Given data:
* The speed of the bus is 100 km/hr.
* The speed of the conductor is 6 km/hr.
Solution:
(a). If the bus and conductor are traveling in the same direction.
The net speed of the conductor relative to the road is,
[tex]v_1=v_b+v_c_{}[/tex]where v_1 is the velocity of the conductor relative to the road, v_b is the velocity of the bus, and v_c is the velocity of the conductor inside the bus,
Substituting the known values,
[tex]\begin{gathered} v_1=100+6 \\ v_1=106\text{ km/hr} \end{gathered}[/tex]Thus, the speed of the conductor relative to the road is 106 km/hr.
(b). The speed of the conductor relative to the bus is the speed of the conductor on the bus,
Thus, the speed of the conductor relative to the bus is 6 km/hr.
I need help with this question.The answer choices for each one is eitherAB C D E Any type of help will be appreciated even if it’s just a hint!
Conductor that carries electrons:
From the pic we can conclude that the conductor is the wire, in this case it would be A.
Load that transforms energy:
From the pic, it is the bulb which transform the electric energy into light and heat, so it would be B
Insulation that prevents electrons from flowing:
It is the cable sheath, from the pic it is C
Area of low potential energy for electrons:
It is the negative part of the battery, from the pic it is E
Area of high potential energy for electrons:
It is the positive part of the battery, it is D
part a. A delivery drone, hovering at an altitude of 260. m above the ground, drops a package. ("dropped" means that it was stationary when released ). How long will it take to reach the ground? part b. What will be the velocity of the package the instant before it hits the ground?
Given:
h= 260m (height)
a = 9.8 m/s^2 (acceleration due to gravity )
Apply:
• a)
h = 1/2 a t^2
Replacing:
260 = 1/2 (9.8) t^2
260 / (1/2 * 9.8 ) = t^2
t = √[260 / (1/2 * 9.8 ) ]
t = 7.28 s
• b)
Apply.:
Vf = vi + at
Vf = finale velocity
Vi = initial velocity ( rest ) = 0 m/s
Vf = -9.8 (7.28 ) = -71.34 m/s
A puck is moving on an air hockey table. Relative to an x,y coordinate system at time t =0s, the x
components of the puck's initial velocity and acceleration are Vix=1.0 m/s and ax=2.0 m/s². The y
components of the puck's initial velocity and acceleration are Viy=2.0 m/s and ay=2.0 m/s². Find the
magnitude and direction of the puck's velocity at a time of t=0.50 s. Specify the direction relative to
the x axis. HELPP!!!
The supplied puck is moving at a speed of v0x=+3.4m/s on an air hockey table at time t=0.
What is the meaning of velocity?The direction of the movement of the body or the object is defined by its velocity. Most of the time, speed is a scalar quantity. In its purest form, velocity is a vector quantity. It measures how quickly a distance changes. It is the rate at which displacement is changing.
What does the term "tangential velocity" refer to?Any object traveling in a circular motion has a linear speed known as tangential velocity. On a turntable, a point in the center moves less distance in a full rotation than a point near the outside edge.
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The total mechanical energy of the roller coaster cart below at Point A is 180,000 J. The speed of the cart at Point B is +20 m/s. Assume no energy is lost due to dissipative forces such as friction. A) What is the mass (in kg) of the roller coaster cart? B) What is the potential energy at Point A? C) What is the kinetic energy at Point A?
Mechanical energy (ME) = Potential energy(PE) + kinetic energy (KE)
PE = mgh
m= mass
g= gravity
h= height
KE= 1/2 m v^2
v= speed
Point B
ME = KE + PE
PE = 0 (height = 0 )
KE = 1/2 (m) v^2
180,000 = 1/2 (m) (20)^2
m = 180,000 / (1/2 (20)^2 )
m= 900 kg
Point A.
ME = 180,000 J = PEa + KE a
PEa = m g h = 900 (9.8) (20) = 176,400J
MEa = PEa + KEa
KEa = MEa - PEa = 180,000 - 176,400 J = 3,600 J
A) mass = 900 kg
B) 176,400 J
C) 3,600 J
Convert the following number from scientific notation to standard notation1.9300 x 10^-30.000 193 000.001 93000.001931930
The number converted to the standard notation as
[tex]1.93\times10^{-3}=0.00193[/tex]Hence, the correct answer is 0.00193
What is the energy of a proton accelerated through a potential difference of 500,000 V?
