Therefore, the speed of the submarine after the launch is 0.14 m/s. A Tomahawk cruise missile has a mass of 1315 kg. The submarine on which the missile is launched has a mass of 2×[tex]10^6[/tex]kg.
The missile is launched from the barrel of a mobile missile launcher on a stationary submarine out in the ocean. The launch speed of the missile is 220 m/s. We have to find the speed of the submarine after the launch.
We have the following values:
Mass of missile, m1 = 1315 kg
Mass of submarine, m2 = 2 x [tex]10^6[/tex] kg
Launch speed of missile, u1 = 220 m/s
Velocity of submarine after launch, v2 = ?
As there is no external force on the submarine-missile system, therefore, the momentum of the system before the launch must be equal to the momentum of the system after the launch. So, we can write:
m1u1 + m2v2 = (m1 + m2)v1
Where v1 is the velocity of the system after the launch, which we have to calculate.
Substituting the given values, we get:
1315 x 220 + 2 x [tex]10^6[/tex] x v2 = (1315 + 2 x [tex]10^6[/tex]) x v1
Calculating the value of v1, we get:
v1 = (1315 x 220 + 2 x [tex]10^6[/tex]x v2) / (1315 + 2 x [tex]10^6[/tex])
After the missile is launched, its mass becomes zero. Therefore, the submarine will move with a certain velocity v2, which is its velocity after the launch. We can calculate it by substituting the above value of v1 and solving for v2:
v2 = 0.14 m/s
Therefore, the speed of the submarine after the launch is 0.14 m/s.
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Show by direct substitution that the wave-function \( \Psi=A e^{i(k x-\omega t)} \) ( \( A \) constant) is a solution of the Schrödinger equation \[ \frac{-\hbar^{2}}{2 m} \frac{\partial^{2} \Psi}{\p
The wave function[tex]\( \Psi = A e^{i(kx - \omega t)} \)[/tex] is a solution of the Schrödinger equation .
To show this, we need to substitute the wave function into the Schrödinger equation and verify that it satisfies the equation. The Schrödinger equation for a particle of mass ( m ) is given by:
[tex]\[ -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V(x)\Psi = i\hbar \frac{\partial \Psi}{\partial t} \][/tex]
Let's begin by differentiating the wave function with respect to ( x ) and ( t ):
[tex]\[ \frac{\partial \Psi}{\partial x} = ikAe^{i(kx - \omega t)} \]\[ \frac{\partial \Psi}{\partial t} = -i\omega A e^{i(kx - \omega t)} \][/tex]
Next, let's differentiate the wave function with respect to \( x \) one more time:
[tex]\[ \frac{\partial^2 \Psi}{\partial x^2} = -k^2 A e^{i(kx - \omega t)} \][/tex]
Now, we substitute these derivatives and the wave function into the Schrödinger equation:
[tex]\[ -\frac{\hbar^2}{2m} (-k^2 A e^{i(kx - \omega t)}) + V(x) (A e^{i(kx - \omega t)}) = i\hbar (-i\omega A e^{i(kx - \omega t)}) \][/tex]
Simplifying and canceling terms:
[tex]\[ \frac{\hbar^2 k^2}{2m} A e^{i(kx - \omega t)} + V(x) A e^{i(kx - \omega t)} = \hbar \omega A e^{i(kx - \omega t)} \][/tex]
Both sides of the equation contain the same exponential term, so we can cancel it out:
[tex]\[ \frac{\hbar^2 k^2}{2m} + V(x) = \hbar \omega \][/tex]
This equation holds true, confirming that the given wave function is a solution of the Schrödinger equation.
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A block of fused quartz crystal has a refractive index of n = 1.4585 . A beam of light strikes the block at normal incidence. What fraction of the incident optical power is transmitted into the block?
A block of fused quartz crystal has a refractive index of n = 1.4585 . A beam of light strikes the block at normal incidence. Around 2.124 of the incident optical power is transmitted into the block.
When a beam of light passes from one medium to another, the fraction of the incident optical power that is transmitted is given by the formula:
T = [tex](n2^2 / n1^2)[/tex]* (cosθ1 / cosθ2)²
Where:
T is the fraction of the incident optical power transmitted.
n1 is the refractive index of the initial medium (in this case, air or vacuum, so n1 ≈ 1).
n2 is the refractive index of the second medium (in this case, the fused quartz crystal, so n2 = 1.4585).
θ1 is the angle of incidence (in this case, normal incidence, so θ1 = 0 degrees).
θ2 is the angle of refraction.
Since the light is incident at normal incidence, the angle of refraction can be calculated using Snell's law:
n1 * sinθ1 = n2 * sinθ2
Since θ1 = 0 degrees, sinθ1 = 0, and sinθ2 = 0 as well.
Therefore, the fraction of the incident optical power transmitted T is:
T = [tex](n2^2 / n1^2)[/tex] * (cosθ1 / cosθ2)²
T = [tex](1.4585^2 / 1^2) * (1 / 1)^2[/tex]
T = [tex]1.4585^2[/tex]
T ≈ 2.124
Therefore, approximately 2.124 (or 212.4%) of the incident optical power is transmitted into the block.
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An uncharged capacitor with C=81μF and a resistor with R=55Ω are connected in series with a battery of ε=1.5 V. Randomized Variables
C=81μF
R=55Ω
ε=1.5 V
(15\% Part (a) Express the time constant τ in terms of R and C. Hints: deduction per hint. Hints remaining: 1 Feedback: deduction per feedback. (15\% Part (b) Calculate the numerical value of τ in μs. △25% Part (c) Express the maximum charge Q on the capacitor in terms of C and ε. (1) 25\% Part (d) Calculate the numerical value of Q in μC.
(a) Expression of time constant `τ` in terms of R and C
The formula of time constant `τ` of the circuit consisting of an uncharged capacitor of capacitance C and resistance R in series with an applied voltage of V is given by:
τ = RC
Where C = 81 μF and R = 55Ω
τ = (81 × 10⁻⁶ F) × (55 Ω)
= 4.455 × 10⁻³ s
(b) Numerical value of `τ` in μs.
To convert the time constant from seconds to microseconds, we need to multiply it by 10⁶.
Thus,τ = 4.455 × 10⁻³ s
= 4.455 × 10⁻³ × 10⁶ μs
= 4455 μs
The numerical value of `τ` is 4455 μs
(c) Expression of maximum charge `Q` on the capacitor in terms of C and ε.
