Final answer:
The magnitude of the electric force exerted by the tape on the paper at a distance of 8.0 mm is 3.72 x 10^-7 C. The direction of the force is repulsive as like charges repel each other.
Explanation:
When the tape comes within 8.0 mm of the scrap of paper, the electric force magnitude is great enough to overcome the gravitational force exerted by Earth on the scrap and lift it. To determine the magnitude and direction of the electric force exerted by the tape on the paper at this distance, we can use the equation for the electric force:
F = k * (Q1 * Q2) / r^2
where F is the magnitude of the force, k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the distance between the charges.
We know that the gravitational force is equal to the weight of the paper:
Fg = m * g
where Fg is the gravitational force, m is the mass of the paper (190 mg = 0.190 g), and g is the acceleration due to gravity (9.8 m/s^2).
At the point where the tape comes within 8.0 mm of the paper, the electric force is equal to the gravitational force:
F = Fg
k * (Q1 * Q2) / r^2 = m * g
We can solve this equation for the magnitude of the charge Q1 or Q2:
Q1 * Q2 = (m * g * r^2) / k
Substituting the known values, we get:
Q1 * Q2 = (0.190 g * 9.8 m/s^2 * (8.0 mm)^2) / (8.99 x 10^9 Nm^2/C^2)
Q1 * Q2 = 1.383 x 10^-12 C^2
To find the magnitude of the charge, we can assume that Q1 and Q2 are equal, so:
Q1 = Q2 = sqrt(1.383 x 10^-12 C^2) = 3.72 x 10^-7 C
The magnitude of the electric force exerted by the tape on the paper at this distance is 3.72 x 10^-7 C. The direction of the force is repulsive because like charges repel each other.
where in your environment can you observe some of the general properties of wave motion
Answer:
Air, water, rock and soil
Explanation:
The environment where the general properties of wave motion can be observed is the wave medium which is any substance or particle that carries the wave, or through which the wave travels.
ocean waves are carried by watersound waves are carried by airthe seismic waves of an earthquake are carried by rock and soilWhat must be the distance between point charge q1 = 28.0 μC and point charge q2 = −57.0 μC for the electrostatic force between them to have a magnitude of 5.05 N?
Answer:
1.686 m
Explanation:
From coulomb's law,
F = kq1q2/r² ...................................... Equation 1
Where F = electrostatic force between the two charges, q1 = first charge, q2 = second charge, r = distance between the charges.
making r the subject of the equation,
r = √(kq1q2/F).......................... Equation 2
Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C
Constant: k = 9.0×10⁹ Nm²/C².
Substituting into equation 2
r = √(9.0×10⁹×28×10⁻⁶×57.0×10⁻⁶/5.05)
r = √(14364×10⁻³/5.05)
r = √(14.364/5.05)
r = √2.844
r = 1.686 m
r = 1.686 m.
Thus the distance must be 1.686 m
What is the magnitude of the electric force of attraction between an iron nucleus (q=+26e)(q=+26e) and its innermost electron if the distance between them is 1.5×10−12m1.5×10−12m?
The magnitude of the electric force of attraction between the iron nucleus and its innermost electron is approximately 2.645 × 10^-3 Newtons.
To calculate the magnitude of the electric force of attraction between an iron nucleus and its innermost electron, we can use Coulomb's law:
Given:
Charge of the iron nucleus, q1 = +26e
Charge of the electron, q2 = -e (where e is the elementary charge, 1.6 × 10^-19 C)
Distance between them, r = 1.5 × 10^-12 m
Convert the charge of the nucleus from elementary charges to coulombs:
q1 = +26e × 1.6 × 10^-19 C/e = +4.16 × 10^-18 C
Calculate the electric force using Coulomb's law:
F = (8.9875 × 10^9 N m^2/C^2) × |(+4.16 × 10^-18 C) × (-1.6 × 10^-19 C)| / (1.5 × 10^-12 m)^2
F ≈ (8.9875 × 10^9) × | -6.656 × 10^-37 | / (2.25 × 10^-24)
F ≈ (8.9875 × 10^9) × 2.947 × 10^-13
F ≈ 2.645 × 10^-3 N
So, the magnitude of the electric force of attraction between the iron nucleus and its innermost electron is approximately 2.645 × 10^-3 Newtons.
A point charge Q is located a short distance from a point charge 3Q, and no other charges are present. If the electrical force on Q is F, what is the electrical force on 3Q?a. 3Fb. √3Fc. F/√3d. Fe. F/3
Answer:
d) F
Explanation:
According to columb's law:
"The magnitude of electrostatic force between two charges is directly proportional to the product of magnitude of two charges and inversly proportional to separation between them."
If q₁ and q₂ are magnitude of two charges, d is distance between them and k is dielectric constant, then force F is given by
[tex]F=\frac{kq_{1}q_{2}}{d^2}[/tex]
According to this force exerted on point charge Q is same as that of 3Q, so force point 3Q charge experience is also F
An oscillator creates periodic waves on two strings made ofthe same material. The tension is the same in both strings.If the strings have different thicknesses,which of the following parameters, if any, will be different in thetwo strings?Check all that apply.a. wave frequencyb. wave speedc. wavelengthd. none of the above
The wave speed and wavelength will differ across strings of different thicknesses held at the same tension; the wave frequency remains the same.
When two strings of different thicknesses are made of the same material and held at the same tension, the wave speed and wavelength will differ, whereas the wave frequency will remain the same. This is because the speed of a wave on a string is determined by the tension (T) and the linear mass density (μ), where v = (T/μ)¹/2; as the tension is constant and the material is the same, it is the difference in thickness (hence, different densities) that causes a variance in wave speed. Since wave frequency (f) is related to the speed (v) and wavelength (λ) by the equation v = fλ, and the oscillator creates waves at a fixed frequency, a change in wave speed inherently impacts the wavelength.
Suppose the mass is pulled down to where the spring's length is 72 cm . When it is released, it begins to oscillate. What is the amplitude of the oscillation?
