Answers
Answer:
a) [tex]t=5\ s[/tex]
b) [tex]s=61.5\ m[/tex]
Explanation:
Given:
acceleration of the car, [tex]a=-4.92\ m.s^{-1}[/tex]
initial velocity of the car, [tex]u=24.6\ m.s^{-1}[/tex]
final velocity of the car, [tex]v=0\ m.s^{-1}[/tex]
a)
Using eq. of motion:
[tex]v=u+a.t[/tex]
[tex]0=24.6-4.92\times t[/tex]
[tex]t=5\ s[/tex]
b)
Distance travelled before stopping:
[tex]s=u.t+\frac{1}{2} a.t^2[/tex]
[tex]s=24.6\times 5-0.5\times 4.92\times 5^2[/tex]
[tex]s=61.5\ m[/tex]
c)
The car takes deceleration in 5 seconds to stop and travels a distance of 61.5 meters.
a) To find the time it takes for the car to stop, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the deceleration, and t is the time elapsed. Rearranging the equation to solve for t, we have t = (v - u) / a. Substituting the given values, we get t = (0 - 24.6) / -4.92 = 5 seconds.
b) To find the distance traveled during this time, we can use the equation s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the deceleration, and t is the time elapsed. Substituting the given values, we have s = 24.6(5) + (1/2)(-4.92)(5)^2 = 61.5 meters.
c) The graph of x vs t would be a straight line with a negative slope, representing the car's distance decreasing over time. The graph of v vs t would also be a straight line with a negative slope, representing the car's velocity decreasing over time.
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Related Questions
A cathode-ray tube (CRT) is an evacuated glass tube. Electrons are produced at one end, usually by the heating of a metal. After being focused electromagnetically into a beam, they are accelerated through a potential difference, called the accelerating potential. The electrons then strike a coated screen, where they transfer their energy to the coating through collisions, causing it to glow. CRTs are found in oscilloscopes and computer monitors, as well as in earlier versions of television screens.If the accelerating potential is 95.0 V, how fast will the electrons be moving when they hit the screen?
Answers
Answer: v = 5.79 * 10^10m/s.
Explanation: By using the work-energy theorem, we know that the work done on the electron by the potential difference equals the kinetic energy of the electrons.
Mathematically, we have that
qV = 1/2mv²
q= magnitude of an electronic charge = 1.609*10^-16c
V= potential difference = 95v
m = mass of an electronic charge = 9.11* 10^-31kg.
v = velocity of electron.
Let us substitute the parameters, we have that
1.609*10^-16 * 95 = (9.11*10^-31 * v²) /2
1.609*10^-16 * 95 * 2 = 9.11*10^-31 * v²
305.71 * 10^-16 = 9.11 * 10^-31 * v²
v² = 305.71 * 10^-16/ 9.11 * 10^-31
v² = 3.355 * 10^21.
v = √3.355 * 10^21
v = 5.79 * 10^10 m/s
Final answer:
Using the principle of energy conservation and the equation KE = qV = ½ mv^2, the velocity of the electrons in a CRT with an accelerating potential of 95.0 V is calculated to be approximately 5.8 x 10^6 m/s.
Explanation:
To determine how fast electrons will be moving when they hit the screen in a Cathode-Ray Tube (CRT) with an accelerating potential of 95.0 V, one can use the principle of energy conservation. The amount of kinetic energy (KE) gained by an electron when it is accelerated through a potential difference (V) is equal to the change in electrical potential energy (PE), which is given by the equation KE = qV, where q is the charge of the electron (1.602 x 10-19 Coulombs) and V is the potential difference.
The formula for kinetic energy is KE = ½ mv^2, with 'm' representing the mass of the electron (9.109 x 10^-31 kg) and 'v' being the velocity of the electron. Setting these equations equal gives us qV = ½ mv^2. Solving for 'v' and plugging in the values for q, V, and m, we can find the velocity of the electrons when they hit the screen.
Calculation:
qV = ½ mv^2
2qV = mv^2
v = √(2qV/m)
v = √(2 * 1.602 x 10^-19 C * 95.0 V / 9.10^9 x 10^-31 kg)
v ≈ 5.8 x 10^6 m/s
Thus, the electrons would be traveling at approximately 5.8 x 10^6 m/s when they hit the screen.
A physics demo launches a ball horizontally while dropping a second ball vertically at exactly the same time. Trajectories of two balls. The balls are initially placed at the same height above the ground. The first ball moves vertically downward, and hits the ground directly below the initial point. The second ball is launched horizontally, and follows a parabolic trajectory until it hits the ground at some horizontal distance from the initial point. Which ball hits the ground first?
Answers
Answer:
The two balls will hit the ground at the same time neglecting air resistance.
Explanation:
The vertical motion of the two ball are independent of each other. Also, the balls are falling from the same height in the same time. The ball projected horizontal neglecting air resistance is traveling with a constant velocity ( the same distance in the equal time) has only force of gravity acting on it. They will therefore hit the ground at the same time because they are acted upon by the same acceleration in the same time from the same height.
The vertical and horizontal motions of an object are independent. Hence, despite different trajectories, a ball dropped vertically and a ball launched horizontally from the same height will hit the ground simultaneously, as exemplified by Galileo's thought experiment.
Explanation:In the realm of Physics, an interesting question has been asked: Which ball hits the ground first - a ball that is dropped vertically or a ball that is launched horizontally? According to the principle of independence of motion, the horizontal motion and the vertical motion of an object are independent of each other. It implies that the horizontal velocity of a body has no effect on its vertical velocity. Thus, the ball dropped directly and the ball launched horizontally would strike the ground at the same time. Both balls are subjected to the same gravitational acceleration (g), without any initial vertical velocity.
This concept is a perfect illustration of Galileo's thought experiment, whereby he proved that the time for the balls to hit the ground is determined by their vertical motion alone and that horizontal motion does not affect the descent time.
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A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 18.0 cm , giving it a charge of-29.0 μC .
A Find the electric field just inside the paint layer.
Express your answer with the appropriate units. Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward.
B Find the electric field just outside the paint layer.