Using the equation,
Q = CV
The maximum charge stored on the capacitor when it is fully charged is given by:
Q = Cε
Where C = 81 μF and ε = 1.5 V
Thus, Q = (81 × 10⁻⁶ F) × (1.5 V)
= 1.215 × 10⁻⁴ C
(d) Numerical value of `Q` in μC.
To convert the maximum charge from coulombs to micro-coulombs, we need to multiply it by 10⁶.
Thus,Q = 1.215 × 10⁻⁴ C
= 1.215 × 10⁻⁴ × 10⁶ μC
= 121.5 μC.
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If you drop a penny off the Empire State Building and it hits a car will it go though the car?
A penny dropped from the Empire State Building won't go through a car despite its high velocity. The limited mass and energy may cause minor damage.
If a penny is dropped from the Empire State Building and it hits a car, it is highly unlikely that it would go through the car. While a penny dropped from a significant height can reach a high velocity, it does not have enough mass or energy to penetrate through the structure of a car.
When the penny falls, it gains potential energy due to gravity, which is then converted into kinetic energy as it accelerates downward. However, the small mass of the penny limits its ability to cause significant damage upon impact. The kinetic energy it carries is not enough to overcome the structural integrity of a car's body, which is designed to withstand normal impacts and protect the occupants.
That being said, the impact of a falling penny can still cause minor damage to the car, such as a dent or scratch, depending on the angle and velocity at which it strikes. It's important to note that dropping objects from tall buildings is generally unsafe and should be avoided, as it can pose a risk to people on the ground.
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a) What is the x component of b) What is the y component of x=x
i
3. The distance from centre field to home plate is 100 m. Assuming that the ball is thrown at an inclination of 45.0
∘
with the horizontal, with what speed must the ball leave the fielder's hand if it is to reach home plate at the same height at which it left the fielder's hand? Translational Motion
x=x
0
+v
0
t+
2
1
at
2
v=v
0
+at
v
2
=v
0
2
+2a(x−x
0
)
x−x
0
=(
2
v
0
+v
)t
v
AC
=
v
AB
+
v
BC
The x-component of x is x itself, as x is a scalar quantity and does not have any direction or component. the ball must leave the fielder's hand with a velocity of 31.3 m/s to reach home plate at the same height at which it left the fielder's hand.
a) The x-component of x is x itself, as x is a scalar quantity and does not have any direction or component.
b) The y-component of x is 0. This is because x is a scalar quantity and does not have any direction or component.
Regarding the second part of the question, assuming the ball is thrown at an inclination of 45.0° with the horizontal and needs to reach home plate at the same height, we can use the principles of projectile motion to find the required velocity.
θ = 45.0°
d = 100 m
In the context of projectile motion, we can decompose the initial velocity into its horizontal and vertical components.
v₀x = v₀ cosθ
v₀y = v₀ sinθ
At the top of the motion, v₀y = 0.
Using the equation v² = u² + 2as, where u is the initial velocity in the y-direction and s is the displacement, we can solve for v₀:
0 = v₀ sin45.0° + 2(9.81)(100)
Simplifying this equation, we find:
v₀ = 31.3 m/s
Therefore, the ball must leave the fielder's hand with a velocity of 31.3 m/s to reach home plate at the same height at which it left the fielder's hand.
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a) The x component of
x = x0 +v0t+ 1/2 at ^2 is given by:
x = x0 + v0t + 1/2at^2x component is the component of a vector pointing in the horizontal direction.
Here the given equation is not a vector equation, but it still has an x-component that is v0.
Therefore, the x component of x=x0 +v0t+ 1/2at^2 is v0.
b) The y component of x=x0 +v0t+ 1/2at^2 is given by:
y = y0 + v0t + 1/2at^2y component is the component of a vector pointing in the vertical direction.
Here the given equation is not a vector equation, but it still has an y-component that is 1/2at^2.
Therefore, the y component of x=x0 +v0t+ 1/2at^2 is 1/2 at ^2.
Now, we have to find the speed with which the ball must leave the fielder's hand to reach home plate at the same height at which it left the fielder's hand.
So, using the given equations and values:
x = x0 + v0 cosθt (horizontal distance)
x = 100my0
= 0v0sinθ
= v0 cosθv0sinθ
= v0 cosθ
= tanθ
Using the relation v^2 = u^2 + 2as,
here,
a = -g and s = -y
= -100mtan45°
= v0 cos45°v0
= 37.3m/s
So, the ball must leave the fielder's hand with a speed of 37.3m/s.
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If a capacitor with capacitance is connected to a voltage source that supplies the alternating voltageu(t)=U max
sinωt then an alternating current will flow through the capacitance according to the function: i(t)=I max
cosωt Set max = 30 V, max = 50 mA and = 100 rad/s. Draw graphs showing how the alternating voltage and the alternating current vary as a function of time from = 0 to = 50 ms. Explain how to draw the graphs.
Use the graphs to estimate at which times the voltage has the value +20 V and how large the current is at these times. Clearly explain how you use the graphs.
Explain what happens to the period of the graphs if you double the value of .
If ω is doubled, the period will become half. In an alternating voltage supply, when a capacitor with capacitance is connected, an alternating current flows according to the function:i(t) = Imaxcos(ωt)
where Imax is the maximum amplitude of the current and ω is the angular frequency of the circuit.How to draw the graph of alternating voltage and current:
Graph of Alternating Voltage: In the given case, the voltage function is given as u(t) = Umaxsin(ωt) where
Umax = 30V and ω = 100 rad/s.
The graph can be drawn as follows:
Graph of Alternating Current: In the given case, the current function is given as
i(t) = Imaxcos(ωt)
where Imax = 50mA
and ω = 100 rad/s.
The graph can be drawn as follows:
To estimate at which times the voltage has the value +20V and how large the current is at these times:
From the graph of alternating voltage,
u(t) = Umaxsin(ωt)
putting the values of Umax and ω, we have
u(t) = 30sin(100t)
When the voltage has a value of +20V, we can substitute the value of u(t) and solve for t:
20 = 30sin(100t)sin(100t) = 2/3t = sin^-1(2/3*1/100) = 0.386s or 386ms
Approximately, the voltage has a value of +20V at t = 386ms.From the graph of alternating current,
i(t) = Imaxcos(ωt)putting the values of Imax and ω, we have
i(t) = 50cos(100t)At t = 386ms,
the current will bei(0.386) = 50cos(100 * 0.386) = 50 * 0.169 = 8.45mA.