ANSWER
[tex]8.01\cdot10^{-14}J[/tex]EXPLANATION
We want to find the energy of the proton accelerated through the given potential.
To do this, apply the relationship between energy and potential:
[tex]V=\frac{E}{q}[/tex]where q = charge
V = potential
The charge of a proton is:
[tex]1.602\cdot10^{-19}C[/tex]Therefore, we have that the energy of the proton is:
[tex]\begin{gathered} E=V\cdot q \\ E=500000\cdot1.602\cdot10^{-19} \\ E=8.01\cdot10^{-14}J \end{gathered}[/tex]That is the answer.
1.In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law the current through the circuit is 2 Amps.2.If a second identical lamp is connected in series, the 6-volt battery must push a charge through a total resistance of 6Ω. The current in the circuit is then 1 Amps.3.If a third identical lamp is connected in series, the total resistance is now 9Ω.4.The current through all three lamps in series is now _________ Amps. The current through each individual lamp is __________ Amps.
ANSWER
The current through all three lamps in series is now 0.67 Amps. The current through each individual lamp is 0.67 Amps.
EXPLANATION
There are three lamps connected in series, each with a resistance of 3 Ω, resulting in a total resistance of 9 Ω.
By Ohm's law, if the voltage from the battery is 6 V, then the current through all three lamps - i.e. the total current in the circuit is,
[tex]I=\frac{V}{R_{eq}}=\frac{6V}{9\Omega}=\frac{2}{3}Amps\approx0.67Amps[/tex]And, since the three lamps are connected in series - which means there are no dividing paths, the current through each individual lamp is the same as the total current of the circuit, 0.67 Amps.
Hence, the current through all three lamps and through each individual lamp is 0.67 Amps, rounded to the nearest hundredth.
10. A boy of mass 55kg runs at 12m/s and hops on a 15kg skateboard that was at rest. What is thevelocity of the boy on the skateboard afterwards?
M = mass of the boy = 55kg
V = initial velocity of the boy = 12 m/s
m= mass of stationary skateboard = 15kg
v= velocity os stationary sketeboard= 0 m/s
V' = velocity of the boy on the skateboard after collision
Conservation of momentum:
MV + mv = (M + m) V'
Replacing:
55 kg * 12 m/s + 15 kg *0 = (55 kg+ 15 ) V'
Solve for V´'
660 = 70 V´'
660/70 = V'
V'= 9.42 m/s
Which of the following statements about air is TRUE?
A. Air is not a source of resistance.
B. Air has mass, but not inertia.
C. Air is not affected by human movement.
D. None of these statements are true.
The true statement among the following is that the air is not affected by human movement. Hence, option C is correct.
What is Air?Air relates to the atmosphere of the planet. Several gases and minute dust particles make up the air. Living organisms breathe and thrive in this pure gas. Its shape and volume are ill-defined. Considering that it is matter, it has mass and weight. Atmospheric pressure is generated by air weight. The space vacuum lacks air.
About 78% of the gas within air is nitrogen, 21% of the gas is oxygen, 0.9% of the gas is argon, 0.04% of the gas is co2, and very little other gas is present.
A typical amount of water vapor is around 1%.
Since respiration requires oxygen, animals must breathe it to survive. The lungs transfer back carbon dioxide back into the atmosphere when breathing, putting oxygen into the body.
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The chart shows data for four moving objects.
Object
W
X
Y
Z
Which object has the greatest acceleration?
W
Initial Velocity
(m/s)
Ο Ζ
11
10
12
20
Final Velocity
(m/s)
29
34
40
28
Change in
Time (s)
6
12
7
8
Object Y has the greatest acceleration = 4m/s2
What is an acceleration ?
acceleration: the rate at which the speed and direction of a moving object vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates. Even if the speed is constant, motion on a circle accelerates because the direction is always shifting. Both effects contribute to the acceleration for all other motions.
