(A). The number of Coulombs in one fundamental charge of an electron is,
[tex]1e\text{ = 1.}6\times10^{-19}\text{ C}[/tex](B). The charge in terms of the number of electrons is,
[tex]q=ne[/tex]where q is the net charge, e is the fundamental charge and n is the number of electrons,
For the net charge of q = 1 Coulomb, the number of electrons is,
[tex]\begin{gathered} 1\text{ C=n}\times1.6\times10^{-19}\text{ C} \\ n=\frac{1}{1.6\times10^{-19}} \\ n=0.625\times10^{19} \\ n=6.25\times10^{18}\text{ } \end{gathered}[/tex]Thus, the number of electrons in the 1 Coulomb of charge is,
[tex]6.25\times10^{18}[/tex]
One coulomb (1 C) is equal to approximately 6.24 x 10 18 elementary charges. Thus, an elementary charge is approximately 1.60 x 10 -19 C.
The unit of electrical charge quantity in the International System of Units is the coulomb. Coulomb is the SI unit of electric charge which is equal to the amount of charge transported by a current of one ampere in one second. It can be also, a property of a matter due to which electrical and magnetic effects are produced. It is denoted by C. Mathematically, 1 Coulomb = 1 Ampere × 1 second.
To illustrate the magnitude of 1 Coulomb, an object would need an excess of 6.25 x 1018 electrons to have a total charge of -1 C. And of course, an object with a shortage of 6.25 x 1018 electrons would have a total charge of +1 C. The charge on a single electron is -1.6 x 10 -19 Coulomb.
Charge on an electron(e)= 1.6∗10^−19 C approximately.
By quantization property of charge(Smallest unit is e),
Charge (Q) = 1C where n-number of electrons.
Electron (e) = 1.6× 10^-19
A number of electrons (n) =?
Q= ne
1 C = n × 1.6×10^-19 C
n = 1/(1.6×10^-19)
Which gives, n= 6.25*10^18 electrons
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what force does the ground exert on a 1,000 kg car as it moves at 15 m/s through a dip in the road with a path radius of 30 m?
Answer:
Science; Physics; Physics questions and answers; 3. What force does the ground exert on a 1,000 kg car as it moves at 15 m/s through a dip in the road with a path radius of 30 m.
Explanation:
Glade 2 help :)
A person pulls a block 3 m along a horizontal surface by a constant force F=20 N. Determine the work done by force F acting on the block.
Take into account that the work done is given by the following formula:
[tex]W=F\cdot d[/tex]where,
F: applied force = 20N
d: distance = 3m
Replace the previous values into the formula for W:
[tex]W=(20N)(3m)=60J[/tex]Hence, the work done by the person is 60J
A bowling ball of mass 7.29 kg and radius 11.0 cm rolls without slipping down a lane at 3.00 m/s. Calculate the total kinetic energy.
We have the next information
m=7.29 kg
v=3 m/s
r=11cm=0.11m
We can find the kinetic energy using the next formula
[tex]KE=\frac{1}{2}mv^2+\frac{1}{2}Iw^2[/tex]I is the moment of inertia
w is the angular velocity
First, we need to calculate the moment of inertia
[tex]I=\frac{2}{5}mr^2[/tex][tex]I=\frac{2}{5}(7.29)(0.11)^2=0.0353kgm^2[/tex]Then for the angular velocity
[tex]\omega=\frac{v}{r}[/tex][tex]\omega=\frac{3}{0.11}=27.28[/tex]then we will substitute the values in the kinetic energy formula
[tex]KE=\frac{1}{2}(7.29)(3)^2+\frac{1}{2}(0.0353)(27.28)^2[/tex][tex]KE=45.9J[/tex]The total kinetic energy is 45.9 J
What is the resistance of an electric frying pan that draws 12 amperes of current when connected to 120 Volt circuit
10 ohms
Explanation
the resistance of an electric circuit is given by:
[tex]\begin{gathered} R=\frac{V}{I} \\ where\text{ R is the resitance} \\ Vi\text{s the voltage} \\ I\text{ is the current} \end{gathered}[/tex]so
Step 1
a) let
[tex]\begin{gathered} R=R \\ V=120\text{ volts} \\ I=12\text{ Amperes} \end{gathered}[/tex]b) now, replace in the expression
[tex]\begin{gathered} R=\frac{V}{I} \\ R=\frac{120\text{ volt}}{12\text{ Amp}} \\ R=10\text{ ohms} \end{gathered}[/tex]therefore, the answer is
10 ohms
I hope this helps you
A baseball is hit from theground straight up into the airwith a speed of 14.6 m/s. Howlong is the ball in air (time inseconds from ground-to-ground)?