Express your answer with the appropriate units. Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward
C Find the electric field 6.00
cm outside the surface of the paint layer.
Express your answer with the appropriate units. Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward.
Answers
A) The electric field inside the paint layer is zero
B) The electric field just outside the paint layer is [tex]3.2\cdot 10^7 N/C[/tex] (radially inward)
C) The electric field at 6.00 cm from the surface is [tex]1.2\cdot 10^7 N/C[/tex] (radially inward)
Explanation:
A)
We can solve the problem by applying Gauss Law, which states that the electric flux through a Gaussian surface must be equal to the charge contained in the surface divided by the vacuum permittivity:
[tex]\int EdS = \frac{q}{\epsilon_0}[/tex]
where
E is the magnitude of the electric field
dS is the element of the surface
q is the charge contained within the surface
[tex]\epsilon_0[/tex] is the vacuum permittivity
By taking a sphere centered in the origin,
[tex]\int E dS = E \cdot 4\pi r^2[/tex]
where [tex]4\pi r^2[/tex] is the surface of the Gaussian sphere of radius r.
In this problem, we want to find the electric field just inside the paint layer, so we take a value of r smaller than
[tex]R=9.0 cm = 0.09 m[/tex] (radius of the plastic sphere is half of the diameter)
Since the charge is all distributed over the plastic sphere, the charge contained within the Gaussian sphere is zero:
[tex]q=0[/tex]
And therefore,
[tex]E4\pi r^2 = 0\\\rightarrow E = 0[/tex]
So, the electric field inside the plastic sphere is zero.
B)
Here we apply again Gauss Law:
[tex]E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex]
In this case, we want to calculate the electric field just outside the paint layer: this means that we take r as the radius of the plastic sphere, so
[tex]r=R=0.18 m[/tex]
The charge contained within the Gaussian sphere is therefore
[tex]q=-29.0 \mu C = -29.0\cdot 10^{-6}C[/tex]
Therefore, the electric field is
[tex]E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.09)^2}=-3.2\cdot 10^7 N/C[/tex]
And the negative sign indicates that the direction of the field is radially inward (because the charge that generates the field is negative). However, the text of the question says "Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward", so the answer to this part is
[tex]E=3.2\cdot 10^7 N/C[/tex]
C)
For this part again, we apply Gauss Law:
[tex]E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex]
In this case, we want to calculate the field at a point 6.00 cm outside the surface of the paint layer; this means that the radius of the Gaussian sphere must be
r = 9 cm + 6 cm = 15 cm = 0.15 m
While the charge contained within the sphere is again
[tex]q=-29.0 \mu C = -29.0\cdot 10^{-6}C[/tex]
Therefore, the electric field in this case is
[tex]E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.15)^2}=-1.2\cdot 10^7 N/C[/tex]
And again, this is radially inward, so according to the sign convention asked in the problem,
[tex]E=1.2\cdot 10^7 N/C[/tex]
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To find the electric field just inside the paint layer, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of the medium.
A) Inside the paint layer:
Since the paint is spread in a very thin uniform layer over the surface of the plastic sphere, we can consider a Gaussian surface just inside the paint layer, with a radius slightly smaller than the sphere's radius (let's call it r). The charge enclosed within this Gaussian surface is the charge on the sphere, which is -29.0 μC.
Gauss's law equation can be written as:
Φ = Q_enclosed / ε₀
The electric field just inside the paint layer (E₁) is given by:
E₁ * A = Q_enclosed / ε₀
E₁ * 4πr² = -29.0 μC / ε₀
E₁ = (-29.0 μC / ε₀) / (4πr²)
The direction of the electric field just inside the paint layer will be radially outward, opposite to the direction of the charge.
B) Just outside the paint layer:
The electric field just outside the paint layer (E₂) will be the same as the electric field inside the sphere when it was uncharged. This is because the charge on the paint layer is spread on the outer surface of the sphere.
So, E₂ = E₁ (since it's radially outward)
C) 6.00 cm outside the surface of the paint layer:
To find the electric field 6.00 cm outside the surface of the paint layer, we need to calculate the electric field due to the charge on the sphere.
E₃ = k * Q / r²
where k is the Coulomb's constant (k = 9.0 x 10^9 N m²/C²), Q is the charge on the sphere (-29.0 μC), and r is the distance from the center of the sphere to the point where we want to find the electric field (6.00 cm = 0.06 m + radius of the sphere).
Remember that the electric field due to a charged object decreases with the square of the distance.
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A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/s. A 1.0-kg stone is thrown from the basket with an initial velocity of 15.0 m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. That person sees the stone hit the ground 5.00 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 m/s. (a) How high is the balloon when the rock is thrown? (b) How high is the balloon when the rock hits the ground? (c) At the instant the rock hits the ground, how far is it from the basket? (d) Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer (i) at rest in the basket and (ii) at rest on the ground.
Answers
Answer:
a)-296 m/s
b)-176 m
c) 198 m
Explanation:
given data:
[tex]v_{oy}\\[/tex]=-20 m/s
[tex]v_{x}[/tex]= 15 m/s
t = 6 s
To find
a)[tex]H_{1}[/tex]
b)[tex]H_{2}[/tex]
c) d
d) find the velocity of observer at rest in basket and rest on ground
a) according to kinematic equation the displacement is given by:
[tex]H_{1}[/tex]=[tex]v_{oy}t\\[/tex]-[tex]\frac{1}{2} gt^{2}[/tex]..............(1)
putting values in 1
[tex]H_{1}[/tex]=-296 m/s
negative sign is due to direction and also we choose to take off point to be the origin
b) while the stone was travelling [tex]H_{1}[/tex] for 6 s so the ballon was also travelling displacement [tex]H_{b}[/tex] with [tex]v_{oy}\\[/tex]
[tex]H_{b}[/tex] =[tex]v_{oy}t\\[/tex]
=-6*20
=-120 m
the height above the earth is given by
[tex]H_{2}[/tex]=[tex]H_{1}[/tex]-[tex]H_{2}[/tex]= -176 m
c) the horizontal displacement is given by
x=[tex]v_{x}t\\[/tex]=15*6=90 m
so the distance between balloon and stone is
d=[tex]\sqrt{H_{1}^{2}+x^{2} }[/tex]=198 m
d) the velocity component of the balloon :
1) relative to person rest in the balloon are given by
The vertical component:
[tex]v_{yr} =v_{by} +v_{py}[/tex]=[tex](-gt)-0=9.8*6=-58.8 m/s[/tex]
The horizontal component:
[tex]v_{xr} =v_{bx} +v_{px}= 15-0=15m/s[/tex]
2) relative to person rest in the earth are given by
The vertical component:
[tex]v_{yr} =v_{by} -v_{py}=(-gt)-20=9.8*6-20=-78.8 m/s[/tex]
The horizontal component:
[tex]v_{xr} =v_{bx} -v_{px}= 15-0=15m/s[/tex]
A student decides to measure the muzzle velocity of a pellet shot from his gun. He points the gun horizontally. He places a target on a vertical wall a distance x away from the gun. The pellet hits the target a vertical distance y below the gun.
