temperature when the lead and water reach thermal equilibrium is [tex]23.14^{o} C[/tex]
Explanation:
Using the law of conservation of energy the heat lost by the lead ball is gained by the water.
[tex]-q_{L} =+q_{w}[/tex] .............................1
where
[tex]q_{L}[/tex] is the heat energy of the lead ball;
[tex]q_{w}[/tex] is the heat energy of water;
but q = msΔ
T...............................2
substituting expression in equation 2 into the equation 1;
Δ
T =Δ
T .............................3
where [tex]m_{L}[/tex] is the mass of the lead ball = 5.2 g
[tex]s_{L}[/tex] is the specific heat capacity of the lead ball = 0.128 Joules/g-deg
[tex]m_{w}[/tex] is the mass of water = volume x density = 34.5 ml x 1 g/ml = 34.5 g
[tex]s_{w}[/tex] is the specific heat capacity of water = 4.184 J/g-deg
Δ
T is the temperature changes encountered by the lead ball and water respectively
inputting the values of the parameters into equation 3
5.2 g x 0.128 Joules/g-deg x (183-22.4) = 34.5 g x 4.184 J/g-deg x ([tex]T_{2}[/tex] -22.4)
[tex]T_{2}[/tex] -22.4 = [tex]\frac{106.9}{144.3}[/tex]
[tex]T_{2}[/tex] -22.4 = 0.7408
[tex]T_{2}[/tex] = [tex]23.14^{o} C[/tex]
Therefore the final temperature when the lead and water reach thermal equilibrium is [tex]23.14^{o} C[/tex]
Two 5.0 mm × 5.0 mm electrodes are held 0.10 mm apart and are attached to 7.5 V battery. Without disconnecting the battery, a 0.10-mm-thick sheet of Mylar is inserted between the electrodes.A) What is the capacitor's potential difference before the Mylar is inserted?
B) What is the capacitor's electric field before the Mylar is inserted?
C) What is the capacitor's charge before the Mylar is inserted?
D) What is the capacitor's potential difference after the Mylar is inserted?
E) What is the capacitor's electric field after the Mylar is inserted?
F) What is the capacitor's charge after the Mylar is inserted?
Answer:
A) V = 7.5 V
B) E = 75,000 V/m
C) Q = 16.6 pC
D) V = 7.5 V
E) E = 24,000 V/m
F) Q = 52 pC
Explanation:
Given:
- The Area of plate A = ( 5 x 5 ) mm^2
- The distance between plates d = 0.10 mm
- The thickness of Mylar added t = 0.10 mm
- Voltage supplied by battery V = 7.5 V
Solution:
A) What is the capacitor's potential difference before the Mylar is inserted?
- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.
B) What is the capacitor's electric field before the Mylar is inserted?
- The Electric Field E between the capacitor plates is given by:
E = V / k*d
k = 1 (air) E = 7.5 / 0.10*10^-3
E = 75,000 V/m
C) What is the capacitor's charge Q before the Mylar is inserted?
C = k*A*ε / d
k = 1 (air) C = ( 0.005^2 * 8.85*10^-12 ) / 0.0001
C = 2.213 pF
Q = C*V
Q = 7.5*(2.213)
Q = 16.6 pC
D) What is the capacitor's potential difference after the Mylar is inserted?
- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.
E) What is the capacitor's electric field after the Mylar is inserted?
- The Electric Field E between the capacitor plates is given by:
E = V / k*d
k = 3.13 E = 7.5 / (3.13)0.10*10^-3
E = 24,000 V/m
F) What is the capacitor's charge after the Mylar is inserted?
C = k*A*ε / d
k = 3.13 C = 3.13*( 0.005^2 * 8.85*10^-12 ) / 0.0001
C = 6.927 pF
Q = C*V
Q = 7.5*(6.927)
Q = 52 pC
Here, we are required to evaluate parameters relating to the capacitor.
