If an electron travels 0.300 m from an electron gun to a TV screen in 43.9 ns. The voltage used to accelerate the electron is approximately 6,849 volts.
To determine the voltage used to accelerate the electron, we can use the equation for the electric potential energy gained by the electron, which is equal to the work done on the electron.
Calculate the time in seconds:
Given that the time is given in nanoseconds (ns), we need to convert it to seconds. 1 ns is equal to 1 × 10^-9 seconds. Therefore, 43.9 ns is equal to 43.9 × 10^-9 seconds.
Calculate the speed of the electron:
Using the distance traveled and the time taken, we can calculate the speed of the electron using the formula speed = distance / time.
Calculate the kinetic energy of the electron:
The kinetic energy of the electron is given by the equation KE = (1/2)mv^2, where m is the mass of the electron and v is its velocity (speed).
Calculate the electric potential energy:
The electric potential energy gained by the electron is equal to its kinetic energy. This is because the electric potential energy gained is converted into kinetic energy as the electron accelerates.
Calculate the voltage:
The electric potential energy gained by the electron is equal to the product of its charge (e) and the voltage (V). Thus, we can set up the equation PE = eV and solve for V.
Substitute the values and solve:
Substitute the known values into the equation and solve for the voltage.
Therefore, the voltage used to accelerate the electron is approximately 6,849 volts.
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A merry-go-round has a mass of \( 1640 \mathrm{~kg} \) and a radius of \( 7.5 \mathrm{~m} \). Approximating the merry-go-round as a solid cylinder. how much work is required to accelerate it from rest
183750 J of work is required to accelerate the merry-go-round from rest.
[tex]\( KE = \frac{1}{2} I \omega^2 \)[/tex]
Where:
- KE is the rotational kinetic energy
- I is the moment of inertia
-[tex]\( \omega \) \\[/tex] is the angular velocity
In this case, we can approximate the merry-go-round as a solid cylinder, which has a moment of inertia given by:
[tex]\( KE = \frac{1}{2} I \omega^2 \)[/tex]
Where:
- m is the mass of the merry-go-round
- r is the radius of the merry-go-round
First, let's calculate the moment of inertia:
[tex]\( I = \frac{1}{2} \times 1640 \, \mathrm{kg} \times (7.5 \, \mathrm{m})^2 \)\( I = 183750 \, \mathrm{kg \cdot m^2} \)[/tex]
Next, we need to find the final angular velocity. Since the merry-go-round is starting from rest, the initial angular velocity is 0. The final angular velocity can be calculated using the following equation:
[tex]\( \omega_f = \sqrt{\frac{2 \cdot \text{work}}{I}} \)\( \text{work} = \frac{1}{2} I \omega_f^2 \)[/tex]
Now, substitute the values we have:
[tex]\( \text{work} = \frac{1}{2} \times 183750 \, \mathrm{kg \cdot m^2} \times \left(\sqrt{\frac{2 \cdot \text{work}}{183750 \, \mathrm{kg \cdot m^2}}}\right)^2 \)[/tex]
Simplifying the equation, we get:
[tex]\( \text{work} = \frac{183750}{2} \times 2 = 183750 \, \mathrm{J} \)[/tex]
Therefore, approximately 183750 J of work is required to accelerate the merry-go-round from rest.
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Suppose you are trying to hammer a nail in a wall in order to hang something, only unbeknownst to you the spot you have chosen has a metal beam just on the other side of the wall. When you hit the nail with the hammer, the nail barely moves while the hammer recoils backwards hard. How does the force the hammer exerts on the nail compare to the force the nail exerts on the hammer? The amount the hammer forces the nail is less than the amount the nail forces the hammer. The amount the hammer forces the nail is larger than the amount the nail forces the hammer. The amount the hammer forces the nail is the same as the amount the nail forces the hammer. Not enough information is given to know for sure how the forces compare. Which of the following statements about normal force are true? Surface friction is a type of normal force. The normal force is always vertical. Normal forces keep objects from moving through surfaces. Normal force is always a push, never a pull. All objects always have a normal force acting on them. Question 4 1 pts Which of the following are true of weight/mass? When in orbit around the Earth, objects have essentially no weight. Objects have less mass the higher they are off the surface of the Earth. All of the other statements are false. Mass and weight are equivalent terms for the same concept. The Earth pulls on you more than you pull on the Earth. When in orbit around the Earth, objects have essentially no mass.
Answer:
The correct statements are:
- The amount the hammer forces the nail is the same as the amount the nail forces the hammer.
- Normal forces keep objects from moving through surfaces.
- All objects always have a normal force acting on them.
- When in orbit around the Earth, objects have essentially no weight.
Explanation:
The force the hammer exerts on the nail is the same as the force the nail exerts on the hammer, according to Newton's third law of motion: "For every action, there is an equal and opposite reaction." Therefore, the amount the hammer forces the nail is the same as the amount the nail forces the hammer.
Regarding the statements about the normal force:
- Surface friction is a type of force, but not a type of normal force. So, the statement "Surface friction is a type of normal force" is false.
- The normal force is not always vertical. It acts perpendicular to the surface with which an object is in contact. So, the statement "The normal force is always vertical" is false.
- Normal forces can act to prevent objects from moving through surfaces, so the statement "Normal forces keep objects from moving through surfaces" is true.
- The normal force can be a push or a pull, depending on the situation. So, the statement "Normal force is always a push, never a pull" is false.
- All objects have a normal force acting on them when they are in contact with a surface, so the statement "All objects always have a normal force acting on them" is true.
Regarding weight/mass:
- When in orbit around the Earth, objects experience weightlessness because they are in freefall, but they still have mass. So, the statement "When in orbit around the Earth, objects have essentially no weight" is true.
- The mass of an object remains the same regardless of its height above the Earth's surface. So, the statement "Objects have less mass the higher they are off the surface of the Earth" is false.
- Mass and weight are different concepts. Mass is a measure of the amount of matter in an object, while weight is the force exerted on an object due to gravity. So, the statement "Mass and weight are equivalent terms for the same concept" is false.
- According to Newton's third law, the force of gravity between the Earth and an object is equal in magnitude but opposite in direction. So, the Earth pulls on an object with the same force as the object pulls on the Earth. Thus, the statement "The Earth pulls on you more than you pull on the Earth" is false.
- Objects in orbit around the Earth still have mass; it is just that they are in a state of continuous freefall, resulting in the feeling of weightlessness. So, the statement "When in orbit around the Earth, objects have essentially no mass" is false.
Therefore, the correct statements are:
- The amount the hammer forces the nail is the same as the amount the nail forces the hammer.
- Normal forces keep objects from moving through surfaces.
- All objects always have a normal force acting on them.
