Answer:
.
Explanation:
.
What is displacement?
a. The distance an object travels.
b. The distance between the starting point and the ending point of an object's
journey.
C. The amount of time it takes an object to travel to a destination.
d. The path in which an object travels.
Answer:
displacement is the distance between the starting point and the ending point of an object's journey
I don’t even understand anyone help please.
Answer:
a) A:170572.5 J
C: 55794.9J
b) 170572.5 J
c) 41.4413265306m
d) 2.7455874717m/s
Explanation:
a) Kinetic energy = 0.5*m*v²
KE at A = 0.5*420*28.5² = 170572.5 J
KE at C = 0.5*420*16.3² = 55794.9 J
b) Mechanical energy is the total kinetic energy plus potential energy at a point on the system. There is no potential energy at A.
ANSWER: 170572.5 J
c) v²=u²+2as
28.5²=2(9.8)s
812.25/19.6=s
s=41.4413265306m
d) h=height from part c, r=radius of loop
v²=u²+2as
v²=gr or a=v²/r
Ei=Ef
mgh=0.5mv²+mg(2r)
gh=0.5v²+2gr
h=0.5r+2r
h=5/2r
r=2/5h=(2/5)(41.4413265306)=16.5765306122
F=ma
mg=m(v²/r)
g=v²/r
v²=(9.8)(16.5765306122)
v=√162.45
=12.7455874717m/s
What are the major properties of the sun?
PLEASE HELP!!
Answer:
Explanation:
Properties of the sun
Introduction to Stars A. The Sun 1. ...
Layers of the sun (Atmosphere) a. Photosphere: Layer that emits the radiation we see. ...
Layers of the sun (Interior) a. ...
Solar Wind: a. ...
The sun in X-Rays: a. ...
Luminosity: The measure of the energy put out by the sun. ...
Luminosity continued: c. ...
Luminosity continued: e.
The heat of the Sun's surface is so large that neither solid nor liquid can reside there. The heat is the physical property of the sun.
What is the sun?The Sun is a billion star at the core of our universe, a hot. It is an incandescent ball of hydrogen and helium.
The Sun is millions of kilometers away from the land, and life is not possible without its energy.
The major properties of the sun are as follows;
1. The mass of the sun is 1.98892 x 10³⁰ kg.
2 The density of the sun is 1.622 x 10⁵ kg/m³
3. The surface gravity of the sun is 27.94 g.
4. The diameter is 1,391,000 kilometers.
5. The volume of the Sun is 1.412 x 1018 km³
The heat of the Sun's surface is so hot that neither solid nor liquid can reside there instead, the component materials are mostly gaseous atoms with a few molecules. As a result, no permanent surface exists.
Hence heat, density, and mass are the major properties of the sun.
To learn more about the sun refer to the link;
https://brainly.com/question/2526507
Like charges will exert a force of
a. positive
b. negative
c. attraction
d. repulsion
e. neutral
Answer:
D- Repulsion
Explanation:
A positively charged object will exert a repulsive force upon a second positively charged object.
calculate ine gravitational potential energy of the ball using pe=m×g×h.(use g=9.8 n/kg)
A 4.0-kilogram ball held 1.5 meters above the floor has ________ joules of potential energy
Answer:
58.8J
Explanation:
Given parameters;
Mass of ball = 4kg
Height above the floor = 1.5m
g = 9.8n/kg
Unknown:
Potential energy = ?
Solution:
The potential energy of a body is the energy due to the position of the body.
It is mathematically expressed as:
Potential energy = mass x acceleration due to gravity x height
Potential energy = 4 x 9.8 x 1.5 = 58.8J
25 points!
A 6 kg object accelerates from 5 m•s to 25 m•s in 30 seconds. What was the net force acting on the
object? Give your answer in Newtons to one significant figure and without a unit.
(Show Work)
Answer:
6N
Explanation:
Given parameters:
Mass of object = 6kg
Initial velocity = 5m/s
Final velocity = 25m/s
Time = 30s
Unknown:
Net force acting on the object = ?