A) V = 7.5 VB) E = 75,000 V/mC) Q = 16.6 pCD) V = 7.5 VE) E = 24,000 V/mF) Q = 52 pCData:
The Area of plate A = ( 5 x 5 ) mm^2 = 2.5 × 10^(-5)m²The distance between plates d = 0.10 mm = 0.1 × 10^(-3) The thickness of Mylar added t = 0.10 mm = 0.1 × 10^(-3) Voltage supplied by battery V = 7.5 VA) The potential difference across the two plates is equal to the voltage provided by the battery prior to the insertion of the Mylar, V = 7.5 V which remains constant throughout.
The potential difference across the two plates is equal to the voltage provided by the battery prior to the insertion of the Mylar, V = 7.5 V which remains constant throughout.B) The capacitor's Electric Field E between the capacitor plates is given as:
The potential difference across the two plates is equal to the voltage provided by the battery prior to the insertion of the Mylar, V = 7.5 V which remains constant throughout.B) The capacitor's Electric Field E between the capacitor plates is given as: E = V / k*dThe potential difference across the two plates is equal to the voltage provided by the battery prior to the insertion of the Mylar, V = 7.5 V which remains constant throughout.B) The capacitor's Electric Field E between the capacitor plates is given as: E = V / k*dwhere dielectric constant , k(air) = 1The potential difference across the two plates is equal to the voltage provided by the battery prior to the insertion of the Mylar, V = 7.5 V which remains constant throughout.B) The capacitor's Electric Field E between the capacitor plates is given as: E = V / k*dwhere dielectric constant , k(air) = 1 E = 7.5 /(1 × 0.10*10^-3) E = 75,000 V/mC) The capacitor's charge Q before the Mylar is inserted can be evaluated as follows;
C = k × A × ε /dε /d
ε /d C =ε /d C = (1×2.5×10^(-5)×8.85*10^-12)/0.0001
ε /d C = (1×2.5×10^(-5)×8.85*10^-12)/0.0001 C = 2.213 × 10^(-12) = 2.213 pFε /d C = (1×2.5×10^(-5)×8.85*10^-12)/0.0001 C = 2.213 × 10^(-12) = 2.213 pF Q = C × V Q = 7.5*(2.213) Q = 16.6 pCD) The potential difference across the two plates, after the mylar has been inserted, V = 7.5 V which remains constant throughout.
E) The Electric Field, E between the capacitor plates is given by:
where the dielectric constant of the mylar = 3.13
E = V / k × d E = 7.5 / (3.13)0.10*10^-3 E = 24,000 V/mF) C = k×A×ε / d
C = 3.13×( 0.005^2 × 8.85×10^-12 ) / 0.0001 C = 6.927pC Q = C × V Q = 7.5 × (6.927) Q = 52 pCRead more:
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If we instead connect this headlight and starter in series with the 12.0 V battery, how much total power in W would the headlight and starter consume? (You may neglect any other resistance in the circuit and any change in resistance in the two devices.)
Answer:
If [tex]\large{P_{H}}[/tex] and [tex]\large{P_{S}}[/tex] be the power consumed by the headlight and starter respectively, then the total power consumed will be [tex]\large{P = \dfrac{P_{H}P_{S}}{P_{H} + P_{S}}}[/tex]
Explanation:
We know that electric power ([tex]P[/tex]) is given by
[tex]\large{P = \dfrac{V^{2}}{R}}[/tex]
where '[tex]V[/tex]' is the applied voltage and '[tex]R[/tex]' is the resistance.
