Answer:
t = 12.6 seconds
Explanation:
Given that,
Initial velocity, u = 0
Acceleration of an electric vehicle, a = 2.3 m/s²
Final velocity, v = 29 m/s
We need to find the time elapses from start to stop. The acceleration of an object is given by the relation as follows :
[tex]a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{29-0}{2.3}\\\\t=12.6\ s[/tex]
So, 12.6 seconds is elapsed from start to stop.
Micro-bats use a form of radar called echolocation to navigate and find their prey such as flying insects. They locate the surrounding objects by bouncing sound wave pulses off these objects and detecting the time delay between the emitted pulses and the reflected pulses. Determine the time delay between the pulse emitted by the micro-bat and the detected pulse reflected from an insect located 10 m away from the micro-bat. Assume the approximate speed of sound waves to be 340 m-s-1
Answer:
t = 5.88 10⁻² s
Explanation:
The speed of the sound wave after it is emitted by the bat is constant, so we can use the uniform motion relationships
v = [tex]\frac{x}{t}[/tex]
t = [tex]\frac{x}{v}[/tex]
in this case the distance is that of the sound in going from the bat to the insect and back
x = 2d
x = 2 10
x = 20 m
let's calculate
t = 20/340
t = 5.88 10⁻² s
We can see that the time is very short, so the distance traveled by the two animals has little influence on the result.
During a storm, a car traveling on a level horizontal road comes upon a bridge that has washed out. The driver must get to the other side, so he decides to try leaping the river with his car. The side of the road the car is on is18.0 m above the river, while the opposite side is only1.8 m above the river. The river itself is a raging torrent69.0 m wide.Part AHow fast should the car be traveling at the time it leaves the road in order just to clear the river and land safely on the opposite side?Part BWhat is the speed of the car just before it lands on the other side?
Answer:
Part A: The speed the car should be travelling when leaping the river is approximately 37.948 m/s
Part B: The speed of the car just before it lands is approximately 41.92345 m/s
Explanation:
The parameters of the car attempting leaping the river are;
The height of the car over the river = 18 m
The height of the opposite side of the bridge above the river = 1.8 m
The width of the river, x = 69.0 m
Part A
The time, 't' it would take the car to fall from 18 m above the river to 1.8 m above the river is given as follows;
t = √(2·h/g)
Where;
h = The height of the fall = 18 m - 1.8 m = 16.2 m
g = The acceleration due to gravity = 9.8 m/s²
∴ t = √(2×16.2 m/(9.8 m/s²)) = (9/7)·√2 s
The horizontal speed, 'vₓ', with which the car should be travelling at the time it leaves the road in order just to clear the river and land safely on the opposite side is given as follows;
vₓ = x/t = 69.0 m/((9/7)·√2 s) = (161/6)·√2 m/s ≈ 37.948 m/s
The horizontal speed the car should be travelling when leaping the river, vₓ ≈ 37.948 m/s
Part B;
The vertical velocity of the car is given as follows;
[tex]v_y[/tex]² = [tex]u_y[/tex]² + 2·g·h
∴ [tex]v_y[/tex]² =2·g·h = 2 × 9.8 m/s² × 16.200 m = 317.52m²/s²
[tex]v_y[/tex] = √(317.52 m²/s²) = (63/5)·√2 ≈ 17.819 m/s
The magnitude of the speed of the car, 'v', just before it lands is given using Pythagoras' theorem for resultant vectors as follows;
v = √([tex]v_y[/tex]² + vₓ²) = √(317.52 m²/s² +((161/6)·√2 m/s)²) ≈ 41.92345 m/s
The speed of the car just before it lands, v ≈ 41.92345 m/s.
The kinetic theory states that the higher the temperature, the faster the
Answer: the higher the kinetic energy
Explanation:
The superheroine Xanaxa, who has a mass of 65.1 kg , is pursuing the 78.7 kg archvillain Lexlax. She leaps from the ground to the top of a 153 m high building then dives off it and comes to rest at the bottom of a 17.5 m deep excavation where she finds Lexlax and neutralizes him. Does all this bring about a net gain or a net loss of gravitational potential energy
Answer:
There is net loss of gravitational energy .