- When in orbit around the Earth, objects have essentially no weight.
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Find the rms (a) electric and (b) magnetic fields at a point 5.00 m from a lightbulb that radiates 90.0 W of light uniformly in all directions.
To find the RMS electric and magnetic fields at a distance of 5.00 m from a lightbulb radiating 90.0 W of light uniformly in all directions, we can use the formulas relating power to electric and magnetic fields in electromagnetic radiation.
(a) RMS Electric Field (E):
The formula relating power to the electric field is:
P = (1/2) * ε₀ * c * E²
where ε₀ is the vacuum permittivity and c is the speed of light in a vacuum.
To find the RMS electric field (E), we rearrange the equation:
E = √(2 * P / (ε₀ * c))
Substituting the given values:
E = √(2 * 90.0 W / (8.854 × 10⁻¹² F/m * 3.00 × 10⁸ m/s))
Calculate the value of E using the above formula.
(b) RMS Magnetic Field (B):
The relationship between the electric and magnetic fields in electromagnetic radiation is given by:
B = E / c
where B is the RMS magnetic field and c is the speed of light in a vacuum.
Substituting the value of E calculated above, we can find B.
Calculate the value of B using the above formula.
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"A block with mass m1 = 7.9 kg rests on the
surface of a horizontal table which has a coefficient of kinetic
friction of μk = 0.63. A second block with a
mass m2 = 9.9 kg is connected to the first by"
The acceleration of the blocks is determined as 0.62 m/s².
What is the acceleration of the blocks?The acceleration of the blocks is calculated by applying Newton's second law of motion as follows;
F(net) = ma
where;
m is the total mass of the blocksa is the acceleration of the blocks.F(net) = m₂g - μm₁g
F(net) = (9.9 x 9.8) - (0.63 x 7.9 x 9.8)
F(net) = 48.25 N
The acceleration of the blocks is calculated as;
a = F(net) / m
a = ( 48.25 N ) / (7.9 + 9.9)
a = 0.62 m/s²
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The complete question is below:
A block with mass m1 = 7.9 kg rests on the
surface of a horizontal table which has a coefficient of kinetic
friction of μk = 0.63. A second block with a
mass m2 = 9.9 kg is connected to the first by a pulley. Find the acceleration of the two blocks.
When a charge is accelerated through a potential difference of 500 V, its kinetic energy increases from 2.010^−5J to 6.010^−5J. What is the magnitude of the accelerated charge?
A charge undergoes acceleration through a potential difference of 500 V, resulting in an increase in its kinetic energy from 2.010^(-5) J to 6.010^(-5) J. The question asks for the magnitude of the accelerated charge.
The change in kinetic energy of a charged particle accelerated through a potential difference can be calculated using the formula ΔK = qΔV, where ΔK is the change in kinetic energy, q is the magnitude of the charge, and ΔV is the potential difference.
Given that the initial kinetic energy (K1) is 2.010^(-5) J and the final kinetic energy (K2) is 6.010^(-5) J, the change in kinetic energy (ΔK) is (6.010^(-5) J) - (2.010^(-5) J) = 4.010^(-5) J.
We also know that the potential difference (ΔV) is 500 V.
Using the formula ΔK = qΔV, we can rearrange the equation to solve for q:
q = ΔK / ΔV = (4.010^(-5) J) / (500 V) = 8.010^(-8) C.
Therefore, the magnitude of the accelerated charge is 8.010^(-8) C.
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The forces on a small airplane (mass 1160 kg) in horizontal flight heading eastward are as follows: weight = 16.0 kN downward, lift = 16.0 kN upward, thrust = 1.800 kN eastward, and drag = 1.400 kN westward. At t = 0, the plane’s speed is 90.0 m/s. If the forces remain constant, how far does the plane travel in the next 42.0 s?
The plane will travel approximately 3.78 kilometers (km) in the next 42.0 seconds, assuming the forces remain constant.
To calculate the distance traveled by the plane, we need to determine the net force acting on the plane and then use kinematic equations to find the distance traveled.
The net force acting on the plane is the vector sum of thrust and drag forces: net force = thrust - drag. Substituting the given values, we have net force = 1.800 kN - 1.400 kN = 0.400 kN eastward.
Using Newton's second law (F = ma), we can find the acceleration of the plane. The net force acting on the plane is equal to the mass of the plane multiplied by its acceleration: 0.400 kN = 1160 kg * a. Solving for a, we find the acceleration to be approximately 0.00034 m/s².
Next, we can use the kinematic equation s = ut + (1/2)at² to calculate the distance traveled by the plane. Since the initial velocity (u) is given as 90.0 m/s and the time (t) is 42.0 seconds, we substitute these values along with the acceleration into the equation. Plugging in the values, we get s = (90.0 m/s) * (42.0 s) + (1/2) * (0.00034 m/s²) * (42.0 s)².
Calculating the expression, we find that the plane will travel approximately 3.78 km in the next 42.0 seconds, assuming the forces remain constant.
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The distance between two telephone poles is 54.4 m. When a 1.00-kg bird lands on the telephone wire midway between the poles, the wire sags 0.190 m. Draw a free-body diagram of the bird. How much tension does the bird produce in the wire? Ignore the weight of the wire.
The free-body diagram of a bird that lands on a wire midway between two telephone poles is shown below.
Hence, the tension that the bird produces in the wire is approximately 97.3 N.
First, let's draw the free-body diagram of the bird shown below.FBD:If we consider the wire on which the bird lands, the tension T exists in both directions, one of which is pulling up and the other is pulling down.
The gravitational force of the earth (Fg) acts on the bird and pulls it down. The bird is in equilibrium, so the two forces (T and Fg) are equal and opposite, as shown in the free-body diagram above.The distance between two telephone poles is 54.4 m.
The sag of the wire, or how much the wire stretches when the bird lands on the wire midway between the poles, is 0.190 m.The wire is considered to be in a uniform gravitational field since it is nearly parallel to the earth's surface. Therefore, we may make the following calculations based on the wire's gravitational force:Fg = mg = (1.00 kg)(9.80 m/s²) = 9.80 N
The tension in the wire may be calculated using the following formula:T = Fg/2 + Fg·x²/2L²where L is the span length of the wire and x is the sag of the wire. Substituting the values we get:T = (9.80 N/2) + (9.80 N)(0.190 m)²/[2(54.4 m)]≈97.3 N
Thus, the tension that the bird produces in the wire is approximately 97.3 N.