Acceleration = (final velocity - initial velocity ) /time
For object W
Acceleration = ( 29-11)/6 = 3m/s2
For object X
Acceleration = (34-10)/12 = 2m/s2
For object Y
Acceleration = ( 40-12)/7 = 4m/s2
For object Z
Acceleration = (28-20)/8 = 1m/s2
Object Y has the greatest acceleration = 4m/s2
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The amount of work done by two boys who apply 300 N of force in an unsuccessful attempt to move a stalled car is:1. 600 N · m.2. 300 N.3. 300 N · m.4. 600 N.5. 0.
The work done by a force can be calculated with the formula below:
[tex]W=F\cdot d[/tex]Where W is the work in J, F is the force in N and d is the distance in meters.
Since in this case, the attempt was unsuccessful, the car didn't move, so the distance is zero.
Therefore the work is zero, and the correct option is 5 (work is zero).
A 1500 kg car traveling at 10 m/s suddenly runs out of gaswhile approaching the valley shown in FIGURE EX10.11. The alertdriver immediately puts the car in neutral so that it will roll.What will be the car’s speed as it coasts into the gas station onthe other side of the valley?
If the car does not brake, its mechanical energy will be conserved along all its trajectory.
Then, the sum of the kinetic energy (K) of the car and its gravitational potential energy (U) will be the same for any two points in the trajectory:
[tex]K_1+U_1=K_2+U_2[/tex]The kinetic energy of an object with mass m and speed v is:
[tex]K=\frac{1}{2}mv^2[/tex]The gravitational potential energy of an object with mass m at a height h above a reference level, is:
[tex]U=mgh[/tex]Where g is the acceleration of gravity.
In the beginning, the car has a height of 10m and a speed of 10m/s. In the end, the car reaches the gas station at a height 15m and a unknown speed v.
Then, the initial speed and height are known, as well as the final height. Use the equation for the conservation of mechanical energy to isolate v_2, the unknown final speed of the car:
[tex]\begin{gathered} K_2+U_2=K_1+U_1 \\ \Rightarrow\frac{1}{2}mv_2^2+mgh_2=\frac{1}{2}mv_1^2+mgh_1 \\ \Rightarrow\frac{1}{2}v_2^2+gh_2=\frac{1}{2}v_1^2+gh_1 \\ \Rightarrow\frac{1}{2}v_2^2=\frac{1}{2}v_1^2+gh_1-gh_2 \\ \Rightarrow\frac{1}{2}v_2^2=\frac{1}{2}v_1^2+g\mleft(h_1-h_2\mright) \\ \Rightarrow v_2^2=v_1^2+2g\mleft(h_1-h_2\mright) \\ \\ \therefore v_2=\sqrt{v_1^2+2g(h_1-h_2)} \end{gathered}[/tex]Replace v_1=10m/s, h_1=10m, h_2=15m and g=9.8m/s^2 to find the speed of the car as it reaches the gas station:
[tex]\begin{gathered} v_2=\sqrt{(10\frac{m}{s})^2+2(9.8\frac{m}{s^2})(10m-15m)} \\ =1.4142...\frac{m}{s} \\ \approx1.4\frac{m}{s} \end{gathered}[/tex]Therefore, the car's speed as it coasts into the gas station on the other side of the valley will be approximately 1.4 meters per second.
For a convex lens to form a virtual image the object must be located at some distance less than the focal length. Is this true or false?
For a convex lens to form a virtua
wat is the mass of the car that has kinetic energy of 2400J and is moving with a speed of 20 m\s
Given,
The kinetic energy of the car, E=2400 J
The speed of the car, v=20 m/s
Kinetic energy is the energy that is possessed by an object due to its motion.
It is given by,
[tex]E=\frac{1}{2}mv^2[/tex]Where m is the mass of the car.
On substituting the known values in the above equation,
[tex]\begin{gathered} 2400=\frac{1}{2}\times m\times20^2 \\ m=\frac{2\times2400}{20^2} \\ =\frac{4800}{400} \\ =12\text{ kg} \end{gathered}[/tex]Thus the mass of the car is 12 kg
A car is going at a speed of 25m/s when the driver puts her foot on the gas pedal. The carfeels a net force of 2000N for 50m. The car's mass is 1000kg.How much kinetic energy does the car have initially?