Explanation:
When Xanaxa is on the ground , her potential energy is assumed to be zero . When she leaps to a height of 153 m , she gains gravitational energy . When she dives and reaches the surface , she loses potential energy and on reaching the ground her potential energy becomes zero . When she further goes down inside ground to a depth of 17.5 m , she loses potential energy further . Her potential energy becomes less than zero or negative .
Ultimately her potential energy changes from zero to negative in the whole process . So there is net loss of potential energy .
1 example of a conductor and 1 example of a insulator in your EVERYDAY world.
Answer: Examples of conductors include metals, aqueous solutions of salts (i.e., ionic compounds dissolved in water), graphite, and the human body. Examples of insulators include plastics, Styrofoam, paper, rubber, glass and dry air.
how do positive and negative acceleration differ?
1. positive acceleration represents an object speeding up; negative acceleration represents an object slowing down
2. positive acceleration moves North or east; negative acceleration moves south or west
3. positive acceleration occurs when there is more velocity than speed; negative acceleration occurs when there is less velocity than speed.
4. positive acceleration occurs when an object changes its speed but not its direction; negative acceleration occurs when an object changes both its speed and direction
Answer:
1. positive acceleration represents an object speeding up; negative acceleration represents an object slowing down
Explanation:
Acceleration is clearly defined as the rate of change of velocity with time. When are body is speeding up as we say, it is accelerating. When a body is coming to rest, it is decelerating.
Positive acceleration occurs when the speed of a moving continues to increase.
Negative acceleration is when the speed of a moving body reduces drastically.
Heart Disease is the number one cause of preventable death in the United States
A True
B False
Heart disease is the leading cause of death for both men and women. This is the case in the U.S. and worldwide. More than half of all people who die due to heart disease are men.
TRUE
Meandering valleylike features on the Moon's surface are called
Answer:
Meandering valley like features on the Moon's surface are called rilles
Explanation:
NOUN
rilles (plural noun)
a fissure or narrow channel on the moon's surface.
Which object would have the greatest acceleration?
Answer:
D
Explanation:
A and C are balanced, B has a resultant force of 5N right, and D has a resultant force of 20N right.
Which subatomic particle is NOT found in the nucleus of an atom? *
protons
neutrons
electrons
Answer:
Electrons
Explanation:
Only Protons and Neutrons are found in the nucleus
how can you rewrite the force formula (f=ma) to solve the acceleration?
The force formula can be rewritten to solve the acceleration as:
acceleration = force/mass.
What is acceleration?Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
From Newton's 2nd law of motion, we can write that:
Force = mass × acceleration
⇒ acceleration = force/mass.
Hence, the force formula can be rewritten to solve the acceleration as:
acceleration = force/mass.
Learn more about acceleration here:
brainly.com/question/12550364
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Two kilograms of air is contained in a rigid wellinsulated tank with a volume of 0.6 m3 . The tank is fitted with a paddle wheel (stirrer) that transfers energy to the air at a constant rate of 10 W for 1h. If no changes in kinetic or potential energy occur, determine a) The specific volume at the final state, in m3 /kg. b) The energy transfer by work, in kJ. c) The change in specific internal energy of the air, in kJ/kg.