The question is incomplete. This is the complete question: A spring has an unstretched length of 22 cm. A 150 g mass hanging from the spring stretches it to an equilibrium length of 30 cm. Suppose the mass is pulled down to where the spring's length is 38 cm. When it is released, it begins to oscillate. What is the amplitude of the oscillation?
Answer:
The amplitude of the oscillation is 8 cm.
Explanation:
The amplitude of the oscillation, which is the maximum displacement of the stretched spring from equilibrium or rest, can be calculated by subtracting the spring’s length at equilibrium (when being stretched by 150g mass) from the spring’s length when it was pulled down.
Amplitude = A = the spring’s length when it was pulled down before oscillating (i.e., 38cm) — the spring’s length at equilibrium (i.e., 30cm)
Therefore, A = 38cm — 30cm = 8cm.
A -2.37 µC point charge and a 4.74 µC point charge are a distance L apart.
Where should a third point charge be placed so that the electric force on that third charge is zero? (Hint: Solve this problem by first placing the -2.37 µC point charge at the origin and place the 4.74 µC point charge at x = −L.)
Answer:
[tex]x=2L\pm \sqrt{\frac{(-2L)^2-4\times 1\times(-L^2)}{2\times 1} }[/tex]
Explanation:
Given:
charge on first particle, [tex]q_1=-2.37\times 10^{-6}\ C[/tex]charge on the second particle, [tex]q_2=4.74\times 10^{-6}\ C[/tex]distance between the two charges = LNow the third charge must be placed on the line joining the two charges at a distance where the intensity of electric field is same for both the charges that point will not lie between the two charges because they are opposite in nature.
[tex]E_1=E_2[/tex]
[tex]\frac{1}{4\pi\epsilon_0} \times \frac{q_1}{x^2} =\frac{1}{4\pi\epsilon_0} \times \frac{q_2}{(L+x)^2}[/tex]
[tex]\frac{2.37}{x^2} =\frac{4.74}{L^2+x^2+2xL}[/tex]
[tex]2x^2=L^2+x^2+2xL[/tex]
[tex]x^2-2L.x-L^2=0[/tex]
[tex]x=2L\pm \sqrt{\frac{(-2L)^2-4\times 1\times(-L^2)}{2\times 1} }[/tex]
A large centrifuge is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries. At what angular velocity is the centripetal acceleration 10 g if the rider is 15.0 m from the center of rotation
Answer:
ω = 2.55 rad/sec
Explanation:
Assuming no other external forces acting in the horizontal plane, the only force keeping the rider in a circular path of a radius equal to his distance to the center of rotation, is the centripetal force.
According to Newton's 2nd law, in the horizontal direction, we have:
F = Fc = m*a = m*ω²*r
We know that a = ac = 10*g = 98.0 m/s², and that r = 15.0 m.
Replacing these values in (1), and solving for ω, we get:
ω = √98.0m/s²/ 15.0 m = 2.55 rad/sec
The centrifugal force in a rocket.
The centrifugal force is the force that is related to the outwards or away from the body. The larger force is used to expose the aspiring astronomers to accelerations that are the same as those experienced by the rockets that are launched and the air reentries.
Thus the answer is ω = 2.55 rad/sec
The centrifugal force is used to measure the acceleration of the astronauts when they are launched in the air or atmosphere. The angular velocity of the centripetal force that accelerates to 10g if the rider is 15.0 meters from the center.Taking no external forces acting on the plane,As per the Newton's 2nd law, the formulae.F= Fc = m*a = m*ω²*r know that a = ac = 10*g = 98.0 m/s², and that r = 15.0 m. On replacing these values in (1), and solving for ω, we get: ω equal to 2.55 rad/sec.Learn more about the centrifuge is used.
brainly.com/question/26431449.
When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to make this force twice as strong, the distance would have to be changed to
A) √2d
B) d/√2
C) d/4
D) 2d
E) d/2
Answer:b
Explanation:
Given
Force of attraction is F when charges are d distance apart.
Electrostatic force is given by
[tex]F=\frac{kq_1q_2}{d^2}---1[/tex]
where k=constant
[tex]q_1[/tex] and [tex]q_2[/tex] are charges
d=distance between them
In order to double the force i.e. 2F
[tex]2F=\frac{kq_1q_2}{d'^2}----2[/tex]
divide 1 and 2 we get
[tex]\frac{F}{2F}=\frac{d'^2}{d^2}[/tex]
[tex]d'=\frac{d}{\sqrt{2}}[/tex]
The energy required to dissociate KF into neutral atoms is 498 kJ/mol. Given that the first ionization energy for K is 418 kJ/mol, calculate the electron affinity (in kJ/mol) for F. Show your work for all calculations.
Answer: Electron affinity of F equals
275.8kJ/mol
Explanation: Electron affinity is the energy change when an atom gains an electron.
Let's first calculate the energy required -E(r)to dissociate KF into ions not neutral atom which is given.
E(r) = {z1*z2*e²}/{4π*permitivity of space*r}
z1 is -1 for flourine
z2 is +1 for potassium
e is magnitude of charge 1.602*EXP{-9}C
r is ionic bond length of KF(is a constant for KF 0.217nm)
permitivity of free space 8.854*EXP{-12}.
Now let's solve
E(r)= {(-1)*(1)*(1.602*EXP{-9})²} /
{4*3.142*8.854*EXP(-12)*0.217*EXP(-9)
E(r) = - 1.063*EXP{-18}J
But the energy is released out that is exothermic so we find - E(r)
Which is +1.603*EXP{-18}J
Let's now convert this into kJ/mol
By multiplying by Avogadro constant 6.022*EXP(23) for the mole and diving by 1000 for the kilo
So we have,
1.603*EXP(-18) *6.022*EXP(23)/1000
-E(r) = 640.2kJ/mol.
Now let's obtain our electron affinity for F
We use this equation
Energy of dissociation (nuetral atom)= electron affinity of F +(-E(r)) + ionization energy of K.
498kJ/mol
=e affinity of F + 640.2kJ/mol
+(-418kJ/mol)
(Notice the negative sign in ionization energy for K. since it ionize by losing an electron)
Making electron affinity of F subject of formula we have
Electron affinity (F)=498+418-640.2
=275.8kJ/mol.