If we consider that '[tex]\large{P_{H}}[/tex]' and '[tex]\large{P_{S}}[/tex]' be the power consumed by the headlight and starter respectively, then the resistance ([tex]\large{R_{H}}[/tex]) of the headlight and the resistance ([tex]\large{R_{S}}[/tex]) of the starter can be written as
[tex]&&\large{R_{H} = \dfrac{V^{2}}{P_{H}} = \dfrac{12^{2}}{P_{H}}}\\&and,& \large{R_{S} = \dfrac{V^{2}}{P_{S}} = \dfrac{12^{2}}{P_{S}}}[/tex]
If the headlight and the starter in connected in series, then equivalent resistance ([tex]\large{R_{eq}}[/tex]) will be
[tex]\large{R_{eq} = R_{H} + R_{S} = 12^{2}(\dfrac{1}{P_{H} + P_{S}})}[/tex]
So the total power ([tex]P[/tex]) consumed will be given by
[tex]\large{P = \dfrac{12^{2}}{R_{eq}} = \dfrac{1}{(\dfrac{1}{P_{H}} + \dfrac{1}{P_{S}})} = \dfrac{P_{H}P_{S}}{P_{H} + P_{S}}}[/tex]
When the 30.0-W headlight and the 2.40-kW starter are connected in series to a 12.0-V battery, they consume a total power of 2430 W.
The total power consumed by the headlight and starter when connected in series to a 12.0-V battery can be calculated by adding the individual powers.
Power of headlight = 30.0 WPower of starter = 2.40 kW = 2400 WTotal power = 30.0 W + 2400 W = 2430 Wwhy the sun emits most of its energy in the form of visible light?
Human eye could perceive this visible light and used it to look at the objects.
Explanation:
There are 3 forms of radiation emitted by the sun. They are Infrared radiation, Ultra Violet radiation as well as visible radiation.
Visible light is the type of light can perceive by the human eye and utilizes to see the objects.
Since the sun is hotter (5800 K) than the earth, it emits mostly the radiation energy in the form of this visible light.
A good baseball pitcher can throw a baseball toward home plate at 93 mi/h with a spin of 1510 rev/min. How many revolutions does the baseball make on its way to home plate? For simplicity, assume that the 60 ft path is a straight line.
Answer:
[tex]\Delta \theta =11.073\ rev[/tex]
Explanation:
Given,
Speed of baseball = 93 mi/h
spin = 1510 rev/min
[tex]\Delta x = 60\ ft[/tex]
[tex]v =93\times \dfrac{88}{60} = 136.4 ft/s[/tex]
[tex]\omega = 1510\times \dfrac{1}{60}= 25.167\ rad/s[/tex]
[tex]\Delta t = \dfrac{\Delta x}{v}[/tex]
[tex]\Delta t = \dfrac{60}{136.4 }[/tex]
[tex]\Delta t = 0.44\ s[/tex]
Revolutions of ball
[tex]\Delta \theta = \omega t[/tex]
[tex]\Delta \theta = 25.167\times 0.44[/tex]
[tex]\Delta \theta =11.073\ rev[/tex]
Revolution of the ball is equal to [tex]\Delta \theta =11.073\ rev[/tex].
A concrete highway curve of radius 60.0 m is banked at a 19.0 ∘ angle. what is the maximum speed with which a 1400 kg rubber-tired car can take this curve without sliding? (take the static coefficient of friction of rubber on concrete to be 1.0.)
The student's question involves calculating the maximum speed a car can take a banked curve without sliding, using physic principles of circular motion and the coefficient of static friction.
Explanation:The student's question revolves around determining the maximum speed at which a 1400 kg car can navigate a banked highway curve without sliding, given a static coefficient of friction between rubber and concrete. To solve this, we need to use principles of circular motion and friction.
To find the maximum speed (v), we can use the following equation that balances gravitational, frictional, and centripetal forces:
F_{friction} = μ_s × N
N = m × g × cos(θ)
F_{centripetal} = m × v^2 / r
Where μ_s is the static coefficient of friction (1.0), N is the normal force, m is the mass of the car (1400 kg), g is the acceleration due to gravity (9.8 m/s^2), θ is the banking angle (19.0 degrees), v is the maximum speed, and r is the radius of the curve (60.0 m). The forces due to friction and centripetal need to equal for the car to navigate the curve without sliding.
After applying trigonometry and mechanics principles, we can solve for v which gives us the maximum speed without sliding.
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Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.0*109 V and the quantity of charge of the charge transferred I s30 C? (a )What is the charge in energy of that transferred charge? (b) If all the energy released could be used to accelerate a 1000kg car from rest. What would be its final speed?