Answer:
[tex]0.3\ \text{m}^3/\text{kg}[/tex]
[tex]36\ \text{kJ}[/tex]
[tex]18\ \text{kJ/kg}[/tex]
Explanation:
V = Volume of air = [tex]0.6\ \text{m}^3[/tex]
P = Power = 10 W
t = Time = 1 hour
m = Mass of air = 2 kg
Specific volume is given by
[tex]v=\dfrac{V}{m}\\\Rightarrow v=\dfrac{0.6}{2}\\\Rightarrow v=0.3\ \text{m}^3/\text{kg}[/tex]
The specific volume at the final state is [tex]0.3\ \text{m}^3/\text{kg}[/tex]
Work done is given by
[tex]W=Pt\\\Rightarrow W=10\times 60\times 60\\\Rightarrow W=36000\ \text{J}=36\ \text{kJ}[/tex]
The energy transfer by work, is [tex]36\ \text{kJ}[/tex]
Change in specific internal energy is given by
[tex]\Delta u=\dfrac{Q}{m}+\dfrac{W}{m}\\\Rightarrow \Delta u=0+\dfrac{36}{2}\\\Rightarrow \Delta u=18\ \text{kJ/kg}[/tex]
The change in specific internal energy of the air is [tex]18\ \text{kJ/kg}[/tex]
Which formula is used to find an objects acceleration
Answer:
a=∆v/∆t
Explanation:
The definition of Acceleration is the change in velocity in a given time. So this means you first calculate ∆v (Change in velocity), and you calculate ∆t which is the time taken to apply that change in velocity. Then you find a= ∆v/∆t. This gives us the equation of Acceleration.
Answer:
C. a=∆v/∆tExplanation:
Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the power measured at the detector when only the vertical arm is blocked is 2.25 mW, while the power measured at the detector when only the horizontal arm is blocked is only 2.025 mW. Assume initially the intensity is at its maximum. How much would we need to translate the perfect mirror to the right to get a minimum intensity at detector, and what is that minimum intensity
This question is incomplete, the complete question;
you make an interferometer using 50-50 beam splitter and two mirrors, one being a perfect mirror and one which does not reflect all light. The wavelength of the 9 mW incident laser is 400 nm.
Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the power measured at the detector when only the vertical arm is blocked is 2.25 mW, while the power measured at the detector when only the horizontal arm is blocked is only 2.025 mW. Assume initially the intensity is at its maximum. How much would we need to translate the perfect mirror to the right to get a minimum intensity at detector, and what is that minimum intensity
Options;
a) 200 nm; 0.9 mW
b) 100 nm, 0.0059 mW
c) 200 nm; 0 mW
d) 100 nm; 0.9 mW
e) 200 nm; 0.0059 mW
Answer:
the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector and the minimum intensity are;
100 nm; 0.0059 mW
Option b) 100 nm, 0.0059 mW is the correct answer
Explanation:
Given that the instrument here is an interferometer.
Maximum intensity is obtained when the two waves are exactly in phase.
that is the peaks (crusts and troughs) and nodes (zero value points) of the two waves will be at the exact same point when the wave falls on the detector.
The phase factor of this point is taken as ∅ = 0
Now, to get a minimum point, the phase difference between the two waves should be should be ∅ = π
This corresponds to a path difference between the two waves as half of the wavelength. λ/2
The light gets reflected from the mirror.
Hence, when we move the mirror by a length l, the extra/less path the light has to travel is 2l (light is going and coming back)
hence, to get a path difference of λ/2 the mirror should move half of this distance only
so, the mirror should move;
[tex]l[/tex] = λ/4
here, wavelength is 400nm
the length moved by the mirror = 400/4 = 100 nm
The intensity is given by the equation;
[tex]l[/tex] = [tex]l[/tex]1 + [tex]l[/tex]2 + 2√[tex]l[/tex]1[tex]l[/tex]2cos(∅)
where
[tex]l[/tex]1 = 2.25 mW
[tex]l[/tex]2 = 2.025 mW
∅ = π
so we substitute
[tex]l[/tex] = 2.25 + 2.025 - 2√(2.25 × 2.025)
[tex]l[/tex] = 4.275 - 4.26907
[tex]l[/tex] = 0.0059
Therefore; the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector and the minimum intensity are;
100 nm; 0.0059 mW
Option b) 100 nm, 0.0059 mW is the correct answer
Two children, Ferdinand and Isabella, are playing with a water hose on a sunny summer day. Isabella is holding the hose in her hand 1.0 meters above the ground and is trying to spray Ferdinand, who is standing 10.0 meters away. Will Isabella be able to spray Ferdinand if the water is flowing out of the hose at a constant speed v0 of 3.5 meters per second?
Answer:
Isabella will not be able to spray Ferdinand.