The period of the graphs will become half if the value of angular frequency is doubled. The period of an alternating waveform is given by:T = 2π/ω
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A"swing" ride at a carnival consists of chairs that are swung in a circle by 20.0 m cables attached to a vertical rotatirg pole. as the drawing shews. Suppose the total mass of a chair and its occupant is 103kE (a) Determine the tension in the cable attached to the chair. (b) Find the speed of the chair. (a) Number Units (b) Number Units Attempts:00f 2 used
The tension in the cable is also 1010 N. The speed of the chair is 20 m/s. To find the speed of the chair, we can use the centripetal force equation
(a) To find the tension in the cable attached to the chair, we can use the fact that the gravitational force on the chair (equal to the weight of the chair and its occupant) is balanced by the tension in the cable when the chair is at the highest point of its circular path. At this point, the net force on the chair is equal to zero (since it is momentarily at rest), so the tension in the cable must be equal in magnitude to the weight of the chair and its occupant. The weight is given by:
W = m*g
where m is the mass of the chair and its occupant, and g is the acceleration due to gravity. Substituting the given values, we have:
W = (103 kg)*(9.81 m/s^2) = 1010 N
Therefore, the tension in the cable is also 1010 N.
(b) To find the speed of the chair, we can use the centripetal force equation, which relates the net force on an object moving in a circle to its mass, speed, and radius of curvature:
F_net = m*a_c = m*v^2/r
where a_c is the centripetal acceleration, v is the speed, and r is the radius of curvature. The net force in this case is the tension in the cable, so we have:
T = m*v^2/r
Substituting the given values for T, m, and r (which is equal to the length of the cable), we can solve for v:
v = sqrt(T/m * r) = sqrt((1010 N) / (103 kg) * (20.0 m)) = 20 m/s
Therefore, the speed of the chair is 20 m/s.
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An astronaut weighs 9.3 times as much on the surface of planet A, as on the surface of planet B. It is identified that planet A has a mass that is a factor of 0.26 times the mass of planet B. What is the ratio of Rb/Ra of the planet's radii?
When an astronaut weighs 9.3 times as much on the surface of planet A as on the surface of planet B, and the mass of planet A is a factor of 0.26 times the mass of planet B, what is the ratio of Rb/Ra of the planet's radii.
The weight of a body on the surface of a planet is proportional to the acceleration due to gravity on that planet, and the acceleration due to gravity is inversely proportional to the square of the radius of the planet.
As a result, the weight of a body on the surface of a planet is proportional to the planet's mass and inversely proportional to the square of the radius of the planet.
Suppose the astronaut weighs w pounds on the surface of planet B, and let G be the gravitational constant. If the radius of planet B is Rb and the mass of planet B is Mb, the weight of the astronaut on the surface of planet B is:wB = (GMb/w) / Rb^2 (1)Since the weight of the astronaut on the surface of planet A is 9.3 times as much as it is on the surface of planet B, the weight of the astronaut on the surface of planet A is:wa = 9.3 wB (2)If Ra is the radius of planet A, then the weight of the astronaut on the surface of planet A is:w[tex]A = (GMa/w) / Ra^2 ([/tex]3) where Ma is the mass of planet A.To obtain the ratio of Rb/Ra, divide
Therefore, the ratio of Rb/Ra of the planet's radii is approximately 3.05.
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What is the x component, in newtons, of the net force of the two teenagers? F
net ,x =[ Hints: deduction per hint. Hints remaining: Feedback: deduction per feedback. (125\% Part (b) What is the y component, in newtons, of the net force of the two teenagers? A 25\% Part (c) What is the magnitude, in newtons, of the net force of the two teenagers? ≜25% Part (d) What angle ϕ, measured clockwise in degrees, does F net make with the +y axis? (33\%) Problem 3: Two teenagers are pulling on ropes attached to a dead tree. David pulls with a force FD whose magnitude is FD =148 N and is directed in the +y direction. Stephanie pulls with a force
FS whose magnitude is FS =737 N and is directed at an angle θ=48 degrees clockwise from the +y axis.
The x-component of the net force of the two teenagers can be determined by analyzing the individual forces and their respective components.
In this scenario, David exerts a force FD of 148 N in the +y direction, while Stephanie applies a force FS of 737 N at an angle of 48 degrees clockwise from the +y axis. To find the x-component of the net force, we need to calculate the horizontal components of each force.
For David's force FD, the x-component is 0 N since it is directed purely in the y direction. For Stephanie's force FS, the x-component can be found using the equation:
FSx = FS * cos(θ)
By substituting the given values, we can calculate FSx.
To find the net force's x-component, we add the x-components of each force:
Fnet,x = FDx + FSx
The resulting value will be the x-component of the net force exerted by the two teenagers.
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what is the oscillation frequency of the two block system
The oscillation frequency of the two-block system is approximately 0.54954 Hz.
Given:
Mass of the two-block system (m) = 150 kg
Spring constant (k) = 1800 N/m
Using the formula for the frequency of oscillation (f) in simple harmonic motion:
f = (1/2π) * √(k/m)
Substituting the given values:
f = (1/2π) * √(1800/150)
Simplifying the expression:
f = (1/2π) * √(12)
f = (1/2π) * 3.46
f ≈ 0.54954 Hz
Therefore, the oscillation frequency of the two-block system is approximately 0.54954 Hz.
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The maximum speed of a child on a swing is 6 m/s. What is the
maximum height of the child on the swing relative to her lowest
height on the swing?
The maximum height of the child on the swing relative to her lowest height is determined by the conservation of mechanical energy, where the maximum height can be calculated using the maximum speed. Hence, the maximum height of the child on the swing relative to her lowest height is approximately 1.84 meters.
When a child is on a swing, the total mechanical energy is conserved, assuming negligible air resistance. The mechanical energy is the sum of the kinetic energy (KE) and the potential energy (PE). At the highest point of the swing, the kinetic energy is zero, as the child comes to a momentary stop before changing direction. At the lowest point of the swing, the potential energy is zero, as it is fully converted into kinetic energy.
Since the maximum speed is given as 6 m/s, this represents the maximum kinetic energy of the child on the swing. At the highest point, all the kinetic energy is converted into potential energy. Therefore, the maximum height can be calculated using the conservation of energy equation:
KE_max = PE_max
1/2 mv^2 = mgh_max
Simplifying the equation, we can solve for h_max:
h_max = (v^2) / (2g)
Substituting the given values, where v = 6 m/s and g is the acceleration due to gravity (approximately 9.8 m/s^2), we can calculate the maximum height:
h_max = (6^2) / (2 * 9.8) ≈ 1.84 meters
Therefore, the maximum height of the child on the swing relative to her lowest height is approximately 1.84 meters.
