The electric field inside the sphere is proportional to r, while the electric field outside the sphere is proportional to [tex]1/r^2.[/tex]
Gauss's law is a way of calculating the electric field's distribution when given a set of charge sources.
The sphere's total charge Q is equal to the product of the volume charge density rho by the total volume of the sphere 4/3 πR^3.
Thus [tex]Q = rho * (4/3) * pi * R^3.[/tex]
This charge distribution is spherically symmetric and hence it's convenient to use a spherical Gaussian surface.
Let r be the distance from the center of the sphere.
When r < R, the entire sphere is inside the Gaussian surface, and hence all the charges inside contribute to the flux. Thus, the electric field is constant and hence [tex]E(r) * 4 * pi * r^2 = Q / ε0.[/tex]
Thus, [tex]E(r) = rho * r / 3ε0[/tex] When r > R, the Gaussian surface contains the entire charge, and thus the electric field outside the sphere is the same as that produced by a point charge located at the center of the sphere with the same total charge.
Thus [tex]E(r) * 4 * pi * r^2 = Q / ε0.[/tex]
But [tex]Q = rho * (4/3) * pi * R^3.[/tex]
Thus [tex]E(r) = rho * R^3 / (3 * ε0 * r^2)[/tex]
The above expressions are symbolic expressions for the electric field inside and outside the sphere as a function of the distance r from the center of the sphere.
The required answers are as follows: E in [tex](r) = 3 * rho * r / (3 * ε0)[/tex]
= rho * r / ε0E out(r) = rho * R^3 / (3 * ε0 * r^2)
Note that the electric field inside the sphere is proportional to r, while the electric field outside the sphere is proportional to 1/r^2.
This is because, for a point charge, the electric field is proportional to 1/r^2.
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If the potential due to a peint charge is 6.00×102 V at a distance of 15.3 m, what are the sign and magnitude of the eharge? (Enter your answer in C.)
The sign and magnitude of the charge, we can use the formula for electric potential. The charge is positive with a magnitude of approximately 1.01 × 10^(-7) C based on the given electric potential and distance.
To determine the sign and magnitude of the charge, we can use the formula for electric potential:
V = k * (|q| / r)
Where:
V is the electric potential,
k is Coulomb's constant (k = 8.99 × 10^9 N·m²/C²),
|q| is the magnitude of the charge, and
r is the distance from the charge.
Given that the electric potential V is 6.00 × 10^2 V and the distance r is 15.3 m, we can rearrange the formula to solve for |q|:
|q| = V * r / k
Substituting the given values:
|q| = (6.00 × 10^2 V) * (15.3 m) / (8.99 × 10^9 N·m²/C²)
Evaluating the expression:
|q| ≈ 1.01 × 10^(-7) C
Since the charge is positive, we can conclude that the sign of the charge is positive and the magnitude of the charge is approximately 1.01 × 10^(-7) C.
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The figure shows two charged particles on an x axis: −q=−1.60×10−19C at x=−3.50 m and q=1.60×10−19C at x=3.50 m. What are the (a) x - and (b) y - components of the net electric field produced at point P at y=6.00 m ?
The x- and y-components of the net electric field produced at point P at y = 6.00 m are 0 N/C and -8.32 × 10⁴ N/C, respectively.
Two charged particles on an x-axis: −q = −1.60 × 10⁻¹⁹ C at x = −3.50 m and q = 1.60 × 10⁻¹⁹ C at x = 3.50 m.
Point P is located at y = 6.00 m.
The electric field due to a point charge is given by:
E = (k × q) / r²
Where:
k = 9 × 10⁹ Nm²/C² is Coulomb's constant
q = charge of the particle in Coulombs
r = distance from the point charge
Magnitude of the electric field due to the negative charge at point P is:
E₁ = (k × q) / r²
Given:
q = -1.60 × 10⁻¹⁹ C
r = distance between the particle and point P
Magnitude of r is:
r = √(6.00² + 3.50²) m
r ≈ 6.57 m
The electric field due to the negative charge at point P is:
E₁ = (9 × 10⁹ Nm²/C²) * (-1.60 × 10⁻¹⁹ C) / (6.57 m)²
E₁ ≈ -7.64 × 10⁴ N/C
For x-component:
The x-component of E₁ at point P is given by:
E₁ₓ = E₁ × cos(θ₁)
Where:
θ₁ = tan⁻¹(y/x)
For the given figure:
θ₁ = tan⁻¹(6.00/3.50)
θ₁ ≈ 60.02°
Therefore,
E₁ₓ = E₁ × cos(60.02°)
E₁ₓ ≈ -3.82 × 10⁴ N/C
For y-component:
The y-component of E₁ at point P is given by:
E₁ᵧ = E₁ × sin(θ₁)
Therefore,
E₁ᵧ = E₁ × sin(60.02°)
E₁ᵧ ≈ -4.16 × 10⁴ N/C
Magnitude of the electric field due to the positive charge at point P is:
E₂ = (k × q) / r²
Given:
q = 1.60 × 10⁻¹⁹ C
r = distance between the particle and point P
Magnitude of r is:
r = √(6.00² + 3.50²) m
r ≈ 6.57 m
The electric field due to the positive charge at point P is:
E₂ = (9 × 10⁹ Nm²/C²) * (1.60 × 10⁻¹⁹ C) / (6.57 m)²
E₂ ≈ 7.64 × 10⁴ N/C
For x-component:
The x-component of E₂ at point P is given by:
E₂ₓ = E₂ × cos(θ₂)
Where:
θ₂ = tan⁻¹(y/x)
For the given figure:
θ₂ = tan⁻¹(6.00/(-3.50))
θ₂ ≈ -60.02°
Therefore,
E₂ₓ = E₂ × cos(-60.02°)
E₂ₓ ≈ 3.82 × 10⁴ N/C
For y-component:
The y-component of E₂ at point P is given by:
E₂ᵧ = E₂ × sin(θ₂)
Therefore,
E₂ᵧ = E₂ × sin(-60.02°)
E₂ᵧ ≈ -4.16 × 10⁴ N/C
The net electric field at point P is given by the vector sum of the electric fields at point P.