A horizontal plank of mass m and length L is pivoted at one end. The plank’s other end is supported
by a spring of force constant k. The plank is displaced by a
small angle
from its horizontal equilibrium position and
released. (a) Show that it moves with simple harmonic motion
with an angular frequency
= 3k/m . (b) Evaluate the
frequency if the mass is 5.00 kg and the spring has a force
constant of 100 N/m
Final answer:
The horizontal plank on a pivot supported by a spring exhibits simple harmonic motion with angular frequency ω = √(3k/m). When the mass is 5.00 kg and the spring constant is 100 N/m, the frequency is approximately 1.95 Hz.
Explanation:
A horizontal plank of mass m and length L is pivoted at one end and supported by a spring of force constant k at the other. This setup is displaced by a small angle θ from its horizontal equilibrium position and released.
Part A: Derivation of Simple Harmonic Motion
To show the plank moves with simple harmonic motion (SHM) and to find the angular frequency ω, we analyze the forces acting on the plank when displaced. The restoring force F exerted by the spring is F = -kx, where x is the linear displacement of the spring. For small angles, θ, x ≈ Lθ. Thus, F ≈ -kLθ. Applying Newton's second law for rotational motion, τ = Iα, where τ is the torque, I is the moment of inertia of the plank, and α is the angular acceleration. The torque caused by the spring is τ = -kL²θ, and the moment of inertia of the plank about the pivot is I = (1/3)mL². From τ = Iα, we get α = -3k/mθ, indicating SHM with an angular frequency ω = √(3k/m).
Part B: Calculating the Frequency
Given m = 5.00 kg and k = 100 N/m, ω = √(3k/m) = √(3*100/5) = √60. Hence, the frequency f is f = ω/(2π) = √60/(2π) ≈ 1.95 Hz.
Light from a helium-neon laser (λ=633nm) passes through a circular aperture and is observed on a screen 4.0 m behind the aperture. The width of the central maximum is 2.5 cm .What is the diameter (in mm) of the hole?
Answer:
d = 0.247 mm
Explanation:
given,
λ = 633 nm
distance from the hole to the screen = L = 4 m
width of the central maximum = 2.5 cm
2 y = 0.025 m
y = 0.0125 m
For circular aperture
[tex]sin \theta = 1.22\dfrac{\lambda}{d}[/tex]
using small angle approximation
[tex]\theta = \dfrac{y}{D}[/tex]
now,
[tex]\dfrac{y}{D} = 1.22\dfrac{\lambda}{d}[/tex]
[tex]y = 1.22\dfrac{\lambda\ D}{d}[/tex]
[tex]d = 1.22\dfrac{\lambda\ D}{y}[/tex]
[tex]d = 1.22\dfrac{633\times 10^{-9}\times 4}{0.0125}[/tex]
d =0.247 x 10⁻³ m
d = 0.247 mm
the diameter of the hole is equal to 0.247 mm
The diameter of the hole is approximately 0.0201 mm.
Explanation:The width of the central maximum in a diffraction pattern can be determined using the formula:
w = (2 * λ * D) / x
Where w is the width of the central maximum, λ is the wavelength of the light, D is the distance between the aperture and the screen, and x is the diameter of the hole. Rearranging the formula, we can solve for x:
x = (2 * λ * D) / w
Plugging in the given values, we get:
x = (2 * 633 * 10^-9 * 4.0) / 0.025
x ≈ 0.0201 mm
Two vertical springs have identical spring constants, but one has a ball of mass m hanging from it and the other has a ball of mass 2m hanging from it.Part A If the energies of the two systems are the same, what is the ratio of the oscillation amplitudes?
To solve this problem we will start from the definition of energy of a spring mass system based on the simple harmonic movement. Using the relationship of equality and balance between both systems we will find the relationship of the amplitudes in terms of angular velocities. Using the equivalent expressions of angular velocity we will find the final ratio. This is,
The energy of the system having mass m is,
[tex]E_m = \frac{1}{2} m\omega_1^2A_1^2[/tex]
The energy of the system having mass 2m is,
[tex]E_{2m} = \frac{1}{2} (2m)\omega_1^2A_1^2[/tex]
For the two expressions mentioned above remember that the variables mean
m = mass
[tex]\omega =[/tex]Angular velocity
A = Amplitude
The energies of the two system are same then,
[tex]E_m = E_{2m}[/tex]
[tex]\frac{1}{2} m\omega_1^2A_1^2=\frac{1}{2} (2m)\omega_1^2A_1^2[/tex]
[tex]\frac{A_1^2}{A_2^2} = \frac{2\omega_2^2}{\omega_1^2}[/tex]
Remember that
[tex]k = m\omega^2 \rightarrow \omega^2 = k/m[/tex]
Replacing this value we have then
[tex]\frac{A_1}{A_2} = \sqrt{\frac{2(k/m_2)}{(k/m_1)^2}}[/tex]
[tex]\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m_1}{m_1}}[/tex]
But the value of the mass was previously given, then
[tex]\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m}{2m}}[/tex]
[tex]\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{1}{2}}[/tex]
[tex]\frac{A_1}{A_2} = 1[/tex]
Therefore the ratio of the oscillation amplitudes it is the same.
An eight-turn coil encloses an elliptical area having a major axis of 40.0 cm and a minor axis of 30.0 cm. The coil lies in the plane of the page and carries a clockwise current of 6.20 A. If the coil is in a uniform magnetic field of 1.98 10-4 T directed toward the left of the page, what is the magnitude of the torque on the coil? Hint: The area of an ellipse is A = ?ab, where a and b are, respectively, the semimajor and semiminor axes of the ellipse.
Answer:
9.25 x 10^-4 Nm
Explanation:
number of turns, N = 8
major axis = 40 cm
semi major axis, a = 20 cm = 0.2 m
minor axis = 30 cm
semi minor axis, b = 15 cm = 0.15 m
current, i = 6.2 A
Magnetic field, B = 1.98 x 10^-4 T
Angle between the normal and the magnetic field is 90°.