Givens.
• Initial speed = 14.6 m/s.
,• FInal speed = 0 m/s (at highest point).
,• Gravity = 9.8 m/s^2.
First, find the time needed to reach the highest point.
[tex]\begin{gathered} v_f=v_0+gt \\ t=\frac{v_f-v_0}{g} \\ t=\frac{0-14.6\cdot\frac{m}{s}}{-9.8\cdot\frac{m}{s^2}} \\ t\approx1.49\sec \end{gathered}[/tex]It takes 1.49 seconds to reach the highest point.
The time that the baseball takes to reach the ground is double t because the trajectory is symmetrical, that is, it takes the same time to go from ground level to highest point than from highest point to ground level.
[tex]t_{\text{total}}=2\cdot1.49\sec =2.98\sec [/tex]Therefore, the baseball is 2.98 seconds in the air.
determine the total force and absolute pressure on the bottom of a swimming pool 28m by 8.5m whose uniform depth is 1.8. what will be the pressure against the side of the pool near the bottom
Answer:
Explanation:
Let us draw the pool for our visualization:
Pressure is defined as Force / area.
Now, what is the total force on the bottom of the pool?
It will be equal to the force due to the atmospheric pressure + force due to the mass of the water.
If we call p the density of water, then the force is
[tex]F=mg=\text{pVg}[/tex]where V = volume of the pool.
Therefore, the pressure due to water is
[tex]P=\frac{\rho Vg}{A}[/tex]Here remember that V = abh and A = ab; therefore, the above gives
[tex]P=\frac{\rho(abh)g}{ab}[/tex][tex]\boxed{P=\rho gh\text{.}}[/tex]adding the atmospheric pressure gives
[tex]\boxed{P=\rho gh+P_{\text{atm}}}[/tex]This is the pressure exerted on the bottom of the pool. But what about the sides?
Now near the bottom of the pool, the pressure exerted on the walls will be about the same as that exerted on the bottom of the pool. Therefore, we can use the above equation to find the pressure on the walls near the bottom.
SInce p = 1000kg / m^3, g = 9.8 m s^2, P_atm = 101325 Pascals, and h = 1.8 m, the equation gives
[tex]P=(\frac{1000\operatorname{kg}}{m^3})\times(\frac{9.8m}{s^2})\times(1.8m)+101325(N/m)[/tex]which evaluates to give
[tex]P=118965(N/m)[/tex]or in scientific notation, the above is
[tex]\boxed{P=1.19\times10^5(N/m)}[/tex]which is our answer!
Note that we also added to atmospheric pressure. This pressure will of course be canceled with the outside atmospheric pressure, leaving only the pressure due to water in effect.
Magnetic field lines of force will not go through ______A) your hand B) a tree C) a steel plate D) a paper plate E) water
Magnetic field lines can pass through metals and non metals. But it will not pass through super conductors. A steel plate is a superconductor. Thus, the correct option is
C) a steel plate
The rumble feature on a video game controller is driven by a device that turns electrical energy into mechanical energy. This device is best referred to as _________?A. an electric generatorB. an electromagnetC. a solenoidD. a motor
The rumble feature on a video game controller is driven that turns electrical energy into mechanical energy. This device is best referred to as a motor.
An electric motor converts electrical energy into mechanical energy.
Thus, the option (D) is correct.
A 3.10-kg block is moving to the right at 2.60 m/s just before it strikes and sticks to a 1.00-kg block initially at rest. What is the total momentum of the two blocks after the collision? Enter a positive answer if the total momentum is toward right and a negative answer if the total momentum is toward left.