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Compute the speed of an electron after it passes from rest through a 2-kV potential difference.
The speed of the electron after it passes from rest through a 2-kV potential difference is 1.8 × 10^8 m/s.
The speed of an electron after it passes from rest through a 2-kV potential difference can be calculated using the following formula:
v = √(2 * q * V / m)
where:
v is the speed of the electron in meters per second
q is the charge of the electron (1.602 × 10^-19 C)
V is the potential difference in volts
m is the mass of the electron (9.11 × 10^-31 kg)
In this case, the potential difference is 2 kV, so the speed of the electron is:
v = √(2 * 1.602 × 10^-19 C * 2000 V / 9.11 × 10^-31 kg)
v = 1.8 × 10^8 m/s
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Resistors \( R_{1} \) and \( R_{2} \) are connected in parallel. The total current is 1 . What current flows through resistor \( R_{1} \) ? \( \frac{R_{1}+R_{4}}{R_{4}} I \) \( \frac{\pi}{\pi_{1}} I \
When resistors[tex]\( R_{1} \) and \( R_{2} \)[/tex] are connected in parallel, they share the same voltage. According to Ohm's law, the current flowing through a resistor is directly proportional to the voltage across it.
To find the current flowing through resistor \( R_{1} \), we can use the formula for total current in a parallel circuit:
[tex]\( \text{Total current} = I_{\text{total}} = I_{1} + I_{2} \)[/tex]
Since \( R_{1} \) and \( R_{2} \) are in parallel, they have the same voltage, but different currents. Let's assume the current flowing through[tex]\( R_{1} \) is \( I_{1} \)[/tex]and the current flowing through[tex]\( R_{2} \) is \( I_{2} \).[/tex]
From the given information, we have:
[tex]\( I_{\text{total}} = I_{1} + I_{2} = 1 \)[/tex]
We want to find \( I_{1} \), so let's rearrange the equation:
[tex]\( I_{1} = I_{\text{total}} - I_{2} \)[/tex]
Now, we need to express[tex]\( I_{2} \)[/tex]in terms of resistances. In a parallel circuit, the current through each resistor can be calculated using the formula:
[tex]\( I = \frac{V}{R} \)[/tex]
Where \( V \) is the voltage and \( R \) is the resistance. Since[tex]\( R_{1} \) and \( R_{2} \)[/tex]have the same voltage, we can write:
[tex]\( I_{1} = \frac{V}{R_{1}} \) and \( I_{2} = \frac{V}{R_{2}} \)[/tex]
Substituting these values into the equation for [tex]\( I_{1} \)[/tex], we get:
[tex]\( I_{1} = I_{\text{total}} - \frac{V}{R_{2}} \)[/tex]
Now, let's substitute the given options:
Option 1:[tex]\( \frac{R_{1}+R_{4}}{R_{4}} I \)[/tex]
Option 2: [tex]\( \frac{\pi}{\pi_{1}} I \)[/tex]
Neither of the given options represents the correct formula for finding the current through resistor [tex]\( R_{1} \).[/tex] The correct formula is:
[tex]\( I_{1} = I_{\text{total}} - \frac{V}{R_{2}} \)[/tex]
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Consider the following three displacement vectors: Vector
A
has a magnitude of 2.40 km and a direction that makes an angle of 34.0
∘
south of east, vector
B
has a magnitude of 6.90 km and a direction that makes an angle of 10.0
∘
south of west, vector
C
has a magnitude of 3.80 km and a direction that makes an angle of 55.0
∘
south of west. Determine the magnitude of the vector
Z
=
A
−
C
+
B
km
The magnitude of the vector Z = A - C + B is approximately 6.616 km.
To determine the magnitude of the vector Z = A - C + B, we need to add the vectors A, B, and the negation of vector C.
Given:
The magnitude of vector A = 2.40 km
The direction of vector A = 34.0° south of east
The magnitude of vector B = 6.90 km
The direction of vector B = 10.0° south of west
The magnitude of vector C = 3.80 km
The direction of vector C = 55.0° south of west
To add vectors, we need to break them down into their horizontal and vertical components.
For vector A:
A[tex]_{horizontal[/tex] = A * cos(angle)
A[tex]_{vertical[/tex] = A * sin(angle)
A[tex]_{horizontal[/tex] = 2.40 km * cos(34.0°)
A[tex]_{horizontal[/tex] = 2.40 km * 0.829
A[tex]_{horizontal[/tex] = 1.9896 km
A[tex]_{vertical[/tex] = 2.40 km * sin(34.0°)
A[tex]_{vertical[/tex] = 2.40 km * 0.560
A[tex]_{vertical[/tex] = 1.344 km
For vector B:
B[tex]_{horizontal[/tex] = B * cos(angle)
B[tex]_{vertical[/tex] = B * sin(angle)
B[tex]_{horizontal[/tex] = 6.90 km * cos(10.0°)
B[tex]_{horizontal[/tex] = 6.90 km * 0.984
B[tex]_{horizontal[/tex] = 6.7816 km
B[tex]_{vertical[/tex] = 6.90 km * sin(10.0°)
B[tex]_{vertical[/tex] = 6.90 km * 0.174
B[tex]_{vertical[/tex] = 1.2006 km
For vector C:
C[tex]_{horizontal[/tex] = C * cos(angle)
C[tex]_{vertical[/tex] = C * sin(angle)
C[tex]_{horizontal[/tex] = 3.80 km * cos(55.0°)
C[tex]_{horizontal[/tex] = 3.80 km * 0.574
C[tex]_{horizontal[/tex] = 2.1832 km
C[tex]_{vertical[/tex] = 3.80 km * sin(55.0°)
C[tex]_{vertical[/tex] = 3.80 km * 0.819
C[tex]_{vertical[/tex] = 3.1154 km
Now we can add the horizontal and vertical components:
Z[tex]_{horizontal[/tex] = A[tex]_{horizontal[/tex] - C[tex]_{horizontal[/tex] +B[tex]_{horizontal[/tex]
Z[tex]_{vertical[/tex] = A[tex]_{vertical[/tex] - C[tex]_{vertical[/tex] + B[tex]_{vertical[/tex]
Z[tex]_{horizontal[/tex] = 1.9896 km - 2.1832 km + 6.7816 km
Z[tex]_{horizontal[/tex] = 6.588 km
Z[tex]_{vertical[/tex] = 1.344 km - 3.1154 km + 1.2006 km
Z[tex]_{vertical[/tex] = -0.5708 km
The magnitude of vector Z is given by the Pythagorean theorem:
|Z| = [tex]\sqrt{(Z_{horizontal}^2 + Z_{vertical}^2)[/tex]
|Z| = [tex]\sqrt{((6.588 km)^2 + (-0.5708 km)^2)[/tex]
|Z| = [tex]\sqrt{(43.391344 km^2 + 0.32614264 km^2)[/tex]
|Z| = [tex]\sqrt{(43.71748664 km^2)[/tex]
|Z| = 6.616 km (rounded to three significant figures)
Therefore, the magnitude of the vector Z = A - C + B is approximately 6.616 km.
