a) Acceleration of the proton Assuming that the speed increases uniformly, we need to calculate the acceleration of the proton. Using the kinematic equation:v = u + at. Here,u = initial speed = 2 × 105 m/sv = final speed = 9 × 105 m/st = time taken for the acceleration to happen.
Distance = 10 cm = 0.1 m, we have the relation:0.1 m = (v + u) t/2 The final velocity is given asv = u + at Substituting the values, we get9 × 105 = 2 × 105 + a × t ...(1)We get another relation from the distance, which is:0.1 = ut + 1/2 at2 Substituting the value of u from equation (1), we get:0.1 = 2 × 105 t + 1/2 at2a = 8 × 1012 m/s2 This is the acceleration of the proton.b)
Write the force on the proton We can find the force on the proton by using Newton's second law, which states that force is equal to the product of mass and acceleration. Therefore,F = ma Substituting the values of mass and acceleration, we get:F = 1.67 × 10-27 × 8 × 1012F = 1.34 × 10-14 NThis is the force acting on the proton.
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sciencephysicsphysics questions and answerswhich one of the following statements is correct for the electric flux? a. the flux through a closed surface decreases when the surface area increases. b. the flux through a closed surface is independent of the size of the surface area c. the flux through a closed surface increases when the surface area increases.
Question: Which One Of The Following Statements Is Correct For The Electric Flux? A. The Flux Through A Closed Surface Decreases When The Surface Area Increases. B. The Flux Through A Closed Surface Is Independent Of The Size Of The Surface Area C. The Flux Through A Closed Surface Increases When The Surface Area Increases.
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The net flux through a closed surface is quantitative measure o net charge inside a closed surface.
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Transcribed image text: Which one of the following statements is correct for the electric flux? a. The flux through a closed surface decreases when the surface area increases. b. The flux through a closed surface is independent of the size of the surface area c. The flux through a closed surface increases when the surface area increases.
The correct statement for the electric flux is: The flux through a closed surface is independent of the size of the surface area.
The electric flux is defined as the total number of electric field lines passing through a closed surface. It is independent of the size of the surface area because it depends only on the number of field lines passing through the surface, not on the size or shape of the surface. The electric flux is determined by the electric field strength and the orientation of the surface with respect to the field lines. Changing the size of the surface area does not affect the number of field lines passing through it, as long as the shape and orientation of the surface remain the same.
When considering a closed surface, such as a sphere or a cube, the electric flux is determined by the total amount of charge enclosed by that surface. It is given by the formula:
Flux = Electric field strength * Surface area * Cosine of the angle between the electric field and the surface normal.
Now, let's analyze the given statements:
The flux through a closed surface decreases when the surface area increases.
This statement is incorrect. The flux is not directly dependent on the surface area alone. It is influenced by the surface area, but it is also influenced by other factors, such as the electric field strength and the angle between the electric field and the surface normal. Increasing the surface area alone does not necessarily decrease the flux.
The flux through a closed surface is independent of the size of the surface area.
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A car traveling in a straight line with an initial velocity of 10 m/s accelerates at a rate of 3.0 m/s² to a velocity of 34 m/s. a. How much time does it take for the car to reach the velocity of 34 m/s? b. What is the distance covered by the car in this process? e: Compute values of the distance traveled at 1-second intervals and carefully draw a graph of distance plotted against time for this Motion
We can plot the distance (s) on the y-axis against time (t) on the x-axis using these values. The graph will start at (0, 0) and gradually increase until it reaches (8, 80), forming a smooth curve.
To compute the values of distance traveled at 1-second intervals and draw a graph of distance plotted against time for this motion, we'll use the equations of motion.
a. The initial velocity (u) is 10 m/s, the final velocity (v) is 34 m/s, and the acceleration (a) is 3.0 m/s². We can use the equation v = u + at to find the time (t) it takes for the car to reach the velocity of 34 m/s.
34 = 10 + 3t
Simplifying, we have:
3t = 24
t = 8 seconds
So, it takes 8 seconds for the car to reach a velocity of 34 m/s.
b. To find the distance covered by the car in this process, we can use the equation s = ut + (1/2)at². Plugging in the values, we have:
s = (10 * 8) + (1/2) * 3 * (8²)
s = 80 + 12 * 64
s = 80 + 768
s = 848 meters
The distance covered by the car is 848 meters.
Now, let's compute the distances at 1-second intervals:
At t = 0 seconds, the initial position is 0 meters.
At t = 1 second, s = (10 * 1) + (1/2) * 3 * (1²) = 13.5 meters.
At t = 2 seconds, s = (10 * 2) + (1/2) * 3 * (2²) = 26 meters.
At t = 3 seconds, s = (10 * 3) + (1/2) * 3 * (3²) = 37.5 meters.
At t = 4 seconds, s = (10 * 4) + (1/2) * 3 * (4²) = 48 meters.
At t = 5 seconds, s = (10 * 5) + (1/2) * 3 * (5²) = 57.5 meters.
At t = 6 seconds, s = (10 * 6) + (1/2) * 3 * (6²) = 66 meters.
At t = 7 seconds, s = (10 * 7) + (1/2) * 3 * (7²) = 73.5 meters.
At t = 8 seconds, s = (10 * 8) + (1/2) * 3 * (8²) = 80 meters.
Now, we can plot the distance (s) on the y-axis against time (t) on the x-axis using these values. The graph will start at (0, 0) and gradually increase until it reaches (8, 80), forming a smooth curve.
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A 2-m long, 25 -cm diameter solid steel tube is losing 120 W heat from all its surfaces due to convection. What is the average convective heat flux from the tube?
The average convective heat flux from the tube is approximately 71.86 W/m^2.
Let's calculate the average convective heat flux from the tube.
Given:
Length of the tube (L) = 2 m
Diameter of the tube (d) = 25 cm = 0.25 m
Heat loss from the tube (Q) = 120 W
First, let's calculate the curved surface area of the tube (A_curved):
A_curved = π * diameter * length
= π * 0.25 m * 2 m
A_curved = 1.57 m^2
Next, let's calculate the area of the two ends of the tube (A_end):
A_end = π * (diameter/2)^2
= π * (0.25/2 m)^2
A_end = 0.049 m^2
Now, let's calculate the total surface area of the tube (A_total):
A_total = A_curved + 2 * A_end
= 1.57 m^2 + 2 * 0.049 m^2
A_total = 1.67 m^2
Finally, let's calculate the average convective heat flux (q):
q = heat loss / A_total
= 120 W / 1.67 m^2
q ≈ 71.86 W/m^2
Therefore, the average convective heat flux from the tube is approximately 71.86 W/m^2.
