Two pieces of amber are hung from threads. Piece A is charged by rubbing piece A with fur. Piece B is charged by rubbing piece B with silk. Nylon is used to rub and charge a third piece of amber. Piece A and B are both repelled by the third piece of amber. This means:____.
Answer:
ieces A and B must also have the same type of charges
Explanation:
In electrostatics, charges of the same sign repel and charges of different signs attract.
If we apply this to our case, we have that part A and C repel each other, therefore they have the same type of charge.
Also part A and C repel each other, therefore they have the same type of charge.
If we use the transitive property of mathematics, pieces A and B must also have the same type of charges
car is moving at 40 m/s. At 10 meters the driver spots a deer on the road and instantly steps on the brakes. If the car is 400 kg how much force must the breaks exert to stop the car in time?
Answer:
32000 N
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 40 m/s
Distance (s) = 10 m
Final velocity (v) = 0 m/s
Mass (m) of car = 400 Kg
Force (F) =?
Next, we shall determine the acceleration of the the car. This can be obtained as follow:
Initial velocity (u) = 40 m/s
Distance (s) = 10 m
Final velocity (v) = 0 m/s
Acceleration (a) =?
v² = u² + 2as
0² = 40² + (2 × a × 10)
0 = 1600 + 20a
Collect like terms
0 – 1600 = 20a
–1600 = 20a
Divide both side by –1600
a = –1600 / 20
a = –80 m/s²
The negative sign indicate that the car is decelerating i.e coming to rest.
Finally, we shall determine the force needed to stop the car. This can be obtained as follow:
Mass (m) of car = 400 Kg
Acceleration (a) = –80 m/s²
Force (F) =?
F = ma
F = 400 × –80
F = – 32000 N
NOTE: The negative sign indicate that the force is in opposite direction to the motion of the car.
If pressurized air pressure is 350 kPa, atmospheric pressure is 100 kPa, initial atmospheric pressure is 100 kPa, initial acceleration of the water rocket is 0.5g, acceleration of the water rocket is 0.5g, mass of water is 0.5 kg and structural mass of water is 0.5 kg and structural mass is 0.5 kg. Calculate the diameter of mass is 0.5 kg. Calculate the diameter of the nozzle where water is leaving the the nozzle where water is leaving the bottle
Answer:
[tex]d=8.657mm[/tex]
Explanation:
From the question we are told that
Pressurized air pressure is [tex]P_{air}=350 kPa,[/tex]
Atmospheric pressure is [tex]P_a=100 kPa[/tex]
Initial acceleration of the water rocket is [tex]a_i=0.5g.[/tex]
Acceleration of the water rocket is [tex]a_r=0.5g[/tex]
Mass of water is [tex]M_w=0.5 kg[/tex]
Generally total mass is given mathematically given as
[tex]T_M=0.5+0.5=>1kg[/tex]
Generally the tension on the rocket is given mathematically given as
[tex]T=(P_{air}-P_a)A[/tex]
[tex]T=(350-100) \frac{\pi d^2}{4}[/tex]
T is also
[tex]T=\frac{3Mg}{2}[/tex]
Therefore
[tex]T=>(350-100) \frac{\pi d^2}{4}= \frac{3Mg}{2}[/tex]
[tex]T=>(350-100) \frac{\pi d^2}{4}= \frac{3*1*9.81}{2}[/tex]
[tex]d^2= \frac{3*1*9.81*4}{2(350-100) \pi}[/tex]
[tex]d=\sqrt{\frac{3*1*9.81*4}{2(350-100) \pi}}[/tex]
[tex]d=8.657mm[/tex]
therefore diameter of nozzle is mathematically given as
[tex]d=8.657mm[/tex]
What test are included in the Physical Fitness Test?
