Answer:
Gravitational force
Explanation:
If two spheres have equal densities and they are subject only to their mutual gravitational attraction. We need to say that the quantities that must have the same magnitude for both spheres. So, the correct option is (E) i.e. gravitational force.
It is because of Newton's third law of motion. It states that the force due to object 1 to object 2 is same as force due to object 2 to object 1. The two forces act in opposite direction.
Hence, the correct option is (E) "Gravitational force".
The gravitational force has the same magnitude for both spheres.
By the Newton's law of gravitation and the Newton's third law we understand that gravitational force is directly proportional to the masses of the spheres and inversely proportional to the square of distance between centers, also that gravitational force experimented by one sphere has the same magnitude but opposite direction than the gravitational force experimented by the other sphere.
Therefore, we conclude that gravitational force has the same magnitude for both spheres. (Correct choice E)
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A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The bottom of the ladder rests on a floor with a rough surface where the coefficient of static friction is 0.42 . The angle between the horizontal and the ladder is θ . The person wants to climb up the ladder a distance of 0.9 m along the ladder from the ladder’s foot. 16.2 kg 0.9 m 2 m θ b µ = 0.42 µ = 0 What is the minimum angle θmin (between the horizontal and the ladder) so that the person can reach a distance of 0.9 m without having the ladder slip? The acceleration of gravity is 9.8 m/s 2
To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.
The forces in the vertical direction would be,
[tex]\sum F_x = 0[/tex]
[tex]f-N_w = 0[/tex]
[tex]N_w = f[/tex]
The forces in the horizontal direction would be,
[tex]\sum F_y = 0[/tex]
[tex]N_f -W =0[/tex]
[tex]N_f = W[/tex]
The sum of Torques at equilibrium,
[tex]\sum \tau = 0[/tex]
[tex]Wdcos\theta - N_wLsin\theta = 0[/tex]
[tex]WdCos\theta = fLSin\theta[/tex]
[tex]f = \frac{Wd}{Ltan\theta}[/tex]
The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore
[tex]f_{max} = \mu W=\frac{Wd}{Ltan\theta}[/tex]
[tex]\theta = tan^{-1} (\frac{d}{\mu L})[/tex]
Replacing,
[tex]\theta = tan^{-1} (\frac{0.9}{0.42*2})[/tex]
[tex]\theta = 46.975\°[/tex]
Therefore the minimum angle that the person can reach is 46.9°
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 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.
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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]
In a 5.00 km race, one runner runs at a steady 11.4 km/h and another runs at 14.7 km/h . How long does the faster runner have to wait at the finish line to see the slower runner cross?
Answer:
0.0986 h or 5 minutes 55 seconds.
Explanation:
Speed: This can be defined as the rate of change of distance of a body. The S.I unit of speed is m/s. Speed is a scalar quantity, because it can only be represented by magnitude alone.
Mathematically,
Speed = distance/time.
S = d/t ........................... Equation 1
making t the subject of the equation
t = d/S ......................... Equation 2
Form the question,
Time taken for the faster runner to reach the finish line
t₁ = d/S₁................... Equation 3
Where t₁ = time taken for the faster runner to reach the finish line, d = distance, S₁ = speed of the faster runner.
Given: d = 5.0 km, S₁ = 14.7 km/h.
Substituting into equation 3
t₁ = 5/14.7
t₁ = 0.340 h
Also,
t₂ = d/S₂................... Equation 4
Where t₂ = time taken for the slower runner to reached the finished line, d = distance, S₂ = speed of the slower runner.
Given: d = 5 km, S₂ = 11.4 km/h.
Substitute into equation 4,
t₂ = 5/11.4
t₂ = 0.4386 h.
The time the faster runner have to wait at the finish line to see the slower runner cross = t₂ - t₁ = 0.4386-0.340
The time the faster runner have to wait at the finish line to see the slower runner cross = 0.0986 h = 5 mins 55 s.