Torque is given by
τ = N i A B SinФ
Where, A be the area of the coil.
Area of ellipse, A = π ab = 3.14 x 0.20 x 0.15 = 0.0942 m²
τ = 8 x 6.20 x 0.0942 x 1.98 x 10^-4 x Sin 90°
τ = 9.25 x 10^-4 Nm
thus, the torque is 9.25 x 10^-4 Nm.
The magnitude of the torque on the coil is 9.25 x 10⁻⁴ Nm
Torque:According to the question we have the following data:
Number of turns of the coil N = 8
Semi-major axis of the ellipse a = 40/2 cm = 0.2 m
Semi-minor axis, b = 30/2 cm = 0.15 m
Current in the coil, i = 6.2 A
Magnetic field, B = 1.98 x 10⁻⁴ T
The angle between the normal and the magnetic field is 90°.
So the torque on the coil is given by:
τ = NiABsinθ
Now, the area of ellipse:
A = πab
A = 3.14 x 0.20 x 0.15 = 0.0942 m²
Thus,
τ = 8 x 6.20 x 0.0942 x 1.98 x 10⁻⁴ x sin 90°
τ = 9.25 x 10⁻⁴ Nm
Learn more about torque:
https://brainly.com/question/6855614?referrer=searchResults
The position of a particle along a straight-line path is defined by s=(t3−6t2−15t+7) ft, wheret is in seconds.A. Determine the total distance traveled when t = 8.3 s .B. What are the particle's average velocity at the time given in part A?C. What are the particle's average speed at the time given in part A?D. What are the particle's instantaneous velocity at the time given in part A?E. What are the particle's acceleration at the time given in part A?
Answer:
A) The total distance traveled when t = 8.3 s is 234 ft.
B) The average velocity of the particle is 4.1 ft/s.
C) The average speed at t = 8.3 s is 28 ft/s.
D) The instantaneous velocity at t = 8.3 s is 92 ft/s.
E) The acceleration of the particle at t = 8.3 s is 37.8 ft/s²
Explanation:
Hi there!
A) The position of the particle at a time "t", in feet, is given by the function "s":
s(t) = t³ - 6t² - 15t + 7
First, let´s find at which time the particle changes direction. The sign of the instantaneous velocity indicates the direction of the particle. We will consider the right direction as positive. The origin of the frame of reference is located at s = 0 and t = 0 so that the particle at t = 0 is located 7 ft to the right of the origin.
The instaneous velocity (v(t)) of the particle is the first derivative of s(t):
v(t) = ds/dt = 3t² - 12t - 15
The sign of v(t) indicates the direction of the particle. Notice that at t = 0,
v(0) = -15. So, initially, the particle is moving to the left.
So let´s find at which time v(t) is greater than zero:
v(t)>0
3t² - 12t - 15>0
Solving the quadratic equation with the quadratic formula:
For every t > 5 s, v(t) > 0 (the other solution of the quadratic equation is -1. It is discarded because the time can´t be negative).
Then, the particle moves to the left until t = 5 s and, thereafter, it moves to the right.
To find the traveled distance at t= 8.3 s, we have to find how much distance the particle traveled to the left and how much distance it traveled to the right.
So, let´s find the position of the particle at t = 0, at t = 5 and at t = 8.3 s
s(t) = t³ - 6t² - 15t + 7
s(0) = 7 ft
s(5) = 5³ - 6 · 5² - 15 · 5 + 7 = -93 ft
s(8.3) = 8.3³ - 6 · 8.3² - 15 · 8.3 + 7 = 40.9 ft
So from t = 0 to t = 5, the particle traveled (93 + 7) 100 ft to the left, then from t = 5 to t = 8.3 the particle traveled (93 + 40.9) 134 ft to the right. Then, the total distance traveled when t = 8.3 s is (134 ft + 100 ft) 234 ft.
B) The average velocity (AV) is calculated as the displacement over time:
AV = Δs / Δt
Where:
Δs = displacement (final position - initial position).
Δt = elapsed time.
In this case:
final position = s(8.3) = 40.9 ft
initial position = s(0) = 7 ft
Δt = 8.3 s
So:
AV = (s(8.3) - s(0)) / 8.3 s
AV = (40.9 ft - 7 ft) / 8.3 s
AV = 4.1 ft/s
The average velocity of the particle is 4.1 ft/s (since it is positive, it is directed to the right).
C) The average speed is calculated as the traveled distance over time. The traveled distance at t = 8.3 s was already obtained in part A: 234 ft. Then, the average speed (as) will be:
as = distance / time
as = 234 ft / 8.3 s
as = 28 ft/s
The average speed at t = 8.3 s is 28 ft/s
D) The instantaneous velocity at any time t was obtained in part A:
v(t) = 3t² - 12t - 15
at t = 8.3 s
v(8.3) = 3(8.3)² - 12(8.3) - 15
v(8.3) = 92 ft/s
The instantaneous velocity at t = 8.3 s is 92 ft/s.
E) The particle acceleration at any time t, is obtained by derivating the velocity function:
v(t) = 3t² - 12t - 15
dv/dt = a(t) = 6t - 12
Then at t = 8.3 s
a(8.3) = 6(8.3) - 12
a(8.3) = 37.8 ft/s²
The acceleration of the particle at t = 8.3 s is 37.8 ft/s²
Given the missing information, we can only calculate the particle's instantaneous velocity and acceleration at 8.3 s, which can be obtained by taking derivatives of the position function.
Explanation:This is a physics problem involving kinematics and calculus. Let's address each part of the question:
A. To find the total distance, we would need to know the initial position of the particle. Without this information, we cannot accurately calculate the total distance.
B. Average velocity is defined as the displacement divided by the time interval, which is irrelevant in this case because we do not know the displacement.
C. Similar to B, without displacement information, we cannot calculate average speed.
D. Instantaneous velocity is given by the first derivative of the position function. By taking the derivative of s, we can get the velocity function: v(t) = 3t^2 - 12t - 15. Plug in t = 8.3 s, we can then get the instantaneous velocity.