PACER/Mile Run, Push Ups, Curl Ups, Trunk Lift, Back Saver Sit-and-Reach
PACER/Mile Run, Push Ups
O Push Ups, Curl Ups, Trunk Lift, Back Saver Sit-and-Reach
PACER/Mile Run, Long Jump, Push Up,
The test that is included in the Physical Fitness Test are: A. PACER/Mile Run, Push Ups, Curl Ups, Trunk Lift, Back Saver Sit-and-Reach.
What is physical fitness test?Physical fitness test can be defined as a measure of both the physical and mental soundness (wellness) of an individual to engage in sports, work and other day-to-day activities.
The examples of physical fitness test.In Health education, some examples of the test that is included in the Physical Fitness Test are:
PACER/Mile RunPush UpsCurl UpsTrunk LiftBack Saver Sit-and-Reach.Read more on physical fitness here: https://brainly.com/question/1809216
a) What magnitude point charge creates a 12596.37 N/C electric
held at a distance of 0.593 m?
Answer:
[tex]Q = 4.9216 * 10^{-7}C[/tex]
Explanation:
Given
[tex]E = 12596.37 N/C[/tex]
[tex]r = 0.593m[/tex]
Required
Determine the magnitude point charge (Q)
This question will be solved using [tex]the\ magnitude[/tex] of the electric field formula
[tex]E = \frac{kQ}{r^2}[/tex]
Where
[tex]k = 9 * 10^9\ Nm^2 / C^2[/tex]
Make Q the subject in [tex]E = \frac{kQ}{r^2}[/tex]
[tex]E * r^2 = kQ[/tex]
[tex]Q = \frac{E * r^2}{k}[/tex]
Substitute values for E, r and k
[tex]Q = \frac{12596.37 * 0.593^2}{9 * 10^9}[/tex]
[tex]Q = \frac{4429.50}{9 * 10^9}[/tex]
[tex]Q = \frac{492.16}{10^9}[/tex]
[tex]Q = 492.16 * 10^{-9}[/tex]
Express in standard form
[tex]Q = 4.9216 * 10^2 * 10^{-9}[/tex]
[tex]Q = 4.9216 * 10^{2-9}[/tex]
[tex]Q = 4.9216 * 10^{-7}C[/tex]
Suppose that material scientists are testing a new alloy that has remarkably large coefficients of thermal expansion. A sample of the alloy is formed into a thin, uniform, solid disk with diameter 2.00 cm when at room temperature. If the temperature of the disk is increased by an amount ΔT the diameter of the disk increases by an amount ΔD.
a. True
b. False
Answer:
True
Explanation:
When the temperature of solid is increased its size increases. It is due to the increase in internal energy of the solid and then there is expansion in the solid. The temperature gives heat to the solid object and it expands in size. The expansion cannot be linear it can be increased in size by the volume.
As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a proton and an electron are situated 865 nm from each other and you study the forces that the particles exert on each other. As expected, the predictions of Coulomb's law are well confirmed. You find that the forces are attractive and the magnitude of each force is:______
Answer:
force F = 1.66 × [tex]10^{-13}[/tex] N
Explanation:
given data
proton and an electron = 865 nm
solution
we get here force that is express as
force F = k q1 q2 ÷ r² ......................1
put here value and we get
force F = 9 × [tex]10^{9}[/tex] × [tex]\frac{1.6\times (10^{-19})^{2}}{865 \times (10^{-9})^{2}}[/tex]
force F = 1.66 × [tex]10^{-13}[/tex] N
A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the size of our Moon but has the mass of our Sun.
Estimate gravity on the surface on this star.
I know the solution of this question it is as the picture shows but I only need to add *10^3 to the lower part of the division lower part to get the correct answer. But I don't know why I should add it can anyone explain?