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 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
Suppose the entire population of the world gathers in ONE spot and everyone jumps at the sound of a prearranged signal. While everyone is in the air, does the Earth gain momentum in the opposite direction?a) No, the inertial mass of Earth is so large that the planet's change in motion is imperceptibleb) Yes, however the change in momentum of Earth is much less than that of all the jumping people because of Earth's large inertial massc) Yes, Earth recoils like that of a rifle firing a bullet with a change in momentum equal to and opposite of peopled) It depends
Final answer:
Yes, the Earth does gain momentum in the opposite direction due to the conservation of momentum principle when everyone jumps, but the effect is negligible given Earth's massive inertial mass. So the correct option is b.
Explanation:
The question posed is whether the Earth gains momentum in the opposite direction when the entire population of the world jumps and everyone is in the air. According to the conservation of momentum, the answer is yes, but the change in the Earth's momentum is incredibly small to the point of being imperceptible. This is because the inertial mass of Earth is so large compared to the combined mass of all the people that the result of this collective jump would be negligible when it comes to the Earth's momentum.
If we consider a closed system that includes both the Earth and the people jumping, then the total change of momentum for the system must be zero. When people jump, they exert a force on the Earth, and Earth exerts an equal and opposite force on them—this is Newton's third law. However, because of the Earth's substantially greater mass, it experiences an inconsequentially small acceleration in response to this force. While the Earth does indeed recoil much like when a force is applied through a goalpost from a football player hitting it, the Earth's recoil is immeasurably small.
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.
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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.
In the macroscopic world, you know that you can hear but cannot see around corners. Under what conditions does light bend around corners (i.e. diffract) ? Explain why sound diffracts easily around a classroom door. 5. Suppose you added to the single slit an identical slit a distance d=0.25mm away from the first. Draw the resulting interference pattern you might expect on the same screen. What happens when we increase the distance between slits ? What happens in the limit that d becomes arbitrarily large?
Answer:
a much larger slit, the phenomenon of Sound diffraction that slits for light.
this is a series of equally spaced lines giving a diffraction envelope
Explanation:
The diffraction phenomenon is described by the expression
d sin θ = m λ
Where d is the distance of the slit, m the order of diffraction that is an integer and λ the wavelength.
For train the diffraction phenomenon, the d / Lam ratio is decisive if this relation of the gap separation in much greater than the wavelength does not reduce the diffraction phenomenon but the phenomena of geometric optics.
The wavelength range for visible light is 4 10⁻⁷ m to 7 10⁻⁷ m. The wavelength range for sound is 17 m to 1.7 10⁻² m. Therefore, with a much larger slit, the phenomenon of Sound diffraction that slits for light.
When we add a second slit we have the diffraction of each one separated by the distance between them, when the integrals are made we arrive at the result of the interference phenomenon, a this is a series of equally spaced lines giving a diffraction envelope
When I separate the distance between the two slits a lot, the time comes when we see two individual diffraction patterns
A car initially going 61 ft/sec brakes at a constant rate (constant negative acceleration), coming to a stop in 7 seconds.
a) Graph the velocity from t = 0 to t = 7.
(b) How far does the car travel?
(c) How far does the car travel if its initial velocity is doubled, but it brakes at the same constant rate?
Answer:
See the attachment below for the graphics in part (a)
The initial velocity for this time interval is u = 61ft/sec and the final velocity is 0m/s because the car comes to a stop.
This a constant acceleration motion considering the given time interview over which the brakes are applied. So the equals for constant acceleration motion apply here.
Explanation:
The full solution can be found in the attachment below.
Thank you for reading. I hope this post is helpful to you.
A 400 kg satellite orbits the moon at a height of 2000 km above the moon at a speed of 946 m/s. The speed of a 800 kg satellite orbiting the moon at 2000 km above the moon is:
To solve this problem we will apply the concepts related to the balance of Forces, in this case the centripetal force of the body must be equal to the gravitational force exerted by the moon on it.