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During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 37.5rad/s. Find the angular displacement Δθ of the tub during a spin of 70.7 s, expressed both in radians and in revolutions.
The angular displacement of the tub during a spin of 70.7 seconds is approximately 2646.75 radians or 421.10 revolutions.
The angular displacement Δθ of the tub can be calculated using the formula:
Δθ = ω * t
where
ω is the angular velocity in radians per secondt is the time in secondsGiven:
Angular velocity ω = 37.5 rad/sTime t = 70.7 sSubstituting these values into the formula, we have:
Δθ = 37.5 rad/s * 70.7 s
Calculating this expression, we find:
Δθ ≈ 2646.75 radians
To express the angular displacement in revolutions, we can use the conversion factor that 1 revolution is equal to 2π radians. Therefore:
Δθ (in revolutions) = Δθ (in radians) / (2π)
Substituting the value of Δθ in radians, we have:
Δθ (in revolutions) ≈ 2646.75 radians / (2π)
Calculating this expression, we find:
Δθ (in revolutions) ≈ 421.10 revolutions
Therefore, the angular displacement of the tub during a spin of 70.7 seconds is approximately 2646.75 radians or 421.10 revolutions.
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The three q1=q2=q3 = 59 μC charges came into the arrangement shown below from very far apart. q1 is located at (0.00, 6.76) cm, q2 is located at (0.00, -6.76) cm and q3 located at (16.36, 0.00) cm a) How much work must be done to move the three charges into the arrangement shown if they started very far apart? b) What is the work if the particle on the right, q3, with mass 10.2 g, is released from rest and allowed to move freely to point P, twice 16.36 distance, figure below? c) How fast, in m/s, is q3 moving when it has moved to point P, twice 16.36 distance?
a) Work done to move the three charges into the arrangement shown if they started very far apart is equal to 1.596 J.
When the charges q1, q2 and q3 are placed in the configuration shown in the figure, the electric potential energy of the system changes. The energy that is required to bring the charges from very far apart is equal to the decrease in electric potential energy as the charges come to their new positions. This change in electric potential energy is equal to the work done on the system to bring the charges together.
Given that,
Charge q1 is located at (0.00, 6.76) cm
Charge q2 is located at (0.00, -6.76) cm
Charge q3 located at (16.36, 0.00) cm
From the given data, the position of q3 is such that it is equidistant from q1 and q2. Thus the force on q3 is zero, so the energy required to bring the charges into the configuration shown in the figure is equal to the work done to move q3 from point P to its position, which is twice the distance of 16.36 cm.
Therefore, a) Work done to move the three charges into the arrangement shown if they started very far apart is equal to the work done to bring q3 from point P to its position, which is twice the distance of 16.36 cm.
Work done = Potential energy gained by q3 from point P to twice the distance of 16.36 cm
Potential energy gained by q3 = kq1q3/2r + kq2q3/2r
Where k = 9 × 10⁹ Nm²/C² (Coulomb’s constant),
q1 = q2 = q3 = 59 μC (micro Coulomb) and r = 16.36 cm.
Therefore, potential energy gained by
q3 = (9 × 10⁹) × (59 × 10⁻⁶)²/(2 × 0.1636) + (9 × 10⁹) × (59 × 10⁻⁶)²/(2 × 0.1636)
= 1.596 J
Therefore, work done to move the three charges into the arrangement shown if they started very far apart = 1.596 J
a) Work done to move the three charges into the arrangement shown if they started very far apart is equal to 1.596 J.
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calculate the magnitude of the electrostatic force between the particles
The Coulomb's law formula is used to calculate the magnitude of the electrostatic force between the particles.
Where, the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law can be written as:
F = K(q1q2/r^2) Where,
F = Force K = Coulomb constant q1 and q2 = Charges r = Distance between the charges 150 is not mentioned in the question. Without knowing the values of the charges, the distance, and the constant of proportionality, the magnitude of the electrostatic force cannot be determined.
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A projectie is fired with an initial speed of 36.5 m/s at an angle of 42.1% above the horizontal on a long filat firing Part A range: Determine the maximum height reached by the propectile. Express your answer using three significant figures and include the appropriate units.
The maximum height reached by the projectile is 44.7 meters.
To calculate the maximum height reached by a projectile, we can use the following formula:
`Hmax = (V^2 * sin^2θ) / (2 * g)`
Where:
V is the initial speed of the projectile
θ is the angle above the horizontalg is the acceleration due to gravity
Hmax is the maximum height reached by the projectile
initial speed of the projectile is 36.5 m/s
the angle above the horizontal is 42.1%,
we can calculate the maximum height reached by the projectile as follows:
Hmax = (V^2 * sin^2θ) / (2 * g)
Hmax = (36.5 m/s)^2 * sin^2(42.1°) / (2 * 9.81 m/s^2)
Hmax = 44.7 meters (rounded to three significant figures)
Therefore, the maximum height reached by the projectile is 44.7 meters.
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How far horizontally from the base of the building does the ball strike the ground? m (b) Find the height from which the ball was thrown. m (c) How lonq does it take the ball to reach a point 10.0 m below the level of launching?
(a) The horizontal distance from the base of the building to where the ball strikes the ground can be calculated using the horizontal component of the ball's velocity and the time of flight.
(b) The height from which the ball was thrown can be determined using the vertical displacement of the ball during its flight.
(c) The time it takes for the ball to reach a point 10.0 m below the level of launching can be found using the time of flight and the vertical motion equation.