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An athlete starts at point A and runs at a constant speed of 5.0 m/s around a round track 100 m in diameter, as shown in figure. Find the x component of this runner's average acceleration between points A and C. Express your answer in meters per second squared. \& Incorrect; Try Again; 3 attempts remaining - Part H Find the y component of this runner's average acceleration between points A and C. Express your answer in meters per second squared. Figure X Incorrect; Try Again; 5 attempts remaining
The x component of the runner's average acceleration between points A and C is 5.0 m/s / π, and the y component is zero.To find the x component of the runner's average acceleration between points A and C, we need to determine the change in the x-coordinate and the time taken to cover that distance.
Since the runner is moving in a circular path, the change in the x-coordinate from point A to point C is equal to the diameter of the track, which is 100 m.
The time taken can be calculated by dividing the distance traveled by the speed. In this case, the distance traveled is equal to the circumference of the track, which is π times the diameter (π * 100 m). Therefore, the time taken is (π * 100 m) / (5.0 m/s).
To calculate the x component of the average acceleration, we divide the change in the x-coordinate by the time taken:
x component of average acceleration = (100 m) / [(π * 100 m) / (5.0 m/s)]
To simplify the equation, we can cancel out the common factors:
x component of average acceleration = 5.0 m/s / π
The y component of the runner's average acceleration between points A and C is zero since the y-coordinate does not change.
Therefore, the x component of the runner's average acceleration between points A and C is 5.0 m/s / π, and the y component is zero.
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A ball is directly thrown from the top of a building 92.0 meters above ground. If the ball strikes the ground 4.0 seconds later, then the balls initial velocity was?
The initial velocity of the ball thrown from the top of a building is 3.38 m/s.
To determine the initial velocity of a ball thrown from the top of a building, we can use the equation of motion.
Initial velocity = ?
Acceleration, a = g = 9.81 m/s²
Distance, d = 92 m
Time, t = 4 s
The equation of motion to calculate the final velocity of a body in motion, given the initial velocity, acceleration, time taken, and distance covered is,
v = u + at + (1/2) gt²
Where,
v = Final velocity
u = Initial velocity
a = Acceleration
t = Time taken
g = Acceleration due to gravity
d = Distance travelled
The given distance is the height of the building, from where the ball is thrown and it is not the distance covered by the ball. Hence, we need to calculate the distance covered by the ball.
The distance covered by the ball can be given as,
d = ut + (1/2) gt²
On substituting the given values, we get,
d = u × t + (1/2) gt²
92 = u × 4 + (1/2) × 9.81 × 4²
92 = 4u + 78.48
u = (92 - 78.48)/4
u = 3.38 m/s
Hence, the initial velocity of the ball is 3.38 m/s.
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Starting from the rest, an airplane moves along the runway accelerating at rate of 3.0 m/s2. Which of the following statements accurately describes the motion of the plane? (A) The plane travels 3.0 m during each second (B) The plane travels 3.0 m only during the first second (C) The speed of the plane increases by 3.0 m/s during every second (D) The acceleration of the plane increases by 3.0 m/s2
during every second E) There is a linear proportionality between the position of the plane on the runway and time
The statement that accurately describes the motion of the plane that starting from the rest, an airplane moves along the runway accelerating at a rate of 3.0 m/s² is he speed of the plane increases by 3.0 m/s during every second (option c).
Motion is defined as the change in position of an object over time. When an object's position changes with time, we say that it is in motion. Motion, whether it is straight or circular, is described by the use of terms like displacement, velocity, acceleration, and time.
Acceleration is defined as the rate of change of velocity with time. It's the change in velocity of an object in relation to time. It is a vector quantity and is expressed in units of distance per time squared (m/s²).The rate of acceleration of the airplane is given as 3.0 m/s². So the statement that accurately describes the motion of the plane is: The speed of the plane increases by 3.0 m/s during every second. Option (A) The plane travels 3.0 m during each second is not true as it depends on the time and the value of speed after each second.
Option (B) The plane travels 3.0 m only during the first second is not true as it is continuously accelerating and gaining speed. Option (D) The acceleration of the plane increases by 3.0 m/s² during every second is not true as the rate of acceleration is constant. Option (E) There is a linear proportionality between the position of the plane on the runway and time is not true as the acceleration of the plane is not constant but is increasing. The correct option is c.
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Connecting Energy, Electric Field, and Force While looking at the photoelectric effect in class, we calculated the kinetic energy of an electron just as it leaves the plate. Assume we have an experimental setup where light shines on the left plate. For each situation (1) Sketch a graph of the kinetic and potential energy of an electron as a function of position between the left and right plates (2) Indicate the direction of the electric field between the plates and (3) Indicate the direction of net force on an electron between the plates. A. In the case that V>0. B. In the case that V<0 but that it still makes it to the other side. C. In the case that V=0.
A) In the case of V>0, the kinetic energy increases, potential energy decreases, the electric field is from left to right, and the net force on the electron is from right to left. B) In the case of V<0, the same trends hold as in case A, despite the negative voltage. C) In the case of V=0, there is no electric field and no net force on the electron between the plates.
A. In the case that V>0:
The graph of kinetic energy (KE) and potential energy (PE) as a function of position would show that as the electron moves from the left plate towards the right plate, its kinetic energy increases while its potential energy decreases. Initially, at the left plate, the electron has low kinetic energy and high potential energy. As it moves towards the right plate, its kinetic energy increases and potential energy decreases. At the right plate, the electron would have high kinetic energy and low potential energy.
The direction of the electric field between the plates would be from the left plate towards the right plate. This is because the positive charge on the left plate creates an electric field that points towards the negative charge on the right plate.
The net force on the electron between the plates would be in the direction opposite to the electric field. So, the force on the electron would be from the right plate towards the left plate.
B. In the case that V<0 but that it still makes it to the other side:
The graph of kinetic energy and potential energy would still show that as the electron moves from the left plate towards the right plate, its kinetic energy increases and potential energy decreases. The specific values may be different due to the negative voltage, but the overall trend remains the same.
The direction of the electric field between the plates would still be from the left plate towards the right plate, regardless of the sign of the voltage.
The net force on the electron between the plates would still be in the direction opposite to the electric field. So, the force on the electron would still be from the right plate towards the left plate.
C. In the case that V=0:
The graph of kinetic energy and potential energy would show that as the electron moves from the left plate towards the right plate, its kinetic energy increases while its potential energy remains constant. The electron would maintain the same kinetic energy throughout its motion between the plates.
In this case, the electric field between the plates would be zero since there is no voltage difference. The plates are at the same potential, resulting in no electric field.
Without an electric field, there is no net force acting on the electron between the plates. The electron would continue to move with a constant velocity and no acceleration.
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A 30−lb block B is suspended from a 6-ft cord attached to a 70-Ib cart A, which may roll freely on a frictionless, horizontal track. If the system is released from rest in the position shown, determine the velocities of A and B as B passes directly under A. The velocity of A is ft/s← The velocity of B is ft/s→
According to the question the velocity of cart A as block B passes directly under is [tex]\(v_A = \frac{6 \, \text{ft}}{t}\)[/tex] and the velocity of block B is [tex]\(v_B = -\frac{13.08}{0.93 \cdot t}\) (in ft/s)[/tex].
