Artesian wells flow freely without any pumping required because the wells are close to the groundwater recharge area and the elevation of the wells is below the elevation of the groundwater recharge.
Artesian wells are the wells that use confined aquifers to extract water to the surface. They are constructed in such a way that the underground aquifer is tapped in the high-lying area which forces the water to rise up due to the natural pressure exerted by the underground water.
This water is under pressure because the aquifer is confined between impervious rock layers.
Therefore, when the elevation of the wells is below the elevation of the groundwater recharge, and the wells are close to the groundwater recharge area, the artesian wells will flow freely without any pumping required.
This is due to the natural pressure from the underground water system, which forces the water to rise up to the surface.
Thus, the flow of artesian wells depends on the elevation of the well, its proximity to the groundwater recharge area, and the pressure exerted by the underground water system.
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A baseball player hits a fly ball that has an initial velocity for which the horizontal component is 30 m/s and the vertical component is 40 m/s. What is the speed of the ball at the highest point of its flight?
1. 50m/s
2. Zero
3. 30m/s
4. 40 m/s
The speed of the ball at the highest point is equal to the magnitude of its horizontal component of velocity, which is 30 m/s.
At the highest point of its flight, the vertical component of the ball's velocity becomes zero while the horizontal component remains unchanged. The speed of the ball at the highest point can be found by calculating the magnitude of the velocity vector.
Using the Pythagorean theorem, we can calculate the magnitude of the velocity vector:
speed = √((horizontal component)^2 + (vertical component)^2)
speed = √((30 m/s)^2 + (0 m/s)^2)
speed = √(900 m^2/s^2)
speed = 30 m/s.
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A charged particle beam (shot horizontally) moves into a region where there is a constant magnetic field of magnitude 0.00343 T that points straight down. The charged particles in the beam move in a circular path of radius 3.15 cm. If the charged particles in the beam were accelerated through a potential difference of 144 V, determine the charge to mass ratio of the charged particles in the beam. Answer in units of C/kg. 00610.0 points A simple mass spectrometer consists of an accelerating electric potential (so that ions of different mass have different velocities) and a uniform magnetic field (so that different velocity ions have different radii paths). This mass spectrometer is analyzing a beam of singly ionized unknown atoms using the following settings: - the magnetic field is 0.152 T; - the charge of an atom is 1.60218×10
−19
C; - the radius of the orbit is 0.0863 m; and - the potential difference is 179 V. Calculate the mass of the unknown atom. Answer in units of kg
The mass of the unknown atom is 4.65 x 10⁻²³ kg.
The equation for the motion of a charged particle in a magnetic field is given by the equation,
F = Bqv
where:
F = force experienced by the charged particle,
B = magnetic field strength,
q = charge on the particle,
v = velocity of the particle perpendicular to the magnetic field,
r = radius of curvature of the path of the particle, and
m = mass of the charged particle
The beam of charged particles moves in a circular path of radius r = 3.15 cm.
Thus the equation for the radius of the path can be given by,
mv²/r = Bqv
The potential difference of 144 V accelerates the charged particles, which gives them an initial kinetic energy of 144 eV. This can be written as the product of the charge and potential difference,
KE = eΔV
where:
e = charge of an electron and ΔV = potential difference
Thus,
KE = (1.602 x 10⁻¹⁹ C)(144 V)KE = 2.3 x 10⁻¹⁶ Joules
Using the equation of conservation of energy,1/2mv² = eΔVand substituting the value of the velocity of the charged particle from this equation into the first equation, the charge to mass ratio of the charged particles in the beam can be found.
mv²/r = Bqv
Where,
m/e = v/(Br)Charge to mass ratio of the charged particles in the beam can be given by,
e/m = Br²/v
Substitute the given values in the above equation,0.00343 T × (0.0315 m)² = (1.602 x 10⁻¹⁹ C)/(e/m)
Thus,
e/m = 1.7589 x 10¹¹ C/kg
Now, for the second question, the mass of the unknown atom can be found using the equation,
m/e = B²r²/2V
Where,
m = mass of the unknown atom,
e = charge of the unknown atom,
B = magnetic field strength,
r = radius of the path of the unknown atom
V = potential difference
The charge of an atom is 1.60218 x 10⁻¹⁹ C, and the magnetic field is 0.152 T.
The radius of the orbit is 0.0863 m, and the potential difference is 179 V. Substituting these values in the above equation,
m/e = (0.152² x 0.0863²)/(2 x 179)
Thus,
m/e = 2.902 x 10⁻⁴ kg/C
The mass of the unknown atom can be calculated using,
m = e(m/e)
Substituting the known values,
m = (1.602 x 10⁻¹⁹ C)(2.902 x 10⁻⁴ kg/C)
Thus,
m = 4.65 x 10⁻²³ kg
Therefore, the mass of the unknown atom is 4.65 x 10⁻²³ kg.
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Light of wavelength 630 nm falls on two slits and produces an Part A interference pattern in which the third-order bright fringe is 40 mm from the central fringe on a screen 3.0 m away. What is the separation of the two slits? Express your answer using two significant figures. X Incorrect; Try Again; 4 attempts remaining
The separation of the two slits is 0.0158 mm (approx).
Given,Light of wavelength 630 nm falls on two slits and produces an interference pattern in which the third-order bright fringe is 40 mm from the central fringe on a screen 3.0 m away.Concept used,The formula for calculating the separation between the two slits is given by: dsinθ = m λ where, m = 1,2,3,...θ = angle of diffractiond = separation between the slitsλ = wavelength of the lightThe path difference between the waves of light emerging from two slits is given by: d sinθ = mλ where, d = separation between the slitsλ = wavelength of lightθ = angle of diffraction.The third order fringe means m = 3. Hence, d sinθ = 3λHere, λ = 630 nm = 6.3 × 10⁻⁷ m. Therefore,3d sinθ = 6.3 × 10⁻⁷ ...................(1)
We know that angle of diffraction, θ can be given by, θ = tan⁻¹(y/L) where y is the fringe width and L is the distance between the screen and the slits. Here, y = 40 mm = 0.04 mL = 3.0 m Therefore, θ = tan⁻¹ (0.04/3)Now, substitute this value of θ in equation (1), we get:3d × (0.04/3) = 6.3 × 10⁻⁷Or, d = (6.3 × 10⁻⁷ )/0.04The value of d is, d = 1.58 × 10⁻⁵ m or 0.0158 mm (approx).Hence, the separation of the two slits is 0.0158 mm (approx).
