The final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.
Time of motion of the girl
The time of motion of the girl is calculated as follows;
h = vt + ¹/₂gt²
where;
v is initial vertical velocity = 0t is time of motiong is acceleration due to gravitySubstitute the given parameters and solve for time of motion;
50.8 = 0 + ¹/₂(9.8)t²
2(50.8) = 9.8t²
101.6 = 9.8t²
t² = 101.6/9.8
t² = 10.367
t = √10.367
t = 3.22 seconds
Final vertical velocity of the skydivervf = vi + gt
where;
vi is the initial vertical velocity = 0
vf = 0 + 9.8(3.22)
vf = 31.56 m/s
Thus, the final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.
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As mass increases, which of the following is always true?
a. volume decreases
b. density decreases
c. inertia increases
d. all of these
As mass increases, the following which is found to be always true is a corresponding increase in the inertia and is therefore denoted as option C.
What is Inertia?This is defined as the property of a body to resist changes in velocity or motion. Such feature is characterized by the body having the tendency to remain at rest or in uniform motion unless an external force acts on it.
When mass increase, inertia also increases due to the large amount of force which is needed to change the direction or motion of a body in this scenario. On the other hand, when mass decreases, there is a corresponding decrease in the inertia which is present in a given body over a period of time.
This is therefore the reason why mass of a body is directly proportional to its inertia and makes it the most appropriate choice.
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2. Three blocks, A,B and C of mass 2kg. 3kg. 5kg respectively kept side by side with one another are accelerated at 2m/s2 across a smooth horizontal surface by a force F evaluated the magnitude of F and what force the blocks exert on each other.
Answer:
Total mass of combination = 2+3+5 = 10kg.
Acceleration produced = 2m/s^2
hence force =( total mass × acceleration)= (2×10)= 20 N.
Net force on 3kg block = acceleration × mass = (2 × 2 )= 4 N
applied force on 2 kg block = 20N
Force between 2 kg and 3 kg block = (20-4) = 16N. ans
Net force on 3 kg block = 3 × 2 =6N.
Applied force on 3 kg block due to 2 kg block = 16N.
hence, force between 3 kg and 5 kg block = (16-6) = 10N .
answers:-
(a) 20 N
(b) 16N
(c) 10 N
7. Four perpendicular forces, 2 N, 3 N, 9 N and 10 N act on object P as shown in Diagram 2. 10 N A B 9 N 3 N 2 N P Diagram 2 What is the magnitude of the resultant force? 10 N C 18 N 14 N D 20 N
Answer: R=
A
2
+B
2
+2ABcosθ
θ=90
o
cos90
o
=0
R=
A
2
+B
2
=
3N
2
+4N
2
=5N
Explanation:
Three bees are oriented as shown in the figure. B1 has +17 µC of charge, B2 has −5 µC of charge, and B3 has +26 µC of charge.
B1 to b2 5 cm
B2 to b3 10 com
Bee number 3 is stationed at the observation location for this problem, and we want to find the net electric field at B3. We'll do this in a few steps.
(a) What are the x and y components of the electric field
at the observation location (Bee number 3) due to B1? Remember that the components can be positive or negative depending on their directions along the x or y axis.
E1x =
E1y =
(b) What are the x and y components of the electric field
at the observation location (Bee number 3) due to B2? Again, remember that the components can be positive or negative depending on their directions along the x or y axis.
E2x = N/C
E2y =
(c) What are the magnitude and direction of the net electric field at the observation location where B3 is resting?
magnitude
N/C
direction
° counterclockwise from the +x-axis
(d) What is the magnitude of the force on B3 due to this net electric field?
Given the following data:
Charge of B₁ = +17 µC to C = 17 × 10⁻⁶ C.Charge of B₂ = -5 µC to C = -5 × 10⁻⁶ C.Charge of B₃ = +26 µC to C = 26 × 10⁻⁶ C.Radius of B₁ to B₂ = 5 cm to m = 0.05 meter.Radius of B₂ to B₃ = 10 cm to m = 0.1 meter.How to determine the x and y components of the electric field?First of all, we would calculate the electric field due to B₁ to B₃ and the electric field due to B₂ to B₃ respectively.
The electric field due to B₁ to B₃ is given by:
E₁ = kq₁/r₁²
E₁ = (9 × 10⁹ × 17 × 10⁻⁶)/(0.05² + 0.1²)
E₁ = 153 × 10³/0.0125
Electric field, E₁ = 12.24 × 10⁶ N/C.
