To solve this problem we will start from the constants given on the two celestial bodies, that is, the earth and the moon. From there we will seek the balance of forces, where the gravitational force is zero. That distance will remain incognito and will be the one we will seek to find, this is
Given that,
[tex]M_E =5.97*10^24 kg[/tex]
[tex]m_m = 7.36*10^22 kg[/tex]
[tex]r_E = 6370 km[/tex]
[tex]r_m = 1737 km[/tex]
The distance between Earth and Moon
[tex]x = 384403*10^3 m[/tex]
As the net gravitational force is zero we have that
[tex]\frac{GMm'}{d^2} = \frac{Gmm'}{(x -d)^2}[/tex]
[tex]\frac{M}{d^2} = \frac{m}{(x -d)^2}[/tex]
Replacing with our values
[tex]\frac{(5.97*10^{24} )}{d^2} = \frac{(7.36*10^{22})}{(x -d)^2}[/tex]
[tex]\frac{(x- d)^2}{d^2}= 0.0123[/tex]
[tex]x^2 + d^2 - 2xd = 0.0123 d^2[/tex]
Replacing x as the distance betwen moon and earth,
[tex](384403 x 10^3)^2+d^2-2(384403 x 10^3)d=0.0123 d^2[/tex]
Solving the polynomial we have that
[tex]d =4.3387*10^8 \text{ or } 3.4506*10^{8}[/tex]
So the minimum value is [tex]3.4506*10^{8}m[/tex]
The helium balloon you got on your birthday has almost entirely deflated, but there is still a very small amount of 4He inside with a total mass of 160 amu. How many protons and how many electrons are present in the balloon (assuming only 4He is there)?
In a helium balloon with a total mass of 160 amu containing helium-4 atoms, there would be 40 helium atoms, resulting in a total of 80 protons and 80 electrons.
Explanation:The question involves calculating the number of protons and electrons in a helium balloon with a total mass of 160 amu, assuming it contains only 4He (helium-4). A helium-4 atom consists of 2 protons, 2 neutrons, and 2 electrons, with a total atomic mass of 4 amu. To find the number of helium atoms, we divide the total mass of helium in the balloon (160 amu) by the mass of a single helium-4 atom, which is 4 amu. This gives us 40 helium atoms.
Since each helium atom has 2 protons and 2 electrons, to find the total number of protons and electrons in 40 helium atoms, we multiply the number of helium atoms by the number of protons or electrons per atom. Therefore, there are 80 protons and 80 electrons in the helium balloon.
A glow-worm of mass 5.0 g emits red light (650 nm) with a power of 0.10 W entirely in the backward direction. To what speed will it have accelerated after 10 y if released into free space and assumed to live
Answer:
a = Np/10 yrs×[3.50^7 yrs /sec]
Explanation:
The energy of the single photon of frequency f or wave length λ is given as
E = hc / λ
since the glow warm emits energy 0.1 J/sec
that is the number of photons n emitted by the photon per sec will be
n = 0.1 W / E
Thus, the number of photons emitted in 10 years
N = n×3.15×10^7 sec/yr ×10 yr
Now, momentum associated with each photon
p= h / λ
and, momentum associated with N photon particles
P= N(h/λ)
hence the change in the momentum of the glow is = Np in 10 years
Therefore, acceleration of the glow
a = Np/10 yrs×[3.50^7 yrs /sec]
Answer:
The speed is 21.06 m/s.
Explanation:
Given that,
mass of glow worm = 5.0 g
Wavelength = 650 nm
Power = 0.10 W
Time = 10 years
The total energy emitted in a period [tex]\tau[/tex] is [tex]P\tau[/tex]
The energy of single photon of frequency or wavelength is
[tex]E=\dfrac{hc}{\lambda}[/tex]
The total number of photons emitted in a interval [tex]\tau[/tex] is then the total energy divided by the energy per photon.