Answer:
a) [tex]U_{e} = 3 \times 10^{10}\,J[/tex], b) [tex]v \approx 7745.967\,\frac{m}{s}[/tex]
Explanation:
a) The potential energy is:
[tex]U_{e} = Q \cdot \Delta V[/tex]
[tex]U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)[/tex]
[tex]U_{e} = 3 \times 10^{10}\,J[/tex]
b) Maximum final speed:
[tex]U_{e} = \frac{1}{2}\cdot m \cdot v^{2}\\v = \sqrt{\frac{2\cdot U_{e}}{m} }[/tex]
The final speed is:
[tex]v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }[/tex]
[tex]v \approx 7745.967\,\frac{m}{s}[/tex]
The change in the energy transferred or potential energy is 3 x 10¹⁰ J.
The final speed of the charge released is 7745.97 m/s.
Change in energy transferredThe change in the energy transferred or potential energy is calculated as follows;
U = QV
U = 30 x 1 x 10⁹
U = 3 x 10¹⁰ J
Speed of the charge releasedThe speed of the charge released is calculated by applying law of conservation of energy.
[tex]K.E = U\\\\\frac{1}{2} mv^2 = U\\\\v= \sqrt{\frac{2U}{m} } \\\\v = \sqrt{\frac{2\times 3\times 10^{10} }{1000} }\\\\v = 7745.97 \ m/s[/tex]
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If a car is taveling with a speed 6 and comes to a curve in a flat road with radius ???? 13.5 m, what is the minumum value the coefficient of friction must be so the car doesn’t slide of the road?
Answer:
[tex]\mu_s \geq 0.27[/tex]
Explanation:
The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:
[tex]f_s=\frac{mv^{2}}{R}[/tex]
But we know that:
[tex]f_s\leq \mu_s N[/tex]
And the normal force is given by the sum of the forces in the vertical direction:
[tex]N-mg=0 \implies N=mg[/tex]
Finally, we have:
[tex]f_s=\frac{mv^{2}}{R} \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR} \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27[/tex]
So, the minimum value for the coefficient of friction is 0.27.
A uniform cylindrical grindstone has a mass of 15.0 kg and a radius of 18.6 cm. The grindstone reaches a rotational velocity of 1.83 rev/min before its motor turns off. After the grindstone motor is turned off, a knife blade is pressed against the outer edge of the grindstone with a perpendicular force of 8.82 N. The coefficient of kinetic friction between the grindstone and the blade is 0.80. Use the work energy theorem to determine how many turns the grindstone makes before it stops.
Answer:
[tex]n \approx 5.778\times 10^{-4}\,rev[/tex]
Explanation:
The Work-Energy Theorem is applied herein:
[tex]-\frac{1}{4}\cdot m_{cyl}\cdot R^{2}\cdot \omega^{2} = - \mu_{k}\cdot N\cdot r \cdot 2\pi \cdot n[/tex]
The number of turns needed to stop the grindstone is:
[tex]n = \frac{m_{cyl}\cdot R^{2}\cdot \omega^{2}}{4\cdot \mu_{k}\cdot N \cdot r\cdot 2\pi}[/tex]
[tex]n = \frac{(15\,kg)\cdot (0.186\,m)^{2}\cdot [(1.83\,\frac{rev}{min} )\cdot (\frac{2\pi\,rad}{1\,rev} )\cdot (\frac{1\,min}{60\,s} )]^{2}}{4\cdot (0.80)\cdot (8.82\,N)\cdot(0.186\,m)\cdot 2\pi}[/tex]
[tex]n \approx 5.778\times 10^{-4}\,rev[/tex]
A physics student is driving home after class. The car is traveling at 14.7 m/s when it approaches an intersection. The student estimates that he is 20.0 m from the entrance to the intersection when the traffic light changes from green to yellow and the intersection is 10.0 m wide. The light will change from yellow to red in 3.00 s. The maximum safe deceleration of the car is 4.00 m/s2 while the maximum acceleration of the car is 2.00 m/s2. Should the physics student decelerate and stop or accelerate and travel through the intersection?
