What does the statement "all motion is relative"
mean?
A monkey has a bit of a heavy for on the gas pedal. As soon as the light turns green the monkey pushes the gas pedal to the floor and accelerates from 0 mph to 60 mph (0 m/s - 28.82 m/s) in 4.2 seconds, calculate the cars acceleration.
Answer:
[tex]s=6.86m/s^2[/tex]
Explanation:
Hello,
In this case, considering that the acceleration is computed as follows:
[tex]a=\frac{v_{final}-v_{initial}}{t}[/tex]
Whereas the final velocity is 28.82 m/s, the initial one is 0 m/s and the time is 4.2 s. Thus, the acceleration turns out:
[tex]a=\frac{28.82m/s-0m/s}{4.2s}\\ \\s=6.86m/s^2[/tex]
Regards.
explain how ozone in the atmosphere affects visible light on earth
The SI system uses three base units. Question 6 options: True False
Answer:
The answer is false
Explanation:
Though the mostly used SI unit of measurement or the most popular units are the
Length,
Time and
Mass
i.e meter (m), seconds (s), kilogram (kg)
Aside all the above stated units for measurements there are other four basic units which are itemized bellow.
they are
1. Amount of substance - mole (mole)
2. Electric current - ampere (A)
3. Temperature - kelvin (K)
4. Luminous intensity - candela (cd)
In an evironmental system of subsystem, the mass balance equation is:__________.
Answer:
Explanation:
The mass balance is an application of conservation of mass, to the analysis of physical system. This is given in an equation form as
Input = Output + Accumulation
The conservation law that is used in this analysis of the system actually depends on the context of the problem. Nevertheless, they all revolve around conservation of mass. By conservation of mass, I mean that the fact that matter cannot disappear or be created spontaneously.
Which law of motion is this an example of? Newton's 1st Law of Motion
Newton's 2nd Law of Motion
Newton's 3rd Law of Motion
Answer:
first law is your answer, if I'm not wrong
Answer:
i think it's the first law of motion :)
Explanation:
What is the final velocity of a 1400 kg drag racer that’s engine applies 910000 N of Force for 120m
Answer:
468,000,000 m/sExplanation:
According to newtons first law of motion;
Force = mass * acceleration
F = ma
where a = v-u/t
v is the final velocity
u is the initial velocity
t is the time taken
F = m(v-u/t)
Ft = m(v-u)
Given parameters;
F = 910,000N
t = 120m = 120 *60 = 7200secs
m = 1400kg
u = 0m/s
Required parameter
Final velocity v
Substituting the given parameters into the formula to get v;
Ft = m(v-u)
910,000 * 72000 = 1400(v-0)
910,000 * 72000 = 1400v
v = (910,000 * 72000)/1400
v = 910,000 * 720/14
v = 910,000 * 60
v = 468,000,000 m/s
Hence the final velocity of the drag racer is 468,000,000 m/s
How many standards unti “forces and motion” cover ?
A golfer needs to sink an 8m putt. She hits the ball, giving it an initial speed of 1.6 m/s, but it stops 1.5m short of the hole. c. What initial speed did the ball need to just make the putt? Assume the acceleration of the grass is constant. Hint: to just make the putt, the speed of the ball will be zero right at the hole. d. In order to make the 1.5m putt, what initial speed does she need to give the ball?
Answer:
1.78 m/s
Explanation:
Distance to putt = 8 m
Distance the ball stops to the putt = 1.5 m
therefore distance traveled by the ball = 8 - 1.5 = 6.5 m
The ball stops at this point 1.5 m from the putt, therefore its final velocity at this point = 0 m/s
the ball was struck with an initial velocity of 1.6 m/s
Using the equation
[tex]v^2[/tex] = [tex]u^2[/tex] + 2as
where v is the final velocity of the ball = 0 m/s
u is the initial speed of the ball = 1.6 m/s
a is the acceleration of the grass
s is the distance the ball travels = 6.5 m
substituting values, we have
[tex]0^2[/tex] = [tex]1.6^2[/tex] + 2(a x 6.5)
0 = 2.56 + 13a
13a = -2.56
a = -2.56/13 = -0.197 m/s^2
If this acceleration of the grass is assumed to be constant, then to the initial speed needed to make the putt will be calculated from
[tex]v^2[/tex] = [tex]u^2[/tex] + 2as
where
v is the final speed at the putt = 0 m/s
u is the initial speed with which the ball is struck = ?