Answer:
4.384 * 10^13
Explanation:
Given the expression :
[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]
Applying the laws of indices
[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]
13.2733 * 10^19 ÷ 3.0276 * 10^6
(13.2733 / 3.0276) * 10^(19 - 6)
4.3840996 * 10^13
= 4.384 * 10^13
Answer:
[tex]g=4.38*10^{7} m/s^2[/tex]
Explanation:
To solve this, we need to know the mass of the sun, and the radius of the moon
[tex]M_{s} = 1.989*10^{30}kg \\R_{m} = 1737400m[/tex]
Now we can plug our values into our equation:
[tex]g=G*\frac{M_{E} }{r^{2} }[/tex]
This gives us:
[tex]g=6.67*10^{-11}*\frac{1.989*10^{30}}{1737400^{2}}[/tex]
This equals:
[tex]g=4.38*10^{7} m/s^2[/tex]
A heavy book is launched horizontally out a window from the first floor, a height, h, above the ground, with initial velocity, v0, and it hits the ground a horizontal distance X1 away from the window. Another book is similarly launched (same initial velocity) from the second floor window, a height 2h above the ground. Where does the second book land relative to the first book
Answer:
x₂ / x₁ = √2
Explanation:
To solve this exercise we can use the projectile launch ratios, let's find the time it takes for the second book to reach the ground
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
as the book is thrown horizontally v_{oy} = 0, when it reaches the ground its height is zero y= 0
0 = y₀ - ½ g t²
t = [tex]\sqrt{ \frac{2y_o}{ g} }[/tex]
t = \sqrt{ \frac{2 \ 2h}{ g} }
with this time we calculate the horizontal distance traveled
x = v₀ t
x₂ = v₀ [tex]\sqrt{ \frac{4h}{g} }[/tex]
now let's calculate the time it takes him to get to the floor when he leaves from the first floor
t =\sqrt{ \frac{2y_o}{ g} }
the horizontal distance traveled is
x₁ = v₀ [tex]\sqrt{ \frac{2h}{g} }[/tex]
therefore the difference in distance between the two runs is
Δx = x₂-x₁
Δx = v₀ \sqrt{ \frac{4h}{g} } - v₀ \sqrt{ \frac{2h}{g} }
Δx = v₀ \sqrt{ \frac{2h}{g} } √2
Δx =√2 x₁
the relationship between the two distances is
x₂ / x₁ = √2
As electric current moves through a wire, heat generated by resistance is conducted through a layer of insulation and then convected to the surrounding air. The steady-state temperature of the wire can be computed as
Answer:
Hello your question is incomplete attached below is the complete question
answer : ri ( thickness of wire ) = 14.167 m
Explanation:
attached below is a detailed solution
using the given data to determine the thickness of wire
3. What is the SI unit of force? What is this unit equivalent to in terms of fundamental units?
4. Why is force a vector quantity?
Answer:
force = mass * acceleration
therefore the SI unit is kg*m/s2 or newton's
it's a vector quantity because it has both direction(acceleration) and size (mass)
What is the difference between a wave and a medium?
Answer:
Mediums in which the speed of sound is different generally have differing acoustic impedances, so that, when a sound wave strikes an interface between
Explanation:The propagation of a wave through a medium will depend on the properties of the medium. For example, waves of different frequencies may travel
A torsional pendulum is formed by attaching a wire to the center of a meter stick with a mass of 5.00 kg. If the resulting period is 4.00 min, what is the torsion constant for the wire
Answer:
The torsion constant for the wire is [tex]2.856\times 10^{-4}\,N\cdot m[/tex].
Explanation:
The angular frequency of the torsional pendulum ([tex]\omega[/tex]), measured in radians per second, is defined by the following expression:
[tex]\omega = \sqrt{\frac{\kappa}{I} }[/tex] (1)
Where:
[tex]\kappa[/tex] - Torsional constant, measured in newton-meters.
[tex]I[/tex] - Moment of inertia, measured in kilogram-square meters.
The angular frequency and the moment of inertia are represented by the following formulas:
[tex]\omega = \frac{2\pi}{T}[/tex] (2)
[tex]I = \frac{m\cdot L^{2}}{12}[/tex] (3)
Where:
[tex]T[/tex] - Period, measured in seconds.
[tex]m[/tex] - Mass of the stick, measured in kilograms.
[tex]L[/tex] - Length of the stick, measured in meters.