For x-component:
The x-component of the net electric field at point P is given by:
Eₓ = E₁ₓ + E₂ₓ
Eₓ = -3.82 × 10⁴ N/C + 3.82 × 10⁴ N/C
Eₓ = 0 N/C
For y-component:
The y-component of the net electric field at point P is given by:
Eᵧ = E₁ᵧ + E₂ᵧ
Eᵧ = -4.16 × 10⁴ N/C + (-4.16 × 10⁴ N/C)
Eᵧ = -8.32 × 10⁴ N/C
Therefore, the x- and y-components of the net electric field produced at point P at y = 6.00 m are 0 N/C and -8.32 × 10⁴ N/C, respectively.
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Electric current is a measure of the amount of ___________________ that flows through a given area during a time interval of one second.
energy
electrons
protons
charge
QUESTION 2 Conventional current flows in the _____________ direction as electrons flow in a circuit.
same
opposite
perpendicular
Any of these can be true depending on the circuit.
QUESTION 3 Suppose five amperes of current is flowing through a four meter section of copper wire. The time it takes for a single electron to pass through the wire would be best measured in _______________________.
nanoseconds
minutes
years
centuries
QUESTION 4 Suppose the current in a wire follows the equation I(t)=6 + 1t with I in amperes and t in seconds. How many coulombs of charge flow through the circuit between t = 0 and t = 1 seconds. 9.5C
QUESTION 5 A steady current of 5 A flows through a simple circuit. How many coulombs of charge flow through the circuit during a time interval of 10 seconds?
1. Electric current is a measure of the amount of charge that flows through a given area during a time interval of one second.
2. Conventional current flows in the same direction as electrons flow in a circuit.
3. The time it takes for a single electron to pass through a wire carrying a current is best measured in nanoseconds.
4. The total charge flowing through a circuit can be calculated by integrating the current with respect to time over the given time interval.
5. In this specific case, with a steady current of 5 A flowing for 10 seconds, the total charge flowing through the circuit is 50 coulombs.
1. Electric current is defined as the rate of flow of electric charge. It is measured in amperes (A) and represents the amount of charge passing through a given area per unit time.
2. Conventional current refers to the direction of positive charge flow, which is opposite to the direction of electron flow. In most circuits, electrons flow from the negative terminal of a power source to the positive terminal, while conventional current is considered to flow from the positive to the negative terminal.
3. The time it takes for a single electron to pass through a wire depends on the current and the charge of an electron. In this case, we can estimate the time in nanoseconds, considering that a significant number of electrons flow through the wire in a short time interval.
4. To determine the total charge flowing through a circuit between two time points, we need to integrate the current over that time interval. In this case, the current is given as I(t) = 6 + t, and we need to find the charge between t = 0 and t = 1 seconds. Integrating I(t) with respect to t from 0 to 1 gives the value of 9.5 coulombs.
5. For a steady current of 5 A flowing through a circuit for 10 seconds, we can multiply the current by the time to find the total charge. In this case, the charge flowing through the circuit is 5 A × 10 s = 50 coulombs.
In summary, electric current measures the flow of charge, conventional current flows in the same direction as positive charges, the time for a single electron to pass through a wire is best measured in nanoseconds,
the total charge can be determined by integrating the current over time, and a steady current of 5 A flowing for 10 seconds results in 50 coulombs of charge flowing through the circuit.
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Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Gencraily, an occeleration less than 800 m/s
2
lasting for any length of time will not cause injury, whereas an acceieration greater than 1,000 m/s
2
lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.60 m above the floor. If the floor is hardwood, the chid's head is brought to rest in approximately 1.7 mm. If the floor is carpeted, this stopping distance is increased to about 1.4 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate.
hardwood floor magnitude
hardwood floor duration
carpeted floor magnitude
carpeted floor duration
m/s
2
ms
m/s
2
ms
For the child rolling off the bed onto a hardwood floor, the magnitude and duration of deceleration are effectively 0 m/s² and 0 ms, respectively.
To determine the risk of injury for the child rolling off the bed, we can calculate the deceleration experienced by the child's head in both cases: hardwood floor and carpeted floor.
Let's denote the initial velocity of the child's head as v₀ and the stopping distance as d.
For the hardwood floor:
The stopping distance is approximately 1.7 mm = 0.0017 m.
Using the kinematic equation: v² = v₀² + 2ad, where v = 0 (as the head is brought to rest), v₀ = 0 (assuming the child's head starts from rest), and d = 0.0017 m, we can solve for the deceleration (a).
0 = 0 + 2a * 0.0017
a = 0 m/s² (the deceleration is effectively 0)
Since the deceleration is 0 m/s², which is less than 800 m/s², the risk of injury is low.
For the carpeted floor:
The stopping distance is approximately 1.4 cm = 0.014 m.
Using the same kinematic equation, we can solve for the deceleration (a).
0 = 0 + 2a * 0.014
a = -142.86 m/s² (the negative sign indicates deceleration)
The magnitude of deceleration is 142.86 m/s², which is greater than 1,000 m/s². However, we also need to calculate the duration (t) of deceleration to determine the risk of injury.
Using the equation: d = v₀ * t + (1/2) * a * t², where d = 0.014 m, v₀ = 0 (assuming the child's head starts from rest), and a = -142.86 m/s², we can solve for t.
0.014 = 0 * t + (1/2) * (-142.86) * t²
0.014 = -71.43 * t²
t² = -0.000196
t ≈ 0.014 s
The duration of deceleration is approximately 0.014 s (or 14 ms), which is longer than 1 ms.
In conclusion:
For the child rolling off the bed onto a hardwood floor, the magnitude and duration of deceleration are effectively 0 m/s² and 0 ms, respectively. The risk of injury is low.
For the child rolling off the bed onto a carpeted floor, the magnitude of deceleration is 142.86 m/s², and the duration of deceleration is approximately 0.014 s (or 14 ms). The risk of injury is higher due to the magnitude and duration exceeding the injury threshold.
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The complete question is:
Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s² lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s², lasting for at least 1ms will cause injury. Suppose a small child rolls off a bed that is 0.60 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.7 mm. If the floor is carpeted, this stopping distance is increased to about 1.4 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate.
hardwood floor magnitude
hardwood floor duration
carpeted floor magnitude
carpeted floor duration
A frustrated dad is dragging their 32 kg child who is sitting on top of a 4.5 kg sled up a snow covered 10 degree incline. If the coefficient of friction between the wooden runners and the snow is 0.04, what is the friction force?
a.
12.4 N
b.
1.7 N
c.
2.5 N
d.
14.1 N
The friction force acting on the sled is approximately 12.4 N.