Given,
The mass of the block moving to the right, M=3.10 kg
The speed of the block, u=2.60 m/s
The mass of the block at rest, m=1.00 kg
The total momentum of the two-block system before the collision is given by
[tex]p_i=Mu_{}[/tex]On substituting the known values,
[tex]\begin{gathered} p_i=3.10\times2.60 \\ =8.06\text{ kg}\cdot\frac{m}{s} \end{gathered}[/tex]From the law of conservation of energy, the total momentum of a system always remains the same.
Thus the momentum after the collision is equal to the total momentum of the blocks before the collision.
Thus the total momentum of the two blocks after the collision is 8.06 kg·m/s
Part 1)If a force of magnitude 125 N is exerted horizontally as shown, find the force exerted by the hammer claws on the nail. (Assume that the force the hammer exerts on the nail is parallel to the nail).Answer in units of N.Part 2)Find the force exerted by the surface on the point of contact with the hammer head. Assume that the force the hammer exerts on the nail is parallel to the nail. Answer in units of N.
Given data:
* The force acting on the hammer in the horizontal direction is F = 125 N.
* The distance of horizontal force from the point of contact is d = 28 cm.
* The angle of the nail with the vertical direction is,
[tex]\theta=26^{^{\circ}}[/tex]* The distance of the nail from the point of contact is x = 5.86 cm.
Solution:
The moment caused by the horizontal force on the hammer is equal to the moment caused by the horizontal component of force at the nail about the point of contact, thus,
[tex]F\times d=F^{\prime}x\text{ sin(}\theta)[/tex]Substituting the known values,
[tex]\begin{gathered} 125\times28=F^{\prime}\times5.86\times\sin (26^{\circ}) \\ 3500=F^{\prime}\times2.57 \\ F^{\prime}=\frac{3500}{2.57} \\ F^{\prime}=1361.87\text{ N} \end{gathered}[/tex]Thus, the force exerted on the nail is 1361.87 N or approximately 1362 N.
A hiker walks 14.91 m, N and 4.40 m, E. What is the magnitude of his resultant displacement?
Givens.
• 14.91 meters North.
,• 4.40 meters East.
First, make a diagram to visualize the vectors and the resultant displacement.
In the figure, the purple vector d represents the resultant displacement, which horizontal component is 4.40m and its vertical component is 14.91m.
Let's use the following formula to find the resultant.
[tex]d=\sqrt[]{(y_{})^2+(x)^2}[/tex]Where y = 14.91 and x = 4.40.
[tex]\begin{gathered} d=\sqrt[]{(14.91m)^2_{}+(4.40m)^2} \\ d=\sqrt[]{222.31m^2+19.36m^2} \\ d=\sqrt[]{241.67m^2} \\ d\approx15.55m \end{gathered}[/tex]Therefore, the magnitude of the resultant displacement is 15.55m.
But, the resultant displacement refers to the vector, which is the following
[tex]d=(4.4i+14.91j)m[/tex]15) A closed, uninsulated system was fitted with a movable piston. The addition of 754.3 J of heat caused the system to expand, doing 424.5 J of work in the process against a constant pressure. What is the value of E for this process?
Given,
Addition of heat, Q=+754.3 J
Work done , W=-424.5 J
The work is done against constant pressure.
To find
The value of E for this process
Explanation
According to the first law of thermodynamics,
[tex]\Delta E=Q+W[/tex]Putting the values,
[tex]\begin{gathered} \Delta E=754.3-424.5 \\ \Rightarrow\Delta E=329.8J \end{gathered}[/tex]Conclusion
The required value is 329.8 J
A particle moving along the x axis has a position given by x = (24t – 2.0t 3) m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero?
The velocity is defined by:
[tex]v=\frac{dx}{dt}[/tex]where x is the position of the particle and t is the time.