E. Acceleration is given by the first derivative of the velocity function or the second derivative of the position function. With our velocity function, v(t), we can take its derivative to find a(t) = 6t - 12. Plugging in t = 8.3 s will give the particle's acceleration at that time.
Learn more about Kinematics here:https://brainly.com/question/35140938
#SPJ3
In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of radius 0.500 cm at the same elevation as the first tube. The pressure differs by DP between the liquid in one tube and the liquid in the second tube. (a) Find the volume flow rate as a function of DP. Evaluate the volume flow rate for (b) DP 5 6.00 kPa and (c) DP 5 12.0 kPa. 49. The Venturi tube discussed
To find the volume flow rate in ideal flow between two horizontal tubes, use the equation A1v1 = A2v2, where A1 and A2 are the cross-sectional areas of the tubes and v1 and v2 are the velocities of the liquid. To find the volume flow rate as a function of DP, substitute the values of A1 and v1 into the equation Q = A1v1. For specific values of DP, substitute the values of A1 and v1 into the equation Q = A1v1.
Explanation:For liquids, the volume flow rate can be determined using the equation A1v1 = A2v2. In this equation, A1 and A2 are the cross-sectional areas of the first and second tubes, and v1 and v2 are the velocities of the liquid flowing through the tubes. Since the first tube is larger in radius, it has a larger cross-sectional area. Therefore, the velocity of the liquid in the first tube is smaller than in the second tube. The volume flow rate can be expressed as:
Q = A1v1
where Q is the volume flow rate and is equal to the product of the cross-sectional area and velocity of the liquid in the first tube.
(a) The volume flow rate as a function of DP can be found by substituting the values of A1 and v1 into the equation Q = A1v1. Since the radii of the tubes are given, the cross-sectional areas can be calculated using the formula A = πr2.
(b) To evaluate the volume flow rate for DP = 6.00 kPa, substitute the values of A1 and v1 into the equation Q = A1v1.
(c) To evaluate the volume flow rate for DP = 12.0 kPa, substitute the values of A1 and v1 into the equation Q = A1v1.
Learn more about liquid here:https://brainly.com/question/34457794
#SPJ2
The electric field everywhere on the surface of a thin, spherical shell of radius 0.770 m is of magnitude 860 N/C and points radially toward the center of the sphere.
(a) What is the net charge within the sphere's surface?
(b) What can you conclude about the nature and distribution of the charge inside the spherical shell?
Answer:
(a) [tex]Q = 7.28\times 10^{14}[/tex]
(b) The charge inside the shell is placed at the center of the sphere and negatively charged.
Explanation:
Gauss’ Law can be used to determine the system.
[tex] \int{\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\
E4\pi r^2 = \frac{Q_{enc}}{\epsilon_0}\\
(860)4\pi(0.77)^2 = \frac{Q_{enc}}{8.8\times 10^{-12}}\\
Q_enc = 7.28\times 10^{14}[/tex]
This is the net charge inside the sphere which causes the Electric field at the surface of the shell. Since the E-field is constant over the shell, then this charge is at the center and negatively charged because the E-field is radially inward.
The negative charge at the center attracts the same amount of positive charge at the surface of the shell.
You are given vectors Aâ = 4.8 i^â 7.0 j^ and Bâ = - 4.0 i^+ 7.3 j^. A third vector Câ lies in the xy-plane. Vector Câ is perpendicular to vector Aâ and the scalar product of Câ with Bâ is 14.0. Find the X and Y components of vector C
Answer:
Vector C = -1.56i^ +1.07j^
This question requires that we use the properties of the scalar product of two vectors to find the required x and y component.
The dot product of two perpendicular vectors is equal to and the product of two vectors that are not parallel is equal to a nonzero value.
These are the properties that have been used in solving this problem alongside solving the simultaneous questions generator.
Explanation:
The full solution can be found in the attachment below.
Thank you for reading and I hope this post is helpful to you.
If He gas has an average kinetic energy of 5930 J/mol under certain conditions, what is the root mean square speed of F2 gas molecules under the same conditions
To solve this problem we will apply the concept related to kinetic energy based on the ideal gas constant and temperature. From there and with the given values we will find the temperature of the system. As the temperature is the same it will be possible to apply the root mean square speed formula that is dependent on the element's molar mass, the ideal gas constant and the temperature, this would be:
[tex]KE = \frac{3}{2} RT[/tex]
Where,
KE = Average kinetic energy of an ideal gas
[tex]R = 8.314JK^{-1}mol^{-1}[/tex]= Ideal gas constant
T = Temperature
Replacing we have,
[tex]KE = \frac{3}{2} RT[/tex]
[tex]5930J/mol = \frac{3}{2}(8.314JK^{-1}mol^{-1})T[/tex]
[tex]T = 475.503K[/tex]
Therefore the temperature is 475.5K
RMS velocity of [tex]F_2[/tex] gas is
[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]
Where,
M = Molar mass of [tex]F_2[/tex]
[tex]M = 38.00g/mol[/tex]
[tex]M = 38.00*10^{-3} kg/mol[/tex]
[tex]T = 475.5K[/tex]
[tex]R = 8.314JK^{-1}mol^{-1}[/tex]
Replacing we have,
[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]
[tex]v_{rms} = \sqrt{\frac{3(8.314JK^{-1}mol^{-1})(475.5K )}{38.00*10^{-3} kg/mol}}[/tex]
[tex]v_{rms} = 558.662m/s[/tex]
Therefore, the RMS velocity of [tex]F_2[/tex] gas is 558.6m/s
For aerodynamic reasons, nearly all modern aircraft feature at leastone large vertical stabilizer. The B-2 "stealth" bomber pictured to theright, however, has no vertical fins at all. Why not?
Answer:
Answer:
The reason why the B-2 "stealth" bomber has no vertical fins (or stabilizers) at all, is because by design, the vertical fins are not required to provide stability; the stability for yaw movement is provided through a computer that controls the B-2 stealth bomber; hence the B-2 stealth bomber does not need a vertical stabilizer in order to fly.