a is the acceleration of the grass = -0.197 m/s^2
s is the distance to the putt = 8 m
substituting values, we have
[tex]0^2[/tex] = [tex]u^2[/tex] + 2(-0.197 x 8)
0 = [tex]u^2[/tex] - 3.152
[tex]u^2[/tex] = 3.152
u = [tex]\sqrt{3.152}[/tex] = 1.78 m/s
What must the charge (sign and magnitude) of a particle of mass 1.50 gg be for it to remain stationary when placed in a downward-directed electric field of magnitude 670 N/CN/C ?
Answer:
q = -2.19 x 10⁻⁵ C
Explanation:
Given;
mass of the particle, m = 1.5 g = 0.0015 kg
magnitude of electric field, E = 670 N/C
Electric field is given by;
[tex]E = \frac{F}{q}[/tex]
where;
q is the magnitude of the
f is the force of the charge
f = mg
[tex]E = \frac{F}{q}\\\\E = \frac{mg}{q}\\\\q = \frac{mg}{E}\\\\q = \frac{0.0015*9.8}{670} \\\\q = 2.19*10^{-5} \ C[/tex]
Since the electric field is acting downward, the force on the charge must be acting upward. Therefore, the charge must be negative
q = -2.19 x 10⁻⁵ C
Why are viruses hard to fight
Answer:Compared to other pathogens, such as bacteria, viruses are minuscule. And because they have none of the hallmarks of living things — a metabolism or the ability to reproduce on their own, for example — they are harder to target with drugs.
Explanation:
A speeder passes a parked police car with a speed of 65 km/h in a 50km/h zone. One second after the speeder has passed the police car, the police begin his pursuit. The police car accelerates with constant acceleration of 2m/s^2.Required:a. How long does it take for the police to catch the speeding car? b. How far did the police car travel before police caught up with the speeder? c. What is the speed of the police car when catches up with the speeder?
Answer:
a) The police will take 18.056 seconds to catch the speedy car, b) The police will travel 326.019 meters before catching the speedy car, c) The speed of the police car when catches up with the speeder is 36.112 meters per second.
Explanation:
Let suppose that speeder moves in a uniform motion, whereas police car has an uniformly accelerated motion.
a) How long does it take for the police to catch the speeding car:
Kinematic equation of each vehicle's position are described:
Speeder
[tex]s_{A} = s_{A,o}+v_{A}\cdot t[/tex]
Police Car
[tex]s_{B} = s_{B,o}+v_{B,o}\cdot t + \frac{1}{2}\cdot a_{B}\cdot t^{2}[/tex]
If [tex]s_{A} = s_{B}[/tex], [tex]s_{A,o} = s_{B,o}[/tex], [tex]v_{A} = 18.056\,\frac{m}{s}[/tex], [tex]v_{B,o} = 0\,\frac{m}{s}[/tex] and [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], the resulting expression is done:
[tex]v_{A} \cdot t = v_{B,o}\cdot t +\frac{1}{2}\cdot a_{B}\cdot t^{2}[/tex]
[tex]\frac{1}{2}\cdot a_{B}\cdot t^{2}+(v_{B,o}-v_{A})\cdot t = 0[/tex]
[tex]t \cdot \left(\frac{1}{2}\cdot a_{B}\cdot t +v_{B,o}-v_{A} \right)= 0[/tex]
[tex]t = 0\,s\,\wedge\, t = \frac{2\cdot (v_{A}-v_{B,o})}{a_{B}}[/tex]
[tex]t = \frac{2\cdot \left(18.056\,\frac{m}{s}-0\,\frac{m}{s} \right)}{2\,\frac{m}{s^{2}} }[/tex]
[tex]t = 18.056\,s[/tex]
The police will take 18.056 seconds to catch the speedy car.
b) How far did the police car travel before police caught up with the speeder?