(a) Show that the position of the pellet when traveling through the air is given by y = Ax^2, where A is a constant.
(b) Express the constant A in terms of the initial velocity v and the free-fall acceleration g.
(c) If x = 3 m and y = 0.21 m, what is the initial speed of the pellet?
Answers
Answer:
a) y = A x² , b) A = - ½ g / v₀², c) v₀ = 15.46 m / s
Explanation:
For this problem of two-dimensional kinematics, we will use that the time to reach the wall is the same
X axis
x = v₀ₓ t
t = x / v₀ₓ
Y Axis
y = [tex]v_{oy}[/tex] t - ½ g t²
As it shoots horizontally the vertical speed is zero
y = - ½ g t²
We replace
y = - ½ g (x / v₀ₓ)²
The initial speed is all horizontal
v₀ₓ = v₀
y = - ½ g / v₀² x²
y = A x²
b) the expression for the constant is
A = - ½ g / v₀²
c) we look for the initial speed
v₀² = - ½ g x² / y
As the object falls below the exit point its height is negative
v₀ = √ (- ½ 9.8 3²/ (-0.21))
v₀ = 15.46 m / s
A dog running in an open field has components of velocity vx = 2.6 m/s and vy = -1.8 m/s at t1 = 10.0 s. For the time interval from t1 = 10.0 s to t2 = 20.0 s, the average acceleration of the dog has magnitude 0.45 m/s2 and direction 31.0o measured from the + x-axis toward the + y-axis. At t2 = 20.0 s, (a) what are the x- and y-components of the dog’s velocity? (b) What are the magnitude and direction of the dog’s velocity? (c) Sketch the velocity vectors at t1 and t2. How do these two vectors differ?
Answers
Answer:
a) vₓ = 6,457 m / s , v_{y} = 0.518 m / s , b) v = 6.478 m / s, θ = 4.9°
Explanation:
a) This is a kinematic problem, let's use trigonometry to find the components of acceleration
sin 31 = [tex]a_{y}[/tex] / a
cos 31 = aₓ = a
a_{y} = a sin31
aₓ = a cos 31
Now let's use the kinematic equation for each axis
X axis
vₓ = v₀ₓ + aₓ (t-t₀)
vₓ = v₀ₓ + a cos 31 (t-t₀)
vₓ = 2.6 + 0.45 cos 31 (20-10)
vₓ = 6,457 m / s
Y Axis
v_{y} = v_{oy} + a_{y} t
v_{y} = v_{oy} + a_{y} sin31 (t-to)
v_{y} = -1.8 + 0.45 sin31 (20-10)
v_{y} = 0.518 m / s
b) let's use Pythagoras' theorem to find the magnitude of velocity
v = √ (vₓ² + v_{y}²)
v = √ (6,457² + 0.518²)
v = √ (41.96)
v = 6.478 m / s
We use trigonometry for direction
tan θ = v_{y} / vₓ
θ = tan⁻¹ v_{y} / vₓ
θ = tan⁻¹ 0.518 / 6.457
θ = 4.9°
c) let's look for the vector at the initial time
v₁ = √ (2.6² + 1.8²)
v₁ = 3.16 m / s
θ₁ = tan⁻¹ (-1.8 / 2.6)
θ₁ = -34.7
We see that the two vectors differ in module and direction, and that the acceleration vector is responsible for this change.
a = (v₂ -v₁) / (t₂-t₁)
As Halley’s comet orbits the sun, its distance from the sun changes dramatically. If the comet’s speed at a distance of 9.8 × 1010 m from the sun is 5.3 × 104 m/s and angular momentum is conserved, what is its speed when it is 3.6 × 1012 m from the sun? Assume the comet can be treated as a point mass. Ignore radial components of momentum
Answers
Answer:
The speed of comet is 1442.77 m/s.
Explanation:
Given that,
Speed of comet's [tex]v_{c}= 5.3\times10^{4}\ m/s[/tex]
Distance from the sun [tex]r_{d}=9.8\times10^{10}\ m[/tex]
Distance [tex]r_{far}=3.6\times10^{12}\ m[/tex]
We need to calculate the speed of comet
Using conservation of angular momentum
[tex]L_{f}=L_{i}[/tex]
[tex]I\omega=I\omega[/tex]
Here. [tex]v = r\omega[/tex]
[tex]\omega=\dfrac{v}{r}[/tex]
[tex]mr_{far}^2\times\dfrac{v_{far}}{r_{far}}=mr_{d}^2\times\dfrac{v_{d}}{r_{d}}[/tex]
[tex]r_{far}\times v_{far}=r_{d}\times v_{d}[/tex]
Put the value into the formula
[tex]3.6\times10^{12}\times v_{far}=9.8\times10^{10}\times5.3\times10^{4}[/tex]
[tex]v_{far}=\dfrac{9.8\times10^{10}\times5.3\times10^{4}}{3.6\times10^{12}}[/tex]
[tex]v_{far}=1442.77\ m/s[/tex]
Hence, The speed of comet is 1442.77 m/s.