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"Three forces are applied to an object with a mass of 2 kg. These
are the only forces acting on the object. F1 has a magnitude of 6N
and points directly to the left. F2 has a magnitude of 5N and
points"
The net force acting on the object is 1 N to the left.
Three forces are applied to an object with a mass of 2 kg. These are the only forces acting on the object. F1 has a magnitude of 6N and points directly to the left. F2 has a magnitude of 5N and points directly upwards. F3 has a magnitude of 4N and points directly downwards.
The first step is to resolve the forces in the horizontal direction since F1 acts horizontally.
Using Pythagoras theorem, we can find that the horizontal component of F2 and the horizontal component of F3 are equal to 3 N. Therefore, the total horizontal force acting on the object is 3N to the left (6N-3N=3N).
Hence, the net force acting on the object is 1 N to the left (3N-2N=1N).
Therefore, the object will move towards the left direction with an acceleration of 0.5 m/s² which is determined using the formula F=ma, where F is the net force acting on the object, m is the mass of the object and a is the acceleration of the object.
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Relative Velocity Sidewalk moves with a steady velocity of 1.1 km/h. How long (in s ) does it take for the passenger to get from one end of the sidewalk to the other, that i to cover the 100 m ? Tries 0/10 How much time does the passenger save by taking the moving sidewalk instead of just walking beside it? Tries 0/10 Through what distance does the passenger walk relative to the moving sidewalk? Tries 0/10 If the passenger's stride is 85 cm, how many steps are taken in going from one end of the moving sidewalk to the other? Tries 0/10
In this scenario of relative velocity, the passenger takes 32.73 seconds to traverse the 100 m distance on the moving sidewalk. By utilizing the sidewalk, the passenger saves 23.84 seconds compared to walking beside it. The passenger walks a distance of 30.92 m relative to the moving sidewalk, and completes this journey in approximately 118 steps, given their stride length of 85 cm. This analysis demonstrates the application of kinematic equations and concepts of relative motion in physics.
Relative Velocity
A passenger walks on a moving sidewalk with a constant speed of 1.1 km/h. The following is required to find:
How long (in seconds) does it take for the passenger to get from one end of the sidewalk to the other, i.e., to cover the 100 m?
The time it will take for the passenger to get from one end of the sidewalk to the other is given as follows:
Velocity = Distance / Time
1.1 km/h = 100 m / Time
Time = 100 / (1.1 × 1000 / 3600)
Time = 32.73 s
Therefore, it will take the passenger 32.73 seconds to get from one end of the sidewalk to the other.
How much time does the passenger save by taking the moving sidewalk instead of just walking beside it?
The velocity of the moving sidewalk is 1.1 km/h. Therefore, if the passenger walks beside it, his velocity will be 4.5 km/h. Let's calculate the time it would take for the passenger to travel 100 m at this velocity.
Velocity = Distance / Time
4.5 km/h = 100 m / Time
Time = 100 / (4.5 × 1000 / 3600)
Time = 8.89 s
The time it will take for the passenger to travel 100 m if he walks beside the moving sidewalk is 8.89 s. Therefore, he saves:
Time Saved = Time Walking - Time on Moving Sidewalk
Time Saved = 32.73 - 8.89
Time Saved = 23.84 s
Therefore, the passenger saves 23.84 seconds by taking the moving sidewalk instead of just walking beside it.
Through what distance does the passenger walk relative to the moving sidewalk?
The velocity of the passenger is 4.5 km/h, while the velocity of the moving sidewalk is 1.1 km/h. Therefore, the velocity of the passenger relative to the moving sidewalk is given as follows:
Relative Velocity = Velocity of Passenger - Velocity of Sidewalk
Relative Velocity = 4.5 - 1.1
Relative Velocity = 3.4 km/h = 0.944 m/s
The distance covered by the passenger relative to the moving sidewalk is given as follows:
Distance = Velocity × Time
Distance = 0.944 m/s × 32.73 s
Distance = 30.92 m
Therefore, the passenger walks a distance of 30.92 m relative to the moving sidewalk.
If the passenger's stride is 85 cm, how many steps are taken in going from one end of the moving sidewalk to the other?
The distance that the passenger walks is 100 m, and his stride length is 85 cm or 0.85 m. Therefore, the number of steps he takes is given as follows:
Number of Steps = Distance Walked / Stride Length
Number of Steps = 100 / 0.85
Number of Steps = 117.64 ≈ 118
Therefore, the passenger takes 118 steps while going from one end of the moving sidewalk to the other.
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A car is travelling at 100 km/hr to the right and slows down to 65 km/hr in 10 seconds. What is the acceleration? How far does it travel in those 10 seconds?
The acceleration of the car is calculated to be -9.72 m/s². And the car travels a distance of 229.2 meters in those 10 seconds.
To calculate the acceleration, we use the formula:
acceleration = (change in velocity) / (change in time).
The change in velocity is the final velocity minus the initial velocity:
change in velocity = 65 km/hr - 100 km/hr = -35 km/hr.
The change in time is given as 10 seconds.
Plugging these values into the formula, we have:
acceleration = (-35 km/hr) / (10 s).
To convert km/hr to m/s, we divide by 3.6:
acceleration = (-35 km/hr) / (10 s) × (1 km / 3600 s) = -9.72 m/s².
Therefore, the acceleration of the car is -9.72 m/s².
To find the distance traveled, we use the formula:
distance = average velocity × time.
The average velocity is the sum of the initial and final velocities divided by 2:
average velocity = (100 km/hr + 65 km/hr) / 2 = 82.5 km/hr.
Converting km/hr to m/s, we divide by 3.6:
average velocity = 82.5 km/hr × (1 km / 3600 s) = 22.92 m/s.
Plugging in the values, we have:
distance = 22.92 m/s × 10 s = 229.2 meters.
Therefore, the car travels a distance of 229.2 meters in those 10 seconds.
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A periodic sawtooth train with fundamental period T
0
, amplitude A, and pulse duration τ
0
, over the fundamental period is expressed as x
1
(t)={
τ
A
(t+
2
τ
)
0
−
2
τ
≤t<
2
τ
,
otherwise
τ
0
Based on the analysis equation, derive the coefficients of the complex exponential Fourier series of the sawtooth train. (Note: The answer for the FS coefficients can be found in the Table 3.2 of the lecture notes.) (15 marks) (b) Consider the periodic signal x
2
(t) with fundamental period T
0
, whose expression over the period is given as x
2
(t)=
⎩
⎨
⎧
τ
2A
(t+
2
τ
)
−
τ
2A
(t−
2
τ
)
0
−
2
τ
≤t<0
0≤t<
2
τ
otherwise
Express x
2
(t) in terms of x
1
(t) and then use the properties of the Fourier series, or otherwise, to find the Fourier series coefficients of the periodic signal x
2
(t). (10 marks) (c) Find the Fourier series of the periodic signal x
3
(t) with period 6 . x
3
(t)=
⎩
⎨
⎧
2(t+2)
2
−2(t−2)
0
−2≤t<−1
−1≤t<1
1≤t<2
otherwise
To derive the coefficients of the complex exponential Fourier series of the sawtooth train x1(t), we need to express it as a sum of complex exponential functions.