The gravitational force is given by the function
[tex]F_g = \frac{GmM}{r^2}[/tex]
Here
G = Gravitational Universal constant
M = Mass of the planet
m = Mass of the satellite
r = Radius(orbit)
Now the centripetal force is given as
[tex]F_c =\frac{mv^2}{r}[/tex]
Here
m = mass of satellite
v = Velocity of satellite
r = Radius (orbit)
Since there must be balance for the satellite to remain in the orbit
[tex]F_c = F_g[/tex]
[tex]\frac{mv^2}{r} = \frac{GmM}{r^2}[/tex]
[tex]v^2= \frac{GM}{r}[/tex]
[tex]v=\sqrt{\frac{GM}{r}}[/tex]
The velocity depends on the mass of the planet and the orbit, and not on the mass, so if the orbit is maintained, the velocity will be the same: 946m/s
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
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.
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.
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A small object is attached to the end of a relaxed, horizontal spring whose opposite end is fixed. The spring rests on a frictionless surface. Let the initial position of the object be defined as x-0. The object is pulled to position x = A and then released, after which it undergoes simple harmonic motion.
In one full cycle of its motion, the total distance traveled by the object is:
1) A
2) A/2
3) 2A
4) 4A
5) A/4
Answer:4A
Explanation:
Given
Mass is displace x= A units from its mean position x=0'
When it is set to free it will oscillate about its mean position with maximum amplitude A i.e. from x=-A to x=A
One cycle is completed when block returns to its original position
so first block will go equilibrium position x=0 and then to x=-A
from x=-A it again moves back to x=0 and finally back to its starting position x=A
so it travels a distance of A+A+A+A=4A
If the accelerating voltage va were 250 volts, what would be the speed of the electrons emerging from the gun?
Answer:
[tex]v=9.34*10^{6}\frac{m}{s}[/tex]
Explanation:
According to the law of conservation of energy, the amount of energy an electron gains after being accelerated is equal to its kinetic energy, that is, the electrical potential energy is converted into kinetic energy :
[tex]U=K\\eV=\frac{mv^2}{2}[/tex]
Solving for v and replacing the given voltage:
[tex]v=\sqrt\frac{2eV}{m}\\v=\sqrt\frac{2(1.6*10^{-19}C)250V}{9.1*10^{-31}kg}\\v=9.34*10^{6}\frac{m}{s}[/tex]
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.
Knowing the P is 1110 N, determine the tension in the rope if the frame is in equilibrium. Express your answer in kN to the nearest 10 N.
Explanation:
It is given that the value of P is 1110 N. And, for pin connection we have only two connections which are [tex]A_{x}[/tex] and [tex]A_{y}[/tex]. Let T be the tension is rope.
So, [tex]\sum F_{y} = 0[/tex] and [tex]A_{y}[/tex] - 1110 = 0
[tex]A_{y} = 1110 N[/tex]
[tex]\sum F_{x}[/tex] = 0
And, [tex]T - A_{x}[/tex] = 0
T = [tex]A_{x}[/tex]
Also, [tex]\sum M_{A}[/tex] = 0
1110(0.75 + 0.75 + 0.75) - T(0.5 + 0.1) = 0
2497.5 - 0.6T = 0
T = 4162.5 N
= 4.16 kN
Therefore, we can conclude that the tension in the rope if the frame is in equilibrium is 4.16 kN.
Knowing that α= 60° and that boom AC exerts on pin C a force directed along line AC, determine (a)the magnitude of t hat force, (b) the tension in cable BC.Figure:Cable BC is making an angle alpha with fixed support.
Answer:
Tbc = 230.69 N ; Fac = 172.31 N
Explanation:
Sum of forces in y direction:
[tex]T_{BC} * sin (35) = 300*sin (70) + F_{AC}*sin (60) .... Eq 1\\[/tex]
Sum of forces in x direction:
[tex]T_{BC} * cos (35) + F_{AC}*cos (60)= 300*cos (70) .... Eq 1\\[/tex]
Solving Eq 1 and Eq 2 simultaneously:
[tex]T_{BC} = 281.9077862 + \sqrt{3} / 2 * F_{AC}\\\\F_{AC} (1.736868124) = 300*cos (70) - 491.4912266*cos (35)\\\\F_{AC} = - \frac{300}{1.736868124}\\\\F_{AC} = - 172.73 N\\\\T_{BC} = 230.69 N[/tex]
Answer: Tbc = 230.69 N ; Fac = 172.31 N
(a) The tension in the cable at AC is -200.67 N.