Final answer:
The speed of Halley's comet when it is 3.6 x 10^12 m away from the sun can be found using the conservation of angular momentum, resulting in a speed of 1.44 x 10^3 m/s.
Explanation:
To find the speed of Halley's comet at a distance of 3.6 × 1012 m from the sun, we will use the conservation of angular momentum. The orbital momentum of an object around the sun is given by L = mrv, where m is the mass of the comet, r is the radius of the orbit (distance from the sun), and v is the velocity of the comet along that radius.
We are given the comet's speed at a distance of 9.8 × 1010 m is 5.3 × 104 m/s. When the comet is at a distance of 3.6 × 1012 m from the sun, we want to find its new speed, v'. Because angular momentum is conserved, the initial and final angular momentum should be equal:
L initial = L final
m r v = m r' v'
Since the mass of the comet, m, cancels out, we are left with:
r v = r' v'
Plugging in the known values, we get:
9.8 × 1010 m × 5.3 × 104 m/s = 3.6 × 1012 m × v'
Therefore:
v' = (9.8 × 1010 × 5.3 × 104) / (3.6 × 1012)
v' = 1.44 × 103 m/s
The speed of Halley's comet at a distance of 3.6 × 1012 m from the sun is thus 1.44 × 103 m/s.
A total charge of 6.3 x 10-8 C is placed on a 2.7 cm radius isolated conducting sphere. The surface charge density is: A. 2.55 x 10-4 C/m3 Explain B. 8.64 x 10-5 C/m2 C. 2.75 x 10-5 C/m2 D. 6.88 x 10-6 C/m2 E. 7.43 x 10-7 C/m
Answers
Answer:
[tex]6.88\times 10^{-6}\ C/m^2[/tex]
Explanation:
Q = Total charge = [tex]6.3\times 10^{-8}\ C[/tex]
[tex]\rho[/tex] = Surface charge density
r = Radius = 2.7 cm
A = Surface area = [tex]4\pi r^2[/tex]
Charge is given by
[tex]Q=A\rho\\\Rightarrow Q=4\pi r^2\times \rho\\\Rightarrow \rho=\dfrac{Q}{4\pi r^2}\\\Rightarrow \rho=\dfrac{6.3\times 10^{-8}}{4\pi (2.7\times 10^{-2})^2}\\\Rightarrow \rho=0.00000687706544224\\\Rightarrow \rho=6.88\times 10^{-6}\ C/m^2[/tex]
The surface charge density is [tex]6.88\times 10^{-6}\ C/m^2[/tex]
Particle physicists use particle track detectors to determine the lifetimeof short-lived particles. A muon has a mean lifetime of 2.2sand makes a track 9.5 cm long before decayinginto an electron and two neutrinos. What was the speed of the muon?
Answers
Final answer:
To calculate the speed of a muon before it decays, the distance the muon travels and the time it takes are used in the formula L = v x t. For a distance of 9.5 cm and a time of 2.20 microseconds, the speed is found to be approximately 43,182 meters per second.
Explanation:
The question is seeking to calculate the speed of a muon before it decays. Assuming the mean lifetime of a muon is 2.20 microseconds (us) and the muon makes a track 9.5 cm long in this time, we can calculate the muon's speed. Let's consider the formula:
L = v x t
where L is the distance traveled, v is the velocity, and t is the time. Given L = 9.5 cm (or 0.095 m) and t = 2.20 x 10-6 s, we plug in the values:
v = L/t
v = 0.095 m / 2.20 x 10-6 s
v = 4.31818 x 104 m/s
The muon's speed was approximately 43,182 m/s.
Millikan is doing his oil drop experiment. He has a droplet with radius 1.6 µm suspended motionless in a uniform electric field of 46 N/C. The density of the oil is 0.085 g/cm3 . Calculate for Millikan the charge on the droplet. Is quantization of charge obeyed within the precision of this experiment?
Answers
Answer:
The charge on the droplet is [tex]3.106\times10^{-16}\ C[/tex].
Yes, quantization of charge is obeyed within experimental error.
Explanation:
Given that,
Radius = 1.6μm
Electric field = 46 N/C
Density of oil = 0.085 g/cm³
We need to calculate the charge on the droplet
Using formula of force
[tex]F= qE[/tex]
[tex]mg=qE[/tex]
[tex]V\times\rho\times g=qE[/tex]
[tex]q=\dfrac{V\times\rho}{E}[/tex]
[tex]q=\dfrac{\dfrac{4}{3}\pi\times r^3\times\rho\times g}{E}[/tex]
Put the value into the formula
[tex]q=\dfrac{\dfrac{4}{3}\times\pi\times(1.6\times10^{-6})^3\times85\times9.8}{46}[/tex]
[tex]q=3.106\times10^{-16}\ C[/tex]
We need to calculate the quantization of charge
Using formula of quantization
[tex]n = \dfrac{q}{e}[/tex]
Put the value into the formula
[tex]n=\dfrac{3.106\times10^{-16}}{1.6\times10^{-19}}[/tex]
[tex]n=1941.25[/tex]
Yes, quantization of charge is obeyed within experimental error.
Hence, The charge on the droplet is [tex]3.106\times10^{-16}\ C[/tex].
Yes, quantization of charge is obeyed within experimental error.
Answer:
3.11 x 10^-16 C
Explanation:
radius of drop, r = 1.6 micro metre = 1.6 x 10^-6 m
Electric field, E = 46 N/C
density of oil, d = 0.085 g/cm³ = 85 kg/m³
Let the charge on the oil drop is q.
So, the weight of the drop is balanced by the electric force on the drop.
mass of drop x g = charge of the drop x electric field
m x g = q E
Volume x density x g = q E
[tex]\frac{4}{3} \pi\times r^{3}\times d\times g = q \times E[/tex]
[tex]\frac{4}{3} \pi\times 1.6^{3}\times 10^{-18}\times 85\times 9.8 = q \times 46[/tex]
q = 3.11 x 10^-16 C
Thus, the charge on the oil drop is 3.11 x 10^-16 C.
t requires1200 kg of coal to produce the energy needed to make 1.0 kg of aluminum metal. If a single soda can requires approximately 15 g of Al, what mass of coal would be needed to produce a 6-pack of cans?