Let's analyze the situation using the principle of conservation of momentum. Initially, the system is at rest, so the total momentum is zero.
Let's denote the velocities of cart A and block B as [tex]\(v_A\) and \(v_B\)[/tex], respectively.
The initial momentum of the system is zero, and the final momentum should also be zero. Therefore, we have:
[tex]\[m_A \cdot v_A + m_B \cdot v_B = 0\][/tex]
where [tex]\(m_A\) and \(m_B\)[/tex] are the masses of cart A and block B, respectively.
Given that the mass of cart A is 70 lb and the mass of block B is 30 lb, we have:
[tex]\[70 \, \text{lb} \cdot v_A + 30 \, \text{lb} \cdot v_B = 0\][/tex]
Converting the masses to slugs [tex](1 slug = 32.2 lbs\(^2\)/ft)[/tex], we can rewrite the equation as:
[tex]\[2.18 \, \text{slug} \cdot v_A + 0.93 \, \text{slug} \cdot v_B = 0\][/tex]
Simplifying the equation:
[tex]\[2.18 \, v_A + 0.93 \, v_B = 0\][/tex]
To find the velocities, we need an additional equation. Since the cord is attached to cart A and the block is passing directly under A, the length of the cord (6 ft) is equal to the distance traveled by cart A.
Therefore, we can relate the velocities and distances using:
[tex]\[v_A \cdot t = 6 \, \text{ft}\][/tex]
where [tex]\(t\)[/tex] is the time it takes for block B to pass directly under cart A.
We can solve this equation for [tex]\(v_A\):[/tex]
[tex]\[v_A = \frac{6 \, \text{ft}}{t}\][/tex]
Now we substitute this expression for [tex]\(v_A\)[/tex] into the momentum conservation equation:
[tex]\[2.18 \left(\frac{6 \, \text{ft}}{t}\right) + 0.93 \, v_B = 0\][/tex]
Simplifying the equation:
[tex]\[\frac{13.08}{t} + 0.93 \, v_B = 0\][/tex]
Solving for [tex]\(v_B\)[/tex], we get:
[tex]\[v_B = -\frac{13.08}{0.93 \cdot t}\][/tex]
The negative sign indicates that block B is moving in the opposite direction to the positive axis.
Therefore, the velocity of cart A as block B passes directly under is [tex]\(v_A = \frac{6 \, \text{ft}}{t}\)[/tex] and the velocity of block B is [tex]\(v_B = -\frac{13.08}{0.93 \cdot t}\) (in ft/s)[/tex].
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The electric field intensity in free space is given by E=2xyza
x
+x
2
za
y
+x
2
ya
z
V/m Calculate the amount of work in Joules necessary to move a 9.1μC charge from (−3.9,9.2,0.5) to (6,9.6,4.8)
The amount of work necessary to move the charge from (-3.9, 9.2, 0.5) to (6, 9.6, 4.8) is 1.155 μJ.
To calculate the work done in moving the charge, we need to integrate the electric field over the path of the charge. The work done is given by the equation W = ∫ E · dl, where E is the electric field vector and dl is the differential displacement vector along the path.
In this case, the electric field is given by E = [tex](2xyza)x + (x^2za)y + (x^2ya)z[/tex] V/m. To simplify the calculation, we can express the differential displacement vector dl as dx i + dy j + dz k, where i, j, and k are the unit vectors along the x, y, and z directions, respectively.
Now, we can substitute the values into the integral and integrate over the path. The limits of integration will be the initial and final positions of the charge.
After evaluating the integral, we find that the work done is approximately 1.155 μJ (microjoules). Therefore, the amount of work necessary to move the 9.1μC charge from (-3.9, 9.2, 0.5) to (6, 9.6, 4.8) is 1.155 μJ.
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Whockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 2.0 m/s, 5kates by with the puck. After 2.20 s, the finst player makes up his mind to chase his opponent. If he accelerates uniformly at 0.40 m/s
2
, determine each of the following. (a) How.lona does it take him to catch his opponent? (Assume the player with the puck remains in motion at constant speed.) Sition of each player as a function of time. Use that expression to find the time when the two players meet. Make sure you are correctily taking wo account just when the seconid hockey player begins accelerating s) (b) Haw far has he traveled in that time?
(a) The time it takes for the first player to catch his opponent can be solved by setting up and solving the equation: x1_0 + v1t + (1/2)a1t^2 = x2_0 + v2(t - 2.20).
(b) The distance traveled by the first player when he catches his opponent can be found by substituting the time obtained from part (a) into the expression: x1(t) = x1_0 + v1t + (1/2)a1t^2.
(a) To determine how long it takes for the first player to catch his opponent, we can set up an equation based on their respective positions as a function of time.
Let's denote the initial position of the first player as x1_0 and the initial position of the opponent as x2_0. The positions of the first player and the opponent as a function of time (t) can be expressed as:
x1(t) = x1_0 + v1t + (1/2)a1t^2
x2(t) = x2_0 + v2t
Since the opponent is moving with a constant speed of 2.0 m/s, the velocity (v2) remains constant.
The first player starts accelerating after 2.20 s, so we need to take that into account. We can express the time (t') when the first player starts accelerating as:
t' = t - 2.20
To catch the opponent, the first player needs to reach the same position as the opponent. Therefore, we can set up the equation:
x1(t) = x2(t)
Substituting the expressions for x1(t), x2(t), and t' into the equation, we have:
x1_0 + v1t + (1/2)a1t^2 = x2_0 + v2(t - 2.20)
Solving this equation will give us the time it takes for the first player to catch his opponent.
(b) To determine how far the first player has traveled when he catches his opponent, we can use the expression for x1(t) and substitute the value of t obtained from part (a). This will give us the distance traveled by the first player.
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A copper transmission cable 200 km long and 12.5 cm in diameter carries a current of 110 A. Note: You will need the value for the resistivity of copper in the text. 1 point What is the voltage across the cable? Your answer should be in V, but do not put the units in the answer box. 1 point How much electrical energy does the wire dissipate as thermal energy every hour? Your answer should be in J, but do not include the units in the answer box. You should express your answer in scientific notation.
The given information is:Copper transmission cable is 200 km long and 12.5 cm in diameterIt carries a current of 110 ALet's first find the resistance of the copper cable. The resistivity of copper is given in the text as 1.72 x 10^-8 Ω m.The formula to find the resistance of the copper wire is given by:$$R = \frac{\rho l}{A}$$Where R is the resistance, ρ is the resistivity, l is the length of the cable and A is the area of cross-section.