First, let's determine the period of x1(t) from the given information.
The fundamental period T0 of the sawtooth train is the duration between two consecutive pulse starts, which is 2τ.
The Fourier series representation of x1(t) can be written as:
[tex]x1(t) = Σ Cn * e^(jnω0t)[/tex]
where Cn are the Fourier series coefficients, [tex]ω0 = 2π/T0[/tex]is the fundamental angular frequency, and [tex]j = √(-1)[/tex] is the imaginary unit.
To find the coefficients Cn, we need to calculate the integral of x1(t) multiplied by the complex conjugate of [tex]e^(jnω0t)[/tex]over a period T0.
Let's compute this integral step by step for the given expression of x1(t):
[tex]∫[τA(t+2τ)] * e^(-jnω0t) dt, -2τ ≤ t < 2τ[/tex]
We can simplify the expression by shifting the variable of integration, let's define [tex]u = t + 2τ[/tex]:
[tex]∫[τA(u)] * e^(-jnω0(u-2τ)) du, 0 ≤ u < 4τ[/tex]
Now, we can expand the exponential term using Euler's formula:
[tex]∫[τA(u)] * [cos(nω0u) - jsin(nω0u)] * e^(jn2π) du, 0 ≤ u < 4τ[/tex]
Since the interval of integration is one period T0, the cosine term integrates to zero:
[tex]∫[τA(u)] * [-jsin(nω0u)] * e^(jn2π) du, 0 ≤ u < 4τ[/tex]
Next, we can distribute the j and the e^(jn2π) term:
[tex]-j * e^(jn2π) * ∫[τA(u) * sin(nω0u)] du, 0 ≤ u < 4τ[/tex]
The term e^(jn2π) is equal to 1, so it can be omitted:
[tex]-j * ∫[τA(u) * sin(nω0u)] du, 0 ≤ u < 4τ[/tex]
Finally, we substitute back t for u:
[tex]-j * ∫[τA(t+2τ) * sin(nω0(t+2τ))] dt, -2τ ≤ t < 2τ[/tex]
The integral of the product of a sine function and a periodic function over one period is zero, so the integral evaluates to zero.
Therefore, for[tex]-2τ ≤ t < 2τ[/tex], the Fourier series coefficient Cn is zero.
For all other values of t, the value of x1(t) is τ0.
Hence, for [tex]-2τ ≤ t < 2τ[/tex], the Fourier series coefficient C0 is τ0.
In summary, the coefficients of the complex exponential Fourier series of the sawtooth train x1(t) are:
[tex]C0 = τ0 (for -2τ ≤ t < 2τ)[/tex]
[tex]Cn = 0 (for all n ≠ 0)[/tex]
The actual coefficients may vary depending on the specific values of τ0 and A.
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The thin plastic rod shown in the figure has length L=11.0 cm and a nonuniform linear charge density λ=cx,wherec=40.6pC/m
2
. Nith V=0 at infinity, find the electric potential at point P
1
on the axis, at distance d=4.10 cm from one end.
The electric potential at point P₁, located at a distance of 4.10 cm from one end of the rod, can be determined by integrating the contributions from all infinitesimally small elements of the rod.
To find the electric potential at point P₁ on the axis, we can use the principle of superposition. We need to consider the contribution to the potential from each infinitesimally small element of the rod and integrate over the entire length.
The electric potential due to an infinitesimally small element of length dx at a distance x from P₁ is given by dV = k * λ * dx / r, where k is the electrostatic constant and r is the distance from the element to P₁.
The linear charge density λ = cx, where c = 40.6 pC/m². Therefore, λ = 0.406x nC/m².
The distance from the element to P₁ is r = sqrt(x² + d²), where d = 4.10 cm = 0.041 m.
The electric potential at P₁ is obtained by integrating the contributions from all the elements:
V = ∫(k * λ * dx / r) from x = 0 to x = L.
V = ∫(k * 0.406x * dx / sqrt(x² + d²)) from x = 0 to x = L.
Solving this integral will give us the electric potential at point P₁.
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ignore the friction force, and determine the acceleration of the barge when each donkey exerts a forme of 408 N on a cable. m/s
2
The acceleration of the barge is directly proportional to the net force and inversely proportional to the mass. The acceleration of the barge is
816 N / m, where m is the mass of the barge.
The net force on the barge is equal to the force exerted by each donkey, so the net force is 2 * 408 N = 816 N.
The mass of the barge is not given, so we can't calculate the acceleration directly. However, we can say that the acceleration is directly proportional to the net force and inversely proportional to the mass.
If we let the acceleration be represented by the variable a, we can write the following equation:
a = 816 N / m
where m is the mass of the barge.
We can't solve this equation for m, but we can say that the acceleration of the barge is 816 N / m.
In other words, the acceleration of the barge depends on the mass of the barge. If the mass of the barge is larger, the acceleration will be smaller. If the mass of the barge is smaller, the acceleration will be larger.
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The impulse response of an LTI filter is given by h(t)=2e
−2t
u(t). (a) Determine the unit step response for this filter, that is find s(t) as the output of the filter when the input is u(t). (b) Determine the output, y(t), of the filter for an input x(t)=u(t+1)−u(t−3).
a. The unit step response s(t) for this filter is: s(t) = -e^(-2t) + 1, for t ≥ 0
b. The output y(t) of the filter for the input x(t) = u(t+1) - u(t-3) is:
y(t) = -e^(-2(t-1)) + 1 - (-e^(-2(t-3)) + 1), for t ≥ 0.
The impulse response of an LTI (Linear Time-Invariant) filter is given by h(t) = 2e^(-2t) u(t), where u(t) is the unit step function.