The student should accelerate and travel through the intersection.
Explanation:
As per the given problem, the velocity with which the car of the student is moving is 14.7 m/s and he is 20 m away from intersection. Also the width of intersection is 10 m. It is also stated that traffic lights will change from yellow to red in 3 s. So totally , the student has to cover a distance of 30 m in 3 s to avoid getting stopped in intersection.
The velocity or speed required by the car to cover 30 m in 3 s is 30/3 = 15 m/s.
And the car is moving with initial velocity of 14.7 m/s. The student has two options either to decelerate at 4m/s² rate or to accelerate at the rate of 2 m/s².
So the velocity or speed of the car in 3 s with acceleration of 2 m/s² will be
Velocity = Acceleration × Time = 2 × 3 =6 m/s.
Thus on accelerating the car by 2 m/s² in 3 s, it will have the speed of 6 m/s and it can cover a distance of 6 × 3 = 18 m.
And the time taken by the car to reach from 20 m to intersection point with speed of 14.7 m/s is 20/14.7 = 1.36 s.
So, the car will take 1.36 s to reach the entrance of intersection with the speed of 14.7 m/s. But if it gets accelerated to 2 m/s², then it will have an increase in speed which will make the car to cover the complete intersection without stopping. So the student should accelerate the car.
1) Steven carefully places a 1.85 kg wooden block on a frictionless ramp, so that the block begins to slide down the ramp from rest. The ramp makes an angle of 59.3° up from the horizontal. Which forces below do non-zero work on the block as it slides down the ramp?a) gravityb) normalc) frictiond) spring2) How much total work has been done on the block after it slides down along the ramp a distance of 1.85 m? 3) Nancy measures the speed of the wooden block after it has gone the 1.85 m down the ramp. Predict what speed she should measure.4) Now, Steven again places the wooden block back at the top of the ramp, but this time he jokingly gives the block a big push before it slides down the ramp. If the block's initial speed is 2.00 m / s and the block again slides down the ramp 1.91 m , what should Nancy measure for the speed of the block this time?
The forces doing non-zero work as the block slides down a frictionless ramp are gravity and any applied forces. The work done can be calculated using the mass, gravitational acceleration, and height. For the pushed block, the initial kinetic energy is combined with gravitational work to predict the final speed.
The forces that do non-zero work on a block as it slides down a frictionless ramp include gravity and any applied forces (like a push or a pull), but not the normal force or friction since the ramp is frictionless. Gravity does positive work as it pulls the block down the plane, increasing its kinetic energy. To calculate the total work done on the 1.85 kg wooden block after sliding down a distance of 1.85 m on the ramp at a 59.3° angle, you can use the formula:
Work = mgh, where m is mass, g is acceleration due to gravity, and h is the height.To predict the speed of the block after sliding down, you can use the conservation of energy principle or kinematic equations that relate distance, acceleration, and velocity. For the scenario where the block is given an initial push, the final speed can be predicted using the same principles by accounting for the initial kinetic energy provided by the push.
Here's an example of calculating work done:
Problem: A block of mass 10 kg slides down through a length of 10 m over an incline of 30°. If the coefficient of kinetic friction is 0.5, then find the work done by the net force on the block. In this example, work done by the net force can be calculated by finding the components of the gravitational force parallel and perpendicular to the incline and applying the kinetic friction force in the opposite direction of motion.Consider two uniform solid spheres where one has twice the mass and twice the diameter of the other. The ratio of the larger moment of inertia to that of the smaller moment of inertia is:_________.a) 2b) 8c) 4d) 10e) 6
Answer:
The ratio of moment of inertia of larger sphere to that of smaller sphere = 4
Explanation:
The moment of inertia of solid sphere is given by I = 2/5MR² where M = mass of sphere and R = radius of sphere.
Radius of smaller sphere = D/2
Radius of larger sphere = 2D/2 = D.