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A football quarterback runs 15.0 m straight down the playing field in 2.90 s. He is then hit and pushed 3.00 m straight backward in 1.71 s. He breaks the tackle and runs straight forward anather 23.0 m in 5.205. (a) Calculate his average velocity ( in m/s) for each of the three intervals. (Aswune the quarterback's inital direction is positive. Indicate the direction with the sign of your answer)
v 1=
v 2=
v 3=
(b) Calculate his average velocity (in m/s ) for the entire motion. (Assume the quartarback's initial direction is positive. Indicate the direction with the sign of your answer) m/s
Average velocity in the first interval, v1 = 5.17 m/s in the positive direction Average velocity in the second interval, v2 = -1.75 m/s in the negative directionAverage velocity in the third interval, v3 = 4.42 m/s in the positive directionAverage velocity for the entire motion, v = 3.56 m/s in the positive direction.
Given, Distance moved by the quarterback in the first interval,
Δx1 = 15.0 m
Time taken in the first interval, Δt1 = 2.90 s
Distance moved by the quarterback in the second interval, Δx2 = -3.00 m (since the quarterback is pushed backward)
Time taken in the second interval, Δt2 = 1.71 s
Distance moved by the quarterback in the third interval, Δx3 = 23.0 m
Time taken in the third interval, Δt3 = 5.205 s(a)
Average velocity in the first interval
,v1 = Δx1 / Δt1
= 15.0 / 2.90
= 5.17 m/s in the positive direction (since the quarterback is moving straight down the playing field)
Average velocity in the second interval,
v2 = Δx2 / Δt2= (-3.00) / 1.71
= -1.75 m/s in the negative direction (since the quarterback is pushed backward)
Average velocity in the third interval,
v3 = Δx3 / Δt3= 23.0 / 5.205
= 4.42 m/s in the positive direction (since the quarterback is moving straight forward)
The average velocity for the entire motion is given as:
v = (Δx1 + Δx2 + Δx3) / (Δt1 + Δt2 + Δt3)
= (15.0 - 3.00 + 23.0) / (2.90 + 1.71 + 5.205)
= 35.0 / 9.815
= 3.56 m/s in the positive direction (since the quarterback's initial direction is positive).
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(a) A circular coil of radius $1 \mathrm{~cm}$ has 200 turns. It is removed from a magnetic field of $2 \mathrm{~T}$ in a time of $0.1 \mathrm{~s}$. The field direction is normal to the plane of the coil.
i. What is the magnitude of the average voltage induced in the coil?
ii. If the coil is removed more slowly in a time of $1 \mathrm{~s}$, what is the magnitude of the voltage induced now?
The magnitude of the voltage induced in the coil is 12.576 V when the coil is removed quickly, and 0.419 V when the coil is removed slowly.
i. Magnitude of the average voltage induced in the coil
The magnitude of the average voltage induced in the coil is given by the following formula:
V = N * dΦ / dt
In this case, the number of turns in the coil is 200, the magnetic field is 2 T, and the time it takes to remove the coil from the magnetic field is 0.1 s. So, the magnitude of the average voltage induced in the coil is:
V = 200 * dΦ / dt = 200 * (2 * B * A) / dt
The area of the coil is pi * R^2, where R is the radius of the coil. In this case, the radius of the coil is 1 cm, so the area of the coil is pi * (0.01)^2 = 3.14 * 10^-4 m^2. So, the magnitude of the average voltage induced in the coil is:
V = 200 * (2 * B * A) / dt = 200 * (2 * 2 * 3.14 * 10^-4) / 0.1
V = 12.576 V
ii. Magnitude of the voltage induced now
If the coil is removed more slowly in a time of 1 s, then the magnitude of the voltage induced in the coil is:
V = 200 * (2 * B * A) / dt = 200 * (2 * 2 * 3.14 * 10^-4) / 1
V = 0.419 V
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The atractive electiostatic force between the poict chirges 366×10
∘
C and Q has a magnitude of Part A 0930 Nibich the separation betwoen the charges it 2.03 th . You may want to rovew? Find the sign and magnitude of the charge O A sphere of radue 4.04 cm and uniform suriace Part A charge densify +13.0μC/m
2
oxerts an eactrostatic force of magntude 4.24×10
−2
N on a poinl charge +1.05 μC Find the separation between the point charge and the centor of the sphere.
The separation between the point charge and the center of the sphere due to attractive force is 3.16 cm.
Charge 1, q1 = 366 × 10^(-6) C, Charge 2, q2 = Q, Separation between charges, r = 2.03 m, Force, F = 0.930 N, Charge density, σ = +13.0 μC/m², Radius of the sphere, R = 4.04 cm = 0.0404 m, Force on point charge, F = 4.24 × 10^(-2) N.
Coulomb's law gives the magnitude of the force, F as:
F = k (q1q2/r²)
where k = 1/4πε₀ = 9 × 10^9 Nm²/C²
0.930 = 9 × 10^9 × (366 × 10^(-6) Q)/(2.03)²0.930 × 2.03²/9 × 10^9
= 366 × 10^(-6) Q
Q = 1.13 × 10^(-6) C.
Charge density is defined as the amount of charge per unit area. For a sphere, it is given as:σ = q/4πR²where q is the charge on the sphere.13.0 × 10^(-6) C/m² = q/4π(0.0404)²q = 6.05 × 10^(-8) C. The force between a point charge and a charged sphere is given by: Coulomb's law: F = k (q1q2/r²)
As the force on the point charge is attractive, the charge on the sphere must be opposite in sign to that of the point charge. Therefore, we take q2 = -6.05 × 10^(-8) C.
4.24 × 10^(-2) = 9 × 10^9 × (1.05 × 10^(-6))(-6.05 × 10^(-8))/r²
r² = (9 × 10^9 × 1.05 × 10^(-6) × 6.05 × 10^(-8))/4.24 × 10^(-2)
r = 0.0316 m = 3.16 cm.
Therefore, the separation between the point charge and the center of the sphere due to attractive force is 3.16 cm.
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a) You have an RC circuit with a time constant of 5.26 s. If the total resistance in the circuit is 229.9 kΩ, what is the capacitance of the circuit (in μF)?
b) You have an RC circuit with a time constant of 1.97 s. If the total capacitance in the circuit is 55.7 μF, what is the resistance of the circuit (in kΩ)?
a) The capacitance of the circuit is approximately 22.9 μF. b) The resistance of the circuit is approximately 35.4 kΩ.