By (2) and (3), (1) is now expanded:
[tex]\frac{2\pi}{T} = \sqrt{\frac{12\cdot \kappa}{m\cdot L^{2}} }[/tex]
[tex]\frac{2\pi}{T} = \frac{2}{L}\cdot \sqrt{\frac{3\cdot \kappa}{m} }[/tex]
[tex]\frac{\pi\cdot L}{T} = \sqrt{\frac{3\cdot \kappa}{m} }[/tex]
[tex]\frac{\pi^{2}\cdot L^{2}}{T^{2}} = \frac{3\cdot \kappa}{m}[/tex]
[tex]\kappa = \frac{\pi^{2}\cdot m\cdot L^{2}}{3\cdot T^{2}}[/tex]
If we know that [tex]m = 5\,kg[/tex], [tex]L = 1\,m[/tex] and [tex]T = 240\,s[/tex], then the torsion constant for the wire is:
[tex]\kappa = \frac{\pi^{2}\cdot (5\,kg)\cdot (1\,m)^{2}}{3\cdot (240\,s)^{2}}[/tex]
[tex]\kappa = 2.856\times 10^{-4}\,N\cdot m[/tex]
The torsion constant for the wire is [tex]2.856\times 10^{-4}\,N\cdot m[/tex].
Suppose you were digging a well into saturated sediments. Why is the sediment’s permeability an important factor in deciding where to put your well?
Answer:
The importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.
Explanation:
When digging a well into saturated sediments, the possibility of the sediment with either little saturation or full saturation being able to provide steady water supply will be limited by how permeable it is. Now, the importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.
Two spherical objects are separated by a distance that is 1.08 x 10-3 m. The objects are initially electrically neutral and are very small compared to the distance between them. Each object acquires the same negative charge due to the addition of electrons. As a result, each object experiences an electrostatic force that has a magnitude of 3.89 x 10-21 N. How many electrons did it take to produce the charge on one of the objects
Answer:
the charge on the object is 71.043×10^-20 C and the number of electron is 4.44
Explanation:
from coulumbs law, The force that is acting over both charge can be computed as
F=( kq1q2)/r^2..............eqn(1)
Where
F=electrostatic force= 3.89 x 10-21 N
k= column constant= 9 x 10^9 Nm^2/C^2
q1 and q2= magnitude of the charges
r= distance between the charges= 1.08 x 10-3 m.
Since both charges are experiencing the same force, eqn(1) can be written as
F=( kq^2)/r^2.
We can make q subject of the formula
q= √(Fr^2)/k
= √[(3.89 x 10^-21× (1.08 x 10^-3)^2]/8.99 x 10^9
q= 71.043×10^-20 C
Hence, the charge is 71.043×10^-20 C
From quantization law, the number of electron can be computed as
N=q/e
Where
N= number of electron
q= charges
=1.6×10^-19C
N=71.043×10^-20/1.6×10^-19
=4.44
Hence, the charge on the object is 71.043×10^-20 C and the number of electron is 4.44
In his novel From the Earth to the Moon (1866), Jules Verne describes a spaceship that is blasted out of 12,000 yards/s. the Columbiad is 900 ft long, but part of it is packed with poweder, so the spaceship accelerates over a distance of only 700 ft. Estimate the acceleration experienced by the occupants of the spaceship during launch. Give your answer in m/s2. (Verne realized that the "travelers would...encounter a violent recoil," but he probably didn't know that people generally lose consciousness if they experience accelerations greater than about 7g ~70 m/s2.)
Answer:
The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.
Explanation:
Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration ([tex]a[/tex]), measured in meters per square second, is estimated by this kinematic formula:
[tex]a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s }[/tex] (1)
Where:
[tex]\Delta s[/tex] - Travelled distance, measured in meters.
[tex]v_{o}[/tex], [tex]v[/tex] - Initial and final speeds of the spaceship, measured in meters.