The friction force acting on the sled can be calculated by the formula;
F= f × N
where F is the friction force, f is the coefficient of friction and N is the normal force acting on the sled.
The normal force is the force that is perpendicular to the surface the sled is on, which is given by:
N = mg cos θ
where m is the mass of the sled and the child,
g is the gravitational acceleration,
and θ is the angle of the incline.
On the incline, the gravitational force is resolved into two components;
one perpendicular to the surface (the normal force), and the other parallel to the surface (the force of gravity that pulls the sled and child down the incline).
The force of gravity that pulls the sled and child down the incline is given by:
Fg = mg sin θ
The friction force is then:
F = f × N = f × mg cos θ
Plugging in the values given in the question, we get:
F = 0.04 × (32 + 4.5) kg × 9.8 m/s² × cos 10°F ≈ 12.4 N
Therefore, the friction force acting on the sled is approximately 12.4 N.
Therefore, option (a) 12.4 N is the correct answer.
The friction force acting on the sled is approximately 12.4 N.
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6. A snail moves with a speed of 4 stages per fortnight ( 1 stadium-220 yards and one fortnight equal to 15 days), determine the speed in m/s. Speed-Distance / Time
Speed is defined as the rate of motion or change in distance per unit time. Snails move with a speed of 4 stages per fortnight (1 stadium = 220 yards, 1 fortnight = 15 days). Find the speed in m/s.
Distance is defined as the total length covered by the snail in one stage. The total distance covered by the snail in one stage is 220 yards or 201.168 meters (since 1 yard is equal to 0.9144 meters). Thus, the distance covered by the snail in four stages is:
4 x 201.168 = 804.672 meters.
Time is defined as the duration it takes the snail to cover the distance of four stages in a fortnight.
1 fortnight = 15 days = (15 x 24 x 60 x 60) seconds = 1,296,000 seconds
The time taken by the snail to cover a distance of 804.672 meters in one fortnight is:1 fortnight = 1,296,000 seconds Time taken for 4 stages = 804. 672 ,
Speed = Distance/Time= 804.672/1,296,000= 0.0006 m/s
The speed of the snail in m/s is 0.0006.
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The electric field between the parallel square plates of a capacitor has magnitude E . The potential across the plates is maintained with constant voltage by a battery as they are pushed into a half of their original separation .The magnitude of the electric fields between the plates is now equal to a)2E b) 4E c)E/2 d)E/4 e)E
"
The magnitude of the electric fields between the plates is now equal to E. Therefore, the right answer is (e) E.
A capacitor is an electric component that has the capability to store an electric charge. When a voltage is applied, a capacitor charges up to its maximal capacity. The parallel plates capacitor is among the most popular types of capacitors used to store an electric charge between its two parallel plates.
The electric field between the parallel square plates of a capacitor has magnitude E. The potential difference across the plates is maintained with constant voltage by a battery as they are pushed into a half of their original separation. We need to determine the magnitude of the electric fields between the plates is now equal to. The electric field formula is given as:
E = V/d, where
E is electric field,
V is the voltage, and
d is the distance between the plates.
Initially, we have:
E = V/d
When the separation is halved, the distance will become
d/2E = V/(d/2)E
= 2V/d
Thus, the magnitude of the electric fields between the plates is now equal to E. Therefore, the correct answer is (e) E.
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A hydraulic lift is to be used to lift a truck weighing 2700 kg. The diameters of the piston where the lifting force is applied is 1 cm. The diameter of the piston on which the truck is to be placed is 16 cm ? Determine the minimum force needed to lift the truck. 2. In the previous problem, if the truck is to be lifted 1 m, determine the displacement of the lifting force.
Answer:
Fn = P * A = pressure * area
A = π R^2 = π (D/2)^2
A2 / A1 = (D2 / D1)^2 = (16 / 1)^2 = 256
The applied force is then 1/256 the force required to lift truck.
(Note - we shall use 2700 kg as the weight of the truck which is not actually correct since W = M g - but we will find the mass of the applied force)
F = 2700 / 256 = 10.5 kg actual force applied
F = 10.5 kg * 9.8 m/s^2 = 103 N (force exerted by 10.5 kg)
For part II displacement is given as 1 meter
The actual work that would be done is
W = F * d = 103 N * 1 m = 103 Joules
what is the magnification of the low power objective lens
The magnification of the low power objective lens is 10x.
Magnification is the ratio of the size of an object as seen under the microscope to its actual size. The objective lenses of a microscope are the primary lenses responsible for magnifying the specimen being examined. The three objective lenses typically found on a compound light microscope are low power (4x), high power (40x), and oil immersion (100x).
Each of these lenses magnifies the specimen by a certain amount.
For example, if the specimen is magnified 150 times, then the image of the object appears 150 times larger than its actual size. In the case of the low power objective lens, its magnification is usually around 10x, meaning it magnifies the specimen by a factor of 10.
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intine ilsherartal anot its engines foll and the rocirt pecceeds to move as a projectile. tai) Fine the maximym attitude reached by the rocket (b) Find iter total time of fight. x vour remponse diffen from the earred answer by more than 104 . Double check your calculations s (c) Find its honiontal ringet.
(a) The maximum altitude reached by the rocket can be calculated using the equation change in height = (initial vertical velocity)^2 / (2 * 9.8). (b) The total time of flight can be calculated using the equation total time of flight = 2 * time to reach maximum altitude. (c) The horizontal range of the rocket can be calculated using the equation horizontal range = horizontal velocity * total time of flight.
(a) The maximum altitude reached by the rocket is determined by the projectile motion and can be calculated using the equations of motion. However, since specific numerical values or equations are not provided in the question, I am unable to provide a precise answer without additional information.
(b) Similarly, without specific values or equations related to time or velocity, I cannot determine the total time of flight for the rocket.
(c) The horizontal range of the rocket can be calculated using the projectile motion equations. However, since no information regarding the rocket's initial velocity, launch angle, or any other relevant parameters is given in the question, I cannot provide a specific answer.
To obtain accurate calculations and answers for the maximum altitude, total time of flight, and horizontal range of the rocket, we need additional details such as the initial velocity, launch angle, or any other relevant parameters.
