You are hired to plan stunts for the next James Bond movie. In one of the film's opening scenes, Bond is going to evade some goons by stepping off a building and landing in the bed of a moving garbage truck. From the top of the building, he will fall 15 meters straight down before landing in the truck, which will be driving by with a constant speed of 22 m/s.
Part A. How fast will the stuntman be moving the moment he hits the truck?Part B. How far from the building should the truck be when you give the stuntman the signal to drop off the ledge?
Answer:
Explanation:
height of the building h = 15 m
initial vertical velocity u = 0
v² = u² + 2 g h
u = 0 , final velocity after falling height h
h = 15
v² = 2 x 9.8 x 15
v = 17.14 m /s
. Time of fall be t
h = 1/2 g t²
15 = 1/2 x 9.8 x t²
t = 1.749 s
This time will be required for truck to come under the building .
speed = distance / time
distance = speed x time
= 22 x 1.749
38.5 m
Truck should be at a distance of 38.5 m . when the stuntman starts falling .
Part A: The time required to hit the truck by a stuntman is 1.749 s.
Part B: The distance between the building and truck is 38.47 m.
How do you calculate the time of hitting the truck and the distance between the building and truck?Given that the height h of the building is 15 m. The constant speed s of driving the truck is 22 m/s.
Part A
The final velocity can be calculated by the formula given below.
[tex]v^2 = u^2 + 2gh[/tex]
Where v is the final velocity, u is the initial velocity, g is the gravitational acceleration and h is the height of the building.
The initial velocity u will be zero and g = 9.8 m/s2.
Substituting the values,
[tex]v^2 = 0 + 2\times 9.8\times 15[/tex]
[tex]v^2 = 294[/tex]
[tex]v = 17.14 \;\rm m/s[/tex]
The time to hit the truck can be calculated as given below.
[tex]h = ut + \dfrac {1}{2}gt^2[/tex]
[tex]15 = 0 + \dfrac {1}{2}\times 9.8 \times t^2[/tex]
[tex]t^2 = 3.06[/tex]
[tex]t = 1.749 \;\rm s[/tex]
Hence we can conclude that the time required to hit the truck by stuntman is 1.749 s.
Part B
The distance between the building and truck is calculated as given below.
[tex]d = s \times t[/tex]
[tex]d = 22 \times 1.749[/tex]
[tex]d =38.47\;\rm m[/tex]
Hence the distance between the building and truck is 38.47 m.
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What mass of aluminum has a total nuclear charge of 2.9 c?
Complete Question
What mass of aluminum has a total nuclear charge of 2.9 C?
Aluminum has atomic number 13. Suppose the aluminum is all of the isotope with 14 neutrons.
Answer:
The mass is [tex]T_m = 6.252 *10^{-5}\ g[/tex]
Explanation:
From the question we are told that
The total nuclear charge is [tex]q = 2.9 \ C[/tex]
The atomic number is [tex]u = 13[/tex]
The number of neutron is [tex]k = 14[/tex]
Generally the number of positive charge is mathematically represented as
[tex]N = \frac{q}{p}[/tex]
here p is the charge on a single proton with value [tex]p = 1.60*10^{-19} \ C[/tex]
So
[tex]N = \frac{2.9}{1.60*10^{-19}}[/tex]
=> [tex]N = 1.813*10^{19} \ protons[/tex]
Now since 1 atom contains 13 proton
The number of atoms present is
[tex]a = \frac{1.813*10^{19}}{13}[/tex]
[tex]a = 1.395 *10^{18} \ atoms[/tex]
Then the number of moles present is mathematically represented as
[tex]n = \frac{a}{N_k}[/tex]
Where N_k is the Boltzmann constant with value
[tex]N_k = 6.023*10^{23}[/tex]
So
[tex]n = \frac{1.395 *10^{18}}{ 6.023*10^{23}}[/tex]
[tex]n = 2.315 *10^{-6}\ moles[/tex]
Generally one mole of aluminum is equal to 27 g
So
The total mass of aluminum is
[tex]T_m = n * 27[/tex]
=> [tex]T_m = 2.315 *10^{-6} * 27[/tex]
=> [tex]T_m = 6.252 *10^{-5}\ g[/tex]
4. Velocity can be calculated using the following equation
v= Ad/t
Mass (kg)
Displacement (m)
Velocity (m/s)
7.58
3.56
35.65
Given the data above from an experiment, calculate the amount of time the object was
in motion
3.56 displacement ( m )
A car travels 60 miles due West first then turns back and travels 120 miles due East in 3 hours. What is...
a) the distance traveled by car?
b) the displacement of the car?