A 3.0 cm × 3.0 cm parallel-plate capacitor has a 3.0 mm spacing. The electric field strength inside the capacitor is 1.2×105 V/m . What is the potential difference across the capacitor? How much charge is on each plate?
To solve this problem it is necessary to apply the concepts related to the voltage depending on the electric field and the distance, as well as the load depending on the capacitance and the voltage. For the first part we will use the first mentioned relationship, for the second part, we will not only define the load as the capacitance by the voltage but also place it in terms of the Area, the permittivity in free space, the voltage and the distance.
PART A ) Voltage in function of electric field and distance can be defined as,
[tex]V = Ed[/tex]
Our values are,
[tex]E = 1.2*10^5 V/m[/tex]
[tex]d = 3.0mm = 3*10^{-3}[/tex]
Replacing,
[tex]V = (1.2*10^5)(3*10^{-3})[/tex]
[tex]V = 360v[/tex]
Therefore the potential difference across the capacitor is 360V
PART B) The charge can be defined as,
[tex]Q = CV = \frac{\epsilon AV}{d}[/tex]
Here,
[tex]\epsilon = 8.85*10^{-12} F/m[/tex], Permittivity of free space
[tex]A = s^2[/tex], area of each capacitor plate
s = Length of capacitor plate
Replacing,
[tex]Q = \frac{\epsilon AV}{d}[/tex]
[tex]Q = \frac{(8.85*10^{-12})(0.03)^2(240)}{2.0*10^{-8}m}[/tex]
[tex]Q = 9.558*10^{-10}C[/tex]
Therefore the charge on each plate is [tex]9.558*10^{-10}C[/tex]
The potential difference across the parallel-plate capacitor is 360 V. The charge on each plate of the capacitor is approximately 0.95 x 10^-8 C.
The potential difference across a parallel-plate capacitor is calculated using the formula V = Ed, where E is the electric field strength and d is the distance (or spacing) between the plates. As given, E is 1.2 x 10^5 V/m, and d is 3.0 mm (or 3.0 x 10^-3 m). Therefore, the potential difference V across the plates is given by V = 1.2 x 10^5 V/m * 3.0 x 10^-3 m = 360 V.
The amount of charge Q on each plate of the capacitor can be found using the formula Q = εEA, where ε is the permittivity of free space (ε = 8.85 x 10^-12 F/m), E is the electric field strength, and A is the area of the plate. Substituting the values given, we have A = 3.0 cm * 3.0 cm = 9 cm^2 = 9 x 10^-4 m^2, E = 1.2 x 10^5 V/m, and ε = 8.85 x 10^-12 F/m. Therefore, Q = εEA = 8.85 x 10^-12 F/m * 1.2 x 10^5 V/m * 9 x 10^-4 m^2 ≈ 0.95 x 10^-8 C.
Learn more about Parallel-Plate Capacitor here:
https://brainly.com/question/30312803
#SPJ6
Ricardo, of mass 80 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 30 kg canoe. When the canoe is at rest in the placid water, they exchange seats, which are 3.0 m apart and symmetrically located with respect to the canoe's center. If the canoe moves 40 cm horizontally relative to a pier post, what is Carmelita's mass?
Answer:
m=57.65 kg
Explanation:
Given Data
Ricardo mass m₁=80 kg
Canoe mass m₂=30 kg
Canoe Length L= 3 m
Canoe moves x=40 cm
When Canoe was at rest the net total torque is zero.
Let the center of mass is at x distance from the canoe center and it will be towards the Ricardo cause. So the toque around the center of mass is given as
[tex]m_{1}(L/2-x)=m_{2}x+m_{2}(L/2-x)[/tex]
We have to find m₂.To find the value of m₂ first we need figure out the value of.As they changed their positions the center of mass moved to other side by distance 2x.
so
2x=40
x=40/2
x=20 cm
Substitute in the above equation we get
[tex]m_{x}=\frac{m_{1}(L/2-x)-m_{2}x }{L/2+x}\\m_{x}=\frac{80(\frac{3}{2}-0.2 )-30*0.2}{3/2+0.2}\\m_{x}=57.65 kg[/tex]
A proton is traveling horizontally to the right at 4.20×10^6m/s.Part A:Find (a)the magnitude and (b) direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.50cm.Part B: counterclockwise from the left directionPart C:How much time does it take the proton to stop after entering the field?Part D:What minimum field ((a)magnitude and (b)direction) would be needed to stop an electron under the conditions of part (a)?
Answer:
2630250 N/C, horizontally left
0
[tex]1.67\times 10^{-8}\ s[/tex]
1434.825 N/C, horizontally left
Explanation:
m = Mass of particle
u = Initial velocity = [tex]4.2\times 10^6\ m/s[/tex]
v = Final velocity = 0
t = Time taken
s = Displacement = 3.5 cm
q = Charge of particle = [tex]1.6\times 10^{-19}\ C[/tex]
Force is given by
[tex]F=qE[/tex]
Acceleration is given by
[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times E}{1.67\times 10^{-27}}\\\Rightarrow a=95808383.23353E[/tex]
[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=2\times 95808383.23353Es\\\Rightarrow E=\dfrac{0^2-(4.2\times 10^6)^2}{2\times 95808383.23353\times 0.035}\\\Rightarrow E=-2630250\ N/C[/tex]
Magnitude of electric field is 2630250 N/C
Direction is horizontally to the left
The angle counterclockwise from left is zero.
[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times -2630250}{1.67\times 10^{-27}}\\\Rightarrow a=-2.52\times 10^{14}\ m/s^2[/tex]
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-4.2\times 10^6}{-2.52\times 10^{14}}\\\Rightarrow t=1.67\times 10^{-8}\ s[/tex]
The time taken is [tex]1.67\times 10^{-8}\ s[/tex]
Acceleration is given by
[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times E}{9.11\times 10^{-31}}\\\Rightarrow a=175631174533.4797E[/tex]
[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=2\times 175631174533.4797Es\\\Rightarrow E=\dfrac{0^2-(4.2\times 10^6)^2}{2\times 175631174533.4797\times 0.035}\\\Rightarrow E=-1434.825\ N/C[/tex]
Magnitude of electric field is 1434.825 N/C
Direction is horizontally to the left
A guitar string is 90 cm long and has a mass of 3.7 g . The distance from the bridge to the support post is L=62cm, and the string is under a tension of 500 N . What are the frequencies of the fundamental and first two overtones?