Moment of inertia of smaller sphere I₁ = 2/5M × D²/4 = MD²/10
Moment of inertia of larger sphere I₂ = 2/5M × D² = 2MD²/5
The ratio of moment of inertia of larger sphere to that of smaller sphere = I₂/I₁ = 2MD²/5 ÷ MD²/10 = 10 × 2/5 = 4
An ice cream machine produced 44 ice creams per minute. After reconditioning, its speed increase to 55 ice creams per minute. By what percent did the speed of the machine increase
Answer:
an ice cream machine produced 44 ice cream per minute.
Now, after reconditioning it speeds up to 55 ice cream per minute.
=> 55 - 44 = 11
=> 11 / 44 = 0.25
=> 0.25 * 100% = 25%.
thus it increased it production to 25%.
Explanation:
Answer:
25%
Explanation:
The way to find this answer is to:
Subtract 44 from 55 (11)
Then, you can put 11 over 44
Simplified, this is equal to 1/4.
1/4 as a percentage is 25%.
Good luck with your RSM work!
A wire of length 8x is bent into the shape of a circle. A. Express the circumference of the circle as a function of x. B. Express the area of the circle as a function of x.
Answer:
[tex]a. C=8X\\\\b. A=\frac{16x^2}{\pi}[/tex]
Explanation:
Given length is [tex]8x[/tex]
Circumference of a circle is given as:
[tex]C=2\pi r[/tex]
a.But the circumference is equal to length of the wire. Therefore circumference as a function of [tex]x[/tex] is:
[tex]C=8x[/tex]
b.Area as function of [tex]x[/tex]
Area is calculated as[tex]A=\pi r^2[/tex]
Rewrite the circumference equation to make r the subject of the formula.
[tex]2\pi r=8x\\r=\frac{4x}{\pi}[/tex]
To express area as a function of [tex]x[/tex]:-
[tex]A=\pi r^2\\=\pi (\frac{4x}{\pi})^2\\=\frac{16x^2}{\pi}[/tex]
Therefore area of circle as a function of [tex]x[/tex] is [tex]\frac{16x^2}{\pi}[/tex]
The circumference of the circle, represented as function of x, is C(x) = 8x. The area of the circle, represented as function of x, is A(x) = (16x²) / π.
Explanation:The given length of the wire bent into a circle is 8x. This is equivalent to the circumference of this circle because a circle's circumference is the distance around it.
So, the function of the circumference, C, in terms of x is C(x) = 8x.
For the area of the circle, we must understand that the formula for the area of a circle is πr² where r is the radius of the circle. The radius is half the diameter, and since the circumference of a circle is πD where D is the diameter, we can find that r = circumference / (2π), or r = 8x / (2π).
So, substituting r into the formula for the area, we get A(x) = π(8x / (2π))² = π(4x / π)² = π((4x)² / π²) = (16x²) / π.
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In Millikan's experiment, an oil drop of radius 2.342 μm and density 0.816 g/cm3 is suspended in chamber C when a downward-pointing electric field of 1.92 ✕ 105 N/C is applied. Find the charge on the drop, in terms of e. (Include the sign of the value in your answer.)
Answer:
-14 e
Explanation:
Assuming oil drop is spherical then V = 4pi.r^3/3
density = 0.816 g/cm^3= 0.816 x 10^-3kg/cm^3= 0.816 x 10^-3/10^-6 kg/m^3= 0.816 x 10^3 kg/m^3
mass =(4 x 3.142 x (2.342 x 10^-6)^3 x 0.816 x 10^3)/3 kg
mass = (12.568 x 12.85 x 10^-18 x 0.816 x 10^3)/3 kg
m= 4.39 x 10^-14 kg
F=qE…. Eqn. 1
F=mg… Eqn. 2
Compare both eqn.1 and eqn. 2
qE=mg
q=mg/E
q = (4.39 x 10^-14 x 9.81)/1.92 x 10^5
q = 2.24 x 10^-18 C
Number of charges, n = 2.24 x 10^-18/1.62 x 10^-19
n=13.827
Charges come in integer values, nearest is 14
Number of charges is 14.