Answer:
A. 180 milesB. 60 milesExplanation:
In this problem, we are required to solve for the total distance that the car travelled. and the displacement
A) the distance travelled by car
this can be gotten by summing all the distances the car has travelled.
i,e total distance= 60 miles+120 miles
total distance= 180 miles
B) the displacement of the car
the displacement can be gotten by subtracting the final distance from the initial distance
final distance = 120 miles
initial distance= 60 miles
displacement= 120-60= 60 miles
A race car has a maximum speed of 0.104 km/s .What is this speed in miles per hour ?
Answer:
232.641374 mph
Explanation:
A race car has a maximum speed of 0.104km/s
Let X represent the speed in miles per hour
Therefore the speed in miles per hour can be calculated as follows
1 km/s = 2,236.936292 mph
0.104km/s = X
X = 0.104 × 2,236.936292
X = 232.641374
Hence the speed in miles per hour is 232.641374 mph
Speed of the car can be defined by the ratio of distance traveled by time elepsed. The speed of the given car is 233.212 miles/hr.
What is speed?
Speed can be defined by the change in position of an object in a given period of time.
[tex]\bold {Speed = \dfrac {Distance }{Time}}[/tex]
Given here,
The speed of the car is 0.104 km/s
Since 1 km = 0.621
So, 0.104 km = 0.0646
and 1 s = 0.000277
So, the speed of the car in miles per hour,
[tex]Speed = \dfrac {0.0646 \rm \ miles }{0.000277}\\\\ Speed =233.212 \rm \ miles/hr[/tex]
Therefore, the speed of the given car is 233.212 miles/hr.
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As an object in motion becomes heavier, its kinetic energy _____. A. increases exponentially B. decreases exponentially C. increases proportionally D. decreases proportionally
Answer:
c
Explanation:
Calculate the speed for a car that went a distance of 125 miles in 2 hours time.
Answer:
65 miles per hour
Explanation:
Speed= Distance/Time
Following this formula, we just need to insert the values.
The distance is 125 miles and the time is 2 hours
125=2=65
65 miles per hour
The speed of the car was exactly 125 mi per 2 hours.
In a more familiar unit, 125mi/2hr = 62.5 mi/hr.
Two violin strings are tuned to the same frequency 294 H. The tension in one string is then decreased by 2.0%. What will be the beat frequency heard when the two strings are played together?
Answer:
Beat frequency together = 2.95 Hz (Approx)
Explanation:
Given:
Frequency (F) = 294 H
Decrease in tension = 2%
Find:
Beat frequency together
Computation:
Tension = (100 - 2) / 100
Tension (T) = 0.98
Beat frequency together = Frequency (F) - (√T × F)
Beat frequency together = 294 - (√0.98 × 294)
Beat frequency together = 2.95 Hz (Approx)
The beat frequency heard when the two strings are played together is 2.95 Hz.
Given data:
The tuning frequency of the violin is, f = 294 Hz.
Decrement in the tension is, 2 %.
Since, tension is reduced at the rate of 2%. Then the new magnitude of tension on the string is,
T = (100 - 2 )/100
T = 0.98
Then the expression for the beat frequency heard when the two strings are played together is given as,
[tex]f_{b}=f -(\sqrt{T \times f})[/tex]
Solving as,
[tex]f_{b}=294-(\sqrt{0.98 \times 294})\\\\f_{b}=2.95\;\rm Hz[/tex]
Thus, we can conclude that the beat frequency heard when the two strings are played together is 2.95 Hz.