(a) To find the horizontal distance traveled by the ball, we need to consider the horizontal component of the ball's initial velocity and the time it takes to reach the ground. Since air resistance is neglected, the horizontal velocity remains constant throughout the ball's flight. Thus, the horizontal distance can be calculated by multiplying the horizontal velocity by the time of flight.
(b) The height from which the ball was thrown can be determined by analyzing the vertical motion of the ball. The ball experiences free fall motion, so we can use the equations of motion to find the height. The vertical displacement can be determined using the formula: Δy = v₀y * t + (1/2) * a * t², where v₀y is the initial vertical velocity, t is the time of flight, and a is the acceleration due to gravity.
(c) To find the time it takes for the ball to reach a point 10.0 m below the level of launching, we can use the equation of motion for vertical displacement. We need to solve the equation Δy = v₀y * t + (1/2) * a * t² for time, where Δy is the vertical displacement (negative value), v₀y is the initial vertical velocity, and a is the acceleration due to gravity. The value of Δy in this case is -10.0 m, as the point is below the level of launching. By rearranging the equation and solving for t, we can find the time it takes for the ball to reach that point.
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A proton is circulating inside lhe ring or a synchrotron. The proton has an orbital fadius of 82 mand a velocity of 1.25×10 ^8m s ^−1What is the magnitude of the magnetic field required to keen the proton in this radial distance? (A) 0.0175 T (B) 0.0185 T (C) 0.0188 T (D) 0.0196 T
The correct option is (C) 0.0188 T.
A proton is moving in a circle inside the ring of a synchrotron.
The proton has an orbital radius of 82 m and a velocity of 1.25 x 108 m/s.
The magnitude of the magnetic field required to keep the proton in this radial distance is determined by the following formula: magnetic field (B) = (mass (m) x velocity (v)) / (charge (q) x radius (r)).
The proton's charge is +1.60 x 10-19 C and its mass is 1.67 x 10-27 kg.
Substituting these values into the formula,
we get: B = (1.67 x 10^-27 kg × 1.25 x 10^8 m/s) / (1.60 x 10^-19 C × 82 m)B = 0.0188 T
Therefore, the magnitude of the magnetic field required to keep the proton in this radial distance is 0.0188 T.
Hence, the correct option is (C) 0.0188 T.
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An electric heater is used to boil small amounts of water and consists of a 11−Ω coil that is immersed directly in the water. It operates from a 120−V socket. How much time is required for the heater to raise the temperature of 0.69 kg of water from 20
∘
C to the normal boiling point? Take the specific heat capacity of water to be 4186 J/(kg⋅C
∘
).
It takes approximately 177.4 seconds (or 2.96 minutes) for the heater to raise the temperature of 0.69 kg of water from 20°C to its normal boiling point.
To calculate the time required for the heater to raise the temperature of the water, we can use the formula:
Q = mcΔT
where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Mass of water (m) = 0.69 kg
Specific heat capacity of water (c) = 4186 J/(kg⋅°C)
Change in temperature (ΔT) = boiling point - initial temperature
The boiling point of water is 100°C, and the initial temperature is 20°C. So, ΔT = 100°C - 20°C = 80°C.
Now, we can calculate the heat energy required to raise the temperature:
Q = mcΔT
Q = (0.69 kg)(4186 J/(kg⋅°C))(80°C)
Q ≈ 232,346.4 J
Next, we need to calculate the power (P) of the electric heater using Ohm's Law:
P = V^2 / R
where V is the voltage and R is the resistance.
Voltage (V) = 120 V
Resistance (R) = 11 Ω
Calculating the power:
P = (120 V)^2 / 11 Ω
P ≈ 1309.09 W
The power is approximately 1309.09 Watts.
Finally, we can calculate the time (t) using the formula:
Q = Pt
t = Q / P
t = 232,346.4 J / 1309.09 W
t ≈ 177.4 seconds
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Two football players collide head-on in midair while chasing a pass. The first player has a 102.5 kg mass and an initial velocity of 6.40 m/s, while the second player has a 112 kg mass and initial velocity of -5.7 m/s. What is their velocity (in m/s) just after impact if they cling together? (Indicate the direction with the sign of your answer.)
The velocity of the players just after impact, when they cling together, is approximately 0.0764 m/s in the direction of the first player's initial velocity. We can apply the principle of conservation of momentum.
To determine the velocity of the football players just after impact, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Given:
Mass of the first player (m1) = 102.5 kg
The initial velocity of the first player (v1i) = 6.40 m/s
Mass of the second player (m2) = 112 kg
The initial velocity of the second player (v2i) = -5.7 m/s
Using the conservation of momentum equation:
(m1 * v1i) + (m2 * v2i) = (m1 + m2) * vf
Substituting the given values:
(102.5 kg * 6.40 m/s) + (112 kg * -5.7 m/s) = (102.5 kg + 112 kg) * vf
(656 kg·m/s) + (-639.6 kg·m/s) = (214.5 kg) * vf
16.4 kg·m/s = 214.5 kg·vf
Dividing both sides by 214.5 kg:
vf = 16.4 kg·m/s / 214.5 kg
vf ≈ 0.0764 m/s
Therefore, the velocity of the players just after impact, when they cling together, is approximately 0.0764 m/s in the direction of the first player's initial velocity.
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A rod 2 meters long is firmly clamped at one end. Calculate the wavelengths of the first three modes for transverse standing waves.
The wavelengths of the first three modes for transverse standing waves on a 2-meter-long rod are 4 meters, 2 meters, and approximately 1.33 meters.
The equation to calculate the wavelengths of the modes for transverse standing waves on a rod is:
λn = 2L/n
where λn is the wavelength of the nth mode, L is the length of the rod, and n is the mode number.
Using this equation, we can calculate the wavelengths of the first three modes for a rod that is 2 meters long:
Mode 1: n = 1
λ1 = 2L/n = 2(2)/1 = 4 meters
Mode 2: n = 2
λ2 = 2L/n = 2(2)/2 = 2 meters
Mode 3: n = 3
λ3 = 2L/n = 2(2)/3 ≈ 1.33 meters
Therefore, the wavelengths of the first three modes for transverse standing waves on a 2-meter-long rod are 4 meters, 2 meters, and approximately 1.33 meters.
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A lens with a focal length of -4.52 cm is placed 7.0 cm in front of an object. A second lens with a focal length of 6.0 cm is then placed on the opposite side of the first lens, with a distance of 9.42 cm between the two lenses. What is the distance between the second lens and the resulting image that it produces?
Provide your answer in cm, correct to 1 decimal place.