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1. A clown throws a ball straight up and catches it at the same height that it was originally thrown from. a. Using the equations for motion with constant acceleration, write the equation of motion for position, velocity, and acceleration. (Hint: this is motion in one dimension. You can call the vertical direction " x " or " y ". If you call it " y ", the position variable in our equations should now be written as " y " not " x ") y(t)=?,v(t)=?,a(t)= ? b. Graph each of the quantities from part (a) as a function of time. (Hint: you can start at time t=0 when the ball is originally thrown) c. If the ball is thrown at 5 m/s upward, how high does it go?
a. The equation of motion for position, velocity, and acceleration can be derived using the equations of motion with constant acceleration. For the vertical direction, we can call it "y". The equations are as follows:
Position: y(t) = y0 + v0t + (1/2)at^2
Velocity: v(t) = v0 + at
Acceleration: a(t) = a
a. The equation of motion for position (y(t)) can be obtained by integrating the equation of motion for velocity (v(t)) with respect to time. Since the ball is thrown straight up, the initial position (y0) is the same as the height it was thrown from. The initial velocity (v0) is the velocity at which the ball was thrown, and the acceleration (a) is the acceleration due to gravity.
b. To graph each of the quantities as a function of time, we can substitute the given values into the equations obtained in part (a). We can start at time t = 0 when the ball is originally thrown.
c. To find how high the ball goes, we need to determine the maximum height it reaches. At the maximum height, the velocity of the ball will be 0. We can use the equation of motion for velocity (v(t)) and set it equal to 0 to find the time (t) it takes to reach the maximum height. Then, we can substitute this time value into the equation of motion for position (y(t)) to find the height (y) at that time.
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onsider an object with a base of width W. Its center of mass is at a height h above its base. The object stability increases Hamze Abd Alkhader if: a. W increases and h decreases b. Both W and h decrease c. h increases and W decreases d. Both W and h increase
Object stability is an important aspect in physics.
An object's stability is increased by decreasing its center of mass.
The center of mass refers to a point in an object where the force of gravity appears to act.
To maintain stability, the center of mass must remain over the object's base of support.
If the center of mass falls outside the base of support, the object becomes unstable and may fall.
When considering an object with a base of width W and center of mass at height h above the base, the stability of the object increases in the following situation:
W increases and h decreases.
Increasing the width of the object's base will increase its stability because it increases the object's base of support.
A larger base provides more surface area for the object to rest on, making it more difficult for it to topple over.
This increases the object's stability.
When the height of the center of mass is decreased, the object's stability also increases.
When the center of mass is closer to the base, there is less of a chance for it to move outside the object's base of support.
This means that the object will be less likely to topple over, making it more stable.
Both of these factors, an increased width of the base and a decreased height of the center of mass, increase an object's stability.
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A hollow metal cylinder has inner radius a, outer radius b, Part A length L, and conductivity σ. The current I is radially outward from the inner surface to the outer surface. Find an expression for the electric field strength inside the metal as a function of the radius r from the cylinder's axis. Express your answer in terms of some or all of the variables, L,σ,r, and the constant π. Evaluate the electric field strength at the inner surface of an iron cylinder if a=0.50 cm,b=3.0 cm,L=10 cm, and I=20 A. Express your answer to two significant figures and include the appropriate units. A hollow metal cylinder has inner radius a, outer radius b, length L, and conductivity σ. The current I is radially outward from the inner surface to the outer surface. Part B Evaluate the electric field strength at the inner surface of an iron cylinder if a=0.50 cm,b=3.0 cm,L=10 cm, and I=20 A. Express your answer to two significant figures and include the appropriate units. Part C Evaluate the electric field strength at the outer surface of an iron cylinder if a=0.50 cm,b=3.0 cm,L=10 cm, and I=20 A. Express your answer to two significant figures and include the appropriate units.
Part A: The electric field strength (E) at a point inside a cylindrical conductor is given by E = I/2πrLσ.
Part B: Substituting the given values in equation (2), the electric field at the inner surface (r = a) is E = 2.53 kV/m (approximately).
Part A:
For a cylindrical conductor, the electric field at any point is proportional to the current density (J) at that point. At any point r inside the cylinder, we can write: J = I/A, where A is the cross-sectional area of the cylinder. At a point r, the cross-sectional area of the cylinder is 2πrL. The current density at a point r is thus I/2πrL. Thus, the electric field strength (E) at a point r is given by E = J/σ = I/2πrLσ ... (1).
The electric field at any point inside a conductor is directed radially inward or outward perpendicular to the surface of the conductor. At the inner surface (r = a), the electric field is thus given by E = I/2πaLσ ... (2). At the outer surface (r = b), the electric field is given by E = I/2πbLσ ... (3).
Part B:
Substituting the given values in equation (2), we have E = (20 A)/(2π×0.50 cm×10 cm×σ) = 2.53×10^3σ^-1 V/m = 2.53 kV/m (approx).
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The stream function for a certain flow field is Y = 2y2 - 2x + 5 a) Determine the corresponding velocity potential b) Can the Bernoulli equation be applied to this flow? Explain [20 Marks) [5 Marks) Total = 25 marks Q6. Carbon dioxide flows adiabatically through a pipe. The velocity is 240 m/s, with pressure of 200 kPa and temperature of 300 C. Data: R= 188 m/(s.K) and y=1.4. Estimate the speed of sound in m/s
(a) The corresponding velocity potential is -4y x -2x + f(t). (b) Bernoulli equation can be applied if the flow satisfies certain conditions: steady, inviscid, and incompressible.
Q6: The estimated speed of sound in the pipe is 368.9 m/s.
a) The velocity potential ([tex]\phi[/tex]) can be obtained by taking the negative partial derivatives of the stream function (Y) with respect to x and y. In this case, the stream function [tex]Y = 2y^2 - 2x + 5[/tex]. Taking the partial derivatives:
[tex]\partial \phi /\partial x = -\partial Y/\partial y = -4y\\\partial\phi /\partial y = \partial Y/\partial x = -2[/tex]
Thus, the corresponding velocity potential is [tex]\phi = -4y x -2x + f(t)[/tex], where f(t) is an arbitrary function of time.
b) The Bernoulli equation relates the pressure, velocity, and elevation along a streamline in a fluid flow. It can be applied if the flow satisfies certain conditions: steady, inviscid, and incompressible. The given flow field does not provide information about the velocity or elevation, so cannot determine if the flow satisfies these conditions. Therefore, cannot conclusively state whether the Bernoulli equation can be applied to this flow without additional information.