Explanation:
We'll begin by calculating the time taken for the water to get to the ground from the hose held at 1 m above the ground. This can be obtained as follow:
Height (h) = 1 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =.?
h = ½gt²
1 = ½ × 9.8 × t²
1 = 4.9 × t²
Divide both side by 4.9
t² = 1/4.9
Take the square root of both side
t = √(1/4.9)
t = 0.45 s
Next, we shall determine the horizontal distance travelled by the water. This can be obtained as follow:
Horizontal velocity (u) = 3.5 m/s
Time (t) = 0.45 s
Horizontal distance (s) =?
s = ut
s = 3.5 × 0.45
s = 1.58 m
Finally, we shall compare the distance travelled by the water and the position to which Ferdinand is located to see if they are the same or not. This is illustrated below:
Ferdinand's position = 10 m
Distance travelled by the water = 1.58 m
From the above, we can see that the position of the water (i.e 1.58 m) and that of Ferdinand (i.e 10 m) are not the same. Thus, Isabella will not be able to spray Ferdinand.
A swimmer, capable of swimming at a speed of 1.60 m/s in still water (i.e., the swimmer can swim with a speed of 1.60 m/s relative to the water), starts to swim directly across a 1.25-km-wide river. However, the current is 0.549 m/s, and it carries the swimmer downstream. (a) How long does it take the swimmer to cross the river
Answer:
t = 781.25 s
Explanation:
This is an exercise in velocity composition, if we set a reference system where the x-axis is perpendicular to the river and the y-axis is parallel to the river.
The swimmer has a velocity on the x axis
vx = 1.60 m / s
a velocity on the y axis, created by the current of the river
vy = 0.549 m / s
time is a scalar, therefore the time it takes to cross the river is the same time it creates the displacement in e; Axis y
X axis
vₓ = x / t
t = x / vₓ
t = 1250 / 1.6
t = 781.25 s
in this time a distance has descended
y = v_y t
y = 0.549 781.25
y = 428.9 m
A ball of mass 0.3 kg flies through the air at low speed, so that air resistance is negligible. (a) What is the net force acting on the ball while it is in motion
Answer:
X axis F=0
Y axis Fg = - 2.94 j ^
Explanation:
The motion of a ball in air where the air residence is indicated to be negligible can be analyzed using Newton's second law.
We set a reference system, where the x-axis is horizontal and the y-axis vertical.
X axis
There are no forces on this axis, therefore the ball goes at constant speed.
Force is zero
Y axis
In this axis it is subjected to the acceleration of gravity that creates a force equal to the weight of the body, in a vertical direction.
Fg = m g
Fg = 0.3 9.8
Fg = 2.94 N
Fg = - 2.94 j ^
the boold are vectors; negative sign indicates that the force eta directed vertically downward
What is the potential energy of an object 20 m in the air with a
mass of 600 kg?
Answer:
Ep = 117600 J
Explanation:
Data:
Mass (m) = 600 kgHeight (h) = 20 mGravity (g) = 9.8 m/s²Potential Energy (Ep) = ?Use formula:
Ep = m * g * hReplace:
Ep = 600 kg * 9.8 m/s² * 20 mMultiply operations, and units:
Ep = 117600 JWhat is the potential energy?
The potential energy is 117600 Joules.
The force of friction occurs primarily because:
A) two surfaces in contact have magnetic forces of attraction.
B) on the microscopic level, two surfaces in contact are rough even if they appear smooth to the touch.
C) two surfaces in contact have a gravitational attraction to one another.
D) both A and B.
Answer:
B
Explanation:
Friction is a force that opposes motion between any surfaces that are touching. Friction occurs because no surface is perfectly smooth Friction produces heat because it causes the molecules on rubbing surfaces to move faster and have more energy.
Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.0988 N when their center-to-center separation is 44.5 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0276 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other
Answer:
(a) The negative charge on one of the charges is -8.79630245 × 10⁻⁷C
(b) The positive charge on one of the other charges is 8.79630245 × 10⁻⁷C
Explanation:
The given parameters are;
The force of attraction between the two spheres = 0.0988 N
The distance between their centers = 44.5 cm = 0.445 m
Therefore, we have;
[tex]F = \dfrac{k \cdot q_1 \cdot q_2}{d^2}[/tex]
Therefore, we have;
[tex]0.0988 \ N = -\dfrac{8.99 \times 10^9 N\cdot m^2 \cdot C^{-2}\cdot q_1 \cdot q_2}{(0.445 \ m)^2}[/tex]
Therefore, we have;
q₁·q₂ = -0.0988 N × (0.445 m)²/(8.99 × 10⁹ N·m²·C⁻²) = -2.17629255 × 10⁻¹² C²
q₁·q₂ = -2.17629255 × 10⁻¹² C²...(1)
When the two charges are connected, we get;
[tex]F_2 = \dfrac{k \cdot \left (\dfrac{q_1 + q_2}{2} \right) ^2}{d^2}[/tex]
Therefore, we have;
[tex]q_1 + q_2 = \sqrt{\dfrac{4 \cdot F_2 \cdot d^2}{k} }[/tex]
[tex]q_1 + q_2 = \sqrt{\dfrac{4 \times 0.0276 \ N \times(0.445 \ m)^2}{8.99 \times 10^9 N\cdot m^2 \cdot C^{-2}} } = 1.59446902743 \times 10^{-6} \ C[/tex]
q₁ + q₂ = 1.59446902743 × 10⁻⁶ C...(2)
From, equation (2), we have;
q₁ = 1.59446902743 × 10⁻⁶ C - q₂
Plugging in the value of q₁ in equation (1) givens;
q₁·q₂ = -2.17629255 × 10⁻¹²
Therefore, we have;
(1.59446902743 × 10⁻⁶ - q₂) × q₂ = -2.17629255 × 10⁻¹²
Which gives;
-q₂² + 1.59446902743 × 10⁻⁶·q₂+2.17629255 × 10⁻¹² = 0
Solving, with a graphing calculator, we get;
q₂ = 2.4741×10⁻⁶ C, or -8.79630245 × 10⁻⁷C
q₁ = 8.79630245 × 10⁻⁷C or -2.4741×10⁻⁶ C
Therefore, we have;
(a) The negative charge on one of the charges = -8.79630245 × 10⁻⁷C
(b) The positive charge on one of the other charges = 8.79630245 × 10⁻⁷C
How does earths magnetic field work
Answer: On Earth, flowing of liquid metal in the outer core of the planet generates electric currents. The rotation of Earth on its axis causes these electric currents to form a magnetic field which extends around the planet.
Explanation:
Answer:
the rotation of earth on its axis causes electric currents to form a magnetic field which extends around the planet
8) A train enters a curved horizontal section of the track at a speed of 100 km/h and slows down with constant deceleration to 50 km/h in 12 seconds. If the total horizontal acceleration of the train is 2 m/s2 when the train is 6 seconds into the curve, calculate the radius of curvature of the track for this instant.