[tex]N=\dfrac{P\tau}{E}[/tex]
[tex]N=\dfrac{P\tau\times\lambda}{hc}[/tex]
[tex]N=\dfrac{P\tau\times\lambda}{hc}[/tex]
We need to calculate the speed
Using de Broglie's relation applies to each photon and thus the total momentum imparted to the glow-worm
[tex]p=\dfrac{Nh}{\lambda}[/tex]
[tex]p=\dfrac{\dfrac{P\tau\times\lambda}{hc}\times h}{\lambda}[/tex]
[tex]p=\dfrac{P\tau}{c}[/tex]
[tex]v=\dfrac{P\tau}{mc}[/tex]
Put the value into the formula
[tex]v=\dfrac{0.10\times3.16\times10^{8}}{5.0\times10^{-3}\times3\times10^{8}}[/tex]
[tex]v=21.06\ m/s[/tex]
Hence, The speed is 21.06 m/s.
In each case indicate whether the quantity in question increased, decreased or stayed the same when the string length is increased. Assume that the tension is unchanged. The function generator is kept at the same frequency, and the string is in resonance in all cases. Part A Number of antinodes ___ Number of antinodes ___ increased. decreased. stayed the same. Request Answer Part B Wavelength ___ Wavelength ___ increased. decreased. stayed the same. Request Answer Part C Fundamental frequency ___ Fundamental frequency ___ increased. decreased. stayed the same. Request Answer Part D Fundamental wavelength ___ Fundamental wavelength ___ increased. decreased. stayed the same. Request Answer Part E Wave speed ___ Wave speed ___ increased. decreased. stayed the same.
Answer:
a) the number of antinodes increases , b) wavelength produced is constant.
, c) fundamental frequency is constant., d) fundamental wavelength does not change, e) the speed of the wave is constant
Explanation:
This is a resonance problem where we have a frequency generator, attached to a chain with a weight in its final part, at the two ends there is a node, point if movement. The condition for resonance of this system is
λ = 2 L / n
Where n is an integer
L = n λ / 2
Let's review the problem questions.
A) As the length of the chain increases the number of wavelengths (Lam / 2) should increase, so the number of antinodes increases
B) when seeing the first equation the wavelength remains the same since the change in the length of the chain and the change in the number between are compensated, therefore for a configuration of generator frequency and weight applied the wavelength produced is constant.
C) the speed of the wave is
v = λ f
In a string the speed is constant for a fixed applied weight, with the wavelength we did not change, therefore the fundamental frequency must also be constant.
D) The value of the fundamental wavelength does not change if the weight does not change, but there is a minimum chain length for this resonance to be observed and corresponds to n = 1
L = λ / 2
E) the speed of the wave depends on the chain tension and its density if they do not change the speed of the wave does not change either
A sphere completely submerged in water is tethered to the bottom with a string. The tension in the string is one-fourth the weight of the sphere.
What is the density of the sphere?
Answer:
ρ = 800 kg/m³
Explanation:
Let the volume of the sphere = V
and the density of the sphere = ρ
density of water , ρ_w = 1000 Kg/m^3
tension in the string is one-fourth the weight of the sphere.
Tension in the rope ,
[tex]T = \rho V \dfrac{g}{4}[/tex]
for the sphere to in equilibrium
T + Weight = buoyant force
[tex]\rho V\dfrac{g}{4} + \rho V g = \rho_w V g[/tex]
[tex] \dfrac{\rho}{4}+\rho= \rho_w[/tex]
[tex] \dfrac{5\rho}{4}= 1000[/tex]
ρ = 800 kg/m³
density of the sphere is equal to 800 kg/m³
Final answer:
To calculate the density of the sphere tethered underwater, we apply Archimedes' principle and equilibrium of forces. Since the tension is one-fourth the sphere's weight, the sphere's density is found to be one-third more than the water's density.
Explanation:
To find the density of the sphere, let's consider the forces acting on the sphere when it is submerged in water. According to Archimedes' principle, the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object.
The weight of the sphere (W) can be expressed as the product of its volume (V), its density (p_sphere), and the acceleration due to gravity (g). It can be written as W = V * p_sphere * g. Similarly, the weight of the water displaced by the sphere, which is also the buoyant force (B), is B = V * p_water * g, where p_water is the density of water.