Answers
Answer:
The mass of coal is 108 kg.
Explanation:
Given that,
Energy of coal = 1200 kg
Mass of aluminum = 1.0 kg
Energy required for single soda can = 15 g of Al
Energy required for 6 pack of cans = [tex]6\times15=90\ g\ of\ Al[/tex]
We need to calculate the mass of coal
Using formula of mass
[tex]\text{mass of coal}=\dfrac{\text{Energy of coal}\times\text{Energy required for 6 pack of cans}}{\text{Mass of aluminum}}[/tex]
Put the value into the formula
[tex]m=\dfrac{1200\times90\times10^{-3}}{1.0}[/tex]
Put the value into the formula
[tex]m=108\ Kg[/tex]
Hence, The mass of coal is 108 kg.
NASA communicates with the Space Shuttle and International Space Station using Ku-band microwave radio. Suppose NASA transmits a microwave signal to the International; Space Station using radio waves with a frequency of 18.0 GHz. What is the wavelength of these radio waves?
Answers
Answer:
0.0167 m
Explanation:
f = Frequency = 18 GHz
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
When the speed of the wave is divided by the freqeuncy we get the wavelength
Wavelength is given by
[tex]\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{18\times 10^9}\\\Rightarrow \lambda=0.0167\ m[/tex]
The wavelength of the radio waves is 0.0167 m
A woman exerts a total force of 7 pounds in a horizontal direction on a box as she pushes it up a ramp that is 10 feet long and inclined at an angle of 30 degrees above the horizontal. Find the work done on the box.
Answers
Answer:
W = 60.62 ft-lbs
Explanation:
given,
Horizontal force = 7 lb
distance of push, d = 10 ft
angle of ramp, θ = 30°
Work done on the box = ?
We know,
Work done is equal to force into displacement.
W = F.d cos θ
W = 7 x 10 x cos 30°
W = 70 x 0.8660
W = 60.62 ft-lbs
Hence, work done on the box is equal to W = 60.62 ft-lbs
The ozone molecule O3 has a permanent dipole moment of 1.8×10−30 Cm. Although the molecule is very slightly bent-which is why it has a dipole moment-it can be modeled as a uniform rod of length 2.5×10−10 m with the dipole moment perpendicular to the axis of the rod. Suppose an ozone molecule is in a 8000 N/C uniform electric field. In equilibrium, the dipole moment is aligned with the electric field. But if the molecule is rotated by a small angle and released, it will oscillate back and forth in simple harmonic motion.
What is the frequency f of oscillation?
Answers
Answer:
934701926.438 Hz
Explanation:
Mass of molecule
[tex]m=3\times 16\times 1.67\times 10^{-27}\ kg[/tex]
Moment of inertia is given by
[tex]I=\dfrac{1}{12}ml^2\\\Rightarrow I=\dfrac{1}{12}\times 3\times 16\times 1.67\times 10^{-27}\times (2.5\times 10^{-10})^2\\\Rightarrow I=4.175\times 10^{-46}\ kgm^2[/tex]
E = Electric field = 8000 N/C
p = Dipole moment = [tex]1.8\times 10^{-30}\ Cm[/tex]
l = Length of rod = [tex]2.5\times 10^{-10}\ m[/tex]
Frequency of oscillations is given by
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{pE}{I}}\\\Rightarrow f=\dfrac{1}{2\pi}\sqrt{\dfrac{1.8\times 10^{-30}\times 8000}{4.175\times 10^{-46}}}\\\Rightarrow f=934701926.438\ Hz[/tex]
The frequency of oscillations is 934701926.438 Hz
A drunken sailor stumbles 600 meters north, 550 meters northeast, then 500 meters northwest. What is the total displacement and the angle of the displacement?
Answers
We will make a graph to better understand the displacement of the individual. From the trigonometric properties we will find the required distances.
[tex]d_1 = \frac{450}{\sqrt{2}} = 318.198[/tex]
[tex]d_2 = \frac{400}{\sqrt{2}} = 282.843[/tex]
[tex]D = 500+d_1+d_2 = 1101.041[/tex]
Displacement ,
[tex]x = \sqrt{1101.041^2+(318.198-282.843)^2}}[/tex]
[tex]x = 1101.608m[/tex]
The angle would be
[tex]\theta = cos^{-1} (\frac{1101.041}{1101.608})[/tex]
[tex]\theta = E 1.838\° N[/tex]
Therefore the displacement was of 1101.608m to a angle of 1.838° from East to North.
The sailor's total displacement is approximately 1118.16 meters, at an angle of approximately 88.13 degrees north of east.
Explanation:This is a problem of physics specifically in the field of mechanics, about displacement and vectors. To find the total displacement, the directions in which the sailor traveled need to be considered. A north direction has an angle of 90 degrees, northeast is 45 degrees, and northwest is 135 degrees.
Using the Pythagorean theorem and some trigonometry, we can calculate the displacement in the x and y directions. The total displacement in the x direction (East-West) is 550*cos(45) - 500*cos(45) = 35.35 meters (approximately). The total displacement in the y direction (North-South) is 600 + 550*sin(45) + 500*sin(45) = 1116.57 meters (approximately).
The total displacement would then be the square root of (35.35 ^2 + 1116.57 ^2) = 1118.16 meters (approximately).
To find the angle of the displacement, we can use the arctangent function (atan function in most calculators). The angle is atan (1116.57 / 35.35) = 88.13 degrees (approximately).
Therefore, the sailor's total displacement is approximately 1118.16 meters, at an angle of approximately 88.13 degrees north of east.
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A motorist is driving at 20m/s when she sees that a traffic light 200m ahead has just turned red. She knows that this light stays red for 15s, and she wants to reach the light just as it turns green again. It takes her 1.0s to step on the brakes and begin slowing. What is her speed as she reaches the light at the instant it turns green?