The direction of the field (downwards) is the direction a positive test charge would move. This means a positive charge would move downwards and repelled by the top plate which must be positive.
Since mg acts downwards then the other force must act up if the oil drop is suspended. This means that the charge is being attracted to the top plate which we have already determined is positive. for the charge to be attracted to the positive top plate then the charge must be negative.
Charge is electron hence the sign is minus.
-14 e
The amount of kinetic energy an object has depends on its mass and its speed. Rank the following sets of oranges and cantaloupes from least kinetic energy to greatest kinetic energy. If two sets have the same amount of kinetic energy, place one on top of the other. 1. mass: m speed: v2. mass: 4 m speed: v3. total mass: 2 m speed: 1/4v4. mass: 4 m : speed: v5. total mass: 4 m speed: 1/2v
The ranking from least to greatest kinetic energy is: Set 3 < Set 1 < Set 5 < Set 2, Set 4.
To rank the sets of oranges and cantaloupes based on their kinetic energy, from the formula for kinetic energy:
Kinetic Energy (KE) = 0.5 × mass × speed²
Calculate the kinetic energy for each set and then rank them from least to greatest kinetic energy:
Set 1: KE = 0.5 × m × v²
Set 2: KE = 0.5 × 4m × v² = 2 × 0.5 × m × v² (Same as Set 1)
Set 3: KE = 0.5 × 2m × (1/4v)² = 0.125 × × v² (Less than Set 1)
Set 4: KE = 0.5 × 4m × v² = 2 × 0.5 × m × v² (Same as Set 1 and 2)
Set 5: KE = 0.5 × 4m × (1/2v)² = 1 × 0.5 × m × v² (Same as Set 1, 2, and 4)
Ranking from least kinetic energy to greatest kinetic energy:
Set 3 (total mass: 2m, speed: 1/4v)
Set 1 (mass: m, speed: v)
Set 5 (total mass: 4m, speed: 1/2v)
Sets 2 and 4 (mass: 4m, speed: v)
Hence, the ranking from least to greatest kinetic energy is: Set 3 < Set 1 < Set 5 < Set 2, Set 4.
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The kinetic energy of an object is dependent on both its mass and speed. The ranking of the sets of oranges and cantaloupes, based on their kinetic energy (from least to greatest), is: KE3, KE5, KE1 = KE4, KE2. These rankings are calculated by substituting the values of mass and speed provided into the formula for kinetic energy.
Explanation:The kinetic energy of an object can be defined as the energy which it possesses due to its motion. It is dependent on both its mass and speed, as outlined by the formula KE = 1/2mv². Where KE is the kinetic energy, m is the mass of the object, and v is the velocity (or speed).
Considering the information provided on the sets of oranges and cantaloupes:
Set 1: KE1 = 1/2m(v)².Set 2: KE2 = 1/2(4m)(v)².Set 3: KE3 = 1/2(2m)((1/4v)²).Set 4: KE4 = 1/2(4m)(v)².Set 5: KE5 = 1/2(4m)((1/2v)²).
Ranking them according to their kinetic energy, from least to greatest, gives us: KE3, KE5, KE1 = KE4, KE2.
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Imagine that you take a beam of monochromatic light and start to reduce the power of this beam of light. As you get the power (intensity) lower and lower, what will happen to your observations of the beam of light with regards to observing individual photons?
Answer:
No change in velocity of photons...
Explanation:
The reason speed of light is constant is that is doesn't depend upon intensity of light as produced by the source. This is the main reason why the wave function (psi)* in Einstein's Photoelectric experiment remains unaffected by intensity. The monochromatic beam of light will glow lower and lower as the power decreases and eventually vanishes from visible range as the emission from the power source has been stopped...
A pole-vaulter just clears the bar at 5.31 m and falls back to the ground. The change in the vaulter's potential energy during the fall is -3900 J. What is his weight?