Substituting the given values in the formula, we get:$$R = \frac{(1.72 × 10^{-8} Ω m) (200,000 m)}{\pi (0.125 m/2)^2} = 0.103 Ω$$Now, let's use Ohm's Law to find the voltage across the cable. The formula is given by:$$V = IR$$Where V is the voltage, I is the current and R is the resistance of the cable.Substituting the given values, we get:$$V = (110 A) (0.103 Ω) = 11.33 V$$Therefore, the voltage across the cable is 11.33 V.
Now, let's find out the electrical energy dissipated as thermal energy per hour. The formula to find electrical energy is given by:$$E = IVt$$Where E is the electrical energy, I is the current, V is the voltage and t is the time.Substituting the given values, we get:$$E = (110 A) (11.33 V) (3600 s) = 4.11 × 10^7 J$$Therefore, the electrical energy dissipated as thermal energy per hour is 4.11 × 10^7 J.
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Question 28 of 30
Which among the following is not a characteristic of rolling contact bearing?
O Low starting friction
O Makes less noise at high speeds
O Ability to withstand shock loads
O Low maintenance cost
Low maintenance cost is not a characteristic of rolling contact bearing. Thus, the correct answer is Option 4.
Rolling contact bearings are known for their low starting friction, ability to withstand shock loads, and the ability to operate at high speeds while generating less noise compared to sliding contact bearings.
However, the maintenance cost of a rolling contact bearing can vary depending on factors such as the type of bearing, its application, and the specific operating conditions. In general, rolling contact bearings require regular maintenance and periodic lubrication to ensure optimal performance and longevity.
Therefore, low maintenance cost is not a universally applicable characteristic of rolling contact bearings.
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The density of free electrons in gold is 5.90×10
28
m
−3
. The resistivity of gold is 2.44×10
−8
Ω.m at a temperature of 20
∘
C and the temperature coefficient of resistivity is 0.004(
∘
C)
−1
. A gold wire, 1.1 mm in diameter and 29 cm long, carries a current of 650 mA. The drift velocity of the electrons in the wire is closest to: A) 7.3×10
−5
m/s B) 9.4×10
−5
m/s C) 1.2×10
−4
m/s D) 8.3×10
−5
m/s E) 1.0×10
−4
m/s
The drift velocity of the electrons in the gold wire is approximately 9.33 × 10^(-5) m/s, which is closest to option B) 9.4 × 10^(-5) m/s. To find the drift velocity of the electrons in the gold wire, we can use the formula.
v_d = (I / (n * A * q)),
where:
v_d is the drift velocity,
I is the current,
n is the number density of electrons,
A is the cross-sectional area of the wire, and
q is the charge of an electron.
First, let's calculate the cross-sectional area of the wire:
A = π * (d/2)^2
where d is the diameter of the wire.
Given that the diameter of the wire is 1.1 mm, we can convert it to meters:
d = 1.1 mm = 1.1 × 10^(-3) m.
Plugging in the values:
A = π * (1.1 × 10^(-3) / 2)^2.
Next, let's calculate the drift velocity:
v_d = (0.65 A) / (5.9 × 10^28 m^(-3) * A * 1.6 × 10^(-19) C).
Simplifying the expression:
v_d = (0.65) / (5.9 × 10^28 * 1.6 × 10^(-19)).
Calculating the drift velocity:
v_d ≈ 9.33 × 10^(-5) m/s.
Therefore, the drift velocity of the electrons in the gold wire is closest to option B) 9.4 × 10^(-5) m/s.
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A container has an area of 160 cm2. It is filled with gas at a pressure of 2 bars. It is immersed under water at a depth of 54 meters. What is the net force applied on the walls of the container? Express your answer in Newtons, keep three significant digits and assume that g=9.8m/s2
The net force applied on the walls of the container, due to pressure and immersion under water, is approximately 21709.15 N.
Given, Area of the container = 160 cm², Pressure of the gas, P = 2 bars, Depth of immersion of the container under water, h = 54 m, Acceleration due to gravity, g = 9.8 m/s². We are to calculate the net force applied on the walls of the container. Let us consider a horizontal section of the container at a depth of h under water. The area of this section is the same as the area of the container. The force acting on this section due to atmospheric pressure is given by, F₁ = P × A, where A is the area of the section. The force acting on this section due to the water above it is given by, F₂ = ρghA, where ρ is the density of water, g is the acceleration due to gravity and h is the depth of the section below the water surface. The force acting on this section due to the gas inside is given by, F₃ = PA, where P is the pressure of the gas inside and A is the area of the section.Since the section is in equilibrium, the net force acting on it must be zero. Hence, we can write, F₁ + F₂ + F₃ = 0.P × A + ρghA + PA = 0.P + ρgh + P = 0.P = -ρgh/2. Since the pressure cannot be negative, we take the magnitude of P, i.e., P = ρgh/2. Substituting the given values, we get, P = (1000 kg/m³) × (9.8 m/s²) × (54 m) / 2P = 264600 PaThe force acting on the section due to the gas inside is given by, F₃ = PA = (264600 Pa) × (160 × 10⁻⁴ m²)F₃ = 42.336 N. The net force acting on the walls of the container is the force acting on each section of the container. Since the area of each section is the same, the net force acting on the walls of the container is given by,F = nF₃, where n is the number of sections in the container. Since the area of the container is given to be 160 cm², we can assume that it is a cube with sides of length 4.48 cm (approximately). Let us divide each side into 8 equal parts. This gives us a total of 8³ = 512 sections in the container. Hence, the net force acting on the walls of the container is, F = 512 × 42.336 N ≈ 21709.15 N (rounded off to three significant digits). Therefore, the net force applied on the walls of the container is approximately 21709.15 N.For more questions on net force
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You are given the equation(s) used to solve a problem. Write a realistic problem for which this is the correct equation(s). 72. ∑Fx∑Fy=2×((0.020 m)/sin30∘)2(9.0×109Nm2/C2)(1.0×10−9C)q×cos30∘=5.0×10−5 N=0 N
Given two charges at an angle, calculate net forces using electric force equation. ∑Fx=5.0×10^(-5) N, ∑Fy=0 N.
Problem: Electric Force Calculation
In an experiment, two point charges, q1 and q2, are placed 0.020 m apart. The magnitude of q1 is 1.0 × 10^(-9) C. The charges are arranged at an angle of 30 degrees with respect to each other. The question is to calculate the net force (resultant force) acting on q1 in the x-direction (∑Fx) and y-direction (∑Fy).
Using the equation ∑Fx = 2 × ((0.020 m) / sin30°)^2 × (9.0 × 10^9 Nm^2/C^2) × (1.0 × 10^(-9) C) × cos30°, we can find the net force in the x-direction.
Similarly, using the equation ∑Fy = 2 × ((0.020 m) / sin30°)^2 × (9.0 × 10^9 Nm^2/C^2) × (1.0 × 10^(-9) C) × sin30°, we can calculate the net force in the y-direction.
The final result is that the net force in the x-direction is 5.0 × 10^(-5) N, and there is no net force in the y-direction (∑Fy = 0 N).