Since the tension (T) in the string is one-fourth the weight of the sphere, we can write T = 1/4 * W. The sphere is in equilibrium, so the sum of the forces equals zero, leading to B = W - T. Substituting in the values, we get (V * p_water * g) = V * p_sphere * g - (1/4 * V * p_sphere * g). From this equation, we can solve for the density of the sphere p_sphere.
After simplifying, p_sphere = 4/3 * p_water, so the sphere's density is one-third more than the density of water.
Which of the following phenomena are due to the electric interaction? (Select all that apply.) surface tension in water friction between tires and pavement the elliptical orbit of comets dissolution of salt in water the binding of protons in an atomic nucleus
Answer:
Surface tension in water
Friction between tires and pavement
Dissolution of salt in water
Explanation:
Surface tension in water: It is due to the electrostatic force of attraction (cohesive force) between water molecules.
Friction between tires and pavement: It is due to the attractive force between tires and pavement.
Dissolution of salt in water: The ions of [tex]Na ^ +[/tex] and [tex]Cl ^ -[/tex] separate due to the strong attraction of water molecules.
Final answer:
Surface tension in water, friction between tires and pavement, and the dissolution of salt in water are phenomena due to electric interaction, which is rooted in the forces of cohesion, adhesion, and intermolecular attractions.
Explanation:
The phenomena that are due to electric interaction include surface tension in water, friction between tires and pavement, and the dissolution of salt in water. The electric interaction plays a crucial role in these phenomena through the forces of cohesion, adhesion, and intermolecular attractions, which are manifestations of the electromagnetic force. While the elliptical orbit of comets is primarily governed by gravitational forces and the binding of protons in an atomic nucleus by the strong nuclear force, surface tension, friction, and dissolution phenomena are deeply rooted in electrical interactions between atoms and molecules.
What is the magnitude of the electric force between a proton and an electron when they are at a distance of 4.09 angstrom from each other?
Answer:
[tex]F=1.38*10^{-9}N[/tex]
Explanation:
According to Coulomb's law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:
[tex]F=\frac{kq_1q_2}{d^2}[/tex]
Here k is the Coulomb constant. In this case, we have [tex]q_1=-e[/tex], [tex]q_2=e[/tex] and [tex]d=4.09*10^-10m[/tex]. Replacing the values:
[tex]F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)(1.6*10^{-19}C)}{(4.09*10^{-10})^2}\\F=-1.38*10^{-9}N[/tex]
The negative sign indicates that it is an attractive force. So, the magnitude of the electric force is:
[tex]F=1.38*10^{-9}N[/tex]
An engine does 25 J of work and exhausts 20 J of waste heat during each cycle. If the cold-reservoir temperature is 30 ∘C, what is the minimum possible temperature in ∘C of the hot reservoir?
Final answer:
The minimum possible temperature of the hot reservoir for an engine that does 25 J of work and exhausts 20 J of waste heat in each cycle, with a cold-reservoir temperature of 30°C is 408.87°C.
Explanation:
To find the minimum possible temperature of the hot reservoir for an engine that does 25 J of work and exhausts 20 J of waste heat during each cycle, with a cold-reservoir temperature of 30 °C, we have to use the concept of efficiency and the Carnot engine.
Firstly, let's convert the cold-reservoir temperature from Celsius to Kelvin:
Temperature in Kelvin (K) = Temperature in Celsius (°C) + 273.15
Tc (cold reservoir) = 30°C + 273.15 = 303.15 K
Next, we can calculate the efficiency (ε) of the engine using the formula:
ε = Work done (W) / Heat absorbed (Qh)
Qh (heat absorbed) = W (work done) + Qc (waste heat) = 25 J + 20 J = 45 J
So, ε = 25 J / 45 J = 0.555... (repeating)
The efficiency of a Carnot engine is also given by:
ε = 1 - (Tc/Th)
Now, we solve for Th (the hot reservoir temperature in Kelvin):
Th = Tc / (1 - ε)
Th = 303.15 K / (1 - 0.555...) = 303.15 K / 0.444... = 682.02 K
Finally, we convert the hot reservoir temperature back to Celsius:
Temperature in Celsius (°C) = Temperature in Kelvin (K) - 273.15
Th (hot reservoir) in °C = 682.02 K - 273.15 = 408.87°C
Thus, the minimum possible temperature of the hot reservoir is 408.87°C.