(b) The tension in the cable at BC is 328.99 N.
Forces resolved along y-axisThe sum of the forces in y-direction is calculated as follows;
T(BC)sin(35) = 400 x sin(65) + F(AC) sin(60) --- (1)
Forces resolved along x-axisT(BC)cos(35) + F(AC) cos(60) = 400 x cos(65) ---(2)
From equation(1);
[tex]T_{BC} = \frac{400 \times sin(65) \ + \ F_{AC} sin(60)}{sin(35)} \\\\T_{BC} = 632 + 1.51F_{AC}[/tex]
From equation (2);
0.82T(BC) + 0.5F(AC) = 169.1
[tex]0.82(632 + 1.51F_A_C) + 0.5F_A_C= 169.1\\\\518.24 + 1.24F_A_C + 0.5F_A_C = 169.1\\\\F_A_C = \frac{-349.14}{1.74} \\\\F_A_C = -200.67 \ N[/tex]
T(BC) = 632 + 1.51(-200.67)
T(BC) = 328.99 N
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An oscillator creates periodic waves on a stretched string.
If the period of the oscillator doubles, what happens to the wavelength and wave speed?
a. The wavelength doubles but the wave speed is unchanged.
b. The wavelength is halved but the wave speed is unchanged.
c. The wavelength is unchanged but the wave speed doubles.
Answer:
A. The wavelength doubles but the wave speed is unchanged
Explanation:
The relationship between the period and wavelength is direct. Doubling the period of the oscillator will correspondingly double the wavelength but the wave speed is unaffected
An oscillator creates periodic waves on a stretched string. If the period of the oscillator doubles, then the wavelength doubles but the wave speed is unchanged. So option A is correct here.
When the period of the oscillator doubles, it means that the time it takes for one complete oscillation or cycle of the wave doubles. The period of a wave is inversely proportional to its frequency. If the period doubles, the frequency is halved. The wavelength of a wave is the distance between two consecutive crests or troughs. The wavelength of a wave is inversely proportional to its frequency. When the frequency is halved, the wavelength doubles to maintain the relationship.
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Two charged small spheres are a distance R apart and exert an electrostatic force F on each other. If the distance is halved to R/5 , the force exerted on each sphere will be..
a) 25F
b) 5F
c) F/5
d) F/25
Answer:
a) 25F
Explanation:
Assuming that the two small spheres can be modeled as point charges, according to Coulomb's law, the magnitude of the electrostatic force is given by:
[tex]F=\frac{kq_1q_2}{R^2}[/tex]
In this case, we have [tex]R'=\frac{R}{5}[/tex]:
[tex]F'=\frac{kq_1q_2}{R'^2}\\F'=\frac{kq_1q_2}{(\frac{R}{5})^2}\\F'=25\frac{kq_1q_2}{R^2}\\F'=25F[/tex]
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.
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.
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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
A 477 g portion of soup is heated in a microwave oven from 25°C to 90°C, using radiation with a wavelength of 1.55 × 10⁻² m. Assuming the heat capacity of the soup is the same as water (4.18 J/°C･g), how many photons are absorbed by the soup?