The electric field due to B₂ to B₃ is given by:
E₂ = kq₂/r₂²
E₂ = (9 × 10⁹ × 5 × 10⁻⁶)/(0.1²)
E₂ = 45 × 10³/0.01
Electric field, E₂ = 4.5 × 10⁶ N/C.
Also, the magnitude of the angle formed is given by tan trigonometry:
Tanθ = 5/10
Tanθ = 0.5
θ = tan⁻¹(0.5)
θ = 26.56°.
Next, we would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₁:
E₁x = E₁cosθ - E₂
E₁x = 12.24 × 10⁶ × cos(26.56) - 4.5 × 10⁶
E₁x = 10.95 × 10⁶ - 4.5 × 10⁶
E₁x = 6.5 × 10⁶ N/C.
Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:
E₁y = -E₁sinθ
E₁y = -12.24 × 10⁶ × sin(26.56)
E₁y = -12.24 × 10⁶ × 0.4471
E₁y = -5.5 × 10⁶ N/C.
Also, the magnitude of the electric field is given by:
Exy = √(E₁x² + E₁y²)
Exy = √(6.5 × 10⁶)² + (-5.5 × 10⁶)²)
Exy = √4.225 × 10¹³ + 3.025 × 10¹³)
Exy = √(7.25 × 10¹³)
Exy = 8.5 × 10⁶ N/C.
Part B.We would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₂:
E₂x = -E₂sinθ
E₂x = -4.5 × 10⁶ × sin(90)
E₂x = -4.5 × 10⁶ N/C.
Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:
E₂y = E₂cosθ
E₂y = -4.5 × 10⁶ × cos(90)
E₂y = 0 N/C.
How to calculate the net electric field at the observation location?The magnitude of the net electric field at B₃ is given by:
Eₙ = √(E₁x + E₂x)² + E₁y²)
Eₙ = √(6.5 × 10⁶ + (-4.5 × 10⁶))² + (-5.5 × 10⁶)²)
Eₙ = √(2.5 × 10⁶)² + (3.025 × 10¹³)
Eₙ = √(6.25 × 10¹²) + (3.025 × 10¹³)
Eₙ = √(3.65 × 10¹³)
Eₙ = 6.04 × 10⁶ N/C.
For the direction, we have:
Tanθ = x/y
Tanθ = 6.5/-5.5
Tanθ = -1.1818
θ = tan⁻¹(-1.1818)
θ = 49.76°.
Part C.The magnitude of the force on B₃ due to this net electric field is given by:
F = B₃ × Eₙ
F = 26 × 10⁻⁶ × 6.04 × 10⁶
F = 157.04 Newton.
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A grinding wheel 0.35 m in diameter rotates at 2600 rpm .
a) Calculate its angular velocity in rad/s .
Express your answer using three significant figures.
b)What is the linear speed of a point on the edge of the grinding wheel?
Express your answer to two significant figures and include the appropriate units.
c)What is the acceleration of a point on the edge of the grinding wheel?
Express your answer to two significant figures and include the appropriate units.
It is calculated that a) The angular velocity of the wheel is 272.13 rad/s,
b) On the edge of the grinding wheel, the linear speed is 47.62 m/s,
and c) On the edge of the grinding wheel, the acceleration is 12958.08 m/s².
Calculation of angular velocity, linear speed & acceleration:
Provided that,
the diameter of the wheel = 0.35 m
So, the radius, r = 0.35/2 = 0.175 m
As 1 revolution = 2π rad
(a) the angular velocity, ω = 2600 rpm = [tex]\frac{2600 * 2\pi }{60}[/tex] rad/s
⇒ω = 272.13 rad/s
So, the angular velocity is 272.13 rad/s.
(b) The linear speed, v = r * ω
⇒v = 0.175 * 272.13
⇒v= 47.62 m/s
(c) The angular acceleration, [tex]a=\frac{v^{2} }{r}[/tex]
[tex]a = \frac{(47.62)^{2} }{0.175}[/tex]
⇒[tex]a[/tex] = 12958.08 m/s²
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The y-position of a damped oscillator as a function of time is shown in the figure.
This function can be described by the y(t) = [tex]A_{0}[/tex][tex]e^{-btx}[/tex]cos(ωt) formula, where [tex]A_{0}[/tex] is the initial amplitude, b is the damping coefficient and ω is the angular frequency.