Solution:
From Newton's second law of motion:
Force = mass x acceleration
Acceleration is the rate of change of velocity with time
Acceleration = [tex]\frac{Final velocity - Initial velocity }{time}[/tex]
Force = mass x [tex]\frac{Final velocity - Initial velocity }{time}[/tex]
So;
Force = 6 x [tex]\frac{25 - 5}{30}[/tex] = 6N
A car enters a 105-m radius flat curve on a rainy day when the coefficient of static friction between its tires and the road is 0.4. What is
the maximum speed which the car can travel around the curve without sliding
Static friction (magnitude Fs) keeps the car on the road, and is the only force acting on it parallel to the road. By Newton's second law,
Fs = m a = W a / g
(a = centripetal acceleration, m = mass, g = acceleration due to gravity)
We have
a = v ² / R
(v = tangential speed, R = radius of the curve)
so that
Fs = W v ² / (g R)
Solving for v gives
v = √(Fs g R / W)
Perpendicular to the road, the car is in equilibrium, so Newton's second law gives
N - W = 0
(N = normal force, W = weight)
so that
N = W
We're given a coefficient of static friction µ = 0.4, so
Fs = µ N = 0.4 W
Substitute this into the equation for v. The factors of W cancel, so we get
v = √((0.4 W) g R / W) = √(0.4 g R) = √(0.4 (9.80 m/s²) (105 m)) ≈ 20.3 m/s
a graduated cylinder.measures 15.3 mL. Convert this measurement to DaL
Answers:
A. 0.0153
B. 0.00153
C. 0.000153
D. 0.153
Answer:
0.000153DaL
Explanation:
We have been given:
15.3mL to convert to DaL
DaL is a unit of volume which indicates a decaliter.
This implies that;
1 Da L = 1 x 10²L
So:
1 mL = 1 x 10⁻³L
So 15.3mL will give 15.3 x 10⁻³L
So;
1 x 10²L = 1 DaL
15.3 x 10⁻³L will give [tex]\frac{15.3 x 10^{-3} }{1 x 10^{2} }[/tex] = 15.3 x 10⁻⁵DaL
Therefore, this is 0.000153DaL
An electric range has a constant current of 10 A entering the positive voltage terminal with a voltage of 110 V. The range is operated for two hours, (a) Find the charge in coulombs that passes through the range, (b) Find the power absorbed by the range, (c) If electric energy costs 12 cents per kilowatt-hour, determine the cost of operating the range for two hours.
Answer:
A. 72000 C
B. 1100 W
C. 26.4 cents.
Explanation:
From the question given above, the following data were obtained:
Current (I) = 10 A
Voltage (V) = 110 V
Time (t) = 2 h
A. Determination of the charge.
We'll begin by converting 2 h to seconds. This can be obtained as follow:
1 h = 3600 s
Therefore,
2 h = 2 h × 3600 s / 1 h
2 h = 7200 s
Thus, 2 h is equivalent to 7200 s.
Finally, we shall determine the charge. This can be obtained as follow:
Current (I) = 10 A
Time (t) = 7200 s
Charge (Q) =?
Q = It
Q = 10 × 7200
Q = 72000 C
B. Determination of the power.
Current (I) = 10 A
Voltage (V) = 110 V
Power (P) =?
P = IV
P = 10 × 110
P = 1100 W
C. Determination of the cost of operation.
We'll begin by converting 1100 W to KW. This can be obtained as follow:
1000 W = 1 KW
Therefore,
1100 W = 1100 W × 1 KW / 1000 W
1100 W = 1.1 KW
Thus, 1100 W is equivalent to 1.1 KW
Next, we shall determine the energy consumption of the range. This can be obtained as follow:
Power (P) = 1.1 KW
Time (t) = 2 h
Energy (E) =?
E = Pt
E = 1.1 × 2
E = 2.2 KWh
Finally, we shall determine the cost of operation. This can be obtained as follow:
1 KWh cost 12 cents.
Therefore, 2.2 KWh will cost = 2.2 × 12
= 26.4 cents.
Thus, the cost of operating the range for 2 h is 26.4 cents.
If a cyclist travels 30 km in 2 h, What is her average speed?
Answer:
15km/h
Explanation:
→ Speed = Distance ÷ Time
30 ÷ 2 = 15km/h
A circus tightrope walker weighing 800 N is standing in the middle of a 15 meter long cable stretched between two posts. The cable was originally horizontal. The lowest point of the cable is now at his feet and is 30 cm below the horizontal. Assume the cable is massless. What is the tension in the cable
Answer:
T = 10010 N
Explanation:
To solve this problem we must use the translational equilibrium relation, let's set a reference frame
X axis
Fₓ-Fₓ = 0
Fₓ = Fₓ
whereby the horizontal components of the tension in the cable cancel
Y Axis
[tex]F_{y} + F_{y} - W =0[/tex]
2[tex]F_{y}[/tex] = W
let's use trigonometry to find the angles
tan θ = y / x
θ = tan⁻¹ (0.30 / 0.50 L)
θ = tan⁻¹ (0.30 / 0.50 15)
θ = 2.29º
the components of stress are
F_{y} = T sin θ
we substitute
2 T sin θ = W
T = W / 2sin θ
T = [tex]\frac{ 800}{ 2sin 2.29}[/tex]
T = 10010 N
While investigating Kirchhoff's Laws, you begin observing a blackbody, such as a star, from Earth using advanced technology that can analyze spectra. While pointing it at the star with nothing between you and the star, you observe a full spectrum. You come back and repeat this same experiment a year later using the same star, except this time you observe an absorption spectrum. What is the most likely explanation for this
Answer:
the second time there is a gas between you and the star,
Explanation:
When you observe the star for the first time you do not have a given between you and the star, therefore you observe the emission spectrum of the same that is formed by lines of different intensity and position that indicate the type and percentage of the atoms that make up the star.