When a third object is brought in contact with the first object (after it gains the electrons), the resulting charge on the third object is 0.9 C. What was its initial charge (in C)?
Answer:
- Contact 1 with 3 , initial charge of 1.8 C.
- contact 1 with 2 and then 1 with 3 , first body should have 3.6 C
Explanation:
The excess charge on a body is distributed evenly throughout the body.
We can have two different configurations:
- Contact 1 with 3
When the third body was touched with the first, the initial charge was distributed between the two, so that when each one separated, it had half the charge, in this configuration the first body should have an initial charge of 1.8 C.
- contact 1 with 2 and then 1 with 3
Another possible configuration of the exercise is that the first body touches the second and the charge decrease to the half and then touches the third where it again decreases by half, so that the first body only gives it every ¼ of its initial load.
Therefore in this configuration if the third body has a load of 0.9C the first body should have 3.6 C
Two identical hard spheres, each of mass m and radius r, are released from rest in otherwise empty space with their centers separated by the distance R. They are allowed to collide under the influence of their gravitational attraction. (a) Find the magnitude of the impulse received by each sphere before they make contact. (b) Find the magnitude of the impulse each receives during their contact if they collide elastically.
Answer:
a) I = √ (2G m³ (1/2r³ - 1/R)), b) I = √ (8 G m³ (1/2r -1/R))
Explanation:
.a) The relation of the Impulse and the moment is
I = Δp = m [tex]v_{f}[/tex] - m v₀
We can use Newton's second law with force the force of universal attraction
F = ma
G m m / r² = m a
dv / dt = G m / r²
Suppose re the direction where the spheres move is x
dv/dx dx/dt = G m / x²
dv/dx v = G m / x²
v dv = G m dx / x²
We integrate
v² / 2 = Gm (-1 / x)
We evaluate this integra from the lower limit v = 0 for x = R to the upper limit, where the spheres v = v and x = 2r are touched
v² / 2-0 = G M (-1 / R + 1 / 2r)
v = √ [2Gm (1 /2r - 1/ R) ]
The impulse on the sphere is
I = m vf - m v₀
I = m vf - 0
I = m √ (2Gm (1 / 2r-1 / R)
I = √ (2G m³ (1/2r³ - 1/R))
b) during the crash each sphere arrives with a velocity v and leaves with a velocity –v, the same magnitude but opposite direction
I = m [tex]v_{f}[/tex]- m v₀
I = m v - m (-v)
I = 2mv
I = 2m √ (2Gm (1 / 2r-1 / R)
I = √ (8 G m³ (1/2r -1/R))
We calculate the impulse received by each sphere before they make contact and during their elastic collision.
Explanation:In this scenario, we have two identical hard spheres separated by a distance R. The spheres are released from rest and allowed to collide under the influence of their gravitational attraction. We need to find the magnitude of the impulse received by each sphere before they make contact and during their elastic collision.
(a) Before the spheres make contact, the magnitude of the impulse received by each sphere can be found using the equation:
Impulse = Change in momentum = Mass x Change in velocity.
Since both spheres are released from rest, their initial velocities are zero. Therefore, the change in velocity is the final velocity. Using the equation:
Final velocity = sqrt(2 x G x m / R).
Substituting the values, we can calculate the magnitude of the impulse received by each sphere before they make contact.
(b) During their elastic collision, the magnitude of the impulse received by each sphere can be found using the equation:
Impulse = Change in momentum = Mass x Change in velocity.
Since the spheres collide elastically, there is no change in velocity. Therefore, the magnitude of the impulse received by each sphere during their contact is zero.
Learn more about Impulse received by spheres here:https://brainly.com/question/31502497
#SPJ12
Consider two waves moving past you at the same speed. The wavelength of wave A is half that of wave B. You then know that ____.