(a) To determine the unit step response for this filter, we need to convolve the impulse response h(t) with the unit step function u(t). The convolution operation is denoted by *, and it is defined as:
s(t) = h(t) * u(t)
In this case, h(t) = 2e^(-2t) u(t) and u(t) = u(t), so the convolution becomes:
s(t) = (2e^(-2t) u(t)) * u(t)
To perform the convolution, we need to integrate the product of h(t) and u(t) over the range from 0 to t:
s(t) = ∫[0,t] (2e^(-2τ) u(τ)) dτ
The unit step function u(τ) is 1 for τ >= 0 and 0 for τ < 0. Therefore, we can simplify the integral by considering two cases:
1. For 0 ≤ τ ≤ t:
s(t) = ∫[0,t] (2e^(-2τ)) dτ
= -e^(-2τ) | [0,t]
= -e^(-2t) + 1
2. For τ > t:
s(t) = ∫[0,t] (2e^(-2τ) u(τ)) dτ + ∫[t,∞] (2e^(-2τ) u(τ)) dτ
= ∫[0,t] (2e^(-2τ)) dτ + ∫[t,∞] 0 dτ
= -e^(-2τ) | [0,t] + 0
= -e^(-2t) + 1
Therefore, the unit step response s(t) for this filter is:
s(t) = -e^(-2t) + 1, for t ≥ 0
(b) To determine the output y(t) of the filter for the input x(t) = u(t+1) - u(t-3), we need to convolve the input signal x(t) with the impulse response h(t):
y(t) = x(t) * h(t)
Substituting the given values of x(t) and h(t) into the convolution equation, we have:
y(t) = (u(t+1) - u(t-3)) * (2e^(-2t) u(t))
Expanding the convolution and simplifying, we can split the integral into two parts:
y(t) = ∫[0,t] (2e^(-2τ) u(t+1-τ)) dτ - ∫[0,t] (2e^(-2τ) u(t-3-τ)) dτ
Considering two cases again:
1. For 0 ≤ τ ≤ t-1:
y(t) = ∫[0,t-1] (2e^(-2τ)) dτ
= -e^(-2τ) | [0,t-1]
= -e^(-2(t-1)) + 1
2. For 0 ≤ τ ≤ t-3:
y(t) = ∫[0,t-3] (2e^(-2τ)) dτ
= -e^(-2τ) | [0,t-3]
= -e^(-2(t-3)) + 1
Therefore, the output y(t) of the filter for the input x(t) = u(t+1) - u(t-3) is:
y(t) = -e^(-2(t-1)) + 1 - (-e^(-2(t-3)) + 1), for t ≥ 0.
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"If a moving object experiences a net zero unbalanced force, then the body:" A. cannot remain at rest B. can be accelerated C. moves with a constant velocity D. will always remain at rest
If a moving object experiences a net zero unbalanced force, it will move with a constant velocity.
If a moving object experiences a net zero unbalanced force, it means that the forces acting on the object are balanced and cancel each other out. In this case, according to Newton's first law of motion, the object will continue to move with a constant velocity. This means that if the object was initially moving, it will keep moving at the same speed and in the same direction without any change in its motion unless acted upon by an external force.
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Let's make another prediction. Suppose that your car is initially at 30 m mark, moving with initial velocity V_i = 10 m/s. If a constant acceleration of −2.0 m/s2 is applied, what would be the car's velocity when it reaches the 25 m mark? Show all your work below. V(25 m)= How long would it take to reach the 25 m mark? Show all your work below.
The time taken to reach the 25 m mark ≈ 0.47 seconds.
find the car's velocity when it reaches the 25 m mark, we can use the kinematic equation:
[tex]V_f^2 = V_i^2[/tex] + 2aΔx
where V_f is the final velocity, V_i is the initial velocity, a is the acceleration, and Δx is the displacement.
V_i = 10 m/s
a = -2.0 [tex]m/s^2[/tex] (negative because it is deceleration)
Δx = 25 m - 30 m = -5 m (negative because the displacement is in the opposite direction of motion)
Plugging in these values into the equation, we have:
[tex]V_f^2 = (10 m/s)^2 + 2(-2.0 m/s^2)(-5 m)[/tex]
[tex]V_f^2 = 100 m^2/s^2 + 20 m^2/s^2[/tex]
[tex]V_f^2 = 120 m^2/s^2[/tex]
Taking the square root of both sides, we get:
[tex]V_f = \sqrt {(120 m^2/s^2)[/tex]
[tex]V_f = 10.95 m/s[/tex]
The car's velocity when it reaches the 25 m mark is approximately 10.95 m/s.
find the time it takes to reach the 25 m mark, we can use the equation:
Δx = V_i * t + (1/2) * a *[tex]t^2[/tex]
Plugging in the known values:
-5 m = (10 m/s) * t + (1/2) *[tex](-2.0 m/s^2) * t^2[/tex]
Simplifying the equation:
-5 m = 10 m/s * t -[tex]1.0 m/s^2 * t^2[/tex]
Rearranging the equation to form a quadratic equation:
[tex]1.0 m/s^2 * t^2[/tex] - 10 m/s * t - 5 m = 0
Solving this quadratic equation, we get two possible solutions:
t ≈ 0.47 s or t ≈ 10.53 s
Since time cannot be negative, the car would take 0.47 seconds to reach the 25 m mark.
The V(25 m) ≈ 10.95 m/s
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A 500g meter stick has a 2kg mass stuck to the end of it so that the center of the 2kg mass is at the end of the stick as shown. How far from the left end of the meter stick is the center of mass of this system in centimeters?
Hint: How far from one end is the center of mass of a meter stick?
The center of mass of this system is 1.1 meters from the left end of the meter stick. Since there are 100 centimeters in a meter, the center of mass is 110 centimeters from the left end.
The center of mass of a uniform meter stick is located at its midpoint. since a 2 kg mass is stuck to the end of the meter stick, the center of mass of the system will be shifted towards the end with the 2 kg mass.
Mass of the meter stick (m₁) = 500 g = 0.5 kg
Mass stuck to the end (m₂) = 2 kg
the position of the center of mass (x) of the system, we can use the formula:
x = (m₁ * d₁ + m₂ * d₂) / (m₁ + m₂)
where d₁ is the distance from the left end of the meter stick to its center of mass, and d₂ is the distance from the left end to the center of mass of the 2 kg mass (which is the length of the meter stick).
Since the center of mass of the meter stick is at its midpoint, we know that d₁ = 0.5 m.
Substituting the values into the formula, we have:
x = (0.5 kg * 0.5 m + 2 kg * 1 m) / (0.5 kg + 2 kg)
x = (0.25 kg + 2 kg) / 2.5 kg
x ≈ 1.1 m
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The starter motor of a car engine draws a current of 180 A from the battery. The copper wire to the motor is 4.60 mm in diameter and 1.2 m long. The starter motor runs for 0.940 s until the car engine starts. How much charge passes through the starter motor? Express your answer with the appropriate units. Part B How far does an electron travel along the wire while the starter motor is on? Express your answer with the appropriate units. X Incorrect; Try Again; 3 attempts remaining
The charge passing through the starter motor is 32.4 C (coulombs), and an electron travels approximately 0.59 cm (centimeters) along the wire during the operation of the starter motor.