How to solve for resistance and capacitance?a) To find the capacitance of the circuit, use the formula for the time constant of an RC circuit:
τ = RC
Given that the time constant (τ) is 5.26 s and the resistance (R) is 229.9 kΩ, rearrange the formula to solve for capacitance (C):
C = τ / R
Substituting the given values:
C = 5.26 s / 229.9 kΩ
Converting kiloohms to ohms:
C = 5.26 s / (229.9 × 10³ Ω)
Converting the result to microfarads:
C ≈ 22.9 μF
Therefore, the capacitance of the circuit is approximately 22.9 μF.
b) To find the resistance of the circuit, use the same formula for the time constant:
τ = RC
Given that the time constant (τ) is 1.97 s and the capacitance (C) is 55.7 μF, rearrange the formula to solve for resistance (R):
R = τ / C
Substituting the given values:
R = 1.97 s / 55.7 μF
Converting microfarads to farads:
R = 1.97 s / (55.7 × 10⁻⁶ F)
Converting the result to kiloohms:
R ≈ 35.4 kΩ
Therefore, the resistance of the circuit is approximately 35.4 kΩ.
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The distance from the Sun to Neptune is 4.5×10 ^{12} m. How long does it take sunlight to reach Neptune? Put your answer in scientific notation. What is the wavelength of a wave that has a speed of 710 m/s and a frequency of 1100 Hz ?
The distance from the Sun to Neptune is 4.5×10 ^{12} m. How long does it take sunlight to reach Neptune? Put your answer in scientific notation.
Sunlight reaches Neptune in approximately 4.5 hours. To convert this to scientific notation, we can write it as
4.5 × 10^0 × 10^12 meters (since 4.5 × 10^12
meters is equal to 4.5 × 10^0 × 10^12 meters).
Then we can divide the time in seconds
(4.5 hours × 3600 seconds/hour = 16,200 seconds)
by this distance: time = distance ÷ speed of light time
= (4.5 × 10^0 × 10^12 m) ÷ (3.00 × 10^8 m/s)
time = 1.50 × 10^4 s Finally, we can convert this to scientific notation
time = 1.50 × 10^4 s = 1.50 × 10^1 × 10^3 s = 1.50 × 10^1 × 10^3 × 10^(-3)
What is the wavelength of a wave that has a speed of 710 m/s and a frequency of 1100 Hz?We know that the velocity of a wave is given by: v = λfwhere v is the velocity, λ is the wavelength and f is the frequency.
Rearranging this equation to solve for wavelengt:
λ = v/f Substituting the given values: v = 710 m/sf = 1100 Hzλ = 710/1100 = 0.646 m
The wavelength of the wave is 0.646 meters.
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An airplane flies 200 km due west from city A to city B and then 345 km in the direction of 34.5° north of west from city B to city C.
(a) In straight-line distance, how far is city C from city A?
(b) Relative to city A, in what direction is city C?
Relative to city A, city C is in the direction of 58.13° north of west.
(a) In straight-line distance, the distance from city A to city C can be determined by using the Pythagorean theorem. This theorem states that in a right-angled triangle, the sum of the squares of the two shorter sides is equal to the square of the longest side, i.e., the hypotenuse. Here, the hypotenuse will be the distance between city A and C.
If AB = 200 km and BC = 345 km, then:
AC = √[(AB)² + (BC)²]
AC = √[(200 km)² + (345 km)²]
AC = √(40,000 km² + 119,025 km²)
AC = √159,025 km²
AC = 398.78 km
So, city C is approximately 398.78 km away from city A.
(b) We can use trigonometry to determine the direction of city C from city A. Let θ be the angle between AC and the westward direction. Then:
tan θ = opposite/adjacent = BC/AB = 345/200
θ = tan⁻¹(345/200)
θ = 58.13° north of west
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do not use rounded intermediate values in vour calculations-including answers submitted in Webassign.) (a) Find the speed \( \{ \) in \( m \) 's \( \} \) at which the ball was launched.
Speed at which the ball was launched = 15.696 m/s (approx).
Given data: Let the initial velocity be u, the final velocity be v, the time taken be t, and the acceleration be a.
Initial velocity (u) = 0 (As the ball is thrown upwards)
Final velocity (v) = ?
Acceleration (a) = g = 9.81 m/s²
Time taken (t) = 1.6 s
Now, the formula for displacement is:v = u + at
Here, u = 0So, v = at
Now, the formula for height or distance traveled by an object is:
s = ut + 1/2 at²
Here, u = 0, so: s = 1/2 at²
This is the formula for height or distance travelled by an object. In our case, the maximum height is reached when the ball is at its highest point and velocity is zero.
Therefore, we can say that: v = 0 (At the highest point)
Now, putting the values of t and g in s = 1/2 at², we get:
s = 1/2 x 9.81 x 1.6²s = 12.55 m
Therefore, the speed at which the ball was launched is: v = at = 9.81 x 1.6
= 15.696 m/s (approx)
Answer: Speed at which the ball was launched = 15.696 m/s (approx).
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Consider the two vectors, A⃗ =A→=-3.14 i+-1.93 j and B⃗ =B→=-1.85 i+-5.64 j.
a)What is the magnitude of C⃗ =A⃗ +B⃗ C→=A→+B→?
b) What is the direction of C⃗ =A⃗ +B⃗ C→=A→+B→ expressed in degrees above to the negative x-axis? Make sure your answer is positive.
c) What is the magnitude of D⃗ =A⃗ −B⃗ D→=A→-B→?
d) What is the direction of D⃗ =A⃗ −B⃗ D→=A→-B→ expressed in degrees below the negative x-axis? Make sure your answer is positive.
a) The magnitude of C⃗ is approximately 9.05. b) The direction of C⃗ is approximately 121.36° above the negative x-axis. c) The magnitude of D⃗ is approximately 3.95. d) The direction of D⃗ is approximately 71.04° below the negative x-axis.
(a) To find the magnitude of C⃗ = A⃗ + B⃗, we can use the formula:
|C⃗| = √(Cₓ² + Cᵧ²)
where Cₓ and Cᵧ are the x and y components of C⃗, respectively.
Given A⃗ = -3.14i - 1.93j and B⃗ = -1.85i - 5.64j, we can add their corresponding components:
Cₓ = Aₓ + Bₓ = -3.14 - 1.85 = -4.99
Cᵧ = Aᵧ + Bᵧ = -1.93 - 5.64 = -7.57
Substituting these values into the magnitude formula:
|C⃗| = √((-4.99)² + (-7.57)²) ≈ 9.05
Therefore, the magnitude of C⃗ is approximately 9.05.
(b) To find the direction of C⃗ expressed in degrees above the negative x-axis, we can use the arctan function:
θ = tan⁻¹(Cᵧ / Cₓ)
Substituting the values from part (a):
θ = tan⁻¹((-7.57) / (-4.99)) ≈ 58.64°
Since we want the answer in degrees above the negative x-axis, we subtract the calculated angle from 180°:
θ = 180° - 58.64° ≈ 121.36°
Therefore, the direction of C⃗ is approximately 121.36° above the negative x-axis.