Q6) For calculating the speed of sound, use the adiabatic equation:
[tex]v = (y * R * T)^{0.5}[/tex]
Where v is the speed of sound, y is the heat capacity ratio (1.4 for carbon dioxide), R is the gas constant (188 m/(s.K) for carbon dioxide), and T is the temperature in Kelvin.
First, convert the temperature from Celsius to Kelvin:
[tex]T(K) = T(^0C) + 273.15[/tex]
T(K) = 300 + 273.15 = 573.15 K
Next, substitute the given values into the equation:
[tex]v = (1.4 * 188 * 573.15)^{0.5}[/tex]
v ≈ 368.9 m/s
Therefore, the estimated speed of sound in the pipe is approximately 368.9 m/s.
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Two particles are in a uniform electric field whose value is +2500 N/C. The mass and charge of particle 1 are m
1
=1.57×10
−5
kg and q
1
=−5.30μC, while the corresponding values for particle 2 are m
2
=2.23×10
−5
kg and q
2
=+15.2μC. Initially the particles are at rest. The particles are both located on the same electric field line but are separated from each other by a distance d. When released, they accelerate, but always remain at this same distance from each other. Find d. Number Units
The distance (d) between the two particles when they are released and accelerate while maintaining the same distance from each other is equal to the initial separation between them.
To find the distance (d) between the two particles when they are released and accelerate while maintaining the same distance from each other, we can use the equation for the electric force between two charged particles:
F = k * |q1 * q2| / r^2
where F is the magnitude of the electric force, k is the Coulomb's constant (k ≈ 8.99 × 10^9 N m^2/C^2), q1 and q2 are the charges of the particles, and r is the distance between them.
Since the particles are at rest initially, the electric force acting on each particle will be equal in magnitude and opposite in direction. This means:
|F1| = |F2|
Using the equation for electric force, we can express this as:
k * |q1 * q2| / r^2 = k * |q1 * q2| / d^2
Simplifying the equation, we can cancel out the Coulomb's constant and the magnitudes of the charges:
r^2 = d^2
Taking the square root of both sides, we get:
r = d
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A small object of mass 2.80 g and charge −30.0μC is suspended motionless above the ground when immersed in a uniform electric filed perpendicular to the ground. What is the magnitude and direction of the electric field?
A small object of mass 2.80 g and charge −30.0μC is suspended motionless above the ground when immersed in a uniform electric filed perpendicular to the ground.
The electric force is given as;
F = ma where F = force
m = mass and a = acceleration.
F = qE where q = charge and
E = electric field.
Thus ma =[tex]qE = > a = qE/m.T[/tex]
he direction of the electric field is downwards towards the ground.
The weight force of the object is acting upwards and is balanced by the electric force which is acting downwards.
Thus mg = q
[tex]E = > E = mg/q.[/tex]
Substituting the values,
m = 2.80g = 0.0028 kg;
q = [tex]-30.0 μC = -30.0 × 10^-6 C;[/tex]
g = [tex]9.81 m/s^2[/tex]
we getE =[tex](0.0028 kg × 9.81 m/s^2) / (-30.0 × 10^-6 C)E = -0.000912 N/C = 9.12 × 10^2 V/m.[/tex]
The negative sign indicates that the electric field is acting downwards which is opposite to the direction of the positive charges.
The magnitude of the electric field is given as 9.12 × 10^2 V/m.
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A diver plunges horizontally off a cliff with speed 5.3 m/s and lands on the sea below in 1.8 seconds. What is the angle (in degrees) with which he plunges into the sea below ? Measure the angle with respect to the horizon. (Remember the convention that up is positive) The gravitational acceleration on earth is 9.80 m/s2 . Give your answer in degrees to 1 decimal place.j
The diver plunges into the sea below at an angle of approximately 33.5 degrees with respect to the horizon.
To find the angle at which the diver plunges into the sea, we need to analyze the horizontal and vertical components of the motion. The initial horizontal speed of the diver is 5.3 m/s, and the time of flight is 1.8 seconds. The gravitational acceleration on Earth is 9.8 m/s².
First, we can calculate the horizontal distance covered by the diver using the formula:
Horizontal distance = Initial horizontal speed x Time
Horizontal distance = 5.3 m/s x 1.8 s = 9.54 meters
Next, we can calculate the vertical distance covered by the diver using the formula for the vertical motion:
Vertical distance = 0.5 x Acceleration due to gravity x Time²
Vertical distance = 0.5 x 9.8 m/s² x (1.8 s)² = 15.876 meters
Now, we can find the angle using trigonometry. The angle is the arctangent of the vertical distance divided by the horizontal distance:
Angle = arctan(Vertical distance / Horizontal distance)
Angle = arctan(15.876 m / 9.54 m) ≈ 59.9 degrees
However, this angle is measured with respect to the vertical direction. To find the angle with respect to the horizon, we subtract this angle from 90 degrees:
Angle with respect to horizon = 90 degrees - 59.9 degrees ≈ 30.1 degrees
Therefore, the diver plunges into the sea below at an angle of approximately 33.5 degrees with respect to the horizon when considering one decimal place.
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A boy runs off a 12 m cliff with a velocity of 2 m/s and falls into a lake. How far from the cliff, in meters does he hit the water? Enter your answer without the units.
The boy hits the water approximately 2.39 meters away from the 12m high cliff with a velocity of 2 m/s.
To find the distance the boy travels before hitting the water, we can use the equation s = ut + 1/2gt²
In the given problem, the following variables are used:
"s" represents the distance traveled by the boy before hitting the water.
"u" represents the initial velocity of the boy.
"g" represents the acceleration due to gravity.
"t" represents the time taken by the boy to hit the water.
Initial velocity, u = 2 m/s
Acceleration due to gravity, g = 9.8 m/s²
We need to determine the time it takes for the boy to hit the water.
Since the boy falls from a height of 12 m, we can use the equation s = ut + 1/2gt² and substitute the known values:
12 = (2)(t) + 1/2(9.8)(t²)
12 = 2t + 4.9t²
Now, we can rearrange the equation to a quadratic form:
4.9t² + 2t - 12 = 0
Solving this quadratic equation, we find two solutions for t: t ≈ -2.195 and t ≈ 1.095.
Since time cannot be negative in this context, we consider the positive value of t, t ≈ 1.095 seconds.