To calculate the charge passing through the starter motor, we can use the formula Q = I * t, where Q represents the charge, I is the current, and t is the time. In this case, the current drawn by the starter motor is 180 A, and it runs for 0.940 s. Plugging these values into the formula, we get Q = 180 A * 0.940 s = 169.2 C. Therefore, approximately 169.2 C or 32.4 C of charge passes through the starter motor.
To find the distance an electron travels along the wire, we need to calculate the length of the wire. The wire's diameter is given as 4.60 mm, and we can use the formula for the circumference of a circle, C = π * d, where C is the circumference and d is the diameter. Substituting the given value, we find C = π * 4.60 mm = 14.45 mm. Converting mm to cm, we get C ≈ 1.445 cm. Since the electron travels along the wire's length, which is 1.2 m or 120 cm, the distance the electron travels is approximately 1.445 cm * (120 cm / 1.445 cm) = 0.59 cm. Therefore, during the operation of the starter motor, an electron travels approximately 0.59 cm along the wire.
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A real object in air is 50 cm away from a lens with a focal power of +5.00. D. What is the image vergence? −3.00D 0 −2.00 D 0 +3.00D 0 +7.00D
Vergence is the degree to which light rays are concentrated at the focal point, which is a physical quantity measured in diopters.
The image vergence is the vergence of light rays that are parallel to the axis of a lens that converge onto the lens and then leave it again. How do you determine the image vergence? The image vergence is determined by the formula:
V′ = V − D where V = the vergence of light incident on the lens and D = the power of the lens in diopters.
Since the object is real, it is located on the opposite side of the lens from the observer, and its image is formed on the same side as the observer. The distance between the lens and the real object is d = -50 cm since it is located on the opposite side of the lens.
The power of the lens in diopters is P = +5.00D. In this case, we have a positive power lens since it is a converging lens. Therefore, we need to use the formula:
V′ = V − D Where, V = the vergence of light incident on the lens and
D = the power of the lens in diopters V = 1/d V = 1/-50 cm V = -0.02 D
Now, we'll substitute the values in the equation: V′ = V − D⇒ V′ = -0.02 - 5⇒ V′ = -5.02D
The image vergence is -5.02 D. Answer: The correct option is -5.02 D.
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Vector
A
has a magnitude of 79 units and points due west, while vector
B
has the same magnitude and points due south. Specify the directions relative to due west. Find the magnitude and direction of
A
−
B
.
The magnitude of A-B is 111.68 units and the direction of A-B is 360° or due west direction.
Relative direction of A: A points towards west side so, it makes 270° with North and its direction is 270° west.
Relative direction of B:B points towards south side so, it makes 180° with North and its direction is 180° south.
The direction of A-B will be such that its direction will be 270° + θ, where θ is the angle between vector A and vector B.
θ is the angle between vector A and B.
Hence it can be given asθ = 180° - 90°θ = 90°
Magnitude of A is 79 Magnitude of B is 79
To find A-B, we can use the head to tail rule of vector addition.
Now the direction of A-B = 270° + θ = 270° + 90° = 360°
The magnitude of A-B is given by[tex]|A-B| = \sqrt{(|A|^{2} +|B|^{2} - 2|A||B| cos(\theta))} =[/tex][tex]\sqrt{(79^{2} + 79^{2} - 2(79)(79)cos(90))} = \sqrt{(2(79)^{2} )} = \sqrt{(2)}\times 79= 111.68[/tex]
So, the magnitude of A-B is 111.68 units and the direction of A-B is 360° or due west direction.
Hence, the answer is:A-B has a magnitude of 111.68 units and points due west.
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Astronomical observations of our Milky Way galaxy indicate that it has a mass of about 8×10
11
solar masses, A star orbiting near the galaxy's periphery is 5.9×10
4
light years from its center. (For your calculations, assume that the galaxy's mass is concentrated near its center.) (a) What should the orbital period of that star be? ve yr (b) If its period is 6.5×10
7
years instead, what is the mass of the galaxy? Such calculations are used to imply the existence of "dark matter" in the universe and have indicated, for example, the existence of very massive black holes at the centers of some galaxies. solar masses:
a) The orbital period of that star is 1.10×10 8 yr.
b) The mass of the galaxy is 2.10×10 12 solar masses.
a) Mass of the Milky Way galaxy = 8×10 11 solar masses.
Distance of star from the center of the galaxy, r = 5.9×10 4 light-years.
Force of attraction between the star and the galaxy,
F = GMm/r ²
Here,
M = mass of the galaxy,
m = mass of the star,
r = distance between the star and the galaxy,
G = gravitational constant
Orbital speed,
v = 2πr/T,
where
T is the orbital period of the star
Using the Third Law of Kepler,
T²/R³= 4π²/GM --------(1)
Where
R is the distance of star from the center of the galaxy
T² = (4π²/GM)×R³ = (4π²/GM)(5.9×10 4 × 9.46×10 15 )³ yr ²...[putting R = 5.9×10 4 light-years = 5.9×10 4 × 9.46×10 15 m]T² = (4π²/GM)(2.09×10 41 ) yr ²
T² = 1.23×10 21 (M/M☉) yr ²
On comparing this with the standard formula,
T² = (4π²/GM)R³
We get,
T² = R³ × (M/M☉) × 1.51×10 - 8 yr ²
We know that for the Sun,
M = M☉ and T = 1 year
So,1 year = R³ × 1.51×10 - 8 yr ²
1 year = R³ × 1.51×10 - 8 yr ²
T = (5.9×10 4 × 9.46×10 15 )³ × 1.51×10 - 8 yr
T = 1.10×10 8 yr
(b) We have,
T² = R³ × (M/M☉) × 1.51×10 - 8 yr ²
T² = (6.5×10 7 )² yr ²
R³ = (5.9×10 4 × 9.46×10 15 )³ m ³
On substituting these values, we get
(6.5×10 7 )² yr ²= (5.9×10 4 × 9.46×10 15 )³ × (M/M☉) × 1.51×10 - 8 yr ²
M = (6.5×10 7 )²/[(5.9×10 4 × 9.46×10 15 )³ × 1.51×10 - 8 ] × M☉
M = 2.10×10 12 solar masses.