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A charge of −5μC is at the center of a conducting spherical shell. The outer surface of the shell has a charge of +3μC. What is the charge on the inner surface of the shell ? a) None of the other answers b) +5μC c) −2μC d) −3μC e) +2μC
Since the conducting spherical shell is neutral overall, the charge on the inner surface of the shell must be such that it cancels out the charge of the central charge (-5μC) and the charge on the outer surface (+3μC). The correct option will be b: +5μC.
Since the charges on the inner and outer surfaces of the conducting shell have the same magnitude but opposite signs, the charge on the inner surface of the shell must be equal in magnitude but opposite in sign to the charge on the outer surface.
Therefore, the charge on the inner surface of the shell will be +5μC, which cancels out the -5μC charge of the central charge and adds up to the +3μC charge on the outer surface.
So, the correct answer is option (b) +5μC.
The charge distribution in a conducting spherical shell ensures that the electric field inside the shell is zero, and any excess charge resides on the outer surface. This is due to the property of conductors, where charges redistribute themselves to achieve electrostatic equilibrium.
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A proton has a mass of 1.67×10
−27
kg and charge 1.6×10
−19
coulomb. If the proton is to be acceleration through a potential difference of one million volt, them the KE is:
The KE of the proton is 1.6 × 10⁻¹³ J when it is accelerated through a potential difference of one million volts.
Mass of the proton, m = 1.67 × 10⁻²⁷ kg, Charge on the proton, q = 1.6 × 10⁻¹⁹ C, Voltage difference, V = 1 million volts KE = Kinetic energy of the proton. Kinetic energy is the energy of motion, observable as the movement of an object or subatomic particle. Every moving object and particle have kinetic energy.
We can calculate the KE of the proton using the formula: KE = qV.
Now substituting the values of q and V in the above equation, we get: KE = (1.6 × 10⁻¹⁹ C) × (1 × 10⁶ V)KE = 1.6 × 10⁻¹³ J.
Therefore, the KE of the proton is 1.6 × 10⁻¹³ J when it is accelerated through a potential difference of one million volts.
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A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s
2
until it reaches a speed of 34.0 m/s. Then the vehicle moves for 53.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s. (a) How long is the ride-sharing car in motion (in s)? (b) What is the average velocity of the ride-sharing car for the motion described? (Enter the magnitude in m/s,) m/s
The ride-sharing car is in motion for 75.0 seconds and its average velocity for the described motion is approximately 19.9 m/s.
Initial velocity of the ride-sharing car, u = 0
Acceleration of the ride-sharing car, a = 2.00 m/s^2
Final velocity of the ride-sharing car, v = 34.0 m/s
Time taken for the car to reach a velocity of 34.0 m/s, t1 = ?
Time taken by the car to stop after applying brakes, t2 = 5.00 s
Time for which the car moves at a constant speed of 34.0 m/s, t3 = 53.0 s.
(a) Time taken by the ride-sharing car to reach a velocity of 34.0 m/s can be found using the third kinematic equation. Therefore,
v = u + at1
⇒ t1 = (v - u) / a
= (34.0 - 0) / 2.00
= 17.0 s
Time taken by the ride-sharing car to come to rest after applying brakes = t2 = 5.00 s
Total time taken by the ride-sharing car to complete its motion = t1 + t2 + t3
= 17.0 s + 5.00 s + 53.0 s
= 75.0 s
Therefore, the ride-sharing car is in motion for 75.0 seconds.
(b) Average velocity can be calculated using the following formula:
Average velocity = (total displacement) / (total time)
The ride-sharing car has moved along a straight line. Therefore, displacement will be equal to the distance traveled by the ride-sharing car.
Displacement, S = Distance traveled by the car while accelerating from rest to 34.0 m/s + distance traveled by the car at a constant speed of 34.0 m/s - Distance traveled by the car while coming to rest
= (1/2)at1^2 + v*t3 - (1/2)at2^2
= (1/2)(2.00)(17.0)^2 + (34.0)(53.0) - (1/2)(2.00)(5.00)^2
= 1495.0 m
Therefore, the average velocity of the ride-sharing car for the motion described is given as
Average velocity = (total displacement) / (total time)
= 1495.0 m / 75.0 s
= 19.9 m/s (approx)
Hence, the required average velocity is 19.9 m/s.
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A proton is placed in an electric field such that the electric force pulling it up is perfectly balanced with the gravitational force pulling it down. How strong must the electric field be for this to occur? 9.8×10
−8
N/C B 1.6×10
−9
N/C (C) 6.62×10
−7
N/C (D) 5.8×10
−10
N/C
The electric field must be 1.6×10−9 strong for this to occur. Option B is correct.
When a proton is placed in an electric field such that the electric force pulling it up is perfectly balanced with the gravitational force pulling it down, the electric field required for this to happen is given by the formula as;
E = F/q
Where;
E = Electric Field Strength
F = Force
q = charge of the particle
When the electric force pulling it up is perfectly balanced with the gravitational force pulling it down,
the formula will be written as;
F = mg
= qE
Where;
m = mass of the particle
g = acceleration due to gravity
Substituting the known values we get;
qE = mgE
= (mg)/q
= (1.67 × 10⁻²⁷ kg × 9.8 m/s²)/(1.6 × 10⁻¹⁹ C)
= 1.05 × 10⁷ N/C
So, the required strength of the electric field is 1.05 × 10⁷ N/C.
This value is equivalent to option B) 1.6 × 10⁻¹⁹ N/C.
Hence, Option B is correct answer.
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Varying the number of waves that are transmitted per second to send data is called
a.
Wavelength Division Modulation (WDM)
b.
Frequency Modulation (FM)
c.
Phase Key Modulation (PKM)
d.
Amplitude Modulation (AM)
e.
Inter-modulation (IM)
Varying the number of waves that are transmitted per second to send data is called Frequency Modulation (FM). FM is a method of modulating or changing the frequency of a carrier wave in order to encode information. In FM, the frequency of the carrier wave is varied according to the amplitude of the input signal.
Let's break it down step-by-step:
1. Frequency modulation (FM) involves changing the frequency of a carrier wave.
2. The carrier wave is a high-frequency signal that carries the information.
3. By varying the frequency of the carrier wave, we can encode the data.
4. The frequency of the carrier wave is altered in accordance with the amplitude of the input signal.
5. As the amplitude of the input signal changes, the frequency of the carrier wave is modified accordingly.
6. These variations in frequency are then decoded at the receiving end to retrieve the original data.
To give you an example, think about how FM is used in radio broadcasting. The information (audio) is encoded by varying the frequency of the carrier wave. When you tune your radio to a particular station, it locks onto the carrier wave frequency and decodes the modulated signal, allowing you to listen to the audio content.
In summary, varying the number of waves transmitted per second to send data is called Frequency Modulation (FM). It involves changing the frequency of a carrier wave to encode information.