Answer:
the radius of curvature of the track for this instant is 266 m
Explanation:
Given that;
The Initial Velocity u = 100 km/h = 100 × [tex]\frac{5}{18}[/tex] = 27.77 m/s
velocity of the train at t=12 s is;
[tex]V_{t=12}[/tex] = 50 km/h = 50 × [tex]\frac{5}{18}[/tex] = 13.89 m/s
now, we calculate the deceleration of the train
[tex]V_{t=12}[/tex] = u + at
13.89 = 27.77 + [tex]a_{t}[/tex]12
[tex]a_{t}[/tex] = (13.89 - 27.77) / 12
[tex]a_{t}[/tex] = -13.88 / 12
[tex]a_{t}[/tex] = - 1.1566 m/s²
Now, the velocity of the train at 6 seconds is;
[tex]V_{t=6}[/tex] = u + at
[tex]V_{t=6}[/tex] = 27.77 + ( - 1.1566 m/s²)6
[tex]V_{t=6}[/tex] = 27.77 - 6.9396
[tex]V_{t=6}[/tex] = 20.83 m/s
The acceleration at t=6 s is;
a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]
a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]
we substitute
2m/s² = √[ (- 1.15 )² + ([tex]a_{n}[/tex])²]
4 = (- 1.1566 )² + ([tex]a_{n}[/tex])²
4 = 1.3377 + ([tex]a_{n}[/tex])²
([tex]a_{n}[/tex])² = 4 - 1.3377
([tex]a_{n}[/tex])² = 2.6623
[tex]a_{n}[/tex] = √2.6623
[tex]a_{n}[/tex] = 1.6316 m/s²
Now the radius of curve is;
a = V² / p
[tex]p_{t=6}[/tex] = ( [tex]V_{t=6}[/tex] )² / [tex]a_{n}[/tex]
[tex]p_{t=6}[/tex] = ( 20.83 m/s )² / 1.6316 m/s²
[tex]p_{t=6}[/tex] = 433.8889 / 1.6316
[tex]p_{t=6}[/tex] = 265.9 m ≈ 266 m
Therefore; the radius of curvature of the track for this instant is 266 m
6th grade science I mark as brainliest !
A motorcycle reaches the speed of 40 m / s, how far does it travel in 10 seconds?
Answer:
d = 200 m
Explanation:
Data:
Initial Velocity (Vo) = 0 m/s Final Velocity (Vf) = 40 m/s Time (t) = 10 s Distance (d) = ?Use formula:
[tex]\boxed{d=\frac{Vf+Vo}{2}*t}[/tex]Replace:
[tex]\boxed{d=\frac{40\frac{m}{s}+0\frac{m}{s}}{2}*10s}[/tex]Sum in the numerator:
[tex]\boxed{d=\frac{40\frac{m}{s}}{2}*10s}[/tex]It divides:
[tex]\boxed{d=20\frac{m}{s}*10s}[/tex]Simplify the seconds (s), and multiply:
[tex]\boxed{d=200\ m}[/tex]How far does it go?
Travel a distance of 200 meters.
A star can give off white light. Why is this evidence that a star is a blackbody
radiator?
A. White light is made up of many different wavelengths of light.
B. The star reflects the white light.
C. The star absorbs the white light.
D. White light is only one wavelength of light.
O
Answer:
It's A. White light is made up of many different wavelengths of light.
which is the product of cellular respiration? A. ATTP B. light C. oxygen D.sugar
Bungee jumping is an example of
A. wind resistance and insanity
B. gravitational and air pressure energy
C. gravitational and elastic energy
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton's law of cooling. If the coffee has a temperature of 205 degrees Fahrenheit when freshly poured, and 2.5 minutes later has cooled to 195 degrees in a room at 70 degrees, determine when the coffee reaches a temperature of 160 degrees.
Answer:
Tt = 70 + 135e^-0.031t
13 minutes
Explanation:
Given that :
Initial temperature, Ti = 205°
Temperature after 2.5 minutes = 195°
Temperature of room, Ts= 70
Using the relation :
Tt = Ts + Ce^-kt
Temperature after time, t
When freshly poured, t = 0
205 = 70 + Ce^-0k
205 = 70 + C
C = 205 - 70 = 135°
T after 2.5 minutes to find proportionality constant, k
Tt = Ts + Ce^-kt
195 = 70 + 135e^-2.5k
125 = 135e^-2.5k
125 / 135 = e^-2.5k
0.9259 = e^-2.5k
Take In of both sides :
−0.076989 = - 2.5k
k = −0.076989 / - 2.5
k = 0.031
Equation becomes :
Tt = 70 + 135e^-0.031t
t when Tt = 160
160 = 70 + 135e^-0.031k
90 = 135e^-0.031t
90/135 = e^-0.031t
0.6667 = e^-0.031t
In(0.6667) = - 0.031t
−0.405465 = - 0.031t
t = 0.405465/ 0.031
t = 13.071
t = 13 minutes
Fusion probability is greatly enhanced when appropriate nuclei are brought close together, but mutual Coulomb repulsion must be overcome. This can be done using the kinetic energy of high-temperature gas ions or by accelerating the nuclei toward one another. Calculate the potential energy of two singly charged nuclei separated by 1.00 x 10-12 m by finding the voltage of one at that distance and multiplying by the charge of the other.