If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v = 10968\,\frac{m}{s}[/tex] and [tex]\Delta s = 212.8\,m[/tex], then the acceleration experimented by the spaceship is:
[tex]a = \frac{\left(10968\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (212.8\,m)}[/tex]
[tex]a = 282652.782\,\frac{m}{s^{2}}[/tex]
The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.
describe why people are better off not consuming an additional good or service if the marginal cost is greater than the marginal benefit.
Ellie is shooting a free throw. At the instant the ball leaves her hand the ball is traveling at a velocity of 9 m/sec, at an angle above the horizontal of 29.5 degrees. If you are standing beneath the basketball rim, videotaping, you can only see the vertical motion of the ball, since its horizontal motion is straight towards you. What is the magnitude of the ball's initial vertical velocity (if you measured it using your video data, in m/sec)
Answer:
Vy = 4.43 m/s
Explanation:
given data
velocity = 9 m/sec
angle horizontal = 29.5 degrees
solution
we get here ball initial vertical velocity will be here
Vy = v sinθ ...........................1
put here value and we get
Vy = 9 × sin(29.5 )
Vy = 4.43 m/s
differentiate between computer and computer system
A computer is a programmable device that can automatically perform a sequence of calculations or other operations on data once programmed for the task. It can store, retrieve, and process data according to internal instructions. A computer may be either digital, analog, or hybrid, although most in operation today are digital. Digital computers express variables as numbers, usually in the binary system. They are used for general purposes, whereas analog computers are built for specific tasks, typically scientific or technical. The term "computer" is usually synonymous with a digital computers, and computers for business are exclusively digital.
Answer:
The core, computing part of a computer is its central processing unit (CPU), or processor. ... A computer system, therefore, is a computer combined with peripheral equipment and software so that it can perform desired functions.
Explanation:
Hope the answer was helpful
Mary is trying to pull Julie on a sled across a flat snowy field. Mary pulls on the rope attached to the sled. Her pulling force is directed horizontally. Julie weighs 109 pounds. The sled weights 12 pounds. If the coefficient of static friction between the sled runners and the snow is 0.42, how much force must Mary pull with (in lbs) to start moving the sled
Answer: F = 498.04 lbs
Explanation: The forces acting on the sled and Julie are show in the figure below. In it, we notice that, for the sled and Julie to go accross the field, they only need force of friction, because, force of friction is a force that resists the relative motion of surfaces.
Force of friction is given by the formula
[tex]F_{f}=\mu.F_{N}[/tex]
where
μ is coefficient of friction
[tex]F_{N}[/tex] is normal force
Normal force is the force the surface exerts on the object. It is always perpendicular and a force of contact.
In the case of the sled, since it is on a horizontal plane, Normal Force has the same magnitude of Gravitational Force. So
[tex]F_{N}=m.g[/tex]
Coefficient of friction is how much friction exists between two surfaces.
Rearraging friction force is
[tex]F_{f}=\mu.m.g[/tex]
Mass for this system is the sum of Julie and the sled, therefore
m = 109 + 12
m = 121 lb
Calculating Friction Force:
[tex]F_{f}=0.42.121.9.8[/tex]
[tex]F_{f}=[/tex] 498.04 lbs
LBS is a unit of measurement referred as pound by weight.
In conclusion, force Mary needs to start moving the sled is 498.04 lbs
A trolley of mass 5.0 kg is moving at 1.0 ms to the right. A constant force of 25 N acts to the left for 0.75 seconds.
Calculate the change of kinetic energy of the trolley.
(Show Work)
Answer:
change in kinetic energy of the trolley is 53.91 J.
Explanation:
mass of the trolley, m = 5.0 kg
initial velocity of the trolley, u = 1.0 m/s
external force on the trolley, F = 25 N
time of force action, t = 0.75 s
The final velocity of the trolley at the end of 0.75 s is calculated as follows;
[tex]F = \frac{m(v-u)}{t} \\\\25 = \frac{5(v-1)}{0.75}\\\\5(v-1) = 18.75\\\\v-1 = \frac{18.75}{5} \\\\v-1 = 3.75\\\\v = 4.75 \ m/s \ in \ the \ direction \ of \ the \ applied \ force[/tex]
The change in kinetic energy of the trolley is calculated as;
Δ K.E = ¹/₂m(v² - u²)
Δ K.E = ¹/₂ x 5(4.75² - 1²)
Δ K.E = 53.91 J.