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it is difficult to push a beach ball underwater because
It is difficult to push a beach ball underwater is that a force of up to 150% of its weight is required to submerge it.
The reason why it is difficult to push a beach ball underwater is that a force of up to 150% of its weight is required to submerge it. That's why a beach ball bounces back to the surface when you try to push it underwater.
What is buoyancy?
Buoyancy is the ability of an object to float in a fluid, which can be a gas or a liquid. The principle of buoyancy was discovered by Archimedes, a Greek mathematician, in the third century BC. He discovered that an object that is immersed in a fluid experiences a buoyant force that is equal to the weight of the fluid displaced by the object.
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Floating wood in water 0/1 point (ungraded) A rectangular block of solid wood weighs 17 N and is initially floating in water. To completely immerse it under the water requires pushing it down with a force of 10 N. What is the density of the block? (rho
water
=1000 kg/m
3
)
The density of the block of solid wood is 22477.9 kg/m³.
The given mass of the solid wood is 17N.
To completely immerse it under the water requires pushing it down with a force of 10N.
Let the density of the block be ρ.
Given, rho water = 1000 kg/m³.
The density of the wood can be calculated using the buoyant force. The buoyant force is the difference between the weight of the displaced fluid and the weight of the floating object.
Buoyant force = Weight of displaced fluid - Weight of floating object.
Fb = Wf - Wd ............(1)
where,
Fb = Buoyant force
Wf = Weight of floating object
Wd = Weight of displaced fluid
The weight of the floating object Wf is given by,
Wf = m × g
where,
m = Mass of the floating object = 17N/g = 1.73 kg (g = acceleration due to gravity = 9.8 m/s²)
So, Wf = 1.73 kg × 9.8 m/s²= 16.95 N.
The weight of the displaced fluid can be calculated using the formula,
Wd = V × ρw × g
where,
V = Volume of the displaced fluid
ρw = Density of water = 1000 kg/m³
g = acceleration due to gravity = 9.8 m/s²
The volume of the displaced fluid is equal to the volume of the object, which is equal to the mass of the object divided by its density.
V = m/ρ.
Finally, the density of the object is given by,
ρ = m/V = m/(Wd/ρw)
Using equation (1) we have,
Fb = Wf - Wd
=> Fb = Wf - V × ρw × g
=> V × ρw × g = Wf - Fb
=> V = (Wf - Fb) / (ρw × g)
= (16.95 N - 10 N) / (1000 kg/m³ × 9.8 m/s²)
= 0.0007551 m³.
The mass of the object is,
m = ρ × V = ρ × 0.0007551
The given mass of the solid wood is 17N.
Hence,ρ × 0.0007551 = 17Nρ = 22477.9 kg/m³
Therefore, the density of the block is 22477.9 kg/m³.
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A 65.0-kg man is sliding across his kitchen floor. Between socks and polished hardwood, the coefficient of static friction is 0.34 and the coefficient of kinetic friction is 0.25. What is the magnitude of the acceleration of the man as he slows down?
Answer in m/s^2
The magnitude of the acceleration of the man as he slows down is approximately 2.45 m/s², directed opposite to the direction of his motion.
To determine the magnitude of the acceleration of the man as he slows down, we need to consider the forces acting on him and compare them to the frictional forces.
When the man is sliding across the kitchen floor, initially there is static friction between the socks and the polished hardwood. The maximum static frictional force (F_static_max) can be calculated using the formula:
F_static_max = μ_s * N
where μ_s is the coefficient of static friction and N is the normal force.
The normal force (N) is equal to the weight of the man (mg), where m is the mass and g is the acceleration due to gravity.
Next, we need to compare the maximum static frictional force with the applied force on the man. If the applied force is less than or equal to the maximum static frictional force, the man will not start moving, and the acceleration will be zero. However, if the applied force exceeds the maximum static frictional force, the man will start moving, and the frictional force acting on him will change to the kinetic frictional force.
The kinetic frictional force (F_kinetic) can be calculated using the formula:
F_kinetic = μ_k * N
where μ_k is the coefficient of kinetic friction.
Since the man is slowing down, we know that the applied force is less than the maximum static frictional force, and the frictional force acting on him is the kinetic frictional force.
To find the magnitude of the acceleration, we can use Newton's second law of motion:
F_applied - F_kinetic = ma
Given:
Mass of the man (m) = 65.0 kg
Coefficient of static friction (μ_s) = 0.34
Coefficient of kinetic friction (μ_k) = 0.25
Acceleration due to gravity (g) ≈ 9.8 m/s²
First, calculate the maximum static frictional force:
N = mg = 65.0 kg * 9.8 m/s²
F_static_max = μ_s * N
Next, calculate the kinetic frictional force:
F_kinetic = μ_k * N
Finally, calculate the acceleration:
F_applied - F_kinetic = ma
Since the man is slowing down, the applied force is less than the maximum static frictional force, so F_applied = 0.
0 - F_kinetic = ma
Solving for acceleration (a):
a = -F_kinetic / m
Substitute the known values and calculate:
a = -(0.25 * N) / m
a = -(0.25 * N) / 65.0 kg
Substitute the value of N (N = mg):
a = -(0.25 * 65.0 kg * 9.8 m/s²) / 65.0 kg
Simplifying:
a ≈ -2.45 m/s²
Therefore, the magnitude of the acceleration of the man as he slows down is approximately 2.45 m/s², directed opposite to the direction of his motion.
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A mechanic pushes a $3.06 \times 10^3 \mathrm{~kg}$ car from rest to a speed of $v f$, doing $6.12 \times 10^3 \mathrm{~J}$ of work in the process. During this time, the car moves $23.0 \mathrm{~m}$. Neglecting friction between the car and the road, find
(a) The speed, vf
(b) The horizontal force exerted on the car.
The force exerted by the mechanic on the car is 265.217 N (3 significant figures) and the acceleration produced is [tex]0.0867 \mathrm{~ms^{-2}}[/tex](3 significant figures).
We are given;
Mass of car, m = 3.06 × 103 kg
Distance traveled by car, d = 23.0 m
Work done on the car, W = 6.12 × 103 J
Friction is neglected.
Hence, no work is done against the force of friction.
Since the force applied is in the horizontal direction and there is no force in the vertical direction, the net force is equal to the force applied in the horizontal direction.
Therefore, the force applied by the mechanic is given by;
F = [tex]\frac {W}{d}[/tex]
= [tex]\frac {6.12 \times 10^3 \mathrm{~J}}{23.0 \mathrm{~m}}[/tex]
= [tex]265.217\mathrm{~N}[/tex]
The kinetic energy of the car is given by;
K.E. = [tex]\frac {1}{2}mv_f^2[/tex]
Potential energy is given by;
P.E. = mgh
Where, h = 0, because the car is not raised or lowered vertically, it is moved horizontally.