The distance travelled by the police is: ([tex]s_{B,o} = 0\,m[/tex], [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], [tex]v_{B,o} = 0\,\frac{m}{s}[/tex] and [tex]t = 18.056\,s[/tex])
[tex]s_{B} = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (18.056\,s)+\frac{1}{2}\cdot \left(2\,\frac{m}{s^{2}} \right) \cdot (18.056\,s)^{2}[/tex]
[tex]s_{B} = 326.019\,m[/tex]
The police will travel 326.019 meters before catching the speedy car.
c) What is the speed of the police car when catches up with the speeder?
The speed of the police car is represented by the following formula:
[tex]v_{B} = v_{B,o} + a_{B}\cdot t[/tex]
Where [tex]v_{B}[/tex] is the speed of the police car, measured in meters per second.
Given that [tex]v_{B,o} = 0\,\frac{m}{s}[/tex], [tex]t = 18.056\,s[/tex] and [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], the final speed of the police car when catches up with the speeder is:
[tex]v_{B} = 0\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right)\cdot (18.056\,s)[/tex]
[tex]v_{B} = 36.112\,\frac{m}{s}[/tex]
The speed of the police car when catches up with the speeder is 36.112 meters per second.
During a morning run, Juanita averaged a speed of 4.83 m/s for 11.2 minutes. She then averaged a speed of 5.99 m/s for 7.44 minutes. What total distance did Juanita run during her 20.0 minute jog.
Answer:
I believe the answer is 10.82 m/s.
Explanation:
I just saw that the question asked for the TOTAL DISTANCE.. which means to add. And I added 4.83 and 5.99 together which then lead me to get 10.82 m/s. So yeah i think that's the answer. Hope this helps.
Consider an embedded system which uses a battery with a 17.15-Amp-Hour capacity. What is the maximum average current draw (in micro Amps to one decimal place) that your embedded processor can have if you want the battery to last 23 years without being replaced
Answer:
85.12 μAmp
Explanation:
The battery power output = 17.15 Amp-hr
If the battery is to last 23 years, we have to calculate how many hours there are in 23 years
in one year there are 24 hours x 365 day = 8760 hrs
in 23 years there are 23 x 8760 = 201480 hours
maximum current to be drawn from the battery = (17.15 Amp-hr) ÷ (201480 hours) = 85.12 x 10^-6 Amp = 85.12 μA
A substan e has a melting point of 20◦C and a heat of fusion of 3.5 × 104 J/kg. The boiling point is 150◦C and the heat of vaporization is 7.0 × 104 J/kg at a pressure of 1.0 atm. The spe i heats for the solid, liquid, and gaseous phases are 600 J/(kg.K), 1000 J/(kg.K), and 400 J/(kg.K), respe tively. The quantity of heat given up by 0.50 kg of the substan e when it is ooled from 170◦C to 88◦C, at a pressure of 1.0 atmosphere, is losest to (1) 70 kJ (2) 14 kJ (3) 21 kJ (4) 30 kJ (5) None of the above.
Answer: Option (1) 70 kJ
Explanation:
Given that;
melting point of the substance T₁ is 20°C , Boiling point of the substance T₂ is 150°C , heat of fusion L₁ is 3.5 x 10⁴ J/kg , heat of vaporization L₂ is 7 x 10⁴ J/kg , Specific heat in solid state C₁ is 600 J/kg.K , Specific heat in liquid state C₂ is 1000 J/kg.K, Specific heat in gaseous state C₃ is 400 J/(kg.K) , Mass of the substance m is 0.5 kg , Initial temperature of the substance T₃ is 170°C , Final temperature of the substance T₄ is 88°C .