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PLEASE HELP 3 + 2 ∙ 4 = 3 + (2 ∙ 4) True False
The given expression 3 + 2 ∙ 4 = 3 + (2 ∙ 4) is absoulutely correct so it is True.
What is the arithmetic operator?Arithmetic operators are four basic mathematical operations in which summation, subtraction, division, and multiplication involve.,
Summation = addition of two or more numbers or variable
For example = 2 + 8 + 9
Subtraction = Minus of any two or more numbers with each other called subtraction.
For example = 4 - 8
Division = divide any two numbers or variable called division.
For example 4/8
Multiplication = to multiply any two or more numbers or variables called multiplication.
For example 5 × 7.
Given,
3 + 2 ∙ 4 = 3 + (2 ∙ 4)
We know that in every calculation we cannot separate multiplication and need to solve them first
So,
3 + 8 = 3 + 8
11 = 11
Hence, 3 + 2 ∙ 4 = 3 + (2 ∙ 4) is absoulutely correct.
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Usted coloca un recipiente con de mar sobre una bascula y toma nota de la lectura que indica la báscula . Ahora usted suspende la estatua del ejemplo 14.5 en el agua ( figura 14.18 ). Cómo cambia la lectura de la bascula ? ) Se incrementa en 7.84 N; ii) disminuye en 7.84 N: ) permanece igual: ) ninguna de las respuestas anteriores es correcta . cambia la lectura de la báscula cuando la estatua se sumerge en el agua?
Answer:
t the apparent weight of the body decreases when it is submerged in water,
F'_{e} = mg - ρ_liquid g V
Explanation:
When a container is placed on a scale it falls due to the weight of the body and the needle attached to the spring shows the value of the weight that is proportional to the displacement of the spring.
When the body is immersed in a container with water, it exerts a push on the body, so its apparent weight decreases and the compression of the scale is less, therefore it indicates a smaller weight of the body.
Let's write the equilibrium equation for the body is placed on the scale without water
[tex]F_{e}[/tex] - W = 0
F_{e} = W
Let us propose the equilibrium equation for this case
F'_{e} + B - W = 0
where F'_{e} is the elastic force, B the push of the liquid and W the weight of the body
F'_{e} = W -B
The thrust of the liquid is given by the Archimedean principle which says that it is equal to the weight of the dislodged liquid
B = ρ g V
F'_{e} = mg - ρ_liquid g V (1)
we use the density concept
ρ_body = m / V
we substitute
F'_{e}= g V (ρ_body - ρ_agua) (2)
F'_{e} < F_{e}
We see from expression is 1 and 2 that the apparent weight of the body decreases when it is submerged in water,
To know the specific value of the decrease, the weight of the body in the air or its density and volume must be known
How many planets are there?
Answer:
There are eight planets in our Solar System.
Explanation:
Answer:
8
Explanation:
Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune
Einsteins general theory of ————- describes the relationship
between energy and matter, and is major event in the history
of science.
Answer:
Special relativity.
Explanation:
Einstein's general theory of special relativity describes the relationship
between energy and matter, and is major event in the history
of science.
In 1905, Albert Einstein a noble physicist developed a theory that forms part of the basis of modern physics and one of the major event in the history of science, this theory was known as the theory of special relativity.
The theory of special relativity gives the relationship between space and time for any object that is moving at a uniform speed in a straight line, typically at the speed of light.
Simply stated, the theory of special relativity describes how a change in the speed of an object affect or relate with the measurement of space, time and mass. This ultimately implies that, as an object approach the value of light speed, it gains an infinite amount of mass and as such it is unable to travel faster than light.
The theory of special relativity by Albert Einstein gave birth to one of the most famous equation in science;
[tex]E = mc^2[/tex]
The equation illustrates, Energy equals mass multiplied by the square of the speed of light.
Where; E = Energy.
M = Mass.
C = Speed of light.
From the above equation, we can deduce that the mass of an object increases with speed and that mass is equivalent to energy, thus matter (mass) can be interchanged with energy.