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A 10 pole, three phase alternator has 60 slots. Each coil spans 5 slots. If the winding used is half-coil calculate the number of coils per phase.
For a 10-pole, three-phase alternator with 60 slots, each coil spanning 5 slots, and a half-coil winding, there are 12 coils per phase.
The number of coils per phase in a three-phase alternator can be calculated by considering the number of poles and the number of slots. In this case, we have a 10-pole alternator with 60 slots. Each coil spans 5 slots, and the winding used is half-coil.
To calculate the number of coils per phase, we can use the formula:
Number of coils per phase = (Number of slots) / (Number of slots spanned by each coil)
Given that each coil spans 5 slots, we can substitute this value into the formula:
Number of coils per phase = 60 / 5
Simplifying the equation:
Number of coils per phase = 12
Therefore, there are 12 coils per phase in this three-phase alternator.
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According to Equation 20.7, an ac voltage V is given as a function of time t by V=V
o
sin2πft, where V
0
is the peak voltage and f is the frequency (in hertz). For a frequency of 47.7 Hz, what is the smallest value of the time at which the voltage equals one-half of the peakvalue?
The smallest value of time at which the voltage equals one-half of the peak value, for a frequency of 47.7 Hz, is approximately 0.0105 seconds.
According to Equation 20.7, the voltage V as a function of time t is given by V = V0 sin(2πft), where V0 is the peak voltage and f is the frequency. We want to find the smallest value of time at which the voltage equals one-half of the peak value. In other words, we need to solve the equation V = (1/2)V0 for t.
Substituting the given frequency f = 47.7 Hz into the equation, we have:
(1/2)V0 = V0 sin(2π(47.7)t)
Dividing both sides of the equation by V0, we get:
1/2 = sin(2π(47.7)t)
To find the smallest value of time at which the equation is satisfied, we can take the inverse sine (sin^(-1)) of both sides:
sin^(-1)(1/2) = 2π(47.7)t
Simplifying further, we have:
t = sin^(-1)(1/2) / (2π(47.7))
Using a calculator, we can evaluate sin^(-1)(1/2) to be approximately 30 degrees or π/6 radians.
Plugging in this value, we get:
t ≈ (π/6) / (2π(47.7))
Simplifying, we find:
t ≈ 1 / (2(47.7))
t ≈ 0.0105 seconds
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A ball thrown straight up with an initial velocity of +12 m/s. Find its position, velocity, and acceleration at (A) 1.0 s, (B) the maximum height, (C) 2.0 s, and (D) the moment right before it is caught at the same height it was thrown from. d=v
u
t+0.5gt
2
Vf=v
a
+gt where g=−10 m/s/s 13. For the ball in the previous problem, how much time does it take to reach the maximum height? 14. Make a table of the velocities of an object at the end of each second for the first 5 s of free-fall from rest. Assume in this problem that down is the positive direction. Therefore your velocities will all be positive. a. Use the data in your table to plot a velocity time graph below in the grid shown. b. What does the total area under the curve represent? 12. A ball thrown straight up with an initial velocity of +12 m/s. Find its position, velocity, and acceleration at (A) 1.0 s, (B) the maximum height, (C) 2.0 s, and (D) the moment right before it is caught at the same height it was thrown from. d=v
u
t+0.5gt
2
Vf=v
a
+gt where g=−10 m/s/s 13. For the ball in the previous problem, how much time does it take to reach the maximum height? 14. Make a table of the velocities of an object at the end of each second for the first 5 s of free-fall from rest. Assume in this problem that down is the positive direction. Therefore your velocities will all be positive. a. Use the data in your table to plot a velocity time graph below in the grid shown. b. What does the total area under the curve represent?
A ball is thrown straight upwards with an initial velocity of +12 m/s. The velocity, position, and acceleration at different times can be calculated as shown below: At time, t = 1 s, v = u + at= 12 - 10(1) = 2 m/s
A ball is thrown straight upwards with an initial velocity of +12 m/s. The velocity, position, and acceleration at different times can be calculated as shown below: At time, t = 1 s, v = u + at= 12 - 10(1) = 2 m/s
The final velocity,[tex]v_f = v_i[/tex] + at= 12 - 10(1) = 2 m/s
The displacement, s = [tex]v_i[/tex]* t + (1/2) * a * t²= 12(1) + (1/2)(-10)(1)²= 7 m
At the maximum height, v = 0. Therefore, t = [tex]v_f[/tex]/g= 2/-10= 0.2 s
The displacement, s = [tex]v_i[/tex]* t + (1/2) * a * t²= 12(0.2) + (1/2)(-10)(0.2)²= 1.2 m
At time, t = 2 s, v = [tex]v_i[/tex]+ at= 12 - 10(2) = -8 m/s
The final velocity, [tex]v_f = v_i[/tex] + at= 12 - 10(2) = -8 m/s
The displacement, s = [tex]v_i[/tex] * t + (1/2) * a * t²= 12(2) + (1/2)(-10)(2)²= 2 m
At the moment right before it is caught at the same height, v = 0. Therefore, t = [tex]v_f[/tex] /g= -12/-10= 1.2 s
The displacement, s = [tex]v_i[/tex] * t + (1/2) * a * t²= 12(1.2) + (1/2)(-10)(1.2)²= 7.2 m
The velocity of the object at the end of each second for the first 5 s of free-fall from rest can be calculated as shown below: Time, t (s)Velocity, v (m/s)10+0=001+(-10)=9-19+(-10)=8-27+(-10)=7-35+(-10)=6
a) The velocity-time graph is shown below: b) The total area under the curve represents the displacement of the object.
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A rocket has 13653 N of propulsion and experience a constant kinetic friction of 9206 N. The rocket accelerates at a rate of 14 m/s/s. What is the mass of the rocket in kg?
To find the mass of the rocket, we can use Newton's second law of motion, the mass of the rocket is approximately 317.64 kg.
To find the mass of the rocket, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration:
F = m * a
Given:
Propulsion force (F_propulsion) = 13653 N
Kinetic friction force (F_friction) = 9206 N
Acceleration (a) = 14 m/s²
The net force acting on the rocket can be calculated by subtracting the kinetic friction force from the propulsion force:
Net force (F_net) = F_propulsion - F_friction
Substituting the given values:
F_net = 13653 N - 9206 N
= 4447 N
Now, we can use Newton's second law to find the mass (m):
F_net = m * a
4447 N = m * 14 m/s²
Dividing both sides of the equation by 14 m/s²:
m = 4447 N / 14 m/s²
m ≈ 317.64 kg
Therefore, the mass of the rocket is approximately 317.64 kg.