Now Heat given up by the substance to reach boiling point 150°C is
Q₁ = mC₃(T₃ - T₂)
Q₁ = (0.5)(400)(170 - 150)
Q₁ = 4000 J
Heat given up by the substance to turn into liquid from gaseous state at 150°C
Q₂ = mL₂
Q₂ = (0.5)(7x10⁴)
Q₂ = 35000 J
Heat given up by the substance to reach 88°C in liquid state from 150°C
Q₃ = mC₂(T₂ - T₄)
Q₃ = (0.5)(1000)(150 - 88)
Q₃ = 31000 J
Total heat given up by the substance
Q = Q₁ + Q₂ + Q₃
Q = 4000 + 35000 + 31000
Q = 70000 J
Q = 70 kJ
Therefore The quantity of heat given up by 0.50 kg of the substance when it is cooled from 170◦C to 88◦C, at a pressure of 1.0 atmosphere, is closest to 70 kJ
Plz help me :( The average human can run 60.35 meters in 9 seconds, calculate the average speed in meters per second (m/s)
Answer:
6.705555556 m/s
Explanation:
60.35 divide by 9
The element that has an average amu of 40 is represented by the symbol Zr on the Periodic Table of Elements. True False Question 2 (2 points) The element found in Period 2, Group 8 is called Argon. True False Question 3 (2 points) Which element would have 4 valence electrons? a Be b Si c H Question 4 (5 points) Put the following elements in order according to the number of protons they contain: S; Si, Sc; Na Reorder answers 1.Si Reorder answers 2.S Reorder answers 3.Sc Reorder answers 4.Na Question 5 (2 points) Argon is a part of what family in the Periodic Table of Elements? a Halogens b Noble Gases c Alkali Metals Question 6 (4 points) (03.02 MC) Match the element with its description. (4 points) Column A 1. Sodium : a Sodium 2. Silicon : c Silicon 3. Bromine : d Bromine 4. Argon : b Argon Column B a. Malleable, soft, and shiny b. Has properties of both metals and nonmetals c. Highly reactive gas d. Nonreactive gas Question 7 (2 points) A period in the Periodic Table of Elements determines the number of _. a protons b valence electrons c shells/energy levels Question 8 (3 points) (03.02 MC) A neutral atom of chlorine (Cl) has an average mass of 35 amu and 17 electrons. How many neutrons does it have? (3 points) a 18 b 17 c 52 Question 9 (2 points) A balanced/neutral atom will have the same number of protons and neutrons. True False Question 10 (2 points) The element Manganese is represented by _ in the Periodic Table of Elements a Mn b Mg c Ma Question 11 (2 points) Which element family below is best described as being very reactive? a Metalloids b Noble Gases c Alkali Metals Question 12 (2 points) The element Lithium is more reactive than Beryllium. True False Someone answer, please.
Answer:
False
Explanation:
Becuase the average amu of 40 is represented
The standard deviation of Eric’s data is 0.8°C. Martha conducted the same experiment. Her average temperature was 35.1 with a standard deviation of 1.2°C. Her data are more precise than Eric's
Answer:
less precise than
Explanation:
Answer:
less precise than
Explanation:
its right on edge2020
A 130 g copper bowl contains 100 g of water, both at 20.0°C. A very hot 420 g copper cylinder is dropped into the water, causing the water to boil, with 8.63 g being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment. (a) How much energy is transferred to the water as heat? (b) How much to the bowl? (c) What is the original temperature of the cylinder? The specific heat of water is 1 cal/g·K, and of copper is 0.0923 cal/g·K. The latent heat of vaporization of water is 539 Cal/kg.
Answer:
a) 4652 cal
b) 8000 cal
Explanation:
Amount of heat transferred
Q1 = mL(v)
Q1 = 8.63 * 539
Q1 = 4652 cal
Amount of heat transferred to the water
Q2 = mcΔT
Q2 = 100 * 1 * (100 - 20)
Q2 = 8000 cal
Q = Q1 + Q2
Q = 4652 + 8000
Q = 12652 cal
b)
Heat transferred to the copper bowl
Q(b) = m(b) * c(b) * ΔT
Q(b) = 0.13 * 0.0923 * (100 - 20)
Q(b) = 0.96 cal
c)
Original heat of the cylinder
Q(c) = Q + Q(b)
m(c) * c(c) * ΔT = Q + Q(b), making ΔT subject of the formula
ΔT = (Q + Q(b))/ (m(c) * c(c))
ΔT = (12652 + 0.96) / (0.42 * 1)
ΔT = 12652.96/0.42
ΔT = 30126.1
A truck has a mass of 3000 kg and a velocity
of 10 m/s. Calculate momentum!