1. What is the period of the oscillator? Please, notice that the function goes through a grid intersection point.
2. Determine the damping coefficient.
(1) The period of the oscillator is 1 second.
(2) The damping coefficient is 0.93.
What is period of oscillation?
The period of oscillation is the time taken to make one complete cycle.
From the graph, the time taken to make one complete oscillation is 1 second.
Damping coefficientequation of the wave is given as;
y(t) = Ae^(-btx) cos(ωt)
at time, t = 0, y = 3.53.5 = Ae^(-0) cos(0)
3.5 = A x 1
A = 3.5 cm
at time, t = 1 cm, y = - 3cm-3 = 3.5e^(-bx) cos(ω)
-3/3.5 = e^(-bx) cos(ω)
-0.857 = e^(-bx) cos(ω)
-0.857 / cos(ω) = e^(-bx)
ln[-0.857 / cos(ω)] = -bx
ln[-0.857 / cos(ω)] / b = - x ---- (1)
at time, t = 2 cm, y = - 2cm-2 = 3.5e^(-2bx) cos(2ω)
-0.57 = e^(-2bx) cos(2ω)
ln[-0.57 / cos(2ω)] = -2bx
ln[-0.57 / cos(2ω)] /2b = - x ------(2)
solve (1) and (2)
ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b
-0.57 / cos(ω) = 2(-0.857 / cos(ω))
2(-0.857/cosω) = -0.57/cos2ω
-(2 x 0.857) / (-0.57) = cosω/cos 2ω
3 = cosω/cos 2ω
3(cos 2ω) = cosω
3(2cos²ω - 1) = cos ω
6cos²ω - 6 = cosω
6cos²ω - cosω - 6 = 0
let cosω = y
6y² - y - 6 = 0
solve the quadratic equation;
y = 1.1 or -0.92
cosω = -0.92
ω = arc cos(-0.92)
ω = 2.74 rad/s
From equation (1)
ln[-0.857 / cos(ω)] / x = -b ---- (1)
let x = 1
ln(-0.857/cos(2.74) = -b
-0.93 = -b
b = 0.93
Thus, the damping coefficient is 0.93.
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Please help 25 points
Asteroids X, Y, and Z have equal mass of 3.0 kg each. They orbit around a planet with M = 7.20×10^24 kg. The orbits are in the plane of the paper and are drawn to scale.
The three asteroids orbit in the same counterclockwise direction.
The angular velocity of X at i is .... that of Y at i.
The angular velocity of Y at c is .... that at i.
The angular momentum of Z at c is .... that at v.
The angular velocity of X at i is .... that at r.
The angular momentum of Y is .... that of X.
The period of Z is .... that of X.
The period of Y is .... that of X.
(Options are: greater than, less than, equal to)
This is a series of analysis of Angular velocity as relates to an asteroid's orbit. See the explanation below.
What is an asteroid?Asteroids are stony bodies that circle the Sun.
Although asteroids circle the Sun in the same way as planets do, they are far smaller.
Hence, from the information given:
A) The square of the period is proportional to the cube of the semimajor axis, according to Kepler's third law. As a result, the period of Y equals (E) the period of Z.
B) Angular momentum is preserved here, hence it is equivalent (E).
C) As eccentricity increases, so does the angular momentum. In this case, Y and Z have the same period, and both satellites cover the same proportion of the territory in the same length of time.
This indicates that a satellite on Z must cover a lesser area in a given period of time than a satellite on Y. The area swept is approximately 1/2 the radius times the tangential displacement.
Because both satellites have the same "radius" at point y, the satellite on Z must have a lower tangential velocity than the one on Y. As a result, Y has more angular momentum than (G) Z.
D) Using Kepler's third law, X's period is bigger than (G) Z's.
E) In a circular orbit, the angular velocity is constant. As a result, the angular velocity of Y at y equals (E) that at s.
F) Z's angular velocity at c is smaller than (L) at i.
G) Y's angular velocity at y is larger than (G) Z's at y.
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Can we take water instead of clock oil in Milikan oil drop experiment. Explain.
Answer:
No you can't cuz,if you put water instead of clock oil in Millikan oil drop your experiment will fail and it won't turn out the way you wanted it to be
I will run one mile in under 10 minutes 3 months from today.” is an example of a ___________ goal.