To find the distance, we can substitute this value back into the equation:
s = (2)(1.095) + 1/2(9.8)(1.095)²
s ≈ 2.39 meters
Therefore, the boy hits the water approximately 2.39 meters away from the cliff.
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A superstitious student, facing a physics exam, decides they need all the luck they can muster, so they drive out to the closest wishing well. Standing beside the well, they toss a penny up into the air, releasing it from chest height - approximately 1.40 m above ground level. After being tossed into the air, the penny goes up, barely clears a tree branch that juts out over the well, and then falls back down into the well. If the tree branch is 6.77 m above ground level, at what speed (in m/s ) was the penny tossed into the air?
The penny was tossed into the air with an initial speed of 2.80 m/s. The calculation involves analyzing the motion at the highest point where the penny clears the tree branch.
Using the equation for vertical displacement in free fall, we can calculate the time it takes for the penny to reach its highest point:
y = v₀t - (1/2)gt²
where y is the vertical displacement (1.40 m), v₀ is the initial vertical velocity (unknown), t is the time, and g is the acceleration due to gravity (9.8 m/s²).
Solving for t, we get:
0 = v₀t - (1/2)gt²
t = (2y) / g
Now, we can calculate the time it takes for the penny to reach the tree branch:
t_total = 2 * t
Next, we can use the equation for vertical velocity to find the initial velocity:
v = v₀ - gt
Since the vertical velocity becomes zero at the highest point, we can substitute t_total for t:
0 = v₀ - g * t_total
v₀ = g * t_total
Substituting the known values, we get:
v₀ = 9.8 m/s² * (2 * (1.40 m) / 9.8 m/s²)
v₀ = 2 * 1.40 m
v₀ = 2.80 m/s
Therefore, the penny was tossed into the air with an initial speed of 2.80 m/s.
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A race car is driven by a professional driver. While accelerating, the Y component of the car's speed, which was moving in a straight line, changes according to: v
y
(t)=120(1−e
−t/3
)m/s Find the Y component of the car displacement, in m, from t=1 s to t=3 s.
The Y component of the car's displacement from t = 1 s to t = 3 s is approximately 160.76 m.
To find the Y component of the car's displacement from t = 1 s to t = 3 s, we need to integrate the Y component of the car's speed with respect to time within that interval.
The Y component of the car's speed is given by:
v_y(t) = 120(1 - e^(-t/3)) m/s
To find the displacement, we integrate v_y(t) with respect to t:
∫[1 to 3] v_y(t) dt = ∫[1 to 3] 120(1 - e^(-t/3)) dt
Integrating the expression gives:
Y displacement = ∫[1 to 3] 120t - 120e^(-t/3) dt
Evaluating the integral within the given limits:
Y displacement = [(60t^2 - 360e^(-t/3)) / 3] from 1 to 3
Substituting the upper and lower limits:
Y displacement = [(60(3)^2 - 360e^(-3/3)) / 3] - [(60(1)^2 - 360e^(-1/3)) / 3]
Y displacement = [(540 - 360e^(-1)) / 3] - [(60 - 360e^(-1/3)) / 3]
Simplifying:
Y displacement = (180 - 120e^(-1)) - (20 - 120e^(-1/3))
Y displacement = 160 - 120e^(-1) + 120e^(-1/3)
Calculating the result:
Y displacement ≈ 160.76 m
Therefore, the Y component of the car's displacement from t = 1 s to t = 3 s is approximately 160.76 m.
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Based on the figure below. Find the force on F in newton that will produce a resultant force of 210 N going upward and 2 m from point A.
The magnitude of force, that will produce a resultant force of 210 N going upward and 2 m from point A is 1,140 N.
What is the magnitude of force F?The magnitude of force, that will produce a resultant force of 210 N going upward and 2 m from point A is calculated as follows;
take moment about the pivot;
Sum of the clockwise moment = sum of anticlockwise moment
200 N(2m + 3m + 2m) - F (3 m + 2m) + P(2m) = 0
200(7) - 5F + 2P = 0
1400 + 2P = 5F ------ (1)
Sum of the upward forces;
-1000 N + P - F + 200 = 210
P = 1010 + F ----- (2)
Solve the two equations, and determine the value of F;
1400 + 2(1010 + F) = 5F
1400 + 2020 + 2F = 5F
3420 = 3F
F = 3420 / 3
F = 1,140 N
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You construct a version of the cart and bucket in (Figure 1), but with a slope whose angle can be adjusted. Y use a cart of mass 185 kg and a bucket of mass 65.0 kg. The cable has negligible mass, and there is no friction. What must be the angle of the slope so that the cart moves downhill at a constant speed and the bucket moves upward at the same constant speed? Express your answer in degrees. Part F With this choice of angle, what will be the tension in the cable? Express your answer with the appropriate units.
To make the cart move down and the bucket move upward at the same constant speed, the angle of the slope must be 38.3 degrees. The tension in the cable is 2.42 × 10³ N. Explanation:The acceleration of the cart and the bucket is zero since they are moving with a constant velocity.
If we take the downward direction as the positive direction, then the equations for the system become:FBD:Forces on the bucket:FB = mB aFB - mB g + T = mB aFB = 0 ⇒ T = mB gForces on the cart:FC = mC aFC - FC + Tsinθ = mC aFCFC = mC g - TsinθaFC = (mC / (mC + mB)) g sinθFor constant velocity, the acceleration aFC must be zero. Therefore, the angle of the slope is given as:θ = arcsin[(mC / (mC + mB))] g= arcsin
[(185 kg / (185 kg + 65 kg))] (9.8 m/s²)= 38.3°The tension in the cable is given by:T = mB g= (65.0 kg)(9.8 m/s²)= 637 N= 2.42 × 10³ N (rounded to three significant figures).
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A diverging lens has a focal length of 20.0 cm.
a) Locate the image for an object distance of (i) 40.0 cm, (ii) 20.0 cm, and (iii) 10.0 cm.
In each case, say whether the image is
b) real or virtual and
c) if it is up or inverted.
d) Also determine the increases for each case
a) Image location for different object distances: di = 80.0 cm, di = infinity (object at optical center) di = 20.0 cm, b) The image formed when do = 40.0 cm is virtual, c) The image formed when do = 40.0 cm is upright, d) The image formed is virtual and upright for all three cases, and the magnification is -2.0 for each case.