The initial kinetic energy of the car = 312.5 kJ
Explanation:The initial volume of the car, v = 25 m/s
The mass of the car, m = 1000 kg
The initial kinetic energy is given by the formula
[tex]\begin{gathered} KE=\frac{1}{2}mv^2 \\ \end{gathered}[/tex]Substitute m = 1000 kg, and v = 25 m/s into the formula
[tex]\begin{gathered} KE=\frac{1}{2}\times1000\times25^2 \\ KE=500\times625 \\ KE=312500J \\ KE=312.5kJ \end{gathered}[/tex]The initial kinetic energy of the car = 312.5 kJ
Mr.D soars over a large group of zombies and is in the air for a total of 5s. How high did he go?
Kathleen is rowing her canoe northward with an average force of 19.1 N. The wind is blowing directly south with an average force of 14.4 N. What is the net force on the canoe?
4.7 N north
33.5 N south
33.5 N north
4.7 N south
The net force on the canoe rowed northward with an average force of 19.1 N is 4.7 N North if the wind is blowing directly south with an average force of 14.4 N.
Net force = Force due to rowing - Force due to wind
Net force = 19.1 - 14.4
Net force = 4.7 N
The direction is towards North because the force due to rowing is greater than the force due to wind.
If two force act in opposite directions, the resultant force is the difference of these two forces and the resultant direction is the direction in which the greater force is applied.
Therefore, the net force on the canoe is 4.7 N North
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According to the law of conservation of charge which statement can be true?A. A silk cloth gained charge. B. A metal rod lost charge.C. A peice of glass transferred electrons to felt. D. A balloon remains neutrally charged when rubbed.
Option D
Explanation:The law of conservation of charge states that the amount of charge in a system is constant
This means that as time changes, the amount of charge in a system does not change
By careful consideration of the options stated:
Each of options A to C either shows that charge is lost or gained
Only option D typifies the law of conservation of conservation of charges because charges are not lost or gained by the ballon so described.
Projections of a vector make up the components of that vector. Is this true or false?
The given statement 'Projections of a vector make up the components of that vector' is true. The direction of a vector
What is the kinetic energy of a 2.0 kg object moving at 5 m/s?
Let's put the given values into the formula below and get the result;
[tex]K.E=\frac{1}{2}mv^2[/tex]We know the numerical value of speed. We also know the mass. We can jump straight to the conclusion.
[tex]K.E=\frac{1}{2}(2kg)(5m/s)^2[/tex][tex]=\frac{1}{2}(2kg)(25m^2/s^2)[/tex][tex]=25m^2/s^2=25J[/tex]The Kinetic Energy is 25J.
If we want to accelerate an object, we must apply force on it, after applying Force some work has to be done in which energy should be transferred to the object. The energy is known as kinetic energy. The kinetic energy always depends on the mass and the velocity. It is denoted as K and the Si unit is Joules (J).
To calculate the Kinetic Energy,
K= 1/2 mv^2
m = mass
v = velocity
K = kinetic energy
In solving the above equation,
K = 1/2 x2 x 5^2
K = 5x5
K = 25
The Kinetic Energy is 25J.
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A roller coaster has a vertical loop with radius 22.8 m. With what minimum speed should the roller-coaster car be moving at the top of the loop so that the passengers do not lose contact with the seats?
Given,
The radius of the loop of the roller coaster, r=22.8 m
The forces that are acting on the roller coaster when it is at the top of the loop are the centripetal force directed upwards and the weight of the roller coaster including the passengers directed downwards.
For the passengers to stay in the seat, the centripetal force must be, at the least, equal to the weight of the passengers and the rollercoaster.
That is,
[tex]\frac{Mv^2}{r}=Mg[/tex]Where M is the combined mass of the rollercoaster and the passengers, v is the minimum speed of the roller coaster when it is at the top of the loop, and g is the acceleration due to gravity.