Answer:
the Potential Energy is 2.304 × 10⁻¹⁶ J
Explanation:
Given the data in the data in the question;
The expression for the electric potential energy between the charges can be expressed as follows;
PE = qV ------equ 1
where q is the charge and V is the electric potential
Also the formula for electric potential due to point a point in a field is;
V = kq / r -------equ 2
where k is the electrostatic constant and r is the distance form the charged particle
input equation 2 into 1
PE = q × kq / r
PE = kq²/r ------- equ 3
so we substitute into equation 3; 1.00×10⁻¹² for r, 9.00×10⁹ for k( constant ) and 1.60×10⁻¹⁹ for q( charge )
PE = ((9.00×10⁹) (1.60×10⁻¹⁹)²) / 1.00×10⁻¹²
PE = 2.304 × 10⁻²⁸ / 1.00×10⁻¹²
PE = 2.304 × 10⁻¹⁶ J
Therefore, the Potential Energy is 2.304 × 10⁻¹⁶ J
An electron moves from point i to point f, in the direction of a uniform electric field. During this motion:Group of answer choicesthe work done by the field is positive and the potential energy of the electron-field system increasesthe work done by the field is negative and the potential energy of the electron-field system increasesthe work done by the field is positive and the potential energy of the electron-field system decreasesthe work done by the field is negative and the potential energy of the electron-field system decreasesthe work done by the field is positive and the potential energy of the electron-field system does not change
Answer:
the work done by the field is positive and the potential energy of the electron field system decreases
Explanation:
This exercise asks to find the work and the potential energy of an electron in an electric field.
Work is defined by
W = F .d = F d cos θ
the electric force is
F_e = q E
W = q E d cos θ
since the charge of the electron is negative the force is in the opposite direction to the electric field
W = - e E d
we select the direction to the right is positive, point i is to the left of point f,
therefore the work moving from point i to point F has two possibilities
* The electric field lines go from i to f point , so that point i is on the side of the positive charges, so the electron approaches them, This movement is opposite to that indicated
* the field line reaches point i, this implies that the charges are negative, so the electrioc field is then negativeand the electron charge is negative too. The electron moves away from this point, this is in accordance with the indicated movement
In the latter case the electric field lines go from f to i point, therefore the Work is positive
Now let's examine the potential energy
ΔU = - q E .d
so we see that this definition is related to work,
ΔU = -W
Therefore, as the work is positive, the power energy must decrease
When reviewing the different answers, the correct ones are:
the work done by the field is positive and the potential energy of the electron field system decreases
The work done by the electron while moving from point [tex]i[/tex] to point [tex]f[/tex] in the direction of uniform electric field is negative and the potential energy of the electron increases.
An electron moves from point i to point f, in the direction of a uniform electric field, then the potential energy of the electron can be calculated s given below.
[tex]\Delta V=-qEd[/tex]
Where [tex]\Delta V[/tex] is the potential energy, [tex]E[/tex] is the electric field, [tex]q[/tex] is the charge and [tex]d[/tex] is the displacement of the electron.
The work done by the electron in the uniform electric field can be calculated as,
[tex]W = F\times d \times cos\theta[/tex]
Where [tex]W[/tex]is the work done by electron, [tex]F[/tex] is the electric force, [tex]d[/tex] is the displacement of the electron and for uniform electric field, the value of [tex]\theta[/tex] is zero.
Hence [tex]W=F\times d\times 1\\W=F \times d[/tex]
Electric force [tex]F = q E[/tex]
By substituting the value of electric force on the above formula,
[tex]W = qEd[/tex]
Hence, the relation between the work done the electron in an uniform electric field and potential energy of the electron can be given below.
[tex]W = -\Delta V[/tex]
The work done by the electron is negative and the potential energy of the electron increases.
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