(c) To find the magnitude of D⃗ = A⃗ - B⃗, we can use the same magnitude formula:
|D⃗| = √(Dₓ² + Dᵧ²)
Given A⃗ = -3.14i - 1.93j and B⃗ = -1.85i - 5.64j, we subtract their corresponding components:
Dₓ = Aₓ - Bₓ = -3.14 - (-1.85) = -1.29
Dᵧ = Aᵧ - Bᵧ = -1.93 - (-5.64) = 3.71
Substituting these values into the magnitude formula:
|D⃗| = √((-1.29)² + (3.71)²) ≈ 3.95
Therefore, the magnitude of D⃗ is approximately 3.95.
(d) To find the direction of D⃗ expressed in degrees below the negative x-axis, we can again use the arctan function:
θ = tan⁻¹(Dᵧ / Dₓ)
Substituting the values from part (c):
θ = tan⁻¹((3.71) / (-1.29)) ≈ -71.04°
Since we want the answer in degrees below the negative x-axis, we negate the calculated angle:
θ = -(-71.04°) ≈ 71.04°
Therefore, the direction of D⃗ is approximately 71.04° below the negative x-axis.
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what am i doing wrong in the first one? please help me A person is riding a bicycle, and its wheels have an angular velocity of \( 14.8 \mathrm{rad} / 5 \). Then, the brakes are applied and the bike is brought to a uniform stop. During braking, the angula
(a). It takes approximately 4.7 seconds for the bike to come to rest.
(b). The approximate angular acceleration of each wheel is -4.1 rad/s². The negative sign indicates that the wheels are decelerating.
(a) To find the time it takes for the bike to come to rest, we can use the formula:
angular displacement = angular velocity * time
We are given that the angular displacement of each wheel is 14.6 revolutions, which we can convert to radians:
1 revolution = 2π radians So,
14.6 revolutions = 14.6 * 2π radians.
Next, we can rearrange the formula to solve for time:
time = angular displacement / angular velocity
Substituting the given values, we have:
time = (14.6 * 2π radians) / 19.4 rad/s
Simplifying the expression, we get:
time = (29.2π radians) / 19.4 rad/s
The units of radians cancel out, leaving us with:
time = 29.2π / 19.4 s
Now, we can calculate the approximate value of time:
time ≈ 4.7 s
Therefore, the bicycle comes to a stop after 4.7 seconds or thereabouts.
(b) To find the angular acceleration of each wheel, we can use the formula:
angular acceleration = change in angular velocity / time
Since the bike comes to a uniform stop, the final angular velocity is 0 rad/s. The initial angular velocity is given as 19.4 rad/s.
The change in angular velocity is:
change in angular velocity = final angular velocity - initial angular velocity
Substituting the given values, we have:
change in angular velocity = 0 rad/s - 19.4 rad/s
Simplifying the expression, we get:
change in angular velocity = -19.4 rad/s
Now, we can substitute the values into the formula for angular acceleration:
angular acceleration = (-19.4 rad/s) / 4.7 s
Simplifying the expression, we get:
angular acceleration ≈ -4.1 rad/s²
Therefore, each wheel's approximate angular acceleration is -4.1 rad/s². The wheels are slowing down, as shown by the negative sign.
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Complete question is,
A person is riding a bicycle, and its wheels have an angular velocity of 19.4 rad/s. Then, the brakes are applied and the bike is brought to a uniform stop. During braking, the angular displacement of each wheel is 14.6 revolutions. (a) How much time does it take for the bike to come to rest? (b) What is the anguar acceleration (in rad/s2) of each wheel?
A golfer rides in a golf cart at an average speed of 3.59 m/s for 39.3 s. She then gets out of the cart and starts walking at an average speed of 1.46 m/s. For how long (in seconds) must she walk if her average speed for the entire trip, riding and walking, is 1.72 m/s ? Number Units
The golfer does not need to walk for any particular time to maintain an average speed of 1.72 m/s for the entire trip.
To solve this problem, we can use the concept of average speed, which is calculated by dividing the total distance traveled by the total time taken.
Let's start by finding the total distance traveled while riding in the golf cart. We can use the formula:
Distance = Speed × Time
Distance = 3.59 m/s × 39.3 s
Distance = 141.387 m (rounded to three decimal places)
Now, let's assume the golfer walks for time t (in seconds). We can calculate the total distance covered while walking:
Distance = Speed × Time
Distance = 1.46 m/s × t
To maintain an average speed of 1.72 m/s for the entire trip, the total distance traveled while walking and riding should be equal to the distance covered while riding the cart:
141.387 m + 1.46 m/s × t = 141.387 m
Let's solve for t:
1.46 m/s × t = 0
Since the left side of the equation is zero, it means t can be any value. The golfer does not need to walk to maintain the average speed of 1.72 m/s. Therefore, she does not need to walk for any specific duration.
Therefore, the golfer must walk for 1.72 m/s
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Only two horizontal forces act on a 6.8 kg body. One force is 7.1 N, acting due east, and the other is 7.3 N, acting 57° north of west. What is the magnitude of the body's acceleration?
The magnitude of the body's acceleration is 0.522 m/s².
The horizontal component of the 7.3 N force is:
7.3 cos 57° = 3.55 N in the western direction
The net horizontal force acting on the body is:
7.1 N - 3.55 N = 3.55 N to the east
The body's mass is 6.8 kg, and the magnitude of the net horizontal force acting on it is 3.55 N to the east.
Thus, the body's acceleration is given by:
a = F/m
= 3.55 N / 6.8 kg
= 0.522 m/s² (2 significant figures)
Therefore, the magnitude of the body's acceleration is 0.522 m/s².
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A person will hear a sound at a lower pitch if the source Is moving towards them Changes its amplitude Is moving away from them Remains stationary
A person will hear a sound at a lower pitch if the source is moving away from them. The pitch of a sound is determined by its frequency, which is the number of oscillations per unit of time.
The pitch of a sound is determined by its frequency, which is the number of oscillations per unit of time. When a sound source is moving away from an observer, the wavelength of the sound waves appears to increase, resulting in a decrease in frequency. This decrease in frequency leads to a lower perceived pitch of the sound. This phenomenon is known as the Doppler effect. Conversely, if the source is moving towards the observer, the wavelength appears to decrease, resulting in an increase in frequency and a higher perceived pitch. Changes in amplitude or the sound source remaining stationary do not directly affect the pitch of the sound.
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A damped linear oscillator subject to feedback control is described by the following two-dimensional state-space model
dt
dx
=(
0
−1
1
−2
)x+(
0
1
)u,y=(
1
0
)x a) (10 points) Find the transfer function G
yu
(s). b) ( 2 points) Find the steady-state response to a unit step input function. c) (8 points) Find the steady-state response to a sinusoidal input function u= sin(2t)
The transfer function G(y/u) is obtained by transforming the state-space model into the Laplace domain and then manipulating the equations to find the desired response.
a) To find the transfer function G(y/u), we need to first write the state-space model in matrix form.