The distance between the second lens and the resulting image it produces is approximately 4.09 cm.
To solve this problem, we can use the lens formula, which relates the object distance (u), image distance (v), and focal length (f) of a lens:
1/f = 1/v - 1/u
Given:
Focal length of the first lens (f1) = -4.52 cm (negative sign indicates a diverging lens)
Distance of the first lens from the object (u1) = 7.0 cm
We can first find the image distance (v1) produced by the first lens using the lens formula:
1/f1 = 1/v1 - 1/u1
Substituting the values:
1/-4.52 = 1/v1 - 1/7.0
Solving for v1:
1/v1 = 1/-4.52 + 1/7.0
1/v1 = (-7.0 + 4.52) / (-4.52 * 7.0)
1/v1 = -2.48 / (-31.64)
1/v1 ≈ 0.0785
v1 ≈ 12.73 cm
The image produced by the first lens is located approximately 12.73 cm from the lens.
Now, we can consider the second lens with a focal length of 6.0 cm and the distance between the two lenses of 9.42 cm.
Using the lens formula for the second lens:
1/f2 = 1/v2 - 1/u2
Given:
Focal length of the second lens (f2) = 6.0 cm
Distance between the two lenses (d) = 9.42 cm
The object distance for the second lens (u2) is equal to the image distance from the first lens (v1).
Substituting the values:
1/6.0 = 1/v2 - 1/12.73
Solving for v2:
1/v2 = 1/6.0 + 1/12.73
1/v2 = (12.73 + 6.0) / (6.0 * 12.73)
1/v2 = 18.73 / 76.38
1/v2 ≈ 0.2447
v2 ≈ 4.09 cm
Therefore, the distance between the second lens and the resulting image it produces is approximately 4.09 cm.
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a rowing eight is moving at a constant velocity of 6m/s and coming second in its race, in order to try to win, they accelerate at a rate of 0.06t, how fast will they be rowing after accelerating at this rate for 5 seconds?
5) at the start of a race a sprinter accelerates at 2.1m/s^2, for how long will he need to sustain this rate of acceleration in order to reach his maximum speed of 4m/s?
6) an electronic toy truck accelerates at 0.6m/s^2 when switched on, how far will it travel in its first minute?
1. The rowing eight will be moving at a velocity of 7.5 m/s after accelerating at this rate for 5 seconds.
2. The sprinter will need to sustain the rate of acceleration of 2.1m/s^2 for 1.9 seconds to reach his maximum speed of 4m/s.
3. the electronic toy truck will travel a distance of 10,800 meters in its first minute.
1. Initial velocity of the rowing eight = 6 m/s
Acceleration of the rowing eight = 0.06t
Time taken by the rowing eight to accelerate = 5s
Let the final velocity of the rowing eight be Vf. We have to calculate Vf.
Vf = u + at
where u = initial velocity = 6 m/s
a = acceleration = 0.06t = 0.06(5) = 0.3 m/s^2
t = time taken to accelerate = 5s
Vf = 6 + 0.3(5)
Vf = 7.5 m/s
After accelerating at this rate for 5 seconds, the rowing eight will be moving at a velocity of 7.5 m/s.
2. Initial velocity of the sprinter = 0m/s
Acceleration of the sprinter = 2.1m/s^2
Final velocity of the sprinter = 4m/s
Let t be the time taken by the sprinter to reach his maximum speed of 4m/s.
The formula for velocity is v = u + at
4 = 0 + 2.1t
t = 4/2.1
t = 1.9 seconds
The sprinter will need to sustain the rate of acceleration of 2.1m/s^2 for 1.9 seconds to reach his maximum speed of 4m/s.
3. Initial velocity of the electronic toy truck = 0m/s
Acceleration of the electronic toy truck = 0.6m/s^2
Time taken by the electronic toy truck to travel for the first minute = 60s
Let s be the distance covered by the electronic toy truck in its first minute.
The formula for distance is d = ut + 1/2 at^2
where u = initial velocity = 0m/s
a = acceleration = 0.6m/s^2
t = time taken = 60s
Substituting the values in the above equation, we get
d = 0 + 1/2 (0.6) (60)^2
d = 10,800 m
Hence, the electronic toy truck will travel a distance of 10,800 meters in its first minute.
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Three point-like charges are placed at the corners of a rectangle as shown, a = 22.0 cm and b = 58.0 cm. Find the work done by the electric field force when the charge q2 is moved to infinity. q1 = -1.40 uC, q2 = +2.60 uC, q3 = +3.90 uC.
The work done by the electric field force when the charge q2 is moved to infinity is -1.9 J.
The electric field force is the force that acts on a charge in the presence of an electric field. The amount of work done by an electric field force to move a charge between two points in an electric field is the product of the charge and the potential difference between the two points.
Given the three point-like charges are placed at the corners of a rectangle as shown: The work done by the electric field force when the charge q2 is moved to infinity is determined by finding the potential difference between point B and infinity.
Potential difference is the difference in potential energy between two points.
Work done is given as:
W = qΔV
where
q is the charge moved
ΔV is the potential difference
q2 is moved to infinity, the charge on q2 is positive. So, the work done on q2 is negative.
Now,
The distance between B and infinity is equal to a distance between B and the midpoint of AD, which is (1/2)b.
Now,
ΔV = V_B - V_∞V_∞ = 0 as the potential at infinity is zero.
V_B = k[(q_1/d_1) + (q_3/d_3)]
Here, d_1 is the distance between B and q_1 and d_3 is the distance between B and q_3.
We can now solve for ΔV.
k = 1/4πε_0 is the Coulomb's constant and ε_0 is the permittivity of free space.
We have,
d_1 = d_3 = [(a/2)^2 + (b/2)^2]^0.5 = [3382]1/2/2 = 29.18 cm (approx).
So, V_B = (9.0×10^9 Nm^2/C^2){[-1.40 uC/(0.2918 m)] + [3.90 uC/(0.2918 m)]}V_B = 7.4×10^8 Nm^2/C^2.
The work done is:
W = q_2 ΔV = (2.60×10^-6 C)(-7.4×10^8 Nm^2/C^2) = -1.9 J.
So, the work done by the electric field force when the charge q2 is moved to infinity is -1.9 J.
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Vector
A
has a magnitude of 28.0 units and it points in a direction 310
∘
counterclockwise from the positive x-axis. What are the x - and y-components of
A
? A
x
= units A
y
= units
the x- and y-components of vector A are:A_x = 20.84 unitsA_y = -21.46 units
The given vector A has a magnitude of 28.0 units and it points in a direction of 310∘ counterclockwise from the positive x-axis.