Answers
Answer:
5.71428571422 m/s
Explanation:
u = Initial velocity = 20 m/s
v = Final velocity
s = Displacement
a = Acceleration
Time taken = 15-1 = 14 s
Distance traveled in 1 second = [tex]20\times 1=20\ m[/tex]
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 200-20=20\times 14+\frac{1}{2}\times a\times 14^2\\\Rightarrow a=\frac{2(180-20\times14)}{14^2}\\\Rightarrow a=-1.02040816327\ m/s^2[/tex]
[tex]v=u+at\\\Rightarrow v=20-1.02040816327\times 14\\\Rightarrow v=5.71428571422\ m/s[/tex]
The speed as she reaches the light at the instant it turns green is 5.71428571422 m/s
The final speed of the motorist as she reaches the light is 5.72 m/s
To calculate the speed of the motorist as she reaches the light, we need to first find the deceleration of the motorist.
Formula:
S₁ = ut₁.............. Equation 1Where:
S₁ = displacement of the motorist as its slows downu = initial velocityt₁ = time it takes to slow down.Given:
u = 20 m/st₁ = 1Therefore,
S₁ = 20(1) = 20 mThen,
S = 200-20S = 180 m
Where:
S = distance covered by the motorist before the brake is applied.
we use the formula below to calculate the deceleration of the motorist.
S = ut+at²/2........... Equation 2Where:
a = deceleration of the motorist.Given:
u = 20 m/st = 15-1 = 14 sS = 180 mSubstitute these values into equation 2
180 = 20(14)+a(14²)/2180 = 280+98aSolve for a
98a = 180-28098a = -100a = -100/98a = -1.02 m/s²Finally, we use the formula below to get the speed of the motorist as she reaches the light.
v = u+at............ Equation 3Given:
u = 20 m/sa = -1.02 m/s²t = 14 sSubstitute these values into equation 3
v = 20+(-1.02×14)v = 20-14.28v = 5.72 m/s. Hence, The final speed of the motorist as she reaches the light is 5.72 m/s
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A bullet is fired from a rifle that is held 1.60 m above the ground in a horizontal position. The initial speed of the bullet is 1058 m/s. Calculate the time it takes for the bullet to strike the ground.
Answers
Answer:
time takes for the bullet to strike the ground is 0.5711 second
Explanation:
given data
height h = 1.60 m
initial speed u = 1058 m/s
solution
we get here time that is express here as
time = [tex]\sqrt{\frac{2h}{g}}[/tex] ......................1
put here value and we will get here time that is
time = [tex]\sqrt{\frac{2*1.60}{9.81}}[/tex]
time = 0.5711 second
so time takes for the bullet to strike the ground is 0.5711 second
Final answer:
The time it takes for a bullet to strike the ground when fired horizontally from 1.60 m is calculated using the equation t = √(2h/g). Substitute 1.60 m for h and 9.81 m/s² for g to obtain t ≈ 0.571 seconds.
Explanation:
Calculating the Time for a Bullet to Strike the Ground
To calculate the time it takes for a bullet to strike the ground when fired from a horizontal position 1.60 meters above the ground with an initial speed of 1058 m/s, we can use the equations of motion for uniformly accelerated motion (in this case, the acceleration due to gravity). Since the bullet is fired horizontally, there is no initial vertical velocity component, so the vertical motion can be treated as a free-fall problem.
Using the formula for the time of free fall (t) from a height (h):
t = √(2h/g), where g is the acceleration due to gravity (approximately 9.81 m/s²).
Plugging in the given height of 1.60 m and solving for t gives us:
t = √(2 * 1.60 m / 9.81 m/s²) = √(0.3261 s²) ≈ 0.571 seconds.
Therefore, the bullet will take approximately 0.571 seconds to hit the ground.
I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air.
Describe the motion of the ball as I see it. Is the path taken by the ball different than if I was sitting at rest at home and I threw a ball straight up into the air?
Answers
Answer:
Explanation:
I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air. Before , the ball gets separated from my hand , both me the ball will be moving with velocity of 50 m /s in horizontal direction .
As soon as ball is separated from the hand , it acquires addition velocity in upward direction and acceleration in downward direction . This will give relative velocity to the ball with respect to me . So I will see the ball going in upward direction under gravitational acceleration . It appears as if I am sitting at rest and ball is going in upward direction under deceleration . My motion at 50 m/s will have no effect on the motion of ball in upward direction , according to first law of Newton . It is so because ball too will be moving in forward direction with the same speed which will not be visible to me because I too am moving with the same speed.
If I am sitting at rest at home and I threw a ball straight up into the air , I will have the same experience of seeing ball going in similar way as described above.
How and why does a day measured with respect to the Sun differ from a day measured with respect to the stars?
Answers
Answer:
A day measured with respect to the sun differ from a day measured with respect to star by 4 minutes. This is because, it takes 4 minutes for the Earth to rotate the extra amount required for the Sun to return to the same place in the sky.
Explanation:
A day measured with respect to the sun differ from a day measured with respect to star by 4 minutes. This is because, it takes 4 minutes for the Earth to rotate the extra amount required for the Sun to return to the same place in the sky.
Each day that goes by, the Earth needs to turn a little further for the sun to return to the same place in the sky. And that extra time is about 4 minutes or 240 seconds.
Final answer:
A solar day is longer than a sidereal day by about 4 minutes, due to Earth's orbit around the Sun requiring additional rotation to align the Sun to the same position in the sky. The sidereal day, on the other hand, represents the true rotation period of Earth as it measures the time it takes for a distant star to reappear in the same position in the sky.
Explanation:
A day measured with respect to the Sun differs from a day measured with respect to the stars primarily in its length due to Earth's simultaneous rotation on its axis and revolution around the Sun. The solar day is the period during which Earth rotates on its axis so that the Sun appears at the same position in the sky on consecutive days. It is about 4 minutes longer than a sidereal day, which is the time it takes for a distant star to return to the same position in the sky.
This difference occurs because as Earth rotates, it also moves along its orbital path around the Sun, so it has to rotate a little more for the Sun to be in the same position overhead. The sidereal day is actually the true rotation period of Earth, being 235.9 seconds shorter than 24 hours, while the solar day includes an additional 4 minutes to account for Earth's movement in orbit.