The state equation is given by:
[tex]dx/dt = Ax + Bu[/tex]
Where A is the state matrix and B is the input matrix. In this case,
[tex]A = [0 -1; 1 -2] and B = [0; 1].[/tex]
The output equation is given by:
[tex]y = Cx[/tex]
Where C is the output matrix. In this case,
[tex]C = [1 0].[/tex]
To find the transfer function, we need to transform the state-space model into the Laplace domain. The Laplace transform of the state equation is:
[tex]sX(s) - x(0) = AX(s) + BU(s)[/tex]
Rearranging the equation, we get:
[tex](sI - A)X(s) = BU(s) + x(0)[/tex]
Taking the inverse Laplace transform, we get:
[tex]dx/dt = AX + Bu[/tex]
Now, we can substitute the values of A and B to get the transfer function G(s):
[tex]sX(s) - x(0) = [ s 1 ] [ X1(s) ] + [ 0 ] [ U(s) ][/tex] [tex][ X2(s) ][/tex]
[tex]sX1(s) - x1(0) = sX1(s) + X2(s)[/tex]
[tex]sX2(s) - x2(0) = X1(s) - 2X2(s)[/tex]
From the output equation, we have:
[tex]Y(s) = [ 1 0 ] [ X1(s) ] [ X2(s) ][/tex]
By substituting the values, we can obtain the transfer function G(s) = Y(s)/U(s).
b) To find the steady-state response to a unit step input function, we need to set [tex]U(s) = 1/s[/tex] and find the value of Y(s) at
[tex]s = 0[/tex].
c) To find the steady-state response to a sinusoidal input function [tex]u = sin(2t)[/tex], we need to find the frequency response of the system. The frequency response is obtained by substituting the transfer function G(s) and then taking the inverse Laplace transform.
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An ambulance is traveling away from you at 50.0 km/h has a siren that produces a 1,500.0 Hz sound. The outside temperature is 25. ∘C. What is the speed and frequency of the wave that you observe?
As per the Doppler effect, the frequency of a wave appears to change when there is relative motion between the source of the wave and the observer.
When there is motion, the waves in front of the source get compressed and the waves behind the source get elongated. As a result, the frequency of the wave changes.
It is different when the observer is stationary and the source is moving and different when the observer is moving and the source is stationary.In the problem, the ambulance is moving away from the observer.
Therefore, the frequency heard by the observer will be less than the actual frequency produced by the siren. The formula for the apparent frequency heard by the observer is:f′=f(v+v0v−vs)where, f' is the apparent frequency, f is the actual frequency of the source, v is the velocity of sound, v0 is the velocity of the observer, and vs is the velocity of the source.
Substituting the given values,f'=1500(343+0/(343+50))=1414.85 HzTherefore, the frequency observed by the observer is 1414.85 Hz.Speed of wave: The speed of the wave is given by the formula:v=fλwhere, v is the velocity of the wave, f is the frequency of the wave, and λ is the wavelength of the wave.Substituting the given values,v=1414.85(343)=484,919.55 m/hThe speed of the wave is 484,919.55 m/h.Approximating to 1 decimal place, the speed of the wave is 135 m/s.
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a) What loss does laminating the iron core of a transformer reduce? b) Explain why the proportional relationship between the magnetic field strength of an electromagnet and the flux density inside the iron core eventually breakdown as the current continues to increase. C) Draw an equivalent circuit of a transformer with all parameters referred to secondary You can neglect no-load current
Laminating the iron core of a transformer reduces eddy current losses. The proportional relationship between magnetic field strength and flux density in the iron core breaks down as current increases.
a) Laminating the iron core of a transformer reduces a type of loss known as eddy current losses. When an alternating current passes through the windings of a transformer, it induces circulating currents within the iron core. These eddy currents generate heat and consume energy, leading to power losses. By laminating the iron core, which involves dividing it into thin layers insulated from each other, the flow of eddy currents is minimized. This reduces the eddy current losses, making the transformer more efficient.
b) The proportional relationship between magnetic field strength and flux density in the iron core of an electromagnet eventually breaks down as the current continues to increase. Initially, as the current increases, the magnetic field strength and flux density inside the iron core increase in a linear manner.
However, at higher currents, the core starts to approach its magnetic saturation point. At this point, the increase in current does not result in a proportional increase in magnetic field strength. The core becomes saturated, and further increases in current lead to diminishing returns in terms of increased magnetic field strength. This breakdown in the proportional relationship is due to the limitations of the magnetic properties of the core material.
In conclusion, laminating the iron core of a transformer reduces eddy current losses, increasing its efficiency. Additionally, the proportional relationship between magnetic field strength and flux density in an electromagnet's iron core eventually breaks down as current increases due to magnetic saturation.
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A small object has a charge of q=5e, where e is the charge on an electron. (a) Determine the electric potential (in V) due to the charge at a distance r=0.230 cm from the charge. V (b) Determine the electric potential difference (in V ) between a point that is 5r away and this point, that is V(5r)−V(r). V (c) Determine the electric potential difference (in V) between a point that is 5r away and this point, that is V(5r )−V(r). v (d) How would the answers change if the electrons are replaced by protons? The sign of answer (a) would change. The sign of answer (b) would change. The sign of answer (c) would change. The sign of all answers would change. All answers would remain the same.
(a) The electric potential due to a 5e charge at a distance of 0.230 cm is 1.44 × 10^6 V.
(b) The electric potential difference between a point that is 5r away and this point is 3.59 × 10^5 V.
(c) The electric potential difference between a point that is 5r away and this point is -3.59 × 10^5 V.
(d) The sign of all answers would change if the electrons are replaced by protons.
(a) The electric potential due to the charge at a distance r=0.230 cm from the charge is:
V = kq / r
where:
k is the Coulomb constant (8.9875 × 10^9 N⋅m^2/C^2)
q is the charge of the object (5e)
r is the distance from the charge (0.230 cm)
V = 8.9875 × 10^9 N⋅m^2/C^2 * 5e / 0.230 cm = 1.44 × 10^6 V
(b) The electric potential difference between a point that is 5r away and this point is:
V(5r) - V(r) = kq / (5r) - kq / r = 4kq / 5r
V(5r) - V(r) = 4 * 8.9875 × 10^9 N⋅m^2/C^2 * e / (5 * 0.230 cm) = 3.59 × 10^5 V
(c) The electric potential difference between a point that is 5r away and this point is:
V(5r) - V(r) = kq / (5r) - kq / r = -4kq / 5r
Substituting the given values, we get:
V(5r) - V(r) = -4 * 8.9875 × 10^9 N⋅m^2/C^2 * e / (5 * 0.230 cm) = -3.59 × 10^5 V
(d) If the electrons are replaced by protons, the sign of all answers would change.