An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spilled on the arm connected to the line during a laboratory renovation, it is impossible to see the level of the manometer fluid in this arm. During a period when the gas supply is connected to the line but there is no gas flow, a Bourdon gauge connected to the line downstream from the manometer gives a reading of 7.5 psig. The level of mercury in the open arm is 900mm above the lowest part of the manometer.
(a) When the gas is not flowing, the pressure is the same everywhere in the pipe. How high above the bottom of the manometer would the mercury be in the arm connect to the pipe?
(b) When gas is flowing, the mercury level in the visible arm drops by 25mm. What is the gas pressure (psig) at thismoment?
Answer:
A. Using
Pgauge= Pmanometer
And we know that
Pgauge= deta(hpg)
So deta h = Pgauge/density x g
So
= 7.5(6894.76/1psig)/ 13.6*10^3*9.8m/s²)
= 387.9mm
So to find height of pipe connected to the pipe we say
= h -deta h
= 900-387.97mm
=512.02mm
B. We use manometry principle
Pgas+density xg(25*10^3)-PX density{h-(H-0.25)=0
So
Finally Pgas= 6.54psig
a battery of 6v is connected in series with resistors of 0.1 ohm, 0.15 ohm, 0.2 ohm, 0.25 ohm and 6 ohm. how much current would flow through 0.2 ohm resistors
Answer:
0.9 Amps
Explanation:
Ohm's law: Voltage = Current x Resistance
Rearrange equation:
Current =Voltage/Resistance
Voltage=6v
In a series circuit the total resistance is the sum of all the resistors so the total resistance = 0.1+0.15+0.2+0.25+6=6.7 Ohms
6/6.7 = 0.9 (1dp) Amps
Because it is a series circuit, the current is the same throughout the entire circuit, all the resistors have a current of 0.9 Amps.
Hope this helped!
An electron and a proton are separated by a distance of 1.6×10−10m
(roughly the diameter of a single atom). The masses of the electron and proton are mp=1.673×10−27kg
and me=9.11×10−31kg,
respectively. The elementary charge e=1.602×10−19C.
The universal gravitational constant G=6.67×10−11N⋅m2/kg2
and the coulomb constant k=8.988×109N⋅m2/C2
What is the magnitude Fe of the electric force between the electron and the proton? Fe= 7.11×10−9N
What is the magnitude Fg
of the gravitational force between the electron and the proton?
Fg= 31.37×10−49N
In this scenario, how many times stronger is the electric force than the gravitational force?
Answer:
Fe = 9 x 10⁻⁹ N
Fg = 3.97 x 10⁻⁴⁶ N
Fe = 2.26 x 10³⁷ Fg
Explanation:
First we find electric force by Coulomb's Law as follows:
Fe = kq₁q₂/r²
where,
Fe = electric force = ?
k = Coulomb's Constant = 9 x 10⁹ N.m²/C²
q₁ = q₂ = charges on electron and proton = 1.6 x 10⁻¹⁹ C
r = distance between electron and proton = 1.6 x 10⁻¹⁰ m
Therefore,
Fe = (9 x 10⁹ N.m²/C²)(1.6 x 10⁻¹⁹ C)(1.6 x 10⁻¹⁹ C)/(1.6 x 10⁻¹⁰ m)²
Fe = 9 x 10⁻⁹ N
Now we find gravitational force by Newton's Law of Gravitation as follows:
Fg = Gm₁m₂/r²
where,
Fg = gravitational force = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
m₁ = mass of electron = 9.11 x 10⁻³¹ kg
m₂ = mass of proton = 1.673 x 10⁻²⁷ kg
r = distance between electron and proton = 1.6 x 10⁻¹⁰ m
Therefore,
Fg = (6.67 x 10⁻¹¹ N.m²/kg²)(9.11 x 10⁻³¹ kg)(1.673 x 10⁻²⁷ kg)/(1.6 x 10⁻¹⁰ m)²
Fg = 3.97 x 10⁻⁴⁶ N
Dividing both forces:
Fe/Fg = (9 x 10⁻⁹ N)/(3.97 x 10⁻⁴⁶ N)
Fe = 2.26 x 10³⁷ Fg
The electric force is 2.3 × 10^39 times greater than the gravitational force.