To solve this problem we will apply the concept of frequency in a string from the nodes, the tension, the linear density and the length of the string, that is,
[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]
Here
n = Number of node
T = Tension
[tex]\mu[/tex] = Linear density
L = Length
Replacing the values in the frequency and value of n is one for fundamental overtone
[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]
[tex]f = \frac{1}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})[/tex]
[tex]\mathbf{f = 281.2Hz}[/tex]
Similarly plug in 2 for n for first overtone and determine the value of frequency
[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]
[tex]f = \frac{2}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})[/tex]
[tex]\mathbf{f = 562.4Hz}[/tex]
Similarly plug in 3 for n for first overtone and determine the value of frequency
[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]
[tex]f = \frac{3}{2(62*10^{-2})}\bigg (\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}} \bigg)[/tex]
[tex]\mathbf{f= 843.7Hz}[/tex]
How far from the nucleus in angstroms (1 angstrom = 1 × 10–10 m) is the electron in a hydrogen atom if it has an energy of –8.72 × 10–20 J?
Bohr's model of atom postulated that the electrons revolves around the nucleus only in those orbits which have fixed energy and do not lose energy while revolving in them.
According to Bohr's model, the energy at infinite distance is taken to be zero and as it approaches the atom, it starts becoming more negative.
The [tex]n^{th}[/tex] shell of electrons is calculated by
[tex]n^2 = \frac{kZ^2}{E_n}[/tex]
Here
E_n = Energy at [tex]n^{th}[/tex] level
k = Constant
n = Number of shell
Z = Atomic number of the element
Replacing we have that
[tex]n^2 = \frac{-(2.179*10^{-18}J)(1)^2}{-8.72*10^{-20}J}[/tex]
[tex]n = 24.98[/tex]
[tex]n \approx 25[/tex]
Thus
[tex]n = \pm 5[/tex]
Since number of shell cannot be negative we have that n = 5
Now the distance of electron from nucleus is given according to relation
[tex]r = (0.529)(n^2)[/tex]
[tex]r = (0.529)(5^2)[/tex]
[tex]r = 13.225 \AA[/tex]
Therefore the distance of electron from nucleus is 13.225A
The electron in a hydrogen atom is approximately 2.116 angstroms away from the nucleus.
Explanation:To determine the distance of an electron from the nucleus in a hydrogen atom, we can use the equation for the energy of an electron in a hydrogen atom: E = -13.6eV / n^2, where n is the principal quantum number. We can convert the energy from joules to electron volts (eV) by using the conversion factor 1eV = 1.6x10^-19J. Substituting the given energy, we have:
-8.72x10^-20J = -13.6eV / n^2.
By rearranging the equation and solving for n, we find that n≈2. Thus, the electron is in the second energy level. The distance from the nucleus in angstroms (Å) can be calculated using the formula r = 0.529n^2Å, where r is the distance from the nucleus. Substituting n = 2, we get:
r = 0.529 x 2^2 = 2.116Å.
Therefore, the electron is approximately 2.116 angstroms away from the nucleus.
Learn more about distance of electron from nucleus here:https://brainly.com/question/35571736
#SPJ3
A car starting from rest accelerates at a rate of 2 m/sec2 . Find the average speed of this object in the first 10 seconds.
Answer:
The final speed of the car in the first 10 seconds will be 20 m/s.
Explanation:
Given that,
Initial speed of the car, u = 0
Acceleration of the car, [tex]a=2\ m/s^2[/tex]
Time, t = 10 s
We need to find the speed of the car in the first 10 seconds. It can be calculated using first equation of motion. It is given by :
[tex]v=u+at[/tex]
v is the final speed of the car
[tex]v=at[/tex]
[tex]v=2\ m/s^2\times 10\ s[/tex]
v = 20 m/s
So, the final speed of the car in the first 10 seconds will be 20 m/s. Hence, this is the required solution.
A liquid has a specific gravity of 1.1 at room temperature. What is its (a) Density at room temperature in kg/m3 (b) Specific volume at room temperature in ft3/lbm? (c) If the liquid is placed in a 2 L bottle that has a mass of 157 g, how much will the full bottle weigh?
To solve this problem we will proceed to find the density from the specific gravity. Later we will find the specific volume as the inverse of the density. Finally with the data obtained we will find the total weight in the bottle.
a) [tex]\rho = \gamma * 1000[/tex]
Here,
[tex]\rho[/tex] = Density
[tex]\gamma[/tex] = Specific gravity
[tex]\rho = 1.1 * 1000[/tex]
[tex]\rho = 1100 kg/m3[/tex]
b)
[tex]\text{Specific volume}= \frac{1}{\rho}[/tex]
[tex]\upsilon = \frac{1}{1100}[/tex]
[tex]\upsilon = 0.00090909 m^3/kg[/tex]
From the equivalences of meters to feet and kilograms to pounds, we have to
[tex]1m = 3.280839895 ft[/tex]
[tex]1 kg = 2.2046 lbm[/tex]
Converting the previous value to British units:
[tex]\upsilon = 0.00090909 m^3/kg (\frac{3.280839895^3 ft^3}{1m^3} )(\frac{1kg}{2.2046 lbm})[/tex]
[tex]\upsilon= 0.0145757 ft^3 / lbm[/tex]
c)
[tex]V = 2*10^{-3} m^3[/tex]
Mass of the liquid in bottle is
[tex]m = V\rho[/tex]
[tex]m= (2*10^{-3} m^3 )(1100kg/m^3)[/tex]
[tex]m = 2.2kg = 2200g[/tex]
Therefore the Total weight
[tex]W= 157 + 2200 = 2357 g[/tex]
A car starts from rest at a stop sign. It accelerates at 3.8 m/s 2 m/s2 for 6.0 s, coasts for 1.6 ss , and then slows down at a rate of 3.3 m/s 2 m/s2 for the next stop sign.