Therefore, the answers that would change are:
The sign of answer (a) would change.
The sign of answer (b) would change.
The sign of answer (c) would change.
Answer: The answer is (d).
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A uniform electric field of 3.50kN/C points in the +x direction. 1) What is the change in potential energy of a+2.00nC test charge, U
electric, b
−U
electric,
, as it is moved from point a at x=−30.0 cm to point b at x=+50.0 cm ? (Express your answer to two significant figures.) μJ
The potential difference between point a and b in the given scenario is -0.28 kV, and the change in potential energy for a +2.00 nC test charge moving from point a to b is -0.56 µJ. The negative sign indicates a decrease in potential energy during the movement.
Electric field of 3.50 kN/C points in the +x direction. Charge q = +2.00 nC.
The change in potential energy of a test charge as it is moved from point a to b can be calculated using the formula: ΔU = qΔV, where q is the charge of the test charge and ΔV is the change in voltage between the two points. The voltage difference is the work done to move a unit charge from one point to another.
ΔV = -E Δx, where Δx is the distance between the two points.
a) The potential difference between point a and b:
ΔV = -E Δx = (-3.5 kN/C) (80 cm) = -0.28 kV.
Thus, the change in potential energy of a test charge as it is moved from point a to b can be calculated as follows:
ΔU = qΔV = (2.00 nC)(-280 V) = -5.6×10⁻⁴ J = -0.56 µJ.
The change in potential energy of a +2.00nC test charge, ΔUelectric, between points a and b is -0.56 µJ (negative sign indicates a decrease in potential energy).
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A vector has an x-component of −27.0 units and a y-component of 28.0 units. Find the magnitude and direction of the vector. magnitude units direction - (counterclockwise from the +x-axis)
The magnitude of the vector is 38.06 units and the direction of the vector is -46.22°
Vectors are used to represent quantities that have both magnitude and direction, such as velocity, force, or displacement. Vectors are essential in describing physical phenomena, and they are typically represented using arrows or boldface letters.
X-component of vector = -27.0 units
Y-component of vector = 28.0 units
To find: The magnitude and direction of the vector
Formula:The magnitude of vector = √(x² + y²) = √((-27)² + 28²) = 38.06 units
The direction of the vector = tan⁻¹(y/x) = tan⁻¹(28/-27) = -46.22° (counterclockwise from the +x-axis)
Therefore, the magnitude of the vector is 38.06 units and the direction of the vector is -46.22° (counterclockwise from the +x-axis).
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A geologist finds that a Moon rock whose mass is 9.40 kg has an apparent mass of 6.40 kg when submerged in water. Part A What is the density of the rock?
The density of the moon rock is 1000 kg/m³.
Mass of the moon rock = 9.40 kg,
Apparent mass of moon rock in water = 6.40 kg
We can use the Archimedes principle to find the density of the rock.
Archimedes principle states that "Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object."
This means that when an object is submerged in a fluid, there is an upward force acting on the object which is equal to the weight of the fluid displaced by the object.
So, we can find the volume of the rock by using the formula:
Volume of the rock = Volume of water displaced by the rock
Using the apparent mass of the rock, we can find the volume of the rock in water as follows:
Mass of water displaced by the rock = Mass of rock = 9.40 kg
Volume of water displaced by the rock = Mass of water displaced / Density of water
= 9.40 kg / 1000 kg/m³
= 0.0094 m³
Volume of the rock = Volume of water displaced = 0.0094 m³
Now, we can find the density of the rock using the formula:
Density = Mass / Volume
Density of the rock = 9.40 kg / 0.0094 m³
= 1000 kg/m³
So, the density of the moon rock is 1000 kg/m³.
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Part 1 A cart moves on an inclined low-friction track. One end of the track is raised by a block of height H such that the track is inclined by an angle θ. If the length of the track is d=1.00 m and the track is raised by H=9.00 cm, find: a) the angle θ[0.5] b) the acceleration of the cart along the direction of the ramp. Show the relevant diagram. (hint: you have to find the component of the gravitational acceleration g along the incline); Ask your TA, if you have any questions about how to carry this out. [0.5] Part 2 You are now given a stopwatch. Starting from rest, you measure the time t for the cart to slide down the ramp. a) If t=1.50 s, what is the average speed of the cart? [0.5] b) Use the values of t and d to calculate the acceleration of the cart along the direction of the ramp. Is your answer consistent with the acceleration that you calculated in Part 1 (b)? [0.5]
Therefore, the answer is consistent with the acceleration that was calculated in Part 1b.
Part 1a) Since the track is inclined by an angle of θ and the height of the track is raised by H, so the angle is given as:
tan(θ) = H/d = 0.09/1 = 0.09
θ = tan⁻¹0.09
θ = 5.14°
Therefore, the angle of inclination of the track is 5.14°.
b) Now, we need to calculate the acceleration of the cart along the direction of the ramp. Here, we need to find the component of the gravitational acceleration g along the incline. This component is given by:
gsin(θ) = (9.8 m/s²) sin(5.14°) = 0.849 m/s²
Hence, the acceleration of the cart along the direction of the ramp is 0.849 m/s².
Relevant diagram:
Part 2a) The average speed of the cart can be calculated using the formula:
average speed = total distance / total time
Here, the cart moves a distance of 1.00 m. Therefore, the average speed is given by:
average speed = 1.00 m / 1.50 s = 0.67 m/s
b) The acceleration of the cart along the direction of the ramp can be calculated using the formula:
s = ut + 1/2 at²
Here, s = 1.00 m, u = 0 m/s (as the cart starts from rest), t = 1.50 s. Therefore, the acceleration of the cart is given by:
a = 2s / t²
= 2 × 1.00 m / (1.50 s)²
= 0.89 m/s²
The acceleration calculated in Part 1b is 0.849 m/s², whereas the acceleration calculated in Part 2b is 0.89 m/s². Both the values are close, but not exactly the same. The difference can be attributed to experimental errors, such as measurement errors in the stopwatch and the ruler used to measure the length of the track.
Therefore, the answer is consistent with the acceleration that was calculated in Part 1b.
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Suppose you have a 3.7μC charge and a 265 N/C electric field that points due east. What is the magnitude of the force, in newtons, exerted on the charge by the electric field?]
The magnitude of the force exerted on the charge by the electric field is 0.00098 N (rounded to three significant figures).
The electric field equation is given by:
E = F/q
where E is the electric field, F is the electric force acting on a charged particle, and q is the electric charge.