The frequency of wave B is half the frequency of wave A
Explanation:
The wavelength, the speed and the frequency of a wave are related by the wave equation:
[tex]v=f \lambda[/tex]
where
v is the speed of the wave
f is its frequency
[tex]\lambda[/tex] is its wavelength
The equation can also be rewritten as
[tex]f=\frac{v}{\lambda}[/tex]
In this problem, we have wave A with wavelength [tex]\lambda_A[/tex] and speed v, so its frequency is
[tex]f_A=\frac{v}{\lambda_A}[/tex]
Then we have wave B, whose wavelength is twice that of wave A:
[tex]\lambda_B = 2 \lambda_A[/tex]
And its speed is the same; Therefore, its frequency is
[tex]f_B = \frac{v}{\lambda_B}=\frac{v}{2\lambda_A}=\frac{1}{2}(\frac{v}{\lambda_A})=\frac{f_A}{2}[/tex]
So, the frequency of wave B is half that of wave A.
Learn more about wavelength and frequency:
brainly.com/question/5354733
brainly.com/question/9077368
#LearnwithBrainly
There is a 9.0 earthquake that just hit San Diego! There have been multiple aftershocks at a 6.0 magnitude. How much weaker are these aftershocks compared to the original earthquake?
a. 3000x weaker
b. 1000x weaker
c. 100x weaker
d. 10x weaker
The magnitude is used to quantify the size of the earthquakes (measures the energy released during the breakdown of a fault) while the intensity is a qualitative description of the effects of the earthquakes (it involves the perception of people as well as damage material and economic suffered due to the event). The relation between intensity and magnitude is
[tex]\frac{I_1}{I_2} =10^{(M1-M2)}[/tex]
Here,
M = Magnitude
I = Intensity of each one
Given
[tex]M_1 = 9[/tex]
[tex]M_2 =6[/tex]
Replacing,
[tex]\frac{I_1}{I_2} =10^{(9-6)}[/tex]
[tex]\frac{I_1}{I_2} = 10^3[/tex]
[tex]\frac{I_1}{I_2} = 1000[/tex]
So the aftershocks are 1000x weaker compared to the original earthquake. The correct answer is B.
Callisto, one of Jupiter's moons, has an orbital period of 16.69 days and its distance from Jupiter is 1.88*10^6 km. What is Jupiter's mass?
Answer:
The Jupiter´s mass is approximately 1.89*10²⁷ kg.
Explanation:
The only force acting on Calisto while is rotating around Jupiter, is the gravitational force, as defined by the Newton´s Universal Law of Gravitation:
Fg = G*mc*mj / rcj²
where G = 6.67*10⁻¹¹ N*m²/kg², mc= Callisto´s mass, mj= Jupiter´s mass, and rcj = distance from Jupiter for Callisto= 1.88*10⁹ m.
At the same time, there exists a force that keeps Callisto in orbit, which is the centripetal force, that actually is the same gravitational force we have already mentioned.
This centripetal force is related with the period of the orbit, as follows:
Fc = mc*(2*π/T)²*rcj.
In order to be consistent in terms of units, we need to convert the orbital period, from days to seconds, as follows:
T = 16.69 days* 86,400 (sec/day) = 1.44*10⁶ sec.
We have already said that Fg= Fc, so we can write the following equality:
G*mc*mj / rcj² = mc*(2*π/T)²*rcj
Simplifying common terms, and solving for mj, we get:
mj = 4*π²*(1.88*10⁹)³m³ / ((1.44*10⁶)² m²*6.67*10⁻11 N*m²/kg²)
mj = 1.89*10²⁷ kg.
Answer: Mass of Jupiter ~= 1.89 × 10^23 kg
Explanation:
Given:
Period P= 16.69days × 86400s/day= 1442016s
Radius of orbit a = 1.88×10^6km × 1000m/km
r = 1.88 × 10^9 m
Gravitational constant G= 6.67×10^-11 m^3 kg^-1 s^-2
Applying Kepler's third law, which is stated mathematically as;
P^2 = (4π^2a^3)/G(M1+M2) .....1
Where M1 and M2 are the radius of Jupiter and callisto respectively.