Plugging the position function given we have that the velocity is:
[tex]\begin{gathered} v=\frac{dx}{dt} \\ =\frac{d}{dt}(24t-2t^3) \\ =24-6t^2 \end{gathered}[/tex]Hence the velocity is given by the function:
[tex]v=24-6t^2[/tex]to determine the isntant when the velocity is zero we equate its expression to zero and solve for t:
[tex]\begin{gathered} 24-6t^2=0 \\ 6t^2=24 \\ t^2=\frac{24}{6} \\ t^2=4 \\ t=\pm\sqrt[]{4} \\ t=\pm2 \end{gathered}[/tex]Since time is always positive we conclude that the velocity is zero at t=2 s.
Now that we know at which instant the velocity is zero we need to remember that the acceleration is defined as:
[tex]a=\frac{dv}{dt}[/tex]then we have that:
[tex]\begin{gathered} \frac{dv}{dt}=\frac{d}{dt}(24-6t^2) \\ =-12t \end{gathered}[/tex]hence the acceleration is:
[tex]a=-12t[/tex]Plugging the value we found for the time we have that:
[tex]a(2)=-12(2)=-24[/tex]Therefore the acceleration of the particle when its velocity is zero is -24 meters per second per second.
what is the mass of a 2000 pound elephant in kg?
ANSWER:
907.2 kilograms
STEP-BY-STEP EXPLANATION:
To convert pounds to kilograms we must resort to the conversion factor between both units.
Where 1 pound is equal to 0.4536 kilograms.
Therefore:
[tex]2000\text{ pounds}\cdot\frac{0.4536\text{ kg}}{1\text{ pound}}=907.2\text{ kg}[/tex]The mass of the elephant is 907.2 kilograms
Suppose your on the bus and walking from the back to the front while the bus is moving at 10 mph. If you walk at 2 mph, what is the speed in mph?
Given data:
* The speed of the bus is 10 mph.
* The speed of the person in the bus is 2 mph.
Solution:
As the lecture hall is in state of rest with respect to Bus,
Thus, the speed of the person walking inside the bus with respect to the lecture hall is,
[tex]\begin{gathered} S=\text{speed of bus+spe}ed\text{ during walk} \\ S=10\text{ mph + 2 mph} \\ S=12\text{ mph} \end{gathered}[/tex]Thus, the speed relative to the lecture hall is 12 mph.
17. What distinguishes single, double, and triple
covalent bonds?
Answer:
The number of shared electrons is the major distinction between single double and triple bonds. A single bond is formed when two atoms share one pair of electrons, whereas a double bond is formed when two atoms share two pairs (four electrons). Three pairs of electrons (six atoms) are shared to form triple bonds.
Answer
In a single bond one pair of electrons is shared, with one electron being contributed from each of the atoms. Double bonds share two pairs of electrons and triple bonds share three pairs of electrons. Bonds sharing more than one pair of electrons are called multiple covalent bonds.
Explanation:
Rate at which an object with a direction?
Rate at which an object changes its direction is velocity.
So the answer is velocity.
True or False: Wave-particle duality describes how some small particles act like both waves and particles.
Answer:
False
Explanation:
The wave-particle duality says that every particle can act sometimes like a wave and sometimes like a particle. This applies to all particles, so the statement: wave-particle duality describes how some small particles act as both waves and particles is False.
3. Free Fall: A ball is dropped from a window that is 35 meters above gound level.How fast is the ball traveling when it reaches the ground?How long does the ball take to reach the ground?
We are given that a ball is dropped from a height of 35 meters in a free fall and we are asked to determine its final velocity. To do this we will use the following equation of motion:
[tex]v^2_f-v^2_0=2ah[/tex]Where:
[tex]\begin{gathered} v_f\text{ = final velocity} \\ v_0=\text{ initial velocity} \\ a=\text{ acceleration} \\ h=\text{ height} \end{gathered}[/tex]Since the ball is dropped this means that the initial velocity is zero. Also, since we have a free-fall motion the acceleration is the same as the acceleration of gravity. Replacing these values we get:
[tex]v^2_f=2gh[/tex]Now we solve for the velocity by taking the square root to both sides:
[tex]v_f=\sqrt[]{2gh}[/tex]Now we replace the values for the gravity and height:
[tex]v_f=\sqrt[]{2(9.8\frac{m}{s^2})(35m)}[/tex]Solving the operations:
[tex]v_f=26.19\frac{m}{s}[/tex]Therefore, the final velocity is 26.19 meters per second.