To determine the x- and y-components of vector A, we need to use the following trigonometric ratios:cos θ = adjacent/hypotenuse and sin θ = opposite/hypotenusewhere θ is the angle between the vector and the x-axis.
Using the given information, we have:θ = 310∘ - 360∘ (since it is counterclockwise from the positive x-axis)= -50∘A = 28.0 unitsTherefore, the x-component of vector A is given by:A_x = A cos θA_x = 28.0 cos (-50∘)A_x = 20.84 units (rounded to two decimal places)The negative value of cos (-50∘) implies that vector A is in the negative x-direction (to the left of the origin).
Similarly, the y-component of vector A is given by:A_y = A sin θA_y = 28.0 sin (-50∘)A_y = -21.46 units (rounded to two decimal places)The negative value of sin (-50∘) implies that vector A is in the negative y-direction (below the origin).
Therefore, the x- and y-components of vector A are:A_x = 20.84 unitsA_y = -21.46 units
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Sketch the layout of PIM Capacitor having 1pF capacitance
To sketch the layout of a PIM capacitor with a capacitance of 1pF, we need to understand the basic structure of a capacitor and its symbols.
A capacitor consists of two conductive plates separated by an insulating material called the dielectric. The capacitance of a capacitor determines its ability to store electric charge.
To represent a capacitor in a circuit diagram, we use the symbol "C" with two parallel lines representing the conductive plates, and the value of the capacitance is usually specified next to the symbol.
In the case of a PIM (Passive Intermodulation) capacitor, the layout will depend on the specific design and application requirements. However, we can represent a generic PIM capacitor layout with the symbol mentioned above.
To sketch the layout, follow these steps:
1. Start by drawing two parallel lines representing the conductive plates of the capacitor.
2. Label one of the lines with a positive sign (+) and the other with a negative sign (-) to indicate the polarity.
3. Next to the symbol, write the value of the capacitance, which is 1pF in this case.
4. Ensure that the spacing between the lines represents the insulating material or dielectric used in the capacitor.
Here is an example of how the sketch might look:
```
_______
+ _____| |______ -
1pF
```
Please note that this is a simplified representation, and the actual layout may vary depending on the specific design and application requirements of the PIM capacitor. It is always advisable to refer to the manufacturer's datasheet or consult a professional for the accurate layout details.
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When spiking a volleyball, a player changes the Part A velocity of the ball from 4.1 m/s to −29 m/s along a certain direction. If the impulse delivered to the ball by the player is −9.3 kg⋅m/s, what is the mass of the volleyball? Express your answer using two significant figures.
The mass of the volleyball is approximately 0.281 kg.
To find the mass of the volleyball, we can use the equation for impulse: impulse = change in momentum. Impulse is given as -9.3 kg⋅m/s.
The change in momentum can be calculated by subtracting the initial momentum from the final momentum.
Momentum is the product of mass and velocity, so we can rewrite the equation as:
impulse = (mass × final velocity) - (mass × initial velocity).
We know that the initial velocity is 4.1 m/s and the final velocity is -29 m/s. Substituting these values into the equation, we get:
-9.3 = (mass × -29) - (mass × 4.1).
Simplifying, we have: -9.3 = -29mass - 4.1mass.
Combining like terms, we get: -9.3 = -33.1mass.
Now we can solve for the mass by dividing both sides of the equation by -33.1: mass = -9.3 / -33.1.
Calculating this gives us:
mass = 0.281 kg.
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Explain each of the following using the action and reaction forces: a) A hockey player bounces back off the boards after running into them. (3 marks) b) A runner can speed up. ( 3 marks) 11. Two students are standing on skateboards facing each other at rest. Student A (mass of 50.0 kg ) pushes on student B (mass of 60.0 kg ) with a force of 150 N[ W ). You may assume there is no friction. a) Draw the FBD of each skateboarder. (6 marks) b) Find the acceleration of each skateboarder. ( 6 marks) 12. A helicopter of mass 4500 kg is hovering at rest above the ground. a) Draw an FBD of the hovering helicopter. ( 2 marks) b) Calculate the forces acting on the helicopter while it is hovering. ( 4 marks) c) Explain what causes the upward force on the helicopter. (3 marks) d) Find the acceleration of the helicopter if the propeller blades apply a force of 49000 N [down] on the air.
a) A hockey player bounces back off the boards after running into them:
When a hockey player bounces back off the boards after running into them, the action force is the force the player exerts on the board, the boards exert an equal and opposite force on the player which causes the player to bounce back off the boards.
b) A runner can speed up: When a runner speeds up, the action force is the force that the runner exerts on the ground with their feet. The reaction force is the force that the ground exerts back on the runner.
Two students are standing on skateboards facing each other at rest.
Student A (mass of 50.0 kg) pushes on student B (mass of 60.0 kg) with a force of 150 N [W).
You may assume there is no friction.
a) FBD of each skateboarder:
(The diagram below shows the FBD of each skateboarder. )
b) Acceleration of each skateboarder:
For skateboarder A:
Acceleration =[tex]Force / mass = 150 N / 50 kg = 3.0 m/s^2 [W[/tex]]For skateboarder B:
Acceleration = [tex]Force / mass = 150 N / 60 kg = 2.5 m/s^2 [E][/tex]12.
A helicopter of mass 4500 kg is hovering at rest above the ground.
a) FBD of the hovering helicopter: (The diagram below shows the FBD of the hovering helicopter. )
b) Forces acting on the helicopter while it is hovering: The forces acting on the helicopter while it is hovering are the weight of the helicopter, which acts downwards, and the upward force from the rotors, which cancels out the weight of the helicopter to keep it hovering.
c) Cause of the upward force on the helicopter: The upward force on the helicopter is caused by the rotors pushing air downwards, according to Newton's third law, which states that every action has an equal and opposite B. the rotors cause an upward force on the helicopter by pushing air downwards.
d) Acceleration of the helicopter if the propeller blades apply a force of 49000 N [down] on the air:
The acceleration of the helicopter if the propeller blades apply a force of 49000 N [down] on the air can be found using Newton's second law:
Force = mass x acceleration Rearranging this equation to solve for acceleration:
Acceleration = [tex]Force / mass Acceleration = 49000 N / 4500 kg Acceleration = 10.9 m/s^2 [down][/tex]
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