Since M1 >> M1
M1+M2 ~= M1
Equation 1 becomes;
P^2 = (4π^2a^3)/G(M1)
M1 = (4π^2a^3)/GP^2 .....3
Substituting the values into equation 3 above
M1 = (4 × π^2 × (1.88 × 10^9)^3)/(6.67×10^-11 × 1442016^2)
M1 = 1.89 × 10^27 kg
All valid equations in physics have consistent units. Are all equations that have consistent units valid?
a. No. Any equation can be made to have consistent units through unit conversion.
b. Yes. Consistent units indicate that the equation was derived correctly.
c. No. An equation may have consistent units but still be numerically invaid.
d. Yes. In physics, consistent units guarantee that both sides of an equation represent the same physical quanity
Answer:
c. No. An equation may have consistent units but still be numerically invaid.
Explanation:
For an equation to be corrected, it should have consistent units and also be numerically correct.
Most equation are of the form;
(Actual quantity) = (dimensionless constant) × (dimensionally correct quantity)
From the above, without the dimensionless constant the equation would be numerically wrong.
For example; Kinetic energy equation.
KE = 0.5(mv^2)
Without the dimensionless constant '0.5' the equation would be dimensionally correct but numerically wrong.
Answer:
No. An equation may have consistent units but still be numerically invalid.
Explanation:
A 2,537-kg truck moving at 14 m/s strikes a car waiting at a traffic light, hooking bumpers. The two continue to move together at 8 m/s. What was the mass (in kg) of the struck car
Answer:
1902.75 kg
Explanation:
From Law of conservation of momentum,
m₁u₁ + m₂u₂ = V (m₁ + m₂).................... Equation 1
make m₂ the subject of the equation,
m₂ = (m₁V - m₁u₁)/(u₂-V)..................... Equation 2
Where m₁ = mass of the truck, m₂ = mass of the car, u₁ initial velocity of the truck, u₂ = initial velocity of the car V = common velocity
Given: m₁ = 2537 kg, u₁ = 14, V= 8 m/s, u₂ = 0 m/s ( as the car was at rest waiting at a traffic light)
Substituting into equation 2.
m₂ =[2537(8) - 2537(14)]/(0-8)
m₂ = (20296-35518)/-8
m₂ = -15222/-8
m₂ = 1902.75 kg.
Thus the mass of the car = 1902.75 kg
At 20 m from a localized sound source you measure the intensity level as 75 dB. How far away must you be for theperceived loud-ness to drop in half [i.e., to an intensity level of 65 dB)?
Answer: 21.48m
Explanation:
Using the inverse square law, the intensity of sound is inversely proportional to the square of the distance.
Intensity ∝ 1/distance²
let l represent intensity and d represent distanc
l₁/l₂ = d₂²/d₁² ......1
given;
l₁ = 75dB
l₂ = 65dB
d₁ = 20 m
substituting into eqn 1, we have;
75/65 = d₂²/20²
d₂² = 75 × 20²/65
d₂² = 461.54
d₂ = √461.54
d₂ = 21.48m
When an object such as a plastic comb is charged by rubbing it with a cloth, the net charge is typically a few microcoulombs. If that charge is 4.0 ?C , by what percentage does the mass of a 33 g comb change during charging?
The concept required to solve this problem is quantization of charge.
First the number of electrons will be calculated and then the total mass of the charge.
With these data it will be possible to calculate the percentage of load in the mass.