Now we will determine the time it takes the ball to reach the ground. To do that we will use the following equation of motion:
[tex]v_f=v_0+gt[/tex]Since the initial velocity is zero the equation simplifies to:
[tex]v_f=gt[/tex]Now we divide by the acceleration of gravity:
[tex]\frac{v_f}{g}=t[/tex]Now we replace the values:
[tex]\frac{26.19\frac{m}{s}}{9.8\frac{m}{s^2}}=t[/tex]Solving the operations we get:
[tex]2.67s=t[/tex]Therefore, it takes 2.67s for the ball to reach the ground in free-fall.
What is the height intercept?(b) What is the slope of the line?185.5 cm/hours(c) What is the height intercept?(d) Assuming Laura started with an empty pool at time = 0 hours, you would expect the height intercept to be zero. Unfortunately due to error in measurements theintercept was not zero. Use the 5% Rule to calculate the height intercept error. What is the the height intercept error?(e) What would you expect the height of the water to be after 14 hours?
C) the height intercept would be 0 given the pool has no water at the beggining.
D) To find the vertical axis intercept error, we will need to use the formula
[tex]Error=\lvert{\frac{vertical\text{ }Axis}{Largest\text{ }Value}}\rvert *100[/tex]A cube 6.0 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.5 cm in diameter all the way through and perpendicular to one face, you find that the cube weighs 6.30 N. 1. What is the density of the metal? 2. What did the cube weigh before you drilled the hole in it?
1.
First, let's calculate the volume of the cube:
[tex]V_1=6^3=216\text{ cm^^b3}[/tex]Now, let's calculate the volume of the cylinder drilled:
[tex]\begin{gathered} V_2=\frac{\pi d^2h}{4}\\ \\ V_2=\frac{3.14159\cdot2.5^2\cdot6}{4}\\ \\ V_2=29.45\text{ cm^^b3} \end{gathered}[/tex]So the volume of the cube after being drilled is:
[tex]\begin{gathered} V=V_1-V_2\\ \\ V=216-29.45\\ \\ V=186.55\text{ cm^^b3} \end{gathered}[/tex]If the weight is 6.3 N, let's find the mass:
[tex]\begin{gathered} W=m\cdot g\\ \\ 6.3=m\cdot9.8\\ \\ m=\frac{6.3}{9.8}\\ \\ m=0.643\text{ kg} \end{gathered}[/tex]And the density of the metal is:
[tex]\begin{gathered} d=\frac{m}{V}\\ \\ d=\frac{0.643}{186.55}\\ \\ d=0.0034468\text{ kg/cm^^b3}\\ \\ d=3.4468\text{ g/cm^^b3} \end{gathered}[/tex]2.
To find the weight of the cube before being drilled, let's use the following rule of three:
[tex]\begin{gathered} volume\rightarrow weight\\ \\ 216\text{ cm^^b3}\rightarrow x\text{ N}\\ \\ 186.55\text{ cm^^b3}\rightarrow6.3\text{ N}\\ \\ \\ \\ \frac{216}{186.55}=\frac{x}{6.3}\\ \\ x=\frac{216\cdot6.3}{186.55}\\ \\ x=7.295\text{ N} \end{gathered}[/tex]A car moves 25km north and then another 25km west after that it moves 40km north and then 20km west. Find the displacement
Given:
The distance traveled by car;
d₁=25 km to the north
d₂=25 km to the west.
d₃=40 km to the north.
d₄=20 km to the west.
To find:
The displacement of the car.
Explanation:
The displacement of an object is the shortest distance between its initial and final position.
Referring to the diagram, D is the displacement of the car.