Charge 9 × 10−18 C is on the y axis a distance 5 m from the origin and charge 9 × 10−18 C is on the x axis a distance d from the origin. What is the value of d for which the x component of the force on 9 × 10−18 C is the greatest? The Coulomb constant is 8.98755 × 109 N · m2 /C 2 .
Answers
Answer:
d = 3.53 m
Explanation:
The Coulomb Force is given as
[tex]\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r[/tex]
The x-component of the force is equal to
[tex]F_x = F\cos(\theta) = F\frac{x}{\sqrt{x^2 + y^2}} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{(5^2 + d^2)}\frac{d}{\sqrt{5^2 + d^2}} = \frac{1}{4\pi\epsilon_0}\frac{dq_1q_2}{(5^2 + d^2)^{3/2}}[/tex]
This is basically a function of (d). So, the maximum value of this function is the point where its derivative with respect to d is equal to zero.
[tex]\frac{dF_x}{dd} = \frac{kq_1q_2}{(d^2 + 5^2)^{3/2}} - \frac{3d^2kq_1q_2}{(d^2 + 5^2)^{5/2}} = 0\\3d^2 = d^2 + 5^2\\2d^2 = 25\\d = 3.53~m[/tex]
The value of d for which the x component of the force on the charge is the greatest, in accordance to Coulomb's Law, is when d equals 5 meters. At this distance, the x and y components of the force are equal, thus maximizing the x component.
Explanation:The scenario you described involves the concept of Coulomb's Law which states the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Looking at your question, the value of d for which the x component of the force on 9×10−18 C is the greatest would be when d equals 5 meters. The charges would then form an equilateral triangle with the origin, meaning the x component and y component of the force would be equal, hence maximizing the x component.
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If a mass of 0.5 kg is displaced by 30 cm using a spring with a spring constant of 2 N/m, find the following:a) The angular frequencyb) The period of oscillationc) The position at t = 2 secondsd) The velocity at t = 2 secondse) The acceleration at t = 2 seconds
Answers
Answer
given,
mass of the block, m = 0.5 Kg
displacement, x = 30 cm = 0.3 m
Spring constant, k = 2 N/m
a) Angular frequency
[tex]\omega = \sqrt{\dfrac{k}{m}}[/tex]
[tex]\omega = \sqrt{\dfrac{2}{0.5}}[/tex]
[tex]\omega = 2\ rad/s[/tex]
b) Period of oscillation
[tex]T=\dfrac{2\pi}{\omega}[/tex]
[tex]T=\dfrac{2\pi}{2}[/tex]
T = 3.14 s
c) Position at t = 2
x = -A cos ω t
A = 0.3 ω = 2 rad/s
x = -0.3 cos (2 x 2)
x = 0.196 m
d) velocity at t= 2
v = A ω sin ω t
v = 0.3 x 2 x sin 4
v = -0.454 m/s
e) acceleration at t= 2
a = A ω² cos ω t
a = 0.3 x 2² cos 4
a = -0.784 m/s²
If the plane is flying in a horizontal path at an altitude of 98.0 m above the ground and with a speed of 73.0 m/s, at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.
Answers
Explanation:
The given data is as follows.
height (h) = 98.0 m, speed (v) = 73.0 m/s,
Formula of height in vertical direction is as follows.
h = [tex]\frac{gt^{2}}{2}[/tex],
or, t = [tex]\sqrt{\frac{2h}{g}}[/tex]
Now, formula for the required distance (d) is as follows.
d = vt
= [tex]v \sqrt{\frac{2h}{g}}[/tex]
= [tex]73.0 m/s \sqrt{\frac{2 \times 98.0 m}{9.8 m/s^{2}}}[/tex]
= 326.5 m
Thus, we can conclude that 326.5 m is the horizontal distance from the target from where should the pilot release the canister.
The efficiency of a squeaky pulley system is 73 percent. The pulleys are usedto raise a mass to a certain height. What force is exerted on the machine if arope is pulled 18.0 m in order to raise a 58 kg mass a height of 3.0 m
Answers
The efficiency of the machine is defined as
[tex]\eta = \frac{W_{out}}{W_{in}}[/tex]
Here
Work out is the work output and Work in is the work input
To find the Work in we have then
[tex]W_{in} = \frac{W_{out}}{\eta}[/tex]
[tex]W_{in} = \frac{mgh}{\eta}[/tex]
Replacing with our values
[tex]W_{in} = \frac{(58)(9.8)(3)}{73\%}[/tex]
[tex]W_{in} = 2335.89J[/tex]
The work done by the applied force is
W = Fd
Here,
F = Force
d = Distnace
Rearranging to find F,
[tex]F = \frac{W}{d}[/tex]
[tex]F = \frac{2335.89J }{18}[/tex]
F = 129.77N
Therefore the force exerted on the machine after rounding off to two significant figures is 130N
The force exerted on the pulley system when a rope is pulled 18.0 m in order to raise a 58 kg mass a height of 3.0 m with an efficiency of 73 percent is about 129.42 Newtons.
Explanation:To solve this problem, we need to understand the concept of machine efficiency and work. The efficiency of a machine is the ratio of output work to input work.
In this case, the efficiency of the pulley system is given as 73%. Meaning output work is 73% of the input work. The force exerted on the machine is equivalent to the input work divided by the distance pulled.
The output work (W_out) can be determined using the formula W_out = mass * gravity * height = 58 kg * 9.8 m/s^2 * 3.0 m = 1701.6 Joules.
Then, we can find the input work (W_in) using the efficiency formula: W_in = W_out / efficiency = 1701.6 J / 0.73 = 2329.59 Joules.
Finally, we can find the force exerted on the pulley system (F_in) by dividing the input work by the distance pulled: F_in = W_in / distance = 2329.59 J / 18.0 m = 129.42 Newtons.
So, the force exerted on the machine would be approximately 129.42 Newtons.
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Which of the cities that we examined experienced the least significant temperature increase (smallest slopes)?A. Seattle, WA B. Springfield, IL C. Phoenix, AZ D. New York, NY
Answers
Springfield has the smallest slope.
Answer: Option B.