Answer:
Weight = 734.46 N
Explanation:
Given:
Initial height of the pole-vaulter is, [tex]h_i=5.31\ m[/tex]
Final height of the pole-vaulter is, [tex]h_f=0\ m[/tex]
Change in the vaulter's potential energy is, [tex]\Delta U=-3900\ J[/tex]
We know that, change in potential energy is given as:
[tex]\Delta U=mg(h_f-h_i)[/tex]
Where, 'm' is the mass of the object, 'g' is acceleration due to gravity and has a value of 9.8 m/s².
Now, weight of the object is given as the product of its mass and acceleration due to gravity. So, replace 'mg' by weight 'w'. So, the equation becomes,
[tex]\Delta U = w(h_f-h_i)[/tex]
Now, rewriting in terms of 'w', we get:
[tex]w=\dfrac{\Delta U}{(h_f-h_i)}[/tex]
Now, plug in all the given values and solve for 'w'. This gives,
[tex]w=\dfrac{-3900\ J}{(0-5.31)\ m}\\\\\\w=\dfrac{3900}{5.31}\ N\\\\\\w=734.46\ N[/tex]
Therefore, weight of the vaulter is 734.46 N.
The weight of the pole-vaulter is calculated using the principle of conservation of mechanical energy and the equation for gravitational potential energy, resulting in a value of approximately 734.46 Newtons.
Explanation:The question asked concerns the calculation of the weight of a pole-vaulter given a change in potential energy as the vaulter falls. The calculation can be made using the principle of conservation of mechanical energy, and the equation for gravitational potential energy: PE = mgh, where PE is potential energy, m is mass, g is the acceleration due to gravity, and h is the change in height.
The change in the vaulter's potential energy is given in the problem as -3900 J, which is negative because potential energy decreases as height decreases. Assuming that the entire change in potential energy is converted into kinetic energy, we can arrange the equation: -PE = mgh = weight * h for the weight of the vaulter, giving: weight = -PE / h.
Substituting the given values into this equation gives weight = (-(-3900 J)) / (5.31 m) = 734.46 N. So, the weight of the pole-vaulter is approximately 734.46 Newtons.
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Three identical balls are thrown from the top of a building, all with the same initial speed but at different angles. Neglecting air resistance, rank the speeds of the balls at the instant each hits the ground.
Answer:
The speed of each ball would be same at the instant each hits the ground.
Explanation:
From the principle of conservation of energy we know that
PE₁ + KE₁ = PE₂ + PE₂
where PE = mgh
The initial and final potential energies would be same for each ball since all the balls are thrown from the same height.
Now what about KE ?
We know that each ball is thrown with the same initial speed so their initial kinetic energies would be same and in turn their final kinetic energies must be same in order to satisfy th e conservation of energy principle. Therefore, we can conclude that the speed of each ball would be same at the instant they hit the ground. The initial angle of ball doesn't have any impact on the speed of the ball.
Because the block is not moving, the sum of the y components of the forces acting on the block must be zero. Find an expression for the sum of the y components of the forces acting on the block, using coordinate system b.
Answer:
Incomplete questions
Completed question, check attachment for diagram
Because the block is not moving, the sum of the y components of the forces acting on the block must be zero. Find an expression for the sum of the y components of the forces acting on the block, using coordinate system shown. Express your answer in terms of some or all of the variables
Fn, Ff, Fw and θ.
Explanation:
Check attachment for solution
Final answer:
The sum of the y components of forces on a stationary block must be zero due to Newton's second law, which results in the normal force being equal to the weight of the block.
Explanation:
In physics, when analyzing the forces acting on a stationary block, we apply Newton's second law, which implies that the net force in the y-component must be zero because the block is not moving vertically. An expression for the sum of the y components of the forces acting on the block can be deduced by considering all the vertical forces, including the weight of the block (acting downward) and the normal force (acting upward).
For a block on a frictionless surface attached to a spring, we consider the spring force and gravitational force. If the block is in equilibrium, these forces satisfy the equation mg = k Δy, where m is the mass of the block, g is the acceleration due to gravity, k is the spring constant, and Δy is the displacement from the equilibrium position. Therefore, the sum of the y components equates to zero: normal force - weight = 0, which also indicates that the normal force is equal to the weight.