When you observe the same phenomenon for the second time there is a gas between you and the star, this gas absorbs the wavelengths of the star that has the same energies and the atomisms and molecular gas, therefore these lines are not observed by seeing a series of dark bands,
The information obtained from the two spectra is the same, the type of atoms that make up the star
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]
When monochromatic light passes through the interface between two unknown materials at an angle θ where 0∘<θ<90∘, no changes in the direction of propagation of light are observed. What can be said about the two materials? Check all that apply. View Available Hint(s) Check all that apply. The two materials have matching indexes of refraction. The second material through which light propagates has a lower index of refraction. The second material through which light propagates has a higher index of refraction. The two materials are identical.
Answer:
the correct one is the first, the refractive index of the two materials must be the same
Explanation:
When a beam of light passes through two materials, it must comply with the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
where n₁ and n₂ are the refractive indices of each medium.
In this case, it indicates that the light does not change direction, so the input and output angle of the interface must be the same,
θ₁ = θ₂ = θ
substituting
n₁ = n₂
therefore the refractive index of the two materials must be the same
When reviewing the answers, the correct one is the first
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 4.0 rev/s in 9.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 15.0 s. Through how many revolutions does the tub turn during this 24 s interval
Answer:
48 rev
Explanation:
a) we can calculate the distance covered by the tube using the formula:
θ = (ω + ωo)t/2
where ω is the final angular speed, θ is the distance covered, t is the time taken, ωo is the initial angular speed.
Firstly, we calculate the distance covered from 0 to 9 s then from 9s to 24 s.
within 9s, the tub runs from rest (0) to 4 rev/s, hence:
t = 9s, wo = 0, w = 4 rev/s = (4 * 2π) rad/s = 8π rad/s. Hence:
θ = (ω + ωo)t/2 = (0 + 8π)9 / 2 = 36π rad
θ = 36π rad = (36π)/2π rev = 18 rev
Also, within 15 s, the tub runs from 4 rev/s to rest, hence:
t = 15 s, wo = 4 rev/s = 8π rad/s, w = 0 rad/s. Hence:
θ = (ω + ωo)t/2 = (8π + 0)15 / 2 = 60π rad
θ = 60π rad = (60π)/2π rev = 30 rev
Therefore the total revolutions by the tube during 24 s interval = 30 rev + 18 rev = 48 rev
A student is provided with a battery-powered toy car that the manufacturer claims will always operate at a constant speed. The student must design an experiment in order to test the validity of the claim. Which of the following measuring tools can the student use to test the validity of the claim?
a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on.
b. A meterstick to measure the distance of the track that the car travels on.
c. A motion detector that is oriented perpendicular to the direction that the car travels.
d. A mass balance to determine the mass of the car
Answer:
a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on.
b. A meterstick to measure the distance of the track that the car travels on.
Explanation:
Physics can be defined as the field or branch of science that typically deals with nature and properties of matter, motion and energy with respect to space, force and time.
In this scenario, a student is provided with a battery-powered toy car that the manufacturer claims will always operate at a constant speed. The student must design an experiment in order to test the validity of the claim.
Therefore, to test the validity of the claim, the student should use the following measuring tools;
a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on. This device is typically used to measure time with respect to the rate of change of the interruption or block of an infra-red beam.
b. A meterstick to measure the distance of the track that the car travels on.
Hence, with these two devices the student can effectively measure or determine the validity of the claim.