Explanation:
Springfield is a city in the United States of America. This city is in the states of Massachusetts. It is besides the river Connecticut river which adds to the beauty of this state.
Springfield is the nick name of the city "The queen of the Ozarks". The other name of this city is "the cultural center of the Ozarks". It has a beautiful scenery and this scenic beauty adds to the tourist attraction to a lot of people all around the world.
Design an op-amp circuit to provide an output vO =−[2v1 + (v2/2)]. Choose relatively low values of resistors but ones for which the input current (from each input signal source) does not exceed 50 μA for 1-V input signals.
Answers
Answer:
Explanation: see attachment
An astronaut is in space with a baseball and a bowling ball. The astronaut gives both objects an equal push in the same direction. Does the baseball have the same inertia as the bowling ball? Why? Does the baseball have the same acceleration as the bowling ball from the push? Why? If both balls are traveling at the same speed, does the baseball have the same momentum as the bowling ball?
Answers
Answer:
Explanation:
Given
Astronaut in space gives an equal push to baseball and bowling ball.
Since the mass of the bowling ball is more than the baseball so inertia associated with a bowling ball is more as compared to baseball
Force applied on baseball and bowling ball is the same so the acceleration of baseball will be more because the mass of baseball is less.
[tex]Force=mass\times acceleration[/tex]
If both are traveling with the same speed then momentum associated with them is given by-product of mass and velocity
Since the mass of the bowling ball is more, therefore, the momentum of bowling ball is more as compared to baseball
According to the Stefan-Boltzmann law, how much energy is radiated into space per unit time by each square meter of the Sun’s surface? If the Sun’s radius is 696,000 km, what is the total power output of the Sun?
Answers
Answer:
[tex]3.8469943828\times 10^{26}\ W[/tex]
Explanation:
[tex]\sigma[/tex] = Stefan-Boltzmann constant = [tex]5.67\times 10^{-8}\ W/m^2K^4[/tex]
T = Surface temperature of the Sun = 5778 K
r = Radius of the Sun = 696000 km
From Stefan-Boltzmann law
[tex]F=\sigma T^4\\\Rightarrow F=5.67\times 10^{-8}\times 5778^4\\\Rightarrow F=63196526.546\ W/m^2K[/tex]
Power is given by
[tex]P=F4\pi r^2\\\Rightarrow P=63196526.546\times 4\pi\times (696000000)^2\\\Rightarrow P=3.8469943828\times 10^{26}\ W[/tex]
The power output of the Sun is [tex]3.8469943828\times 10^{26}\ W[/tex]
If the potential due to a point charge is 490 V at a distance of 10 m, what are the sign and magnitude of the charge?
Answers
Answer:
+5.4×10⁻⁷ C
Explanation:
Electric potential: This can be defined as the work done in bringing a unit charge from infinity to that point against the action of the field. The S.I unit of potential is volt (V)
The formula for potential is
V = kq/r............................ Equation 1
Where V = electric potential, k = proportionality constant, q = charge, r = distance.
making q the subject of the equation,
q = Vr/k............................ Equation 2
Given: V = 490 V, r = 10 m,
Constant: k = 9×10⁹ Nm²/C²
Substitute into equation 2
q = 490(10)/(9×10⁹)
q = 5.4×10⁻⁷ C
q = +5.4×10⁻⁷ C
Hence the charge is +5.4×10⁻⁷ C
Acetone, a component of some types of fingernail polish, has a boiling point of 56°C. What is its boiling point in units of kelvin? Report your answer to the correct number of significant figures.
Answers
Answer:
The boiling point of Acetone is 329K (in 3 significant figures)
Explanation:
Boiling point of Acetone = 56°C = 56 + 273K = 329K (in 3 significant figures)
Answer: using the formula 0°C + 273.15 = 273.15K the boiling point in units of kelvin to significant figures is 329.15k.
Explanation: The boiling point of a substance ( acetone) is the temperature at which the vapour pressure of the liquid substance equals the pressure surrounding it. The boiling point of acetone serves as it's characteristic physical properties. This is measured in degree Celsius (°C ) which can be converted to units of Fahrenheit or kelvin. To convert degree Celsius to kelvin this formula is used: 0°C + 273.15 = 273.15K . Given that acetone has boiling point of 56°C,from the formula 0°C is substituted for 56°C. This gives us:
56°C + 273.15= 319.15k.
Also,measurements given in Kelvin will always be larger numbers than in Celsius and the Kelvin temperature scale does not use the degree (°) symbol because Kelvin is an absolute scale, based on absolute zero, while the zero on the Celsius scale is based on the properties of water. I hope this helps. Thanks
Four point charges each having charge Q are located at the corners of a square having sides of length a. (a) Find an expression for the total electric potential at the center of the square due to the four charges. (Use any variable or symbol stated above along with the following as necessary: ke.)
Answers
Answer:
[tex]\displaystyle V_t=36\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]
Explanation:
Electric Potential of Point Charges
The electric potential from a point charge Q at a distance r from the charge is
[tex]\displaystyle V=k\frac{Q}{r}[/tex]
Where k is the Coulomb's constant. The total electric potential for a system of point charges is equal to the scalar sum of their individual potentials. The potential is not a vector, so there is no direction or vectors to deal with.
We are required to compute the total electric potential in the center of the square. We need to know the distance from each corner to the center. The diagonal of the square is
[tex]d=\sqrt2 a[/tex]
where a is the length of the side.
The distance from any corner to the center is half that diagonal, thus
[tex]\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}[/tex]
The total potential in the center is
[tex]V_t=V_1+V_2+V_3+V_4[/tex]
Please note all the potentials must be calculated including the sign of the charges. Since all the charges are equal to Q, and the distances are the same, the total potential is 4 times the individual potential of each charge.
[tex]V_t=4\times V[/tex]
[tex]\displaystyle V=9\times 10^9 \frac{Q}{\frac{a}{\sqrt{2}}}[/tex]
Operating
[tex]\displaystyle V=9\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]
Thus:
[tex]\displaystyle V_t=4\times 9\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]
[tex]\boxed{\displaystyle V_t=36\sqrt{2}\times 10^9 \frac{Q}{a}}[/tex]