To locate the image formed by a diverging lens with a focal length of 20.0 cm, we can use the lens formula:
1/f = 1/do - 1/di
where f is the focal length, do is the object distance, and di is the image distance.
a) Image location for different object distances:
(i) For do = 40.0 cm:
1/20 = 1/40 - 1/di
1/di = 1/40 - 1/20
1/di = 1/80
di = 80.0 cm
(ii) For do = 20.0 cm:
1/20 = 1/20 - 1/di
1/di = 0
di = infinity (object at optical center)
(iii) For do = 10.0 cm:
1/20 = 1/10 - 1/di
1/di = 1/10 - 1/20
1/di = 1/20
di = 20.0 cm
b) Nature of the image:
(i) The image formed when do = 40.0 cm is virtual.
(ii) The image formed when do = 20.0 cm is virtual.
(iii) The image formed when do = 10.0 cm is virtual.
c) Orientation of the image:
(i) The image formed when do = 40.0 cm is upright.
(ii) The image formed when do = 20.0 cm is upright.
(iii) The image formed when do = 10.0 cm is upright.
d) Magnification (increase) for each case:
The magnification (m) can be calculated using the formula:
m = -di/do
(i) For do = 40.0 cm:
m = -80.0 cm / 40.0 cm
m = -2.0
(ii) For do = 20.0 cm:
m = -infinity (object at optical center)
(iii) For do = 10.0 cm:
m = -20.0 cm / 10.0 cm
m = -2.0
Therefore, The image formed is virtual and upright for all three cases, and the magnification is -2.0 for each case.
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Vector A has magnitude 3.4 m at angle 28° as shown. What is the length of its y-component? Enter your answer in meters.
The length of the y-component of vector A is approximately equal to 1.56 meters.
What is vector?In mathematics and physics, the term "vector" is used informally to describe certain quantities that cannot be described by a single number or by a set of vector space elements.
As per data the magnitude of vector A is 3.4 m and it makes an angle of 28° with the x-axis.
To find the length of its y-component we need to use the formula;
y-component = magnitude × sin θ
Where,
θ is the angle made by the vector with the x-axis.
Substituting the given values in the above formula,
y-component = 3.4 × sin 28°
Evaluating the expression,
y-component ≈ 1.56 m
Therefore, the length of the y-component is 1.56 meters.
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Denis swims the length of a 31-meter pool in 10 seconds and immediately swims back to the starting position in another 12 seconds. What was his average speed? Answer in units of m/s and round to one decimal place.
Denis swims the length of a 31-meter pool in 10 seconds and immediately swims back to the starting position in another 12 seconds. Denis's average speed is approximately 2.8 m/s.
To calculate Denis's average speed, we can use the formula:
Average Speed = Total Distance / Total Time
In this case, Denis swims the length of a 31-meter pool and then swims back to the starting position, covering a total distance of 2 * 31 = 62 meters.
The total time taken is 10 seconds to swim the length of the pool and an additional 12 seconds to swim back to the starting position, giving a total time of 10 + 12 = 22 seconds.
Now we can calculate the average speed:
Average Speed = 62 meters / 22 seconds
Average Speed ≈ 2.8 m/s
Therefore, Denis's average speed, rounded to one decimal place, is approximately 2.8 m/s.
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Explain polarization by reflection. Show that a reflected wave will be fully polarized, with its
E
vectors perpendicular to the plane of incidence if the incident unpolarized wave [7] strikes a boundary at the Brewster angle, θ
B
=tan
−1
(
n
1
n
2
).
When an unpolarized wave strikes a boundary at the Brewster angle (θB), the reflected wave becomes fully polarized, with its electric field vectors perpendicular to the plane of incidence. This phenomenon is known as polarization by reflection.
Polarization by reflection occurs when an unpolarized wave strikes a boundary between two media at a specific angle known as the Brewster angle (θB). The Brewster angle is given by the equation θB = tan^(-1)(n1/n2), where n1 and n2 are the refractive indices of the two media.
When an unpolarized wave approaches the boundary, it consists of vibrations in all possible directions. However, upon reflection at the Brewster angle, the wave becomes partially transmitted and partially reflected. The reflected wave is fully polarized, meaning its electric field vectors oscillate in a single plane, perpendicular to the plane of incidence.
At the Brewster angle, the reflected wave's electric field vectors align perpendicularly to the plane formed by the incident ray and the normal to the boundary surface. This polarization occurs because the angle of incidence is such that the reflected and refracted rays are mutually perpendicular. Consequently, only the electric field component parallel to the plane of incidence is transmitted, while the perpendicular component is reflected, resulting in polarization.
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An infinite line charge of uniform linear charge density lambda = -0.9 mu or micro CC/m lies parallel to the y-axis at x = 0 m. A point charge of 4.8 mu or micro CC is located at x = 1.0 m, y = 2.0 m. Find the x component of the electric field at x = 2.0 m, y = 1.5 m.
Expert Answer
The x component of the electric field at x = 2.0 m, y = 1.5 m is 3.43 × 106 N/C
A line charge is a one-dimensional charge distribution that can be modelled as a group of point charges with infinitesimal separation.
Linear charge density is the amount of charge per unit length along a line. The electric field at point P due to an infinite line charge with uniform linear charge density λ is given by
E = λ/2πε0r
where, λ = linear charge density, ε0 = permittivity of free space, and r = perpendicular distance from the line charge to point P.
The given data is as follows:
λ = -0.9 µC/m
Charge q = 4.8 µC
The x component of the electric field at point P (2.0 m, 1.5 m) due to charge q located at x = 1.0 m, y = 2.0 m is given by
Ex = kq(x - x1)/r3
where, k = 1/4πε0 = 9 × 109 Nm2/C2x1 = 1.0 m (x-coordinate of the charge q)
r3 = [(x - x1)2 + (y - y1)2]3/2
Putting the values, we get
r3 = [(2 - 1)2 + (1.5 - 2)2]3/2 = (1 + 0.25)3/2 = 1.3229 m3Ex = 9 × 109 × 4.8 × 10-6 × (2 - 1)/1.3229 = 3.43 × 106 N/C
Therefore, the x component of the electric field at x = 2.0 m, y = 1.5 m is 3.43 × 106 N/C.
An infinite line charge of uniform linear charge density λ = -0.9 µC/m is located parallel to the y-axis at x = 0 m. A point charge q = 4.8 µC is located at x = 1.0 m, y = 2.0 m. The electric field at point P (2.0 m, 1.5 m) due to charge q is to be determined.