Long-term
Short-term
SMART
Both A and B
Answer:
long-term
Explanation:
usually a short term goal is able to be accomplished in a week or two. the question gives a 3 month time frame for the person to build up to the end goal.
A light bulb in a camper’s flashlight is labeled 4.1 V, 0.40 A. Find the equivalent current if three of these light bulbs are connected in parallel to a standard C size 1.5 V battery.
Explanation:
first rule of parallel circuits : the voltage is the same in the whole circuit, no matter where we measure.
so, we have 1.5V at every bulb.
second rule of parallel circuits : the total current is the sum of the individual branch currents.
so, if the battery really allows it, then we have 3 times 0.4 A = 3×0.4 = 1.2 A as current in the circuit.
accelerates uniformly at 2.0 ms2 for 10.0s. Calculate its final velocity
Answer:
The distance is
=
7
m
Explanation:
Apply the equation of motion
s
(
t
)
=
u
t
+
1
2
a
t
2
The initial velocity is
u
=
0
m
s
−
1
The acceleration is
a
=
2
m
s
−
2
Therefore, when
t
=
3
s
, we get
s
(
3
)
=
0
+
1
2
⋅
2
⋅
3
2
=
9
m
and when
t
=
4
s
s
(
4
)
=
0
+
1
2
⋅
2
⋅
4
2
=
16
m
Therefore,
The distance travelled in the fourth second is
d
=
s
(
4
)
−
s
(
3
)
=
16
−
9
=
7
m
8- A bottle of water weighs 25 pounds. If the bottle weighs 5 pounds, how many gal does it contain ?
A bottle of water weighs 25 pounds. If the bottle weighs 5 pounds, then it contains 0.599132136584 gal .
A bottle weighs 25 pounds ,
we know that 1 Gallon = 8.34 Lb
so, 25 lb = 2.995660682922 gal
and if the bottle weighs 5 pounds then ,
5 lb = 0.599132136584 gal
hence, 2.4 gal less.
Define pound.
a. a unit of weight equivalent to l6 ounces avoirdupois (453.59237 grams), the fundamental unit of weight in the FPS system;
b. a unit of weight equal to 12 ounces troy or 12 ounces apothecaries' (373.2418 grams).
It is an imperial unit of mass or weight measurement.
Define gallons.
In both imperial and US customary units, the gallon is a unit of volume.
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Find the orbital speed of an ice cube in the rings of Saturn. The mass of Saturn is 5.68 x 1026 kg, and use an orbital radius of 3.00 x 105 km. (G = 6.67 × 10-11 N ∙ m2/kg2)
19.5 km/s
27.5 km/s
11.2 km/s
20.5 km/s
The orbital speed of an ice cube in the rings of Saturn is 355358.97m/s
Law of gravitationAccording to the gravitation law, the force of gravitation is directly proportional to the product of the masses and inversely proportional to the distance between them. Mathematically;
F = GMm/r²
where
m = mass of ice cube and
s = Gm1/r^2
Hence,
F = sm2
On rearranging,
s = m2/F
let V = orbital speed
centripetal acceleration = V^2/r
Such that;
V²/r = Gm/r²
V² = Gm/r
V = √Gm/r
Substitute the given parameters
V = √6.67×10^-11 * 5.68 x 10^26 / 3.00 x 10^5
V = 355358.97m/s
Hence the orbital speed of an ice cube in the rings of Saturn is 355358.97m/s
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How much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 10^5 km?
The amount of time required for reflected sunlight to travel from the moon to the earth according to the given distance between then and the speed of light is; 1.28 seconds.
How long does it take reflected sunlight to each the earth from the moon?It follows from the task content that the time required for the reflected sunlight to reach the earth can be evaluated if the speed of light is known in which case, the speed of light, c = 3× 10⁸ m/s.
Consequently, we have time taken, t evaluated as follows;
Time taken, t = distance/speed
t = 3.85 ×10⁵ × 10³/3× 10⁸
t = 3.85/3
t = 1.28 seconds.
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Please help, Class ends today!! will give brainiest if correct.
Fill in the blanks with agree or disagree
A list of planet-star radius ratios is an example of evidence.
I ____________ with this statement
Explain your answer, using an example from your investigation about habitable worlds.
A list of planet-star radius ratios is an example of evidence. I agree with this statement.