On simplifying the above equation,
[tex]v=\sqrt[]{gr}[/tex]On substituting the known values,
[tex]\begin{gathered} v=\sqrt[]{9.8\times22.8} \\ =14.95\text{ m/s} \end{gathered}[/tex]Thus the minimum speed that the roller coaster must have when it is at the top of the loop so that the passengers stay in contact with the seats is 14.95 m/s.
A 5 cm spring is suspended with a mass of 1 g attached to it which extends the spring by 3.2 cm. The same spring is placed on a frictionless flat surface and charged beads are attached to each end of the spring. With the charged beads attached to the spring, the spring's extension is 0.01 m. What are the charges, in micro-Coulombs, of the beads?
First, we need to find the spring constant
Force of gravity = force of tension from spring
mg = kx
(1x10^-3)(9.8) = k (3.2x10^-2)
k = .30625
Now we can look at the other situation
Since the spring was moved 0.01 meters, we can find the force
F = kx = (.01)(.30625) = .0030625 N
Now we can set the electric force equal to the force of the beads
kq^2/r^2 = .0030625 N
q = 292 microCoulombs
Find the direction of the vector A⃗ = (5.1 m )x^ + (-1.5 m )y^.Find the direction of the vector B⃗ = (-1.5 m )x^ + (5.5 m )y^Find the magnitude of the vector A⃗ +B⃗ .Find the direction of the vector A⃗ +B⃗ .
The angle between the components x and y of a vector is given by:
[tex]\theta=\tan ^{-1}(\frac{v_x_{}_{}}{v_y})[/tex]once we know this we need to find in which quadrant the vector lies so we know how to calculate the correct direction.
Vector A lies in the fourth quadrant this means that we need to subtract theta to 360° in order to get the direction of the vector, then we have:
[tex]360-\tan ^{-1}(\frac{1.5}{5.1})=343.61[/tex]Therefore the direction of vector A is 343.61°
Vector B lies in the second quadrant, this means that we need to subtract theta (given by the first equation) to 180° in order to get the direction, then we have:
[tex]180-\tan ^{-1}(\frac{5.5}{1.5})=105.26[/tex]Therefore the direction of vector B is 105.26°
Let's find vector A+B:
[tex]\begin{gathered} \vec{A}+\vec{B}=\langle5.1,-1.5\rangle+\langle-1.5,5.5\rangle \\ =\langle5.1-1.5,-1.5+5.5\rangle \\ =\langle3.6,4\rangle \end{gathered}[/tex]Then we have that:
[tex]\vec{A}+\vec{B}=\langle3.6,4\rangle[/tex]To find its magnitude we have to remember that the magnitude of any vector is given by:
[tex]\lvert\vec{v}\rvert=\sqrt[]{v^2_x+v^2_y}[/tex]Then for vector A+B we have:
[tex]\begin{gathered} \lvert\vec{A}+\vec{B}\rvert=\sqrt[]{(3.6)^2+(4)^2} \\ =5.38 \end{gathered}[/tex]Therefore the magnitude of vector A+B is 5.38 meters.
Vector A+B lies in the first quadrant, then its direction is given by the expression for theta, then we have:
[tex]\tan ^{-1}(\frac{4}{3.6})=48.01[/tex]Therefore the direction of vector A+B is 48.01°
If the atomic mass of carbon is 12 amu, how much mass of carbon would be needed to have anAvogadro's number (1 mole) of carbon atoms?A. 12 mgB. 12 gC. 12 kgD. 12 lbs
Given:
Atomic mass of carbon = 12 amu
Let's find how much mass of carbon would be needed to have an Avogadro's number (1 mole) of carbon atoms.
Given that the atomic mass of carbon is 12 atomic mass unit (amu), the amount of mass that would be need for 1 mole of carbon atom is = 12 g
ANSWER:
12 g
7. Which of the following measurement tools would you need to
determine the temperature of boiling water?
Answer:Laboratory thermometer
Explanation:
A 2.5 kg canoe is traveling up the mississippi river at a velocity of 10m/s, north. What is its kinetc energy
Answer:
Explanation:
Given:
m = 2.5 kg
V = 10 m/s
__________
Wk - ?