Therefore, the potential energy is equal to zero.
The net work done on the car is equal to the change in kinetic energy.
Therefore;
Net work done = \frac {1}{2}m{v_f}^2-06.12 \times 10^3\mathrm{~J}
= \frac {1}{2} \times 3.06 \times 10^3 \mathrm{~kg} \times {v_f}^2
On solving the above equation we get;
[tex]{v_f}^2 = \frac {6.12 \times 10^3\mathrm{~J} \times 2}{3.06 \times 10^3 \mathrm{~kg}}{v_f}^2[/tex]
= [tex]4\mathrm{~m^2s^{-2}}[/tex]
Hence, vf = 2 m/s (2 significant figures)
The force exerted by the mechanic is given by;
F = ma265.217 N
= (3.06 × 103 kg)a
On solving the above equation we get;
a = [tex]0.0867 \mathrm{~ms^{-2}}[/tex]
The force exerted by the mechanic on the car is 265.217 N (3 significant figures) and the acceleration produced is [tex]0.0867 \mathrm{~ms^{-2}}[/tex] (3 significant figures).
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Using your knowledge of Op-Amps, design a circuit using non-inverting or inverting amplifiers that can implement the following relationship: Vout = ½ Vin1 + 12Vin2 - 10Vin3
This circuit design utilizes a combination of non-inverting and inverting amplifiers, as well as a summing amplifier, to implement the given equation. By carefully selecting resistor values and configuring the amplifiers correctly,
we can achieve the desired relationship between the inputs and output.
To design a circuit that implements the relationship Vout = ½ Vin1 + 12Vin2 - 10Vin3, we can use a combination of non-inverting and inverting amplifiers.
1. Begin by breaking down the equation into separate amplifiers:
- ½ Vin1 requires a non-inverting amplifier with a gain of 0.5.
- 12Vin2 requires an inverting amplifier with a gain of -12.
- -10Vin3 requires an inverting amplifier with a gain of -10.
2. Start with the non-inverting amplifier for ½ Vin1:
- Connect Vin1 to the positive terminal of the op-amp and ground the negative terminal.
- Connect a resistor, R1, from the positive terminal to the output of the op-amp.
- Connect a feedback resistor, Rf1, from the output of the op-amp to the negative terminal.
- The gain of the amplifier can be calculated using the formula: Gain = 1 + (Rf1 / R1).
- Choose appropriate resistor values for R1 and Rf1 to achieve a gain of 0.5.
3. Next, move on to the inverting amplifier for 12Vin2:
- Connect Vin2 to the negative terminal of the op-amp and ground the positive terminal.
- Connect a resistor, R2, from the negative terminal to the output of the op-amp.
- Connect a feedback resistor, Rf2, from the output of the op-amp to the negative terminal.
- The gain of the amplifier can be calculated using the formula: Gain = -(Rf2 / R2).
- Choose appropriate resistor values for R2 and Rf2 to achieve a gain of -12.
4. Lastly, design the inverting amplifier for -10Vin3:
- Connect Vin3 to the negative terminal of the op-amp and ground the positive terminal.
- Connect a resistor, R3, from the negative terminal to the output of the op-amp.
- Connect a feedback resistor, Rf3, from the output of the op-amp to the negative terminal.
- The gain of the amplifier can be calculated using the formula: Gain = -(Rf3 / R3).
- Choose appropriate resistor values for R3 and Rf3 to achieve a gain of -10.
5. Finally, sum the outputs of the three amplifiers to obtain the desired Vout.
- Connect the output of the non-inverting amplifier to one terminal of a summing amplifier.
- Connect the outputs of the two inverting amplifiers to the other terminals of the summing amplifier.
- The summing amplifier can be implemented using an op-amp with an inverting configuration.
- Connect resistors, Rf4, Rf5, and Rf6, from each input to the output of the summing amplifier.
- Choose appropriate resistor values for Rf4, Rf5, and Rf6 to weigh each input according to the desired equation.
- The output of the summing amplifier will be the desired Vout = ½ Vin1 + 12Vin2 - 10Vin3.
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A jet with mass m = 6 × 104 kg jet accelerates down the runway for takeoff at 1.9 m/s2.
1)
What is the net horizontal force on the airplane as it accelerates for takeoff?
N
2)
What is the net vertical force on the airplane as it accelerates for takeoff?
N
3)
Once off the ground, the plane climbs upward for 20 seconds. During this time, the vertical speed increases from zero to 18 m/s, while the horizontal speed increases from 80 m/s to 98 m/s.
What is the net horizontal force on the airplane as it climbs upward?
N
4)
What is the net vertical force on the airplane as it climbs upward?
N
5)
After reaching cruising altitude, the plane levels off, keeping the horizontal speed constant, but smoothly reducing the vertical speed to zero, in 15 seconds.
What is the net horizontal force on the airplane as it levels off?
N
6)
What is the net vertical force on the airplane as it levels off?
N
PLEASE I NEED THESE
1. Net horizontal force during takeoff: 1.14 × [tex]10^5[/tex] N ,2. Net vertical force during takeoff: 5.88 × [tex]10^5[/tex] N ,3. Negligible net horizontal force during upward climb ,4. Net vertical force depends on lift and weight ,5. No net horizontal force when leveling off ,6. Net vertical force depends on lift balancing weight.
The net horizontal force on the airplane as it accelerates for takeoff can be calculated using Newton's second law, which states that force (F) is equal to mass (m) multiplied by acceleration (a).
Thus, the net horizontal force is F = m * a. Plugging in the given values, we have F =[tex](6 * 10^4 kg) * (1.9 m/s^2)[/tex], resulting in the net horizontal force of [tex]1.14 * 10^5[/tex] N.
The net vertical force on the airplane as it accelerates for takeoff can be determined using the same principle as in the previous question. The net vertical force is also F = m * a, where the acceleration is the same as before, [tex]1.9 m/s^2[/tex].
Thus, the net vertical force is F =[tex](6 * 10^4 kg) * (1.9 m/s^2)[/tex], which equals [tex]1.14 * 10^5 N[/tex].
As the airplane climbs upward, the net horizontal force becomes zero because there is no horizontal acceleration. Therefore, the net horizontal force on the airplane as it climbs upward is 0 N.
The net vertical force on the airplane as it climbs upward can be calculated using the same method as before. The acceleration in the vertical direction can be determined by dividing the change in vertical speed (18 m/s - 0 m/s) by the time taken (20 s).
Thus, the net vertical force is F = [tex](6 * 10^4 kg) * (0.9 m/s^2[/tex]), resulting in the net vertical force of [tex]5.4 * 10^4[/tex] N.
When the airplane levels off, the vertical speed is reduced to zero, indicating zero vertical acceleration. Therefore, the net vertical force on the airplane as it levels off is 0 N.
The net horizontal force on the airplane as it levels off remains the same as during the climb, as there is no horizontal acceleration. Hence, the net horizontal force is 0 N.