The gravitational force between the proton and the electron is obtained from;
Fg = Gmemp/r^2
Where;
me = mass of the electron
mp = mass of the proton
r = distance of separation
Fg = 6.67 × 10^−11 × 9.11 × 10^−31 × 1.673 × 10^−27/(1.6 × 10^−10)^2
Fg = 3.94 × 10^−48 N
For the electric force;
Fe= K e^2/r^2
K = electric constant
e = magnitude of charge on electron and proton
r = distance of separation
Hence;
Fe = 8.988 × 10^9 × (1.602×10^−19)^2/(1.6×10^−10)^2 =
Fe = 9 × 10^−9 N
The number of time the electric force is stronger than the gravitational force = 9 × 10^−9 N/3.94 × 10^−48 N = 2.3 × 10^39 times
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_____ is the bending of waves as they pass by a barrier or through an opening.
A. Interference
B. Coherence
C. Diffraction
D. None of the above
Answer:
C. Diffraction
Explanation:
Diffraction is the bending of waves as they pass by a barrier or through an opening. One of the condition for diffraction is that the size of the aperture or obstacle is comparable to the wavelength of light. Fresnel diffraction and Fraunhofer diffraction are two types of diffraction.
Hence, the correct option is (c) "Diffraction".
A ball is thrown upward with a speed of 28.2 m/s.A. What is its maximum height?B. How long is the ball in the air?C. When does the ball have a speed of 12m/s?
Answer:
(A) The maximum height of the ball is 40.57 m
(B) Time spent by the ball on air is 5.76 s
(C) at 33.23 m the speed will be 12 m/s
Explanation:
Given;
initial velocity of the ball, u = 28.2 m/s
(A) The maximum height
At maximum height, the final velocity, v = 0
v² = u² -2gh
u² = 2gh
[tex]h = \frac{u^2}{2g}\\\\h = \frac{(28.2)^2}{2*9.8}\\\\h = 40.57 \ m[/tex]
(B) Time spent by the ball on air
Time of flight = Time to reach maximum height + time to hit ground.
Time to reach maximum height = time to hit ground.
Time to reach maximum height is given by;
v = u - gt
u = gt
[tex]t = \frac{u}{g}[/tex]
Time of flight, T = 2t
[tex]T = \frac{2u}{g}\\\\ T = \frac{2*28.2}{9.8}\\\\ T = 5.76 \ s[/tex]
(C) the position of the ball at 12 m/s
As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.
v² = u² - 2gh
12² = 28.2² - 2(9.8)h
12² - 28.2² = - 2(9.8)h
-651.24 = -19.6h
h = 651.24 / 19.6
h = 33.23 m
Thus, at 33.23 m the speed will be 12 m/s
What is the approximate height of the average adult in centimeters?a. 100b. 200c. 300
Answer:
Option B
200 cm
Explanation:
From statistical evidence, the average height of an adult is about 200cm. This is like 2m or 6.56 feet.
Despite the fact that there are some exceptions; with some adults reaching heights of about seven feet, and some dwarfs who stand only about 1 m tall, most adults only grow to about 200cm in height before they start shrinking due to old age.
Excess electrons are placed on a small lead sphere with a mass of 7.70 g so that its net charge is −3.35 × 10^−9 C.A) Find the number of excess electrons on the sphere.
B) How many excess electrons are there per lead atom? The atomic number of lead is 82, and its atomic mass is 207 g/mol?