Answer:
[tex] \huge{ \boxed{ \sf{30000 \: kg \: {m/s}^{2}}}}[/tex]
Explanation:
[tex] \text{ \underline{ Given}} : [/tex]
[tex] \star[/tex] Mass of a truck ( m ) = 3000 kg
[tex] \star[/tex] Velocity of a truck ( v ) = 10 m / s
Finding the momentum :
[tex] \boxed{ \sf{momentum \: ( \: p \: ) \: = \: mass \: ∗ \: \: velocity}}[/tex]
[tex] \hookrightarrow{ \sf{momentum \: ( \: p \: ) \: = 3000 \: ∗ \: 10}}[/tex]
[tex] \hookrightarrow{ \sf{ \: momentum \: ( \: p \: ) \: = 30000 \: kg \: m/ {s}^{2} }}[/tex]
Hope I helped!
Best wishes ! :D
~[tex] \sf{TheAnimeGirl}[/tex]
Given:-
Mass of the body (m) = 3000 kgVelocity (v) = 10 m/sTo calculate: Momentum of the body.
We know,
p = mv
where,
p = Momentum,m = Mass &v = Velocity.Thus,
p = (3000 kg)(10 m/s)
→ p = 30000 kg m/s (Ans.)
True or False: If a point charge has electric field lines that point into it, the charge must be positive.
Answer:
False
Explanation:
If a point charge has electric field lines point into it,then charge must be negative because electric lines point into negative charges and point out of positive charges
A particle is moving along a straight line through a fluid medium such that its speed is measured as v = (2t) m/s, where t is in seconds. If it is released from rest at s = 0, determine its positions and acceleration when t = 3 s.
Answer:
The acceleration of the particle is 2 m/s²
The position of the particle is 9m
Explanation:
Given;
particles velocity, v = 2t m/s
The acceleration of the particle is given by;
a = dv/dt
a = d(2t) /dt
a = 2 m/s²
The acceleration of the particle is 2 m/s²
The position of the particle is given by;
s = ut + ¹/₂at²
Since the particle was released from rest, u = 0
s = 0(3) + ¹/₂(2)(3)²
s = 0 + 9 m
The position of the particle when t = 3 s is 9m
If three resistors 2ohm, 3ohm and 4ohm are connected in a circuit. Calculate the equivalent resistance of the combination
You haven't described whether they're connected in series or in parallel. Actually, there are eight (8) different ways they can be arranged, and each way has a different equivalent resistance.
==> All 3 resistors in series. Equivalent resistance = (2+3+4) = 9.000 ohms
==> All 3 resistors in parallel.
Equivalent resistance = 1 / (1/2 + 1/3 + 1/4) = (12/13) ohm or 0.923 ohm
==> The 2 ohm resistor in series with (the 3 and the 4 in parallel)
Equivalent resistance = 2 + 1/(1/3 + 1/4) = 3.714 ohms
==> The 3 ohm resistor in series with (the 2 and the 4 in parallel)
Equivalent resistance = 3 + 1/(1/2 + 1/4) = 4.333 ohms
==> The 4 ohm resistor in series with (the 2 and the 3 in parallel)
Equivalent resistance = 4 + 1/(1/2 + 1/3) = 5.200 ohms
==> The 2 ohm resistor in parallel with (the 3 and the 4 in series)
Equivalent resistance = 1.556 ohms
==> The 3 ohm resistor in parallel with (the 2 and the 4 in series)
Equivalent resistance = 2.000 ohms
==> The 4 ohm resistor in parallel with (the 2 and the 3 in series)
Equivalent resistance = 2.222 ohms
Andy took a bus and then walked from his home to downtown.
For the first 1.6 hour, the bus drove at an average speed of 15
km/h. For the next 0.4 hours, he walked at an average speed
of 4.5 km/h. What was the average speed for the whole
journey?