Therefore, change in kinetic energy of the trolley is 53.91 J.
effects of heat on matter
Answer:
it can melt orcan put them past their boiling point
Explanation:
He throws a second ball (B2) upward with the same initial velocity at the instant that the first ball is at the ceiling. c. How long after the second ball is thrown do the two balls pass each other? d. When the balls pass each other how far are they above the juggler’s hands? e. When they pass each other what are their velocities?
Answer:
hello your question has some missing parts
A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.
answer : c) 0.39 sec
d) 2.25 m
e) 1.92 m/sec
Explanation:
The initial velocity of the first ball = 7.67 m/sec ( calculated )
Time required for first ball to reach ceiling = 0.78 secs ( calculated )
Determine how long after the second ball is thrown do the two balls pass each other
Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 = 9.8t^2 / 2
hence d = 4.9t^2 ----- ( 1 )
Initial speed of second ball = first ball initial speed = 7.67 m/sec
3 - d = 7.67t - 4.9t ---- ( 2 )
equating equation 1 and 2
3 = 7.67t therefore t = 0.39 sec
Determine how far the balls are above the Juggler's hands ( when the balls pass each other )
form equation 1 ;
d = 4.9 t^2 = 4.9 *(0.39)^2 = 0.75 m
therefore the height the balls are above the Juggler's hands is
3 - d = 3 - 0.75 = 2.25 m
determine their velocities when the pass each other
velocity = displacement / time
velocity = d / t = 0.75 / 0.39 sec = 1.92 m/sec
Anyone can help me out with this question ? Just number 2,
Answer:
- 21⁰C .
Explanation:
Speed of jet = 2.05 x 10³ km /h
= 2050 x 1000 / (60 x 60 ) m /s
= 569.44 m / s
Mach no represents times of speed of sound , the speed of jet
1.79 x speed of sound = 569.44
speed of sound = 318.12 m /s
speed of sound at 20⁰C = 343 m /s
Difference = 343 - 318.12 = 24.88⁰C
We know that 1 ⁰C change in temperature changes speed of sound
by .61 m /s
So a change in speed of 24.88 will be produced by a change in temperature of
24.88 / .61
= 41⁰C
temperature = 20 - 41 = - 21⁰C .
Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude of the electric field strength between them, if the potential 7.35 cm from the zero volt plate (and 2.65 cm from the other) is 533 V
Answer:
[tex]E=6.8Kv/m[/tex]
Explanation:
From the question we are told that
Distance b/w plate [tex]d=10cm=>0.1m[/tex]
P_1 Potential at 7.35 [tex]V=533v[/tex]
Generally the equation for electric field at a distance is mathematically given as
[tex]E=\frac{v}{d}[/tex]
[tex]E=\frac{533}{7.85*10^-^2}[/tex]
[tex]E=6789.808917[/tex]
[tex]E=6.8*10^3[/tex]
[tex]E=6.8Kv/m[/tex]
The height above the ground of a stone thrown upwards is given by s(t), where t is measured in seconds. After 1secondnothing,the height of the stone is 47feet above the ground, and after 1.5seconds, the height of the stone is 54feet above the ground. Evaluate s(1)and s(1.5),and then find the average velocity of the stone over the time interval [1,1.5].