In this problem, we are given that the electric field strength is 265 N/C and the charge is 3.7μC.
Thus the electric force can be calculated as:
F = Eq
Substituting the given values:
E = 265 N/Cq = 3.7μ
C= 3.7 x 10^-6
CF = Eq= 265 N/C x 3.7 x 10^-6 C= 0.00098 N (rounded to three significant figures).
Therefore, the magnitude of the force exerted on the charge by the electric field is 0.00098 N (rounded to three significant figures).
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The magnitude of the force exerted on the charge by the electric field is 0.98155 N.
Charge = 3.7 μC
Electric field = 265 N/C
The formula for the electric force acting on a charge due to the electric field is:
F = q * E
Where,
F = Force
q = Charge
E = Electric field
Substituting the given values, we get:
F = 3.7 μC * 265 N/C
F = 981.55 * 10^-6 N
F = 0.98155 N
Therefore, the magnitude of the force exerted on the charge by the electric field is 0.98155 N.
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(a) What vertical component of initial velocity must be given to the ball, such that it barely clears the 1.00 m high net? (b) How far beyond the net will the ball hit the ground?
We need to use the equations of motion for projectile motion. The vertical component of the initial velocity required to barely clear the 1.00 m high net is approximately 1.41 m/s. The ball will hit the ground approximately 5.07 meters beyond the net.
To answer these questions, we need to use the equations of motion for projectile motion.
(a) The vertical component of initial velocity required to barely clear the 1.00 m high net can be determined using the equation:
y = y0 + v0y * t - (1/2) * g * t^2
Where:
y is the vertical displacement (1.00 m)
y0 is the initial vertical position (0 m)
v0y is the vertical component of the initial velocity (what we need to find)
t is the time of the flight
g is the acceleration due to gravity (-9.8 m/s^2)
We can solve for v0y:
1.00 m = 0 + v0y * t - (1/2) * (-9.8 m/s^2) * t^2
Simplifying the equation, we get:
v0y * t = (1/2) * 9.8 m/s^2 * t^2 + 1.00 m
v0y = (1/2) * 9.8 m/s^2 * t + 1.00 m / t
Since the ball barely clears the net, it means that the vertical displacement at the time of flight (t) is equal to 1.00 m. Therefore, we can substitute t = sqrt(2 * y / g) into the equation:
v0y = (1/2) * 9.8 m/s^2 * sqrt(2 * 1.00 m / 9.8 m/s^2) + 1.00 m / sqrt(2 * 1.00 m / 9.8 m/s^2)
Calculating the expression, we get:
v0y ≈ 1.41 m/s
Therefore, the vertical component of the initial velocity required to barely clear the 1.00 m high net is approximately 1.41 m/s.
(b) To determine how far beyond the net the ball will hit the ground, we need to find the horizontal distance traveled by the ball during the time of flight (t). The horizontal distance can be calculated using the equation:
x = v0x * t
Since there is no horizontal acceleration, the horizontal component of the initial velocity (v0x) remains constant throughout the motion.
The initial velocity (v0) can be found using the Pythagorean theorem:
v0 = sqrt(v0x^2 + v0y^2)
Given that the initial speed is 25.0 m/s, we have:
25.0 m/s = sqrt(v0x^2 + (1.41 m/s)^2)
Solving for v0x, we get:
v0x ≈ sqrt((25.0 m/s)^2 - (1.41 m/s)^2)
v0x ≈ 24.99 m/s
Now we can calculate the horizontal distance traveled by the ball:
x = v0x * t = 24.99 m/s * t
Since we already determined t = sqrt(2 * y / g), we can substitute this value into the equation:
x ≈ 24.99 m/s * sqrt(2 * 1.00 m / 9.8 m/s^2)
Calculating the expression, we get:
x ≈ 5.07 m
Therefore, the ball will hit the ground approximately 5.07 meters beyond the net.
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A magazine article states that cheetahs are the fastest sprinters in the animal world and that a cheetah was observed to accelerate from rest to 70.0 km/h in 2.50 s. What average acceleration in m/s2 does this require?: (b) The article also says the cheet ah covered 60.0 m during this 2.50 s interval. How large a constant acceleration is implied by this statement? (c) Accelerations substantially greater than g are difficult for an animal or automobile to attain because there is a tendency to slip even on very rough ground with larger acceleration. Given this information, can you guess if either acceleration is woong in the article
The answers to the given questions are as follows:
(a) The average acceleration required for the cheetah to accelerate from rest to 70.0 km/h in 2.50 s is approximately 7.76 m/s².
(b) The statement that the cheetah covered 60.0 m during the 2.50 s interval implies a constant acceleration of approximately 19.2 m/s².
(c) Considering the constraints of slipping on rough ground, the mentioned accelerations in the article (7.76 m/s² and 19.2 m/s²) are reasonable and unlikely to be wrong.
(a) To find the average acceleration in m/s², we need to convert the speed from km/h to m/s and then use the formula for average acceleration:
Average acceleration = (final velocity - initial velocity) / time
Given:
Initial velocity (u) = 0 m/s (starting from rest)
Final velocity (v) = 70.0 km/h
= (70.0 × 1000) / 3600 m/s
≈ 19.4 m/s
Time (t) = 2.50 s
Average acceleration = (19.4 m/s - 0 m/s) / 2.50 s
Average acceleration ≈ 7.76 m/s²
Therefore, the average acceleration required for the cheetah to accelerate from rest to 70.0 km/h in 2.50 s is approximately 7.76 m/s².
(b) To determine the constant acceleration implied by the statement that the cheetah covered 60.0 m during the 2.50 s interval, we can use the following kinematic equation:
s = ut + (1/2)at²
where:
s is the distance covered (60.0 m),
u is the initial velocity (0 m/s),
t is the time (2.50 s),
a is the constant acceleration (to be determined).
Plugging in the values, we have:
60.0 = 0 + (1/2)a(2.50)²
60.0 = (1/2)(6.25)a
a = (2 × 60.0) / 6.25
a ≈ 19.2 m/s²
Therefore, the implied constant acceleration based on the statement that the cheetah covered 60.0 m in 2.50 s is approximately 19.2 m/s².
(c) Given the information that accelerations substantially greater than g (acceleration due to gravity) are difficult for an animal or automobile to attain due to slipping, it is likely that the acceleration values mentioned in the article are reasonable. The average acceleration of 7.76 m/s² falls within a realistic range for the cheetah's acceleration capability, as does the implied constant acceleration of 19.2 m/s².
Therefore, it is unlikely that the accelerations mentioned in the article are wrong, considering the constraints mentioned regarding slipping on rough ground.
Learn more about Acceleration from the given link:
https://brainly.com/question/30660316
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