How far apart are the stop signs?
To solve this problem we will start by calculating the distance traveled while relating the first acceleration in the given time. From that acceleration we will calculate its final speed with which we will calculate the distance traveled in the second segment. With this speed and the acceleration given, we will proceed to calculate the last leg of its route.
Expression for the first distance is
[tex]s_1 = ut +\frac{1}{2} at^2[/tex]
[tex]s_1 = 0+\frac{1}{2} (3.8)(6)^2[/tex]
[tex]s_1 = 68.4m[/tex]
The expression for the final speed is
[tex]v = v_0 +at[/tex]
[tex]v = 0+(3.8)(6)[/tex]
[tex]v = 22.8m/s[/tex]
Then the distance becomes as follows
[tex]s_2 = vt[/tex]
[tex]s_2 = (22.8)(1.6)[/tex]
[tex]s_2 = 36.48m[/tex]
The expression for the distance at last sop is
[tex]v_1^2=v_0^2 +2as_3[/tex]
[tex]22.8^2 = 0+2(3.3)s_3[/tex]
[tex]s_3 =78.7636m[/tex]
Therefore the required distance between the signs is,
[tex]S = s_1+s_2+s_3[/tex]
[tex]S = 68.4+36.48+78.76[/tex]
[tex]S = 183.64m[/tex]
Therefore the total distance between signs is 183.54m
A lead ball is dropped into a lake from a diving board 6.10 mm above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 4.50 ss after it is released. How deep is the lake?
Answer:
D=1.54489 m
Explanation:
Given data
S=6.10 mm= 0.0061 m
To find
Depth of lake
Solution
To find the depth of lake first we need to find the initial time ball takes to hit the water.To get the value of time use below equation
[tex]S=v_{1}t+(1/2)gt^{2} \\ 0.0061m=(0m/s)t+(1/2)(9.8m/s^{2} )t^{2}\\ t^{2}=\frac{0.0061m}{4.9m/s^{2} }\\ t=\sqrt{1.245*10^{-3} }\\ t=0.035s[/tex]
So ball takes 0.035sec to hit the water
As we have found time Now we need to find the final velocity of ball when it enters the lake.So final velocity is given as
[tex]v_{f}=v_{i}+gt\\v_{f}=0+(9.8m/s^{2} )(0.035s)\\ v_{f}=0.346m/s[/tex]
Since there are (4.50-0.035) seconds left for (ball) it to reach the bottom of the lake
So the depth of lake given as:
[tex]D=|vt|\\D=|0.346m/s*4.465s|\\D=1.54489m[/tex]
Answer: d = 1.54m
The depth of the lake is 1.54m
Explanation:
The final velocity of the ball just before it hit the water can be derived using the equation below;
v^2 = u^2 + 2as ......1
Where ;
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance travelled.
Since the initial velocity is zero, and the acceleration is due to gravity, the equation becomes:
v^2 = 2gs
v = √2gs ......2
g = 9.8m/s^2
s = 6.10mm = 0.0061m
substituting into equation 2
v = √(2 × 9.8× 0.0061)
v = 0.346m/s
The time taken for the ball to hit water from the time of release can be given as:
d = ut + 0.5gt^2
Since u = 0
d = 0.5gt^2
Making t the subject of formula.
t = √(2d/g)
t = √( 2×0.0061/9.8)
t = 0.035s
The time taken for the ball to reach the bottom of the lake from the when it hits water is:
t2 = 4.5s - 0.035s = 4.465s
And since the ball falls for 4.465s to the bottom of the lake at the same velocity as v = 0.346m/s. The depth of the lake can be calculated as;
depth d = velocity × time = 0.346m/s × 4.465s
d = 1.54m
The depth of the lake is 1.54m
75) A river 100 m wide flows 1 m/s due south. A boat that travels 1 m/s relative to the water is pointed due east as it crosses from the west bank. Relative to its starting point, the boat travels
A) 141 m.
B) 100 m.
C) 200 m.
D) more than 200 m.
E) nowhere
Answer: A) 141 m
Explanation:
Given that the boat travels at a speed of 1m/s due east in a river that flows 1m/s due south.
Let north represent positive y axis and east represent positive x axis.
Then we can resolve the resultant velocity of the boat to vector form.
Vr = i - j ( 1 m/s on x axis and -1m/s on y axis)
The time required to travel 100m from west to east at a speed of 1m/s is;
Time t = distance/speed = 100m/1m/s = 100s
Since the boat will use 100s to cross the river, We can now determine the resultant distance after 100s:
Distance = velocity × time = (i - j) × 100 = 100i - 100j
Distance = 100i - 100j (in vector form)
Magnitude of the Resultant distance can be given as:
dr = √(dx^2 + dy^2)
dr = √(100^2 + 100^2)
dr = √(20000)
dr = 141.42m
dr = 141m
A) The boat's overall displacement relative to its starting point is 141 m.
Explanation:To solve this problem, we can break it down into two components: the magnitudes of the boat's eastward displacement and southward displacement. The time it takes for the boat to cross the river can be calculated using the width of the river and the boat's eastward speed. The distance the boat drifts downstream during this time can be calculated using the river's southward speed and the time taken. By combining these two displacements, we can determine the boat's overall displacement relative to its starting point.
The eastward displacement of the boat can be found using the formula: eastward displacement = eastward speed x time.
Plugging in the given values, we get: eastward displacement = 1 m/s x (100 m / 1 m/s) = 100 m.
The southward displacement of the boat can be found using the formula: southward displacement = southward speed x time.
Plugging in the given values, we get: southward displacement = 1 m/s x (100 m / 1 m/s) = 100 m.
Therefore, the boat's overall displacement relative to its starting point, which is the combination of the eastward and southward displacements, is equal to the square root of (eastward displacement squared + southward displacement squared).
Plugging in the calculated values, we get overall displacement = sqrt((100 m)^2 + (100 m)^2) = sqrt(2) x 100 m = 141 m.
Learn more about Boat displacement here:
https://brainly.com/question/36032146
#SPJ3