Answer:
v = 14 ft/s
Explanation:
By definition, the average velocity, is just the rate of change of the position, with respect to time, which can be written as follows:[tex]v_{avg} = \frac{\Delta x}{\Delta t} = \frac{x_{f}-x_{o}}{t_{f} - t_{o}} (1)[/tex]
Defining the vertical position as the y-coordinate, with the origin at ground level, and the upward direction as positive, we can write (1) as follows:[tex]v_{avg} = \frac{\Delta y}{\Delta t} = \frac{y_{f}-y_{o}}{t_{f} - t_{o}} (2)[/tex]
where yf = 54 ft, y₀ = 47 ft, tif = 1.5 s, t₀ = 1s.Replacing in (2) we get:[tex]v_{avg} = \frac{\Delta y}{\Delta t} = \frac{y_{f}-y_{o}}{t_{f} - t_{o}} = \frac{7ft}{0.5s} = 14 ft/s (3)[/tex]
1.18. Which of the following is/are supplementary unit(s)? (1) Kelvin
(II) Newton (III) Second (IV) Radian
A. I and III only
C. I and II only
B. IV only
D. I, II and IV only
Answer:
B. IV only
Explanation:
At baseball practice, Mason and Alfredo both picked up the same bat and neither would let go until one of them had it for himself. Mason pulled the bat with
force of 15 newtons (N) while Alfredo pulled with a force of 20 newtons (N). Why did Alfredo end up with the bat?
A because the force was 5 N in Mason's direction
B. O because the net force was 5 N in Alfredo's direction
c. O because the net force was 15 N in Mason's direction
D.O because the net force was 20 N in Alfredo's direction
Answer:
Option B. O because the net force was 5 N in Alfredo's direction
Explanation:
To know the the correct answer to the question given above, we shall determine the net force acting on the bat. This can be obtained as follow:
Force of pull by Mason (Fₘ) = 15 N
Force of pull by Alfredo (Fₐ) = 20 N
Net force (Fₙ) =?
Fₙ = 20 – 15
Fₙ = 5 N in Alfredo's direction
From the calculation made above, we can see that the net force is 5N in Alfredo's direction. This is the reason why Alfredo end up having the bat.
Suppose that 6 J of work is needed to stretch a spring from its natural length of 26 cm to a length of 39 cm. (a) How much work (in J) is needed to stretch the spring from 30 cm to 35 cm
Answer:
Workdone = 0.89 Joules
Explanation:
Given the following data;
Workdone = 6J
Extension = 39 - 26 = 13cm to meters = 13/100 = 0.13m
The workdone to stretch a string is given by the formula;
Workdone = ½ke²
Where;
k is the constant of elasticity.
e is the extension of the string.
We would solve for string constant, k;
6 = ½*k*0.13²
6 = ½*k*0.0169
Cross-multiplying, we have;
12 = 0.0169k
k = 12/0.0169
k = 710.06
a. To find the workdone when e = 30, 35.
Extension = 35 - 30 = 5 to meters = 5/100 = 0.05m
Workdone = ½*710.06*0.05²
Workdone = 355.03*0.0025
Workdone = 0.89 Joules
Therefore, the amount of work (in J) needed to stretch the spring from 30 cm to 35 cm is 0.89.
) A 1000-nF capacitor with circular parallel plates with a radius of 1cm is accumulating charge at the rate of 52 mC/s at some point in time. What will be the induced magnetic field strength if you are positioned 20 cm radially outward from the center of the plates
Answer:
[tex]B=5.2*10^-^8T[/tex]
Explanation:
From the question we are told that
Capacitor [tex]c=1000nf[/tex]
Radius [tex]r=1cm =>0.001m[/tex]
Charge rate [tex]Q/t=52mC/s[/tex]
Distance [tex]d=20cm =0.2m[/tex]
Generally the rate of charge I is mathematically given as
[tex]I=\frac{dQ}{dt}[/tex]
[tex]I=52Cm/s[/tex]
[tex]I=52*10^-^3C[/tex]
Generally the the magnetic field intensity at distance d is mathematically given as
[tex]B=\frac{\mu I}{2\pi d}[/tex]
[tex]B=\frac{(4 * \pi *10^-^7)(52*10^-^3C)}{2\pi (0.2)}[/tex]
[tex]B=5.2*10^-^8T[/tex]