Answer:
a
[tex]N = 2.094*10^{10} \ electrons[/tex]
b
[tex]O = 9.33*10^{-13} \ electrons[/tex]
Explanation:
From the question we are told that
The mass of the lead sphere is [tex]m = 7.70g = 0.0077 \ kg[/tex]
The net charge is [tex]Q_{net} = -3.35*10^{-9} \ C[/tex]
The atomic number is [tex]u = 82[/tex]
The molar mass is [tex]M = 207 \ g/mol[/tex]
Generally the excess number of electron on the sphere is mathematically represented as
[tex]N = \frac{Q_{net}}{ e }[/tex]
Here e is the charge on the electron is [tex]e = -1.60 *10^{-19} \ C[/tex]
So
[tex]N = \frac{-3.35 *10^{-19}}{ -1.60*10^{-19}}[/tex]
[tex]N = 2.094*10^{10} \ electrons[/tex]
Generally the number of atom present is mathematically represented as
[tex]n = N_a * \frac{m}{ M}[/tex]
Here [tex]N_a[/tex] is the Avogadro's number with value [tex]N_a = 6.0*10^{23} \ atoms[/tex]
[tex]n = 6.03 *10^{23} * \frac{7.70}{ 207}[/tex]
[tex]n = 2.24 *10^{22} \ atoms [/tex]
Generally the electrons are there per lead atom is mathematically represented as
[tex]O = \frac{N}{n}[/tex]
=> [tex]O = \frac{2.24*10^{22}}{2.094*10^{10}}[/tex]
=> [tex]O = 9.33*10^{-13} \ electrons[/tex]
Two parallel wires I and II that are near each other carry currents i and 3i both in the same direction. Compare the forces that the two wires exert on each other. A. The wires exert equal magnitude attractive forces on each other. B. Wire I exerts a stronger force on wire II than II exerts on I.C. Wire II exerts a stronger force on wire I than I exerts on II. D. The wires exert equal magnitude repulsive forces on each other. E. The wires exert no forces on each other.
Answer:
A. The wires exert equal magnitude attractive forces on each other.
Explanation:
Magnetic field due to current i on current 2i
B₁ = 10⁻⁷ x 2 i / r where r is distance between the two wires
Force on wire II due to wire I per unit length
= magnetic field x current in wire II
= B₁ x 2 i
= [ 10⁻⁷ x 2 i / r ] x 2i
= 4 x 10⁻⁷ i² / r
Magnetic field due to current 2i on current i
B₂ = 10⁻⁷ x 4 i / r where r is distance between the two wires
Force on wire I due to wire II per unit length
= magnetic field x current in wire I
= B₂ x i
= [ 10⁻⁷ x 4 i / r ] x i
= 4 x 10⁻⁷ i² / r
So final forces on each wire are same .
This force will be attractive in nature . The direction of force can be known from fleming's right hand rule .
What is a chemical change?
A major advantage of case studies is ________.
An initially neutral glass rod is rubbed with silk. It becomes positively charged by_____.a) electrons are created on the rod.
b) electrons are transferred from the silk to the rod.
c) protons are transferred from the silk to the rod.
d) protons are transferred from the rod to the silk.
e) electrons are transferred from the rod to the silk.
Answer:
e) electrons are transferred from the rod to the silk.
Explanation:
An initially neutral glass rod contains equal number of electron and proton.
If the rod becomes positively charged after being rubbed with silk, then the rod must have lost some its electron to the silk since electrons are more mobile than protons, leaving the rod with excess positive charge (protons), and the silk will be negatively charged (excess electron).
Thus, the rod becomes positively charged by transfer of electrons from rod to the silk.
e) electrons are transferred from the rod to the silk.
Assess platos theory of the forms from Aristotle's standpoint. is he correct? Explain..it’s philosophy
Answer:
The task of philosophy, for Plato, is to discover through reason (“dialectic”) the nature of the Forms, the only true reality, and their interrelations, culminating in an understanding of the most fundamental Form, the Good or the One. Aristotle rejected Plato’s theory of Forms but not the notion of form itself. For Aristotle, forms do not exist independently of things—every form is the form of some thing. A …
Explanation:
Samantha is 1.44 m tall on her eleventh birthday and 1.68 m tall on her twelfth birthday. By what factor has her height increased? By what percentage?
Answer:
Given initial height h¹= 1.44
Final height h²=1.68
So factor increase is 1.68/1.44²= 1.17
Percentage increase will be
1.68-1.44/1.44*100%=16.7%
What can make your computer "sick"?
Answer:
An virus
Explanation:
An computer virus can come from websites that try to steal your Information or apps that's not necessarily scanned and protected so that yor computer can be safe from viruses.