Why do i agree with the statement above?I agree with A list of planet-star radius ratios is an example of evidence because the Sample records that are taken for planet-star radius ratio are known to often gives a kind of an improved approximate of the stellar characteristics in regards to the smallest stars in the Kepler target list.
What does a planets radius mean?The planetary radius is known to be the distance that exist between a what we see as planet's center and its surface.
Therefore, one can say that planetary radius is known to be a measure of a given planet's size.
So, A list of planet-star radius ratios is a sample evidence and I do affirm the above statement.
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A centrifuge in a medical laboratory rotates at an angular speed of 3,750 rev/min. When switched off, it rotates through 48.0 revolutions before coming to rest. Find the constant angular acceleration (in rad/s²) of the centrifuge.
______rad/s²
The constant angular acceleration (in rad/s²) of the centrifuge is 255.66 rad/s².
Constant angular acceleration of the centrifuge
The constant angular acceleration of the centrifuge is calculated as follows;
ωf² = ωi² - 2αθ
where;
ωf is the final angular velocity ωi is initial angular velocityθ is angular displacementα is angular accelerationWhen the centrifuge is switched off, the final angular velocity = 0
Initial angular velocity: ωi = 3,750 rev/min x 2π rad/rev x 1 min/60 s = 392.7 rad/s
angular displacement: θ = 48 rev = 48 rev x 2π rad/rev = 301.6 rad
0 = ωi² - 2αθ
2αθ = ωi²
α = ωi²/2θ
α = (392.7²) / (2 x 301.6)
α = 255.66 rad/s²
Thus, the constant angular acceleration (in rad/s²) of the centrifuge is 255.66 rad/s².
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Suppose that the math man, a super here that fights crime with math and physics, can decelerate the rate of gravity. During a recent fight with a diamond thief, Math man fell from the top of a 500 meter building. The equation D=t^2*.25. How long us he falling from the top of the building to the ground?
Answer:
44.7 seconds
Explanation:
D = 500 m
500 = .25 t^2
500/.25 = t^2
2000 = t^2
t = 44.7 seconds
A dog running to the right at 4 m/s sees a ball and accelerates steadily to catch it. The dog accelerates to the right at a rate of 0.21 m/s^2 and catches the ball after 3.8s. What was the dogs velocity when it caught the ball? *Assume here that running to the right is positive and running to the left is negative.
A- 7.96
B- 10.569
C- 6.782
D- 4.798
E- none
Answer:
D.-4.798m/s
Explanation:
Greetings !
Given values
[tex]u= 4ms \\ a = 0.21ms {}^{2} \\ t = 3.8sec[/tex]
Solve for V of the given expression
Firstly, recall the velocity-time equation
[tex]v = u + at[/tex]
plug in known values to the equation
[tex]v = (4) + (0.21)(3.8)[/tex]
solve for final velocity
[tex]v = 4.792ms[/tex]
Hope it helps!
Explain the changes that occur during upstoke and downstoke of water pump with diagram
Answer: Downstroke: When the handle of a water pump is lifted up, the piston is pushed down. It is downstroke of the piston. In this condition, the volume of part A decreases and water pressure increases in this part. The increased pressure shuts the valve ‘A’ and opens the valve ‘B’ It makes the water of part A into pass part B of the cylinder.
Upstroke: When the handle is pushed down, it lifts the piston up. It is upstroke. During upstroke, the volume of part A increases and water Downstroke Upstroke pressure at that part decreases. It makes valve A opened and valve B closed. Thus, the water from the source enters into part A of the cylinder. The water of part B is lifted up and it comes out from the spout.
Explanation:
what is buoyant force
Answer:
When an object is immersed in water. it is pulled downwards due to gravitational pull of earth. Water exerts upward force on the object. This makes object rise up. This upward force is called buoyancy or upthrust.
Why do planets speed up as they get closer to the sun?:
A. air resistance
B. friction
C. gravity
D. tension
Answer:
C
Explanation:
Gravity is the main reason that make our planets to pull each other
How can i find Air velocity??????????????
Answer: By dividing airflow by duct cross section.
Explanation:
In short, air velocity in the ducts is calculated by dividing airflow by duct cross-section. Airflow is expressed as a simple number. Example: Air conditioner has a max. airflow of 600 CFM.
The magnitude of the net torque
Answer:
The magnitude of a torque about a fixed axis is calculated by finding the lever arm to the point where the force is applied and using the relation |→τ|=r⊥F | τ → | = r ⊥ F , where r⊥ is the perpendicular distance from the axis to the line upon which the force vector lies.