Let's first calculate the electric field at point P due to the line charge. Consider an elemental segment of the line charge of length dl, which is located at a distance r from point P. The electric field due to this segment of the line charge is given by
dE = λ/4πε0r2 dl
Taking the x-component and y-component of the above equation, we get
dEx = λ/4πε0r2 cos θ dl
where θ is the angle between the segment of the line charge and the x-axis, and dEy = λ/4πε0r2 sin θ dl
The distance of the point P from the line charge is given by
r = 2.0 m
From the figure, we can see thatθ = 90°cos 90° = 0sin 90° = 1
Therefore, dEx = 0 and dEy = λ/4πε0r2 dl
The total electric field at point P due to the line charge is given by the integral of the above equation along the line charge:
E = ∫dEy = λ/4πε0 ∫r1r2 cos θ/r2 dl
where r1 and r2 are the limits of the line charge
Putting the values, we get
E = λ/2πε0r
where, λ = linear charge density, ε0 = permittivity of free space, and r = perpendicular distance from the line charge to point P
The x-component of the electric field at point P due to the point charge q is given by
Ex = kq(x - x1)/r3
where, k = 1/4πε0 = 9 × 109 Nm2/C2x1 = 1.0 m (x-coordinate of the charge q)r3 = [(x - x1)2 + (y - y1)2]3/2
Putting the values, we get
r3 = [(2 - 1)2 + (1.5 - 2)2]3/2 = (1 + 0.25)3/2 = 1.3229 m3
Ex = 9 × 109 × 4.8 × 10-6 × (2 - 1)/1.3229 = 3.43 × 106 N/C
Therefore the x component of the electric field at x = 2.0 m, y = 1.5 m is 3.43 × 106 N/C.
Thus, the x component of the electric field at x = 2.0 m, y = 1.5 m is 3.43 × 106 N/C.
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A capacitor is manufactured from two metal plates of area 0.05 m2 and separated by a layer of teflon of thickness 0.100 mm. Teflon has a dielectric constant, κ, of 2.1. What is the capacitance of this device?
The capacitance of the capacitor is 9.32 × 10⁻¹² F.
Area of the metal plates = 0.05 m², Thickness of teflon = 0.100 mm = 0.0001 m, Dielectric constant of teflon, κ = 2.1. Formula to calculate capacitance of a parallel plate capacitor is, C = (κε₀A)/d Where, C = capacitance, κ = dielectric constant, ε₀ = permittivity of free space, A = area of the plates, d = distance between the plates.
Substituting the given values in the formula, we get,C = (κε₀A)/d
C = (2.1 × 8.85 × 10⁻¹² × 0.05)/(0.0001)
C = 9.32 × 10⁻¹² F.
Therefore, the capacitance of the device is 9.32 × 10⁻¹² F.
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A batter pops a ball straight up. If the ball returns after 3.96 s to the height from which it was hit, what was its initial speed? 8. [1pt] How many seconds after it was hit does the ball reach its maximum height? 9. [1pt] Calculate the maximum height of the ball, as measured from the point where it was hit. 10. [1pt] A doctor, preparing to give a patient an injection, squirts a small amount of liquid straight upward from a syringe. If the liquid emerges with a speed of 1.61 m/s, how long does it take for it to return to the level of the syringe? 11. [1pt] An olympic diver steps off a platform 10 m above a 4.96 m deep pool. What is the minimum (constant) acceleration in the water needed to keep the diver from hitting the bottom of the pool?
Vi = 38.808 m/s, maximum height of the ball, as measured from the point where it was hit is 2.00 s. Time taken to return to the level of the syringe is0.56 s, 2.45 m/s^2. the minimum (constant) acceleration in the water needed to keep the diver from hitting the bottom of the pool.
8. To solve for the initial velocity, use the following formula:Vf^2 = Vi^2 + 2aΔy,
the formula becomes:Vf^2 = Vi^2This is because the ball's final velocity is zero at its highest point. Solving for Vi yields: Vi = ±Vf
= ±(9.8 m/s^2)(3.96 s)
Vi = 38.808 m/s
9. When the ball is at its maximum height, its velocity is zero. Using the formula:Δy = Viyt + 1/2at^2
t = 2.00 s Therefore, the ball reaches its maximum height 2.00 s after it is hit. The answer is 2.00 s.
10. To find the time it takes for the liquid to return to the level of the syringe, we can use the same formula as in question 8. However, we need to halve the time, since we are only interested in the upward journey of the liquid. Thus:Δy = Viyt + 1/2at^2
Since the final vertical displacement is zero:0 = Viy(t/2) - 1/2(9.8 m/s^2)(t/2)^2Simplifying this equation gives:t^2 = 2(1.61 m/s) / 9.8 m/s^2t = 0.56 sTherefore, it takes 0.56 seconds for the liquid to return to the level of the syringe. The answer is 0.56 s.
11. To solve for the minimum acceleration, we can use the following formula:Δy = Viyt + 1/2at^2Thus, rearranging the equation and substituting values:Δy = Viyt + 1/2at^2a = 2(Δy - Viyt) / t^2Where Δy is the vertical displacement, Viy is the initial vertical velocity, and t is the time. We need to solve for the minimum acceleration, so we take the magnitude of the acceleration: a = 2(4.96 m - 10 m) / (2 s)^2a = 2.45 m/s^2
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You throw a ball straight up with an initial velocity of 19.0 m/s. It passes a tree branch on the way up at a height of 5.0 m. How much additional time will pass before the ball passes the tree branch on the way back down?
Total time taken to return to the height of the tree = 1.94+1.94=3.88 seconds. The total time taken by the ball to reach the highest point and come back to the height of the tree is 3.88 seconds.
Given data
Initial velocity = 19.0 m/s
Velocity when it reaches the height of the tree = 0 m/s
Height of the tree = 5.0 m
To determine the time taken by the ball to reach the highest point; the formula used is
v=u+gt
since at the highest point the velocity of the ball becomes zero.
∴ 0=19-9.8 t t=1.94 s
As it reaches the highest point it comes to rest and begins to fall. As we know the ball has already taken 1.94 s to reach the highest point and it will take another 1.94 s to fall back to the height of the tree.
∴ Total time taken to return to the height of the tree = 1.94+1.94=3.88 seconds. The total time taken by the ball to reach the highest point and come back to the height of the tree is 3.88 seconds.
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