Answer:
12.9
Explanation:
1.6x15=24
0.4x4.5=1.8
24+1.8=25.8
1.6+0.4=2
25.8/2=12.9
Caroline, a piano tuner, suspects that a piano's B4 key is out of tune. Normally, she would play the key along with her B4 tuning fork and tune the piano to match, but her B4 tuning fork is missing! Instead, she plays the errant key along with her A4 tuning fork (which has a frequency of 440.0 Hz), displays the resulting waveform on a handheld oscilloscope, and measures a beat frequency of 15.9 Hz. Then, she plays the errant key along with her C5 tuning fork (which has a frequency of 523.3 Hz) and measures a beat frequency of 67.4 Hz. What frequency is being played by the out-of-tune key? If the B4 key is supposed to produce a frequency of 493.9, is the frequency of the key lower than it should be ("flat") or higher than it should be ("sharp")?
Answer:
455.9 Hz, Flat
Explanation:
The beat frequency is basically the difference between the frequency of the B4 note and the frequency of the A4 tuning fork. This means the B4 note is 15.9 Hz off of 440Hz, and 67.4 Hz off of 525.3 Hz. As B4 is between the two notes, it would make sense to find its frequency by adding 15.9 to 440 and subtracting 67.4 from 523.3, both if which give us a frequency 455.9 Hz for the B4 key. This is because the note doesn't change for the different turning forks so both differences should result in the same frequency. Because the note should be 493.9 frequency but instead has a frequency of 455.9 Hz, it is flat because the frequency is lower than it is supposed to be.
Hope this helped!
This question deals with the phenomenon of "Beat Frequency".
The frequency being played by the B4 key is "455.9 Hz", which is "lower (flat) than it should be".
When two waves with a slightly different frequency interfere with each other, they produce a pattern known as beat frequency. The beat frequency is the difference between the frequencies of the interfering wave. Since the frequency of B4 key produces beat frequency with both the C5 tuning fork (which has a frequency of 523.3 Hz) and the A4 (which has a frequency of 440 Hz). Hence, the frequency of the B4 key must be between the range of these frequencies, that is, 440 Hz to 523.3 Hz.
ANALYZING THE BEAT FREQUENCY WITH A4 TUNING FORK:
Beat Frequency with A4 tuning fork = Frequency of B4 - Frequency of A4
Frequency of B4 = Frequency of A4 + Beat Frequency with A4 tuning fork
Frequency of B4 = 440 Hz + 15.9 Hz
Frequency of B4 = 455.9 Hz
ANALYZING THE BEAT FREQUENCY WITH C5 TUNING FORK:
Beat Frequency with C5 tuning fork = Frequency of C5 - Frequency of C5
Frequency of B4 = Frequency of C5 - Beat Frequency with C5 tuning fork
Frequency of B4 = 440 Hz - 67.4 Hz
Frequency of B4 = 455.9 Hz
Both the answers validate each other.
Since the expected frequency of the B4 key is 493.9 Hz, which is greater than the actual frequency produced by the B4 key. Hence, we can say that "the frequency of the key lower than it should be ("flat")"
The attached picture shows the general formula of the beat frequency.
Learn more about beat frequency here:
https://brainly.com/question/10703578?referrer=searchResults
A car moves in a straight line at a speed of 72.1 km/h.
How far (in km) will the car move in 7.74 minutes at this speed?
Speed=distance/time.
7.74mins=464.4secs.
72.1=distance/464.4.
distance=464.4×7.21.
=3348.324m.
You and a friend are doing the laundry when you unload the dryer and the discussion comes around to static electricity. Your friend wants to get some idea of the amount of charge that causes static cling. You immediately take two empty soda cans, which each have a mass of 120 grams, from the recycling bin. You tie the cans to the two ends of a string (one to each end) and hang the center of the string over a nail sticking out of the wall. Each can now hangs straight down 30 cm from the nail. You take your flannel shirt from the dryer and touch it to the cans, which are touching each other. The cans move apart until they hang stationary at an angle of 10 degrees from the vertical.