To solve this problem we will use the heat transfer equations, to determine the amount of heat added to the body. Subsequently, through the energy ratio given by Plank, we will calculate the energy of each of the photons. The relationship between total energy and unit energy will allow us to determine the number of photons
The mass of water in the soup is 477g
The change in temperate is
[tex]\Delta T = (90+273K)-(25+273K) = 65K[/tex]
Use the following equation to calculate the heat required to raise the temperature:
[tex]q = mc\Delta T[/tex]
Here,
m = Mass
c = Specific Heat
[tex]q = (477)(4.184)(65)[/tex]
[tex]q = 129724.92J[/tex]
The wavelength of the ration used for heating is [tex]1.55*10^{-2}m[/tex]
The number of photons required is the rate between the total energy and the energy of each proton, then
[tex]\text{Number of photons} = \frac{\text{Total Energy}}{\text{Energy of one Photon}}[/tex]
This energy of the photon is given by the Planck's equation which say:
[tex]E = \frac{hc}{\lambda}[/tex]
Here,
h = Plank's Constant
c = Velocity of light
[tex]\lambda =[/tex] Wavelength
Replacing,
[tex]E = \frac{(6.626*10^{-34})(3*10^8)}{1.55*10^{-2}}[/tex]
[tex]E = 1.28*10^{-23}J[/tex]
Now replacing we have,
[tex]\text{Number of photons} = \frac{82240.7}{1.28*10^{-23}}[/tex]
[tex]\text{Number of photons} = 6.41*10^{27}[/tex]
Therefore the number of photons required for heating is [tex]6.41*10^{27}[/tex]
One day when you come into physics lab you find several plastic hemispheres floating like boats in a tank of fresh water. Each lab group is challenged to determine the heaviest rock that can be placed in the bottom of a plastic boat without sinking it. You get one try. Sinking the boat gets you no points, and the maximum number of points goes to the group that can place the heaviest rock without sinking. You begin by measuring one of the hemispheres, finding that it has a mass of 23 g and a diameter of 8.4 cm . What is the mass of the heaviest rock that, in perfectly still water, won't sink the plastic boat?
To solve this problem we will first proceed to find the volume of the hemisphere, from there we will obtain the mass of the density through the relation of density. Finally the mass of the stone will be given between the difference in the mass given in the statement and the one found, that is
The volume of a Sphere is
[tex]V = \frac{4}{3} \pi r^3[/tex]
Then the volume of a hemisphere is
[tex]V =\frac{1}{2} \frac{4}{3} \pi r^3[/tex]
With the values we have that the Volume is
[tex]V =\frac{1}{2} \frac{4}{3} \pi (8.4/2)^3[/tex]
[tex]V = 155.17cm^3[/tex]
Density of water is
[tex]\rho = 1g/cm^3[/tex]
And we know that
[tex]\text{Mass of water displaced} = \text{Density of water}\times \text{Volume of hemisphere}[/tex]
[tex]m = 1g/cm^3 * 155.17cm^3[/tex]
[tex]m = 155.17g[/tex]
So the net mass is
[tex]\Delta m = m_s-m_w[/tex]
[tex]\Delta m = 155.17-23[/tex]
[tex]\Delta m = 132.17g[/tex]
Therefore the mass of heaviest rock is 132.17g or 0.132kg
The mass of the heaviest rock that, in perfectly still water, won't sink the plastic boat is 155 g.
We know that the density of water = 1 g/cm^3
Volume of the hemisphere = 2/3 πr^2
When diameter = 8.4 cm, radius = 4.2 cm
So, V = 2/3 × 3.14 × (4.2)^3
V = 155 cm^3
Volume of hemisphere = volume of water displaced = 155 cm^3
Mass of water displaced = 155 g
Since the solid displaces its own mass of water, the mass of the heaviest rock that, in perfectly still water, won't sink the plastic boat is 155 g.
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a ferry which runs at 12m/s in still water travels between towns a and b on a river which flows south at 9 m/s. the ferry leavs a and heads due east but reaches the east bank at C. it then heads north to B. whatis the velocity of the ferry as it traverses route ac?
Answer:
= 15 m/s
Explanation:
Considering right side(west) as positive x-axis and south as negative y-axis.
velocity of boat in still water [tex]v_b=12\hat{i}[/tex]
velocity of stream [tex]v_s=-9\hat{j}[/tex]
now relative velocity of boat w.r.t. stream [tex]v_{b/s}=12\hat{i}+9\hat{j}[/tex]
this velocity with which ac distance will be covered.
therefore magnitude of [tex]v_{b/s} =\sqrt{12^2+9^2}[/tex]
= 15 m/s