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
2.19 The drag characteristics of a blimp traveling at 4 m/s are to be studied by experiments in a water tunnel. The prototype is 20 m in diameter and 110 m long. The model is one-twentieth scale. What velocity must the model have for dynamic similarity
Answer:
[tex]Vm=0.894m/s[/tex]
Explanation:
From the question we are told that
Velocity if travel [tex]v=4m/s[/tex]
Diameter of prototype [tex]d_1=20m[/tex] and [tex]d_2=110m[/tex]
Scale ratio=[tex]\frac{1}{20}[/tex]
Generally Velocity of of the model using Froud's model is mathematically given as
[tex]Fm=Fp[/tex]
[tex]\frac{Vm}{\sqrt{Lmg}} =\frac{Vp}{\sqrt{Lpg}}[/tex]
[tex]Vm=Vp*\frac{Vp}{\sqrt{Lpg} }[/tex]
[tex]Vm=4*\frac{1}{\sqrt{20}}[/tex]
[tex]Vm=0.894m/s[/tex]
Give an example of mass making a difference in the amount of gravitational energy. Tell how you know the gravitational energy is different and your example
Please help due today!!
Answer:
The Gravitational potential energy at large distances is directly proportional to the masses and inversely proportional to the distance between them. The gravitational potential energy increases as r increases.
Examples of Gravitational Energy
A raised weight.
Water that is behind a dam.
A car that is parked at the top of a hill.
A yoyo before it is released.
What is the period of an objects motion?
A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water-ice mixture. Fine thermocouple wires suspend the sphere, and the temperature is observed to change from 15 to 14C in 6.35 s. The experiment is repeated with a metallic sphere of the same diameter, but of unknown composition with a mass of 1.263 kg. If the same observed temperature change occurs in 4.59 s, what is the specific heat of the unknown material
Answer:
The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.
Explanation:
Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ([tex]\dot Q[/tex]), measured in watts, that is, joules per second, by the following formula:
[tex]\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}[/tex] (1)
Where:
[tex]m[/tex] - Mass of the sphere, measured in kilograms.
[tex]c[/tex] - Specific heat of the material, measured in joules per kilogram-degree Celsius.
[tex]T_{o}[/tex], [tex]T_{f}[/tex] - Initial and final temperatures of the sphere, measured in degrees Celsius.
[tex]\Delta t[/tex] - Time, measured in seconds.
In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:
[tex]\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}[/tex] (2)
Where:
[tex]m_{I}[/tex], [tex]m_{X}[/tex] - Masses of the iron and unknown spheres, measured in kilograms.
[tex]\Delta t_{I}[/tex], [tex]\Delta t_{X}[/tex] - Times of the iron and unknown spheres, measured in seconds.
[tex]c_{I}[/tex], [tex]c_{X}[/tex] - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.
[tex]c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}[/tex]
If we know that [tex]\Delta t_{I} = 6.35\,s[/tex], [tex]\Delta t_{X} = 4.59\,s[/tex], [tex]m_{I} = 0.515\,kg[/tex], [tex]m_{X} = 1.263\,kg[/tex] and [tex]c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the specific heat of the unknown material is:
[tex]c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)[/tex]
[tex]c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}[/tex]
Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.
1. (6x + 8)(5x - 8)
a. 30x2 + 49x + 20
2. (5x + 6(5x - 5)
b. 24x3 + 8x2 + 6x + 4
3. (6x + 3)(6x - 4)
c. 25x2 + 5x - 30
4. (6x + 5)(5x + 5)
d. 30x2 - 8x - 64
e. 36x2 - 6x - 1
5. (4x + 2) (6x2 - x + 2)
Answer:
form 1 question??????????
On a low-friction track, a 0.66-kg cart initially going at 1.85 m/s to the right collides with a cart of unknown inertia initially going at 2.17 m/s to the left. After the collision, the 0.66-kg cart is going at 1.32 m/s to the left, and the cart of unknown inertia is going at 3.22 m/s to the right. The collision takes 0.010 s.
What is the unknown inertia?
What is the average acceleration of the heavier cart?
What is the average acceleration of the lighter cart?
Answer:
(a) the unknown inertia is 0.388 kg
(b) the average acceleration of the heavier cart is 317 m/s²
(c) the average acceleration of the lighter cart is 539 m/s²
Explanation:
Given;
mass of the first cart, m₁ = 0.66 kg
initial speed of the first cart, u₁ = 1.85 m/s
let the mass of the cart with unknown inertia be m₂
initial velocity of the second cart, u₂ = 2.17 m/s to the left
velocity of the first cart after collision, v₁ = 1.32 m/s to the left
velocity of the second cart after collision, v₂ = 3.22 m/s
time of collision, t = 0.010 s
(a) What is the unknown inertia?