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A motorist is traveling at 14 m/s when he sees a deer in the road 37 m ahead. If the maximum negative acceleration of the vehicle is −7 m/s
2
, what is the maximum reaction time Δt of the motorist that will allow him to avoid hitting the deer? Answer in units of s. Question 6, chap 102, rect 6. part 2 of 210 points If his reaction time is 1.83176 s, how fast will he be traveling when he reaches the deer? Answer in units of m/s.
A motorist is traveling at 14 m/s when he sees a deer in the road 37 m ahead. If the maximum negative acceleration of the vehicle is −7 m/s^2. The maximum reaction time Δt is 2 seconds.(6)The motorist will be traveling at approximately 1.17768 m/s when he reaches the deer.
To calculate the maximum reaction time Δt that will allow the motorist to avoid hitting the deer, we need to determine the time it takes for the vehicle to come to a complete stop from its initial velocity.
Given:
Initial velocity (u) = 14 m/s
Distance to the deer (s) = 37 m
Maximum negative acceleration (a) = -7 m/s^2
We can use the following equation of motion to find the time taken to stop:
s = ut + (1/2)at^2
Since the vehicle comes to a stop, its final velocity (v) will be 0. Plugging in the values:
0 = 14t + (1/2)(-7)t^2
Rearranging the equation:
7t^2 - 14t = 0
Factoring out common terms:
7t(t - 2) = 0
This equation will be true if either t = 0 (which is not the case since the vehicle needs time to react) or t - 2 = 0. Therefore, the maximum reaction time Δt is 2 seconds.
Now, to calculate the final velocity when the motorist reaches the deer after a reaction time of 1.83176 seconds, we can use the equation:
v = u + at
Plugging in the values:
u = 14 m/s
a = -7 m/s^2
t = 1.83176 s
v = 14 + (-7)(1.83176)
v = 14 - 12.82232
v = 1.17768 m/s
Therefore, the motorist will be traveling at approximately 1.17768 m/s when he reaches the deer.
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A0.15 ghoneybee acquires a charge of 19pC while flying. The electric field near the surface of the earth is typically 100 N/C, directed downward.
The electric force experienced by the charged honeybee near the surface of the earth is approximately [tex]$1.9 \, \mu \text{N}$[/tex] , directed upward.
To solve the problem properly, we can use the formula for electric force:
[tex]\[ F = q \cdot E \][/tex]
where [tex]\( F \)[/tex] is the electric force, [tex]\( q \)[/tex] is the charge, and [tex]\( E \)[/tex] is the electric field.
Given:
Charge of the honeybee, [tex]\( q = 19 \, \text{pC} = 19 \times 10^{-12} \, \text{C} \)[/tex]
Electric field near the surface of the earth, [tex]\( E = 100 \, \text{N/C} \)[/tex]
Substituting the values into the formula:
[tex]\[ F = (19 \times 10^{-12} \, \text{C}) \cdot (100 \, \text{N/C}) \][/tex]
Calculating the electric force:
[tex]\[ F = 1.9 \times 10^{-9} \, \text{N} \][/tex]
Therefore, the electric force experienced by the charged honeybee near the surface of the earth is approximately [tex]\( 1.9 \times 10^{-9} \, \text{N} \)[/tex], directed upward.
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A stunt man drives a 1500 kg car at a speed of 25 m/s off a 30 -m-high cliff. The road leading to the cliff is inclined upward at an angle of 20
∘
. How far from the base of the cliff does the car land? Express your answer with the appropriate units. - Part B What is the car's impact speed? Express your answer with the appropriate units
The car lands approximately 44.4 meters from the base of the cliff.
The car's impact speed is approximately 8.54 m/s.
To solve this problem, we can break it down into two parts: finding the horizontal distance traveled by the car and determining the impact speed.
Part A: Finding the horizontal distance traveled by the car
Given:
Mass of the car (m) = 1500 kg
Initial speed of the car (v) = 25 m/s
Height of the cliff (h) = 30 m
Angle of the road (θ) = 20 degrees
First, we need to determine the time it takes for the car to fall from the cliff to the ground. We can use the equation for vertical displacement in free fall:
[tex]h = \frac{1}{2} * g * t^2[/tex]
Solving for time (t):
[tex]t = \sqrt{\frac{2h}{g}}[/tex]
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the known values:
[tex]t = \sqrt{\frac{2 \cdot 30\,}{9.8\, }} \approx 2.19\, \textrm{s}[/tex]
Next, we can find the horizontal distance traveled by the car using the equation:
d = v * cos(θ) * t
Substituting the known values:
d = 25 m/s * cos(20 degrees) * 2.19 s ≈ 44.4 m
Therefore, the car lands approximately 44.4 meters from the base of the cliff.
Part B: Finding the car's impact speed
To determine the impact speed, we can use the equation for horizontal velocity:
v_impact = v * sin(θ)
Substituting the known values:
v_impact = 25 m/s * sin(20 degrees) ≈ 8.54 m/s
Therefore, the car's impact speed is approximately 8.54 m/s.
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Question 4 At a distance of R=8 mm away from the centre of a long straight thin walled conducting tube the electric field strength is E=0.4 V/m. The outer radius of the tube is r=4 mm. Calculate the linear charge density σ on the tube surface in pC/m 2
Given that the distance of R=8 mm away from the center of a long straight thin-walled conducting tube, the electric field strength is E=0.4 V/m.
The outer radius of the tube is r=4 mm. We have to find the linear charge density σ on the tube surface in pC/m2.The formula to calculate the electric field strength outside a uniformly charged thin-walled conducting shell is given by,E = σR / ε0
Here, E = 0.4 V/m, R = 8 mm, and r = 4 mm We know that the electric field strength outside a uniformly charged thin-walled conducting shell is proportional to the charge per unit length on the surface of the shell.
This quantity is given by linear charge density .
σ = q / (2πrL), where L is the length of the tube. Since L is not given in the question, we have to assume that the tube is infinitely long. Therefore, the linear charge density σ is given by,σ
[tex]= (E * ε0 * 2πr) / Lσ = (0.4 * 8.85 × 10^-12 * 2π * 4) / Lσ = 1.118 × 10^-10 /[/tex]LWe need to convert the linear charge density from
[tex]C/m to pC/m2.1 C = 10^12 pC1 m = 10^2 cm∴ 1 C/m = (10^12 pC) / (10^4[/tex]cm^2) = 10^8 pC/m2
σ =[tex](1.118 × 10^-10 / L) * 10^8 pC/m2σ = 11.18 / L p[/tex]C/m2Since L is not given,
we cannot find the exact value of σ. However, we can say that the linear charge density on the tube surface is proportional to 1/L. Therefore, if L is more than 100 times the radius of the tube (i.e., L > 400 mm), then the linear charge density is less than 28 pC/m2.
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