The total displacement of the car is given by,
[tex]D=\sqrt{(d_1+d_3)^2+(d_2+d_4)^2}[/tex]On substituting the known values,
[tex]\begin{gathered} D=\sqrt{(25+40)^2+(25+20)^2} \\ =79.06\text{ m} \end{gathered}[/tex]Final answer:
Thus the total displacement of the car is 79.06 m
# 3. If a wire is connected to row 1, column a , and another to row 1, column c, are the two wires connected ?
In a breadboard, internal connections are made between groups of five holes in each row. That means that holes A, B, C, D, and E in row 1 are all connected, but for example, F and A are not connected.
Since 1A and 1C are in the same row and in the same five-hole group, they are connected.
Answer: The two wires are connected.
As a sports psychologist how would you suggest working with an athlete who suffers from self-
attention in front of the home crowd?
As a sports psychologist my suggestion would be to ensure that they maintain a balanced approach about others opinions and should concentrate on their performance than the opinion of other.
Many athletes are intimidated by or motivated by the support and encouragement of the home audience when competing in front of them. They might make an effort to get people's respect and favor. Although it can improve performance, it can also negatively increase an athlete's pressure to perform or even cause them to feel overconfident. Athletes could feel humiliated if they lose or fail because they couldn't satisfy the expectations of their audience. Additionally, they can lose self-respect, which might have an effect on their morale and future performance.
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A small electric motor produces a force of 3 N that moves a remote-control car 4 m every second. How much power does the motor produce?
f = force = 5N
v = velocity = 4m/s
p = power
p = force * velocity
Replacing:
p= 3N * 4 m/s = 12W
The motor produces 12W of power.
A person walks 15.0 m in 5.00 s and then
walks 12.0 m in 10.00 s. What is the
average speed of the person?
Answer:
Explanation:
Given:
D₁ = 15.0 m
t₁ = 5.00 s
D₂ = 12.0 m
t₂ = 10.00 s
___________
V - ?
The average speed of the person:
V =(D₁ +D₂) / (t₁ + t₂)
V =(15.0 + 12.0) / (5.00 + 10.00) = 27.0 / 15.00 ≈ 1.8 m/s
I have a question of a test that I already took and just want to know why my answer was incorrect, so when I spoke to the prevuios tutor and gave hime the question wuth the possible answer he asked me if it was for a graded test , and I said that it was a graded test and he said to report me , my question is a tutor can't help with an answer of a test that I took and my answer was incorrect, and I just want to understand why is it incorrect
The correct answer is (D)
Strong nuclear forces are responsible for holding together the nucleus of an atom; weak nuclear forces are involved when certain types of atoms break down.
Question 8 of 10Objects with the same charge will:O A. have no effect on each other.B. exert a pulling force on each other.O C. repel each other.O D. attract each other.SU
We will have that:
objects with the same charge will repel each other.
I have a homework problem that I don’t even know where to start or what formulas to use
We will have the following:
a) The velocity after 5 seconds will be:
[tex]v_f=v_o+at[/tex][tex]v=2.79m/s+(9.8m/s^2)(5s)\Rightarrow v=51.79m/s[/tex]So, the velocity after 5 seconds will be 51.79m/s.
b) We will have that the distance below the helicopter will be:
[tex]d=\frac{1}{2}(v_0+v_f)t[/tex][tex]d=\frac{1}{2}(2.79m/s+51.79m/s)(5s)\Rightarrow d=136.45m[/tex]So, it will be 136.45 m below the helicopter.
c) We will have that the velocity and distance given that the helicopter is moving constantly at 2.79m/s will be:
[tex]v=-2.79m/s+(9.8m/s^2)(5s)^2\Rightarrow v=46.21m/s[/tex]So, the velocity would be 46.21 m/s.
[tex]d=\frac{1}{2}(-2.79m/s+46.21m/s)(5s)\Rightarrow d=108.55m[/tex]So, the distance would be 108.55 m below the helicopter.
What does the strength of frictiondepend on?A. The direction of the forces.B. The types of surfaces and how hard theobjects are being pushed.C. The color of surfaces and how hard they push.D. Only how hard the objects are being pushed.
Answer:
B. The types of surfaces and how hard the objects are being pushed.
Explanation:
The force of friction is the result of the