Wk = m·V² / 2
Wk = 2.5·10² / 2 = 125 J
The nearest star to the Earth (other than the Sun) is about 4.0 lightyears away. A lightyear, ly, is the distance light travels in one year. Voyager 1 is traveling at 38,500 miles per hour. How long will it take Voyager 1 to reach the nearest star in years?
First, let's convert the distance of 4 lightyears to miles:
[tex]4\text{ ly}=4\cdot5.879\cdot10^{12}\text{ miles}=23.516\cdot10^{12}\text{ miles}[/tex]Now, let's find the time required:
[tex]\begin{gathered} distance=speed\cdot time\\ \\ 23.516\cdot10^{12}=38500\cdot t\\ \\ t=\frac{23.516\cdot10^{12}}{38500}\\ \\ t=6.10805\cdot10^8\text{ hours} \end{gathered}[/tex]Then, we need to convert from hours to years:
[tex]6.10805\cdot10^8\text{ hours}=\frac{6.10805\cdot10^8}{8760}=6.97266\cdot10^4\text{ years}[/tex]Therefore it will take approximately 69727 years.
Ball A with diameter d and ball B with diameter 2d are dropped from the same height. When the two balls have the same speed, what is the ratio of the drag force on ball A to the drag force on ball B?
Ball A with diameter d and ball B with diameter 2d are dropped from the same height. When the two balls have the same speed, the ratio of the drag force on ball A to the drag force on ball B will be F1 : F2 = 1 : 4
When objects travel through fluids (a gas or a liquid), they will undoubtedly encounter resistive forces called drag forces.
The drag force always acts in the opposite direction to fluid flow. If the body’s motion exists in the fluid-like air, it is called aerodynamic drag.
formula to calculate drag force is = F(d) = 1/2 * C * rho*A * [tex]v^{2}[/tex]
C = drag coefficient
A = area of object
rho = density in which object is moving
v = velocity of object
A = area of the object
F1 ( drag force on ball A ) = 1/2 * C * rho * area of ball A * [tex]v^{2}[/tex]
F2 (drag force on ball A ) = 1/2 * C * rho * area of ball B * [tex]v^{2}[/tex]
since , both the balls have same speed and falling in same environment hence , density and speeds are the same , the only difference is in area of both the balls
F1/F2 = area of ball A / area of ball B = 4 * pi * [tex]r1^{2}[/tex] / 4 * pi * [tex]r2^{2}[/tex]
= [tex]r1^{2}[/tex] / [tex]r2^{2}[/tex]
= [tex](\frac{d}{2} )^{2}[/tex]/ [tex](\frac{2d}{2}) ^{2}[/tex]
= 1/4
F1 : F2 = 1 : 4
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A baseball of mass 1.23 kg is thrown at a speed of 65.8 mi/h. What is its kinetic energy?
Given:
The mass of the ball is
[tex]m=1.23\text{ kg}[/tex]The speed of the ball is
[tex]\begin{gathered} v=65.8\text{ mi/h} \\ \end{gathered}[/tex]Required: calculate the kinetic energy of the baseball
Explanation: to calculate the kinetic energy of a body we will use the formula as
[tex]K.E=\frac{1}{2}mv^2[/tex]first, we convert velocity from mi/h into m/s.
we know that
[tex]1\text{ mi=1609.34 m}[/tex]and
[tex]1\text{ h=3600 sec}[/tex]then the velocity is
[tex]\begin{gathered} v=\frac{65.8\times1602.34\text{ m}}{3600\text{ s}} \\ v=29.29\text{ m/s} \end{gathered}[/tex]now plugging all the values in the above formula, we get
[tex]\begin{gathered} K.E=\frac{1}{2}mv^2 \\ K.E=\frac{1}{2}\times1.23\text{ kg}\times(29.29\text{ m/s})^2 \\ K.E=527.61\text{ J} \end{gathered}[/tex]Thus, the kinetic energy of the baseball is
[tex]527.61\text{ J}[/tex]what happens to the state of a variable if it goes through a two series connected NOT gate.