[tex]Q= ne[/tex]
Here Q is the charge, n is the number of electrons and e is the charge on the electron
[tex]n = \frac{Q}{e}[/tex]
Replacing,
[tex]n = \frac{4*10^{-6}C}{1.6*10^{-19}}[/tex]
[tex]n = 2.5 * 10^{13}77[/tex]
According to the quantization of charge the charge is defined as product of the number of electron and the charge on the electron
The total mass of the charge is
[tex]m= nm_e[/tex]
Here,
m = Mass of the charge
n = Number of electrons
[tex]m_e[/tex] = Mass of the electron
[tex]\text{Percentage change} = \frac{nm_e}{M}*100[/tex]
Replacing we have
[tex]\text{Percentage change} = \frac{(2.5*10^13)(9.1*10^{-28})}{33}*100[/tex]
[tex]\text{Percentage change} = 6.9*10^{-14} \%[/tex]
The percentage change in the mass of the 33 g comb during charging is extremely small, approximately [tex]\(6.91 \times 10^{-16}\%\),[/tex] which is negligible.
To find the percentage change in mass of the comb due to charging, we need to understand the relationship between the amount of charge transferred and the mass of the electrons involved. Here's the step-by-step process:
1. Calculate the number of electrons corresponding to the given charge:
The charge of one electron[tex](\(e\))[/tex] is approximately[tex]\(1.6 \times 10^{-19}\)[/tex] coulombs.
Given:
- Net charge [tex](\(Q\)) = 4.0 μC = \(4.0 \times 10^{-6}\) C[/tex]
The number of electrons n transferred can be calculated using the formula:
[tex]\[ n = \frac{Q}{e} \][/tex]
[tex]\[ n = \frac{4.0 \times 10^{-6} \text{ C}}{1.6 \times 10^{-19} \text{ C/electron}} \][/tex]
[tex]\[ n = 2.5 \times 10^{13} \text{ electrons} \][/tex]
2. Calculate the mass of the transferred electrons:
The mass of one electron [tex](\(m_e\))[/tex] is approximately [tex]\(9.11 \times 10^{-31}\)[/tex] kg.
The total mass[tex](\(m\))[/tex] of the electrons transferred is:
[tex]\[ m = n \times m_e \][/tex]
[tex]\[ m = 2.5 \times 10^{13} \times 9.11 \times 10^{-31} \text{ kg} \][/tex]
[tex]\[ m = 2.28 \times 10^{-17} \text{ kg} \][/tex]
3. Convert the mass of the comb to kilograms:
Given:
- Mass of the comb = 33 g = 0.033 kg
4. Calculate the percentage change in mass:
The percentage change in mass is given by:
[tex]\[ \text{Percentage change} = \left( \frac{\text{Change in mass}}{\text{Original mass}} \right) \times 100\% \][/tex]
[tex]\[ \text{Percentage change} = \left( \frac{2.28 \times 10^{-17} \text{ kg}}{0.033 \text{ kg}} \right) \times 100\% \][/tex]
[tex]\[ \text{Percentage change} = 6.91 \times 10^{-16} \% \][/tex]
In a binary-star system that produces a nova, the white dwarf pulls matter from the companion star. The matter forms an accretion disk that orbits the white dwarf. Then a specific sequence of events must take place for a nova event to occur. Rank the steps leading up to the observed nova event in chronological order from first to last.
As a head-up, it is important to notice that a white dwarf only shines thanks to the stored energy and light, because a white dwarf doesn't have any hydrogen left to perform nuclear fusion.
Now the process:
First, the white dwarf accumulates all the extracted matter from its companion, onto its own surface. This extra matter increases the white dwarf's temperature and density.
After a while, the star reaches about 10 million K, so nuclear fusion can begin. The hydrogen that has been "stolen" from the other star and accumulated in the white dwarf's surface it's used for the fusion, dramatically increasing the star's brightness for a short time, causing what we know as a Nova.
As this fuel its quickly burnt out or blown into space, the star goes back to its natural white dwarf state. Since the white dwarf nor the companion star are destroyed in this process, it can happen countless of times during their lifespan.
Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.
(a) What would be the height of the atmosphere if the air density were constant?
km
(b) What would be the height of the atmosphere if the air density decreased linearly to zero with height?