Assuming that there are equal amounts of charge on each can, you now calculate the amount of charge transferred from your shirt.
Answer:
Q_total = 2 10⁻⁶ C
Explanation:
Let's apply the conditions of static equilibrium to this case, in the adjoint we can see a diagram of the forces.
X axis
Tₓ - Fe = 0
Tₓ = Fe
Y axis
[tex]T_{y}[/tex] - W = 0
T_{y} = W
Let's use trigonometry to find the stress components, the angle is measured with respect to the vertical
sin 10 = Tₓ / T
cos 10 = T_{y} / T
Tₓ = T sin 10
T_{y} = T cos 10
we substitute
T sin 10 = Fe
T cos 10 = W
T = mg / cos10
T = 0.120 9.8 / cos 10
T = 1,194 N
Fe = 1,194 sin 10
Fe = 0.2073 N
the electric force is
F = k q₁q₂ / r²
in this case, as the cans touch, they have the same charge and the distance r is searched for by trigonometry
sin 10 = y / L
y = L sin 10
y = 0.30 sin 10
y = 0.052 m
this is the distance from the vertical to one can the distance between the two cans is
y_total = 2y
y_totlal = 2 0.052 = 0.104 m
Fe = F = k q² / r²
q = √ (F r² / k)
q = √ (0.2073 0.104²/9 10⁹) = √ (24.913 10⁻¹⁴)
q = 4.99 10⁻⁷ C
This is the charge of a can, as the transfer is carried out by contact, the flannel transfers half of its charge to the cans and these when separating face one keeps half of the transferred charge, therefore the total charge of the flannel is
Q_total = 4 q
Q_total = 19.97 10⁻⁷ C
Q_total = 2 10⁻⁶ C
what are 7 examples of potential energy
Answer:
Hewo Otaku Kun Here! (UwU)
Explanation:
1. A rock sitting at the edge of a cliff has potential energy. If the rock falls, the potential energy will be converted to kinetic energy.
2. Tree branches high up in a tree have potential energy because they can fall to the ground.
3. A stick of dynamite has chemical potential energy that would be released when the activation energy from the fuse comes into contact with the chemicals.
4. The food we eat has chemical potential energy because as our body digests it, it provides us with energy for basic metabolism.
5. A stretched spring in a pinball machine has elastic potential energy and can move the steel ball when released.
6. When a crane swings a wrecking ball up to a certain height, it gains more potential energy and has the ability to crash through buildings.
7. A set of double "A" batteries in a remote control car possess chemical potential energy which can supply electricity to run the car.
happy to help!
from: Otaku Kun ^^
Calculate the change in momentum of a 0.5kg ball that strikes the floor at 15 m/s and bounces back up at 12 m/s
Answer:
The change in momentum is: [tex]13.5\,\frac{kg\,m}{s}[/tex]
Explanation:
Let's define that vectors pointing up are positive, and vectors pointing down negative.
Then we express the initial momentum of the ball as the product of its mass times the velocity, and include the negative sign since this momentum is pointing down (velocity vector is pointing down):
[tex]P_i=-0.5\,(15) \frac{kg\,m}{s} =-7.5\, \frac{kg\,m}{s}[/tex]
the final momentum is positive (pointing up) and given by the product:
[tex]P_f=0.5\,(12) \frac{kg\,m}{s} =6\, \frac{kg\,m}{s}[/tex]
Therefore, since the change in momentum is defined as the difference between the final momentum minus the initial one, we get:
[tex]P_f-P_i=6-(-7.5)\,\frac{kg\,m}{s} =13.5\,\frac{kg\,m}{s}[/tex]
A baseball is thrown from the outfield toward the catcher. When the ball reaches its highest point, which statement is true? (A) Its velocity is not zero, but its acceleration is zero. (B) Its velocity and its acceleration are both zero. (C) Its velocity is perpendicular to its acceleration. (D) Its acceleration depends on the angle at which the ball was thrown. (E) None of the above statements are true.
Answer:
C. Its velocity is perpendicular to its acceleration
Explanation:
Because acceleration is always perpendicular to the velocity when the velocity will change direction without change it's magnitude