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The ball is released 2.2 m above the ground at 60.0 degree angle to the horizontal. when the ball leaves his foot it has a speed of 79 km/h. If the midfielder is directly in front of him when the ball drops to the ground how many meters down the field is the midfielder located
The midfielder is located 23.6 meters down the field.Given:Initial velocity of the ball, u = 79 km/h = 79×(5/18) m/s = 22.05 m/sHeight at which the ball is released, h = 2.2 mAngle of projection of the ball, θ = 60°From the given data, we can calculate the horizontal and vertical components of the velocity of the ball:Horizontal component, u_x = u cos θVertical component, u_y = u sin θ
We can use these components of velocity and the acceleration due to gravity, g = 9.8 m/s², to find the time of flight, t, and the range, R, of the ball:Vertical displacement of the ball, y = hRange of the ball, R = u_x tHorizontal displacement of the ball, x = R = u_x tVertical component of velocity at the highest point, u_y' = 0From the first equation of motion, we can find the time of flight of the ball:Vertical displacement of the ball,
y = u_y t - (1/2) g t²y
= h = (u sin θ) t - (1/2) g t²h
= (22.05 sin 60°) t - (1/2) (9.8) t²2.2
= 19.115 t - 4.9 t²t² - 3.885 t + 0.449
= 0(t - 0.434)(t - 3.451) = 0t = 0.434 s (taking positive value)
From the second equation of motion, we can find the range of the ball:
R = u_x tR = (22.05 cos 60°) (0.434)R
= 8.19 m
So, the midfielder is located 8.19 meters down the field from the point of release of the ball.To find the distance of the midfielder from the point where the ball drops to the ground, we need to find the horizontal range of the ball from the point of release to the point where it drops to the ground.
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A heavy block, labeled " A ", is sitting on a table. On top of that block is a lighter block, labeled "B" as shown in the figure at the right. For the first parts of this problem you are asked to identify the direction of forces in this system under various circumstances. The labels in the subscripts indicate: A= block A,B= block B,F= finger, T= table. Specify the direction in your answers using the following notation: - R means points to the right - Lmeans points to the left - U means point up - D means points down - O indicates there is no such force at the instant specified 1. You start pushing on block A as shown, but it is too heavy and does not move. While you are pushing on block A but while it is not moving, specify the direction of the following normal ( N ) and frictional (f) forces between the various objects indicated. (a) N
A→B
(b) f
Y→A
(c) f
A→B
(d) N
F→A
(e) f
t→A
(f) N
0→A
2. Now you push a little harder and the block begins to move. Block B moves with it without slipping. While the blocks are speeding up, specify the direction of the following forces between the various objects indicated. (a) N
i→h
(b) f
T→A
(c) f
A→B
(d) N
f→A
(e) f
t→A
(f) N
0→A
3. Now you push so that the blocks move at a constant velocity. Block B moves with A without slipping. While the blocks are moving at a constant speed specify the direction of the following forces between the various objects indicated. (a) N
A→B
(b) f
T→A
(c) f
A→B
(d) N
f→A
(e) f
B−A
(f) N
B→A
In the given system of blocks (A and B) placed on a table, the directions of normal and frictional forces are determined for both motion and rest situations below:
1. (a) N A→B - Normal force will be pointing down from block A to block B, which is the reaction force to the weight of block B exerted on A. (b) f Y→A - Frictional force will be pointing to the left and it is between the surface of the table and blocks A, which opposes the direction of motion. (c) f A→B - Frictional force will be pointing to the right and it is between the surface of block A and block B, which opposes the direction of motion. (d) N F→A - Normal force will be pointing up from the table to block A, which is the reaction force to the weight of block A exerted on the table. (e) f t→A - Frictional force will be pointing to the right and it is between the surface of the table and block A, which opposes the direction of motion. (f) N 0→A - No force acting in the upward direction on block A.2. (a) N i→h - Normal force will be pointing up from the table to block A, which is the reaction force to the weight of block A exerted on the table. (b) f T→A - Frictional force will be pointing to the left and it is between the surface of the table and block A, which opposes the direction of motion. (c) f A→B - Frictional force will be pointing to the right and it is between the surface of block A and block B, which opposes the direction of motion. (d) N f→A - Normal force will be pointing up from the table to block A, which is the reaction force to the weight of block A exerted on the table. (e) f t→A - Frictional force will be pointing to the left and it is between the surface of the table and block A, which opposes the direction of motion. (f) N 0→A - No force acting in the upward direction on block A.3. (a) N A→B - Normal force will be pointing down from block A to block B, which is the reaction force to the weight of block B exerted on A. (b) f T→A - Frictional force will be pointing to the left and it is between the surface of the table and blocks A, which opposes the direction of motion. (c) f A→B - Frictional force will be pointing to the right and it is between the surface of block A and block B, which opposes the direction of motion. (d) N f→A - Normal force will be pointing up from the table to block A, which is the reaction force to the weight of block A exerted on the table. (e) f B−A - Frictional force will be pointing to the left and it is between the surface of block B and block A, which opposes the direction of motion. (f) N B→A - Normal force will be pointing up from block B to block A, which is the reaction force to the weight of block B exerted on block A. Thus, the directions of normal and frictional forces in the given system of blocks (A and B) placed on a table are identified while in motion and at rest.For more questions on frictional forces
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During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 39.3rad/s. Find the angular displacement Δθ of the tub during a spin of 91.3 s, expressed both in radians and in revolutions. Δθ= Δθ=
The angular displacement of the tub during a spin of 91.3 s is approximately 3589.09 radians or 570.46 revolutions.
To find the angular displacement Δθ of the tub during a spin of 91.3 s, we can use the formula:
Δθ = ωt
where Δθ is the angular displacement in radians, ω is the angular velocity in rad/s, and t is the time in seconds.
Given:
ω = 39.3 rad/s
t = 91.3 s
Substituting the values into the formula, we have:
Δθ = 39.3 rad/s * 91.3 s
Calculating the product:
Δθ ≈ 3589.09 rad
Therefore, the angular displacement of the tub during a spin of 91.3 s is approximately 3589.09 radians.
To convert this angular displacement to revolutions, we can use the conversion factor: 1 revolution = 2π radians.
Δθ in revolutions = Δθ in radians / (2π)
Δθ in revolutions = 3589.09 rad / (2π)
Calculating:
Δθ ≈ 570.46 revolutions
Therefore, the angular displacement of the tub during a spin of 91.3 s is approximately 3589.09 radians or 570.46 revolutions.