Answer
Pressure, P = 1 atm
air density, ρ = 1.3 kg/m³
a) height of the atmosphere when the density is constant
Pressure at sea level = 1 atm = 101300 Pa
we know
P = ρ g h
[tex]h = \dfrac{P}{\rho\ g}[/tex]
[tex]h = \dfrac{101300}{1.3\times 9.8}[/tex]
h = 7951.33 m
height of the atmosphere will be equal to 7951.33 m
b) when air density decreased linearly to zero.
at x = 0 air density = 0
at x= h ρ_l = ρ_sl
assuming density is zero at x - distance
[tex]\rho_x = \dfrac{\rho_{sl}}{h}\times x[/tex]
now, Pressure at depth x
[tex]dP = \rho_x g dx[/tex]
[tex]dP = \dfrac{\rho_{sl}}{h}\times x g dx[/tex]
integrating both side
[tex]P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx[/tex]
[tex] P =\dfrac{\rho_{sl}\times g h}{2}[/tex]
now,
[tex]h=\dfrac{2P}{\rho_{sl}\times g}[/tex]
[tex]h=\dfrac{2\times 101300}{1.3\times 9.8}[/tex]
h = 15902.67 m
height of the atmosphere is equal to 15902.67 m.
Find the Cartesian coordinates for the point whose distance from the origin is 2 m and the angle measured from the positive horizontal axis is 15°
x =
y =
The Cartesian coordinates for the point are approximately: x ≈ 1.93185 m and, y ≈ 0.523599 m.
Here, we have to find the Cartesian coordinates (x, y) for a point given its distance from the origin (r) and the angle (θ) measured from the positive horizontal axis, we can use trigonometric functions.
In this case, r = 2 m and θ = 15°.
The Cartesian coordinates can be calculated as follows:
x = r * cos(θ)
y = r * sin(θ)
Given:
r = 2 m
θ = 15°
First, convert the angle from degrees to radians, since trigonometric functions in most programming languages use radians:
θ_rad = 15° * (π / 180) ≈ 0.261799 radians
Now, calculate the coordinates:
x = r * cos(θ_rad) = 2 m * cos(0.261799) ≈ 1.93185 m
y = r * sin(θ_rad) = 2 m * sin(0.261799) ≈ 0.523599 m
So, the Cartesian coordinates for the point are approximately:
x ≈ 1.93185 m
y ≈ 0.523599 m
learn more on Cartesian coordinates
https://brainly.com/question/13024495
#SPJ12
The Cartesian coordinates for the point with a distance of 2 m from the origin and an angle of 15° from the positive horizontal axis are approximately (1.93 m, 0.52 m), calculated using the cosine and sine of the angle.
Explanation:To find the Cartesian coordinates for a point given its distance from the origin and the angle from the positive horizontal axis, you can use trigonometric functions. Since the distance from the origin is 2 m and the angle is 15°, you can calculate the x and y coordinates using the cosine and sine functions, respectively.
The x-coordinate is found by multiplying the distance by the cosine of the angle: x = 2 m × cos(15°). The y-coordinate is found by multiplying the distance by the sine of the angle: y = 2 m × sin(15°).
Using standard values for cos(15°) and sin(15°), the calculations would yield:
x = 2 m × 0.9659 (approximately) = 1.9318 my = 2 m × 0.2588 (approximately) = 0.5176 mTherefore, the Cartesian coordinates are approximately (1.93 m, 0.52 m).
Jane is pulling a chain of 2 boxes with a massless rope at an angle of 20 degrees above the horizontal with a force of 30N. Jane and the boxes are all on horizontal ground. The boxes are connected horizontally by a length of massless rope. The box closest to Jane is 45 kg and the box farthest from Jane is 30kg.
a. Draw a picture of the situation.
b. Draw a free-body diagram for each box.
c. What is the normal force acting on the 45kg box?
d. What is the acceleration of the boxes?
e. What is the tension in the rope connecting the boxes?
In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, 1.3 m above the water level, onto the surface of the water at a point d = 2.8 m from his foot at the edge of the pool a) Where does the spot of light hit the bottom of the h = 2.1-m-deep pool? Measure from the bottom of the wall beneath his foot.
According to the question,
Length, [tex]l_1 = 2.1 \ m[/tex]Height, [tex]h_1 = 1.