Explanation:
:)
If a body has a velocity 50 m s-1
and the uniform acceleration is 5 m s-2, find the time it takes to travel a distance
of 240 m.
Answer:
Using the relation V^2 = U^2 + 2× a× S
V^2 = 50^2 + (2× 5 × 240 )
V^ 2 = 2500 + 2400
V^ 2 = 4900
V= (4900)^0.5 = 70 m/s
hence, velocity at a distance of 240 m will be 70m/s.
Now using the relation,
V= U + a× t
70 = 50 + 5 × t
20 = 5 × t
t = 20/5 = 4m/s
hence , time required to travel a distance of 240m will be 4s.
A sensor that generates voltages in the range of -2V..+2V for 0..400kg weight values Design a balance circuit with LCD display. ; fosc=4MHz
If the atomic mass of Sodium-18 is 18.02597 u, what is the binding energy?
Answer:
The binding energy of sodium Na=5.407791×10⁹JExplanation:
Greetings !Binding energy, amount of energy required to separate a particle from a system of particles or to disperse all the particles of the system. Binding energy is especially applicable to subatomic particles in atomic nuclei, to electrons bound to nuclei in atoms, and to atoms and ions bound together in crystals.
Formula : Eb=(Δm)c²where:Eb= binding energy .Δm= mass defect(kg) c= speed of light 3.00×10⁸ms¯¹Given valuesm= 18.02597c=3.00×10⁸ms¯¹required valueEb=?Solution:Eb=(Δm)c²Eb=(18.02597)*(3.00*10⁸ms¯¹Eb=5.407791*10⁹JWhat is a small portable device that counts every step taken throughout the day?
A.stethoscope
B.pedometer
C.otoscope
D.thermometer
Answer: B
Explanation: i learned it last year
Pedometer is a small portable device that counts every step taken throughout the day. The correct option is (B).
A pedometer is a small portable device that counts the number of steps taken by an individual throughout the day. It is typically worn on the waist or carried in a pocket and uses a mechanism to detect body motion and count each step.
Pedometers are commonly used for tracking physical activity and encouraging individuals to achieve daily step goals as part of their fitness routines.
Therefore, Pedometer is a small portable device that counts every step taken throughout the day. Choose the option (B).
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A uniform linear charge of 3.0 nC/m is distributed along the y axis from y = -3 m to y=2m. Which of the following integrals is correct for the magnitude of the electric field at y = 4 m on the y axis?
[tex]E = \int\limits^2_{-3} {} \, \frac{27dy}{( 4 - y)^2}[/tex] is the integrals for the magnitude of the electric field.
A characteristic of electricity exists at every location in space when charge of any kind is present. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field.
The formula for the electric field due to a point charge is:
[tex]E = \frac{kq}{r^2}[/tex]
Here q is the charge, r is the distance from the charge, and k = 8.99 × [tex]10^9 Nm^2/ C^2[/tex]
Imagine an infinitely - tiny portion of this charge distribution that has a charge [tex]dq[/tex]. This charge produces an electric field dE with magnitude:
[tex]\int\limits^ {} \, dE = \int\limits^ {} \, \frac{k dq}{r^2}[/tex]
[tex]E = \int\limits^ {} \, \frac{k dq}{r^2}[/tex]
Here, λ = 3.0 nC/m dq
= λ.dy dE = ∫k.dq
From y = -3 m to y = 2 m, there is a 3.0 nC/m uniform linear charge distributed along the y axis.
The integrals for the magnitude of the electric field at y = 4 m on the y axis:
[tex]E = k \lambda \int\limits^2_{-3} {} \, \frac{dy}{( 4 - y)^2}[/tex]
[tex]E = \int\limits^2_{-3} {} \, \frac{27dy}{( 4 - y)^2}[/tex]
Learn more about electric field here:
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When there is a imbalance in the body system and the body can’t maintain homeostasis how might other systems respond
Answer:
the stomach has too much oxygen stored inside
Which of the following is a force acting between objects that do not touch? Pls answer in 5-10 min!
Normal force
Frictional force
Electrical force
Applied force
Answer:
Electrical force
Explanation:
The electrons aren't actually touching each other. Normal is something on something pushed by gravity by they are touching, frictional is rubbing, applied is pushing