Apply the principle of conservation of linear momentum, to determine the unknown inertia.
let leftward direction be negative direction
let rightward direction be positive direction
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.66(1.85) + m₂(-2.17) = 0.66(-1.32) + m₂(3.22)
1.221 - 2.17m₂ = -0.8712 + 3.22m₂
1.221 + 0.8712 = 3.22m₂ + 2.17m₂
2.0922 = 5.39m₂
m₂ = 2.0922 / 5.39
m₂ = 0.388 kg
The unknown inertia is 0.388 kg
(b) the average acceleration of the heavier cart
the heavier cart has a mass of 0.66 kg
[tex]a = \frac{v_1 - u_1}{t} \\\\a = \frac{-1.32 - 1.85}{0.01} \\\\a = -317 \ m/s^2\\\\|a| = 317 \ m/s^2[/tex]
(c) the average acceleration of the lighter cart;
the lighter cart has a mass of 0.388 kg
[tex]a = \frac{v_2 - u_2}{t} \\\\a = \frac{3.22 - (-2.17)}{0.01} \\\\a =\frac{3.22 \ +\ 2.17}{0.01} \\\\a= 539\ m/s^2[/tex]
7. If the impact of the golf club on the ball in the previous question occurs over a time of 2 x 10 seconds, what
force does the ball experience to accelerate from rest to 73 m/s?
Answer:
3.65 x mass
Explanation:
Given parameters:
Time = 20s
Initial velocity = 0m/s
Final velocity = 73m/s
Unknown:
Force the ball experience = ?
Solution:
To solve this problem, we apply the equation from newton's second law of motion:
F = m [tex]\frac{v - u}{t}[/tex]
m is the mass
v is the final velocity
u is the initial velocity
t is the time taken
So;
F = m ([tex]\frac{73 - 0}{20}[/tex] ) = 3.65 x mass
Fill in the blank with the correct word below (from the reading_):
helps you track your progress once you have made a lifestyle
change.
Self-monitoring
Healthy food
Regular xxercise
Goals
Answer:I think it’s self monitoring sorry if wrong
Explanation:
Answer:
It self monitoring
Explanation:
I took the test
state four law of photoelectric effect
Answer:
LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation.
---------------------------------------------
LAW 2: For a given metal, there exists a certain frequency below which the photoelectric emission does not take place. This frequency is called threshold frequency.
-----------------------------------------------
LAW 3: For a frequency greater than the threshold frequency, the kinetic energy of photoelectrons is dependent upon frequency or wavelength but not on the intensity of light.
-----------------------------------------------
LAW 4: Photoelectric emission is an instantaneous process. The time lag between incidence of radiations and emission of electron is 10^-9 seconds.
Explanation:
Answer:
LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation. ... LAW 4: Photoelectric emission is an instantaneous process.
A violin has a string of length
0.320 m, and transmits waves at
622 m/s. At what frequency does
it oscillate?
Answer:
1.9kHz
Explanation:
Given data
wavelength [tex]\lambda= 0.32m[/tex]
velocity [tex]v= 622 m/s[/tex]
We know that
[tex]v= f* \lambda\\\\f= v/ \lambda[/tex]
substitute
[tex]f= 622/ 0.32\\\\f= 1943.75\\\\f= 1.9kHz[/tex]
Hence the frequency is 1.9kHz
Answer:
971.2
Explanation:
It was right on acellus :)
A particle with charge Q and mass M has instantaneous speed uy when it is at a position where the electric potential is V. At a later time, the particle has moved a distance R away to a position where the electric potential is V2 ) Which of the following equations can be used to find the speed uz of the particle at the new position?
a. 1/2M(μ2^2-μ1^2)=Q (v1-v2)
b. 1/2M(μ2^2-μ1^2)^2=Q(v1-v2)
c. 1/2Mμ2^2=Qv1
d. 1/2Mμ2^2=1/4πx0 (Q^2/R)
Answer:
A
Explanation:
Ke = 1/2 MV^2
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
The pickup truck has a changing velocity because the pickup truck
A.can accelerate faster than the other two vehicles
B.is traveling in the opposite direction from the other two vehicles
C.is traveling on a curve in the road
D.needs a large amount of force to move
please get right i need awnser today
Answer:
C. Is traveling on a curve in the road
Hope this helps :3
The pick up truck has a changing velocity because, it is travelling on a curve in the road. A change in direction results in its change in velocity because, velocity is a vector quantity.
What is velocity ?Velocity is a physical quantity that measures the distance covered by an object per unit time. It is a vector quantity, thus having magnitude as well direction.
The rate of change in velocity is called acceleration of the object. Like velocity, acceleration also is a vector quantity. Thus, a change in magnitude or direction or change in both for velocity make the object to accelerate.
Here, all the three vehicles are travelling with the same velocity. But, the truck is moving to a curve on the road. The curvature in the path will make a change in its velocity.
Find more on velocity:
https://brainly.com/question/16379705
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