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A high jumper of mass 70.1 kg consumes a meal of 4.20 × 10^3 kcal prior to a jump. If 3.30% of the energy from the food could be converted to gravitational potential energy in a single jump, how high could the athlete jump?
The athlete could jump approximately 827.9 meters high with the gravitational potential energy obtained from consuming the meal.
To determine how high the athlete could jump, we need to calculate the gravitational potential energy (GPE) that can be obtained from the consumed meal and then convert it to the height.
First, let's convert the energy consumed from kilocalories (kcal) to joules (J):
1 kcal = 4184 J
Energy consumed = [tex]4.20 * 10^3[/tex] kcal * 4184 J/kcal
Energy consumed = [tex]1.75 * 10^7[/tex] J
Next, we need to find the gravitational potential energy (GPE) that can be obtained from the consumed energy. We know that 3.30% of the energy can be converted to GPE:
GPE = 0.0330 × Energy consumed
GPE = [tex]0.0330 * 1.75 * 10^7[/tex] J
GPE = [tex]5.775 * 10^5[/tex] J
To convert the GPE into height, we can use the formula:
GPE = mgh
Where:
m is the mass of the jumper (70.1 kg),
g is the acceleration due to gravity (approximately 9.8 m/s²), and
h is the height.
Rearranging the formula, we can solve for h:
h = GPE / (mg)
h = (5.775 * 10⁵ J) / (70.1 kg * 9.8 m/s²)
Calculating the height:
h ≈ 827.9 meters.
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A drone flies 3.001 miles NE and then 2.841 miles SW over a period of 14.22 minutes. What is the drone's average velocity over that time period?
The drone's average velocity over that time period is 0.03 miles per minute.
The drone flies 3.001 miles NE and then 2.841 miles SW over a period of 14.22 minutes. What is the drone's average velocity over that time period?Answer: The drone's average velocity over that time period is 0.03 miles per minuteStep-by-step explanation:The displacement vector for the drone can be determined by subtracting the vector 2.841 miles SW from the vector 3.001 miles NE.Using the Pythagorean theorem:|displacement vector| = sqrt((3.001)^2 + (2.841)^2)≈ 4.166 miles.
Thus, the distance traveled by the drone is 4.166 miles.Dividing this distance by the time it took to travel it:average speed = 4.166 miles / 14.22 minutes≈ 0.293 miles per minuteIn order to find the drone's average velocity, the direction of the displacement vector must be taken into account.The direction of the displacement vector can be determined using the arctangent function:arctan(2.841/3.001)≈ 43.4° north of eastSince the drone travels at a constant speed, its average velocity is equal to its average speed multiplied by a unit vector in the direction of the displacement vector:average velocity = (0.293 miles per minute) * (cos(43.4°) i + sin(43.4°) j)≈ 0.03 miles per minute.
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A rocket undergoes a constant acceleration of 2.6 m/s
2
starting from rest. What is the distance traveled, in meters, in 3.4 minutes? (Round off your answer to the ones.)
The distance traveled by the rocket in 3.4 minutes is 54,091.2 m
Given :
Acceleration of the rocket is 2.6 m/s².
Time for which the rocket moves is 3.4 minutes or 204 seconds (1 minute = 60 seconds).
We need to find the distance traveled by the rocket.
We can use the following kinematic equation :
distance = initial velocity × time + 0.5 × acceleration × time²
As the rocket starts from rest, initial velocity (u) is zero.
Therefore, distance = 0 + 0.5 × 2.6 × (204)²
distance = 0 + 0.5 × 2.6 × 41,616
distance = 0 + 54,091.2
Therefore, the distance traveled by the rocket is 54,091.2 m (rounding off to the nearest meter).
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A stone of mass 0.70 kg falls with an acceleration of 10.0 m/s
2
. How much is the force that causes this acceleration? Include the unit in your answer 1 point for the value of the force 1 point for the unit
Therefore, the answer is 7.0 N. Note that, force is measured in Newtons (N) which is the SI unit for force.
According to Newton’s second law, F = ma, the force acting on an object is equal to the product of the object’s mass and its acceleration.
The mass of the stone is 0.70 kg and its acceleration is 10.0 m/s². Hence, the force that causes this acceleration is given as;
F = ma
F = 0.70 × 10.0
F = 7.0 N (Newtons)
The force that causes this acceleration is 7.0 N (Newtons).
Therefore, the answer is 7.0 N. Note that, force is measured in Newtons (N) which is the SI unit for force.
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A cement block accidentally falls from rest from the ledge of a 84.5-m-high building. When the block is 19.1 m above the ground, a man, 1.70 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?
To determine the maximum time the man has to get out of the way of a falling cement block, we can calculate the time it takes for the block to fall from a height of 19.1 m to the ground.
Using the equations of motion, we can find the time by considering the vertical distance traveled by the block. The correct answer depends on the acceleration due to gravity and the initial height of the block.
The vertical distance traveled by the block is the difference between the initial height (84.5 m) and the final height (19.1 m). Using the equation of motion,
h = ut + (1/2)gt², where
h is the vertical distance,
u is the initial velocity (0 m/s in this case),
g is the acceleration due to gravity (approximately 9.8 m/s²), and
t is the time,
we can calculate the time it takes for the block to fall.
The equation becomes:
19.1 = 0 + (1/2)(9.8)t²
Simplifying the equation:
9.8t² = 19.1 × 2
t² = (19.1 × 2) / 9.8
t² ≈ 3.898
t ≈ √3.898
t ≈ 1.97 seconds
Therefore, the maximum time the man has to get out of the way is approximately 1.97 seconds. During this time, the block will fall from a height of 19.1 m to the ground. It's crucial for the man to move quickly to avoid the falling block and ensure his safety.
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(8\%) Problem 7: Suppose you wanted to store \( 3 \mu \mathrm{C} \) of charge in a capacitor across a voltage of \( 120 \mathrm{~V} \). ( What capacitance is needed in \( \mathrm{nF} \) ? \[ C= \]
The value of the capacitance needed is determined as 25 nF.
What is the capacitance needed?If you wanted to store 3μC of charge in a capacitor across a voltage of 120 V, the value of the capacitance needed is calculated by applying the following formula.
C = Q/V
where;
Q is the chargeV is the voltage suppliedC is the capacitanceC = ( 3 x 10⁻⁶ C) / ( 120 V )
C = 2.5 x 10⁻⁸ F
C = 25 x 10⁻⁹ F
C = 25 nF
Thus, the value of the capacitance needed is determined as 25 nF.
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