A computer may encounter problems or get "ill" for a variety of reasons. The system can become infected with malware and viruses, which can cause performance issues and data loss.
A failing hard drive or defective RAM are examples of hardware issues that can lead to instability and crashes. Overheating can lead to hardware damage or performance deterioration and is frequently brought on by insufficient cooling or clogged air vents.
Systems may become unstable as a result of software conflicts between out-of-date or incompatible programmes.
Electrical problems or power surges can harm components and interfere with proper operation. Computer issues can also be caused by human error, such as inadvertent file deletion or poor configuration.
Thus, a computer can be kept in good condition with routine maintenance, security precautions, and appropriate usage.
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You leave your home at 1pm. At 3pm, you are 100 km east of your house. What was your average velocity in km/hr
Answer:
50km/h
Explanation:
Average velocity = change in distance (or distance travelled) divided by/ the change in time (or time taken.)
The change in distance has been given as 100km.
The change in time is 3pm-1pm = 2 hours.
Therefore the average velocity was 100/2 = 50km/h (to the east).
Hope this helped!
2. If you are driving at a constant velocity of 5 m/s what is your
acceleration?
Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is at the origin. The third charge, q3 = - 30 μC, is located at x = 2.0 m. What is the force on q2?
(a) 1.65 N in the negative x- direction
(b) 3.15 N in the positive x- direction
(c) 1.50 N in the negative x- direction
(d) 4.80 N in the positive x- direction
(e) 4.65 N in the negative x- direction.
Answer:
3.15 N towards the positive x-axis
Explanation:
first charge has charge q1 = 10 μC = 10 x 10^-6 C
second charge has charge q2 = 20 μC = 20 x 10^-6 C
third charge has charge q3 = -30 μC = -30 x 20^-6 C
According to coulomb's law, force between two charged particle is given as
F = [tex]\frac{-kQq}{r^2}[/tex]
Where
F is the force between the charges
k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.
Q is the magnitude of one charge
q is the magnitude of the other charge
is the distance between these two charges
For the force on q2 due to q1,
distance r between them = 0 - (-1.0) = 1 m
F = [tex]\frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2}[/tex] = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive x-axis)
For the force on q2 due to q3,
distance between them = 2.0 - 0 = 2 m
F = [tex]\frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2}[/tex] = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)
Resultant force on q2 = 1.8 N + 1.35 N = 3.15 N towards the positive x-axis
fter a great many contacts with the charged ball, how is any charge on the rod arranged (when the charged ball is far away)? Select the best description. View Available Hint(s) Select the best description. positive charge on end B and negative charge on end A negative charge spread evenly on both negative charge on end A with end B remaining neutral both ends neutral positive charge spread evenly on both
Answer:
With this analysis the correct answer is
Metal bar. The answer is uniform charge distribute
Non-metallic bar. As we see the charge of the ball, at the ends closest to the ball a charge of the same sign as the ball and at the far end the charge of equal magnitude and opposite sign
Explanation:
When the ball touches the bar, a charge transfer process occurs, leaving half in the bar and the other half in the ball, when the ball bounces and clears it, two dependent processes can occur, the bar is metallic or non-metallic , let's analyze the two cases
* metal bar, the load that is on the bar is distributed evenly throughout the bar, with each bounce of the ball the load on the bar increases and decreases in the balls until the two loads are equal and appear from this rebound no there is more load transfer
* Non-conductive insulating bar, when the ball touches the bar for the first time, it creates a load equal to q / 2, when the ball moves away from the bar, for this non-conductive, the transferred load remains in the same place. Therefore this load induces a load of equal magnitude, but opposite sign at the other end of the bar.
When the ball approaches for the second time, the opening already has a load, therefore the amount of load deposited is less, with between processes in each bounce the bar of the bar sounds, the process continues until the load of the ball and the bar have been equal.
With this analysis the correct answer is
Metal bar. The answer is uniform charge distribute
Non-metallic bar. As we see the charge of the ball, at the ends closest to the ball a charge of the same sign as the ball and at the far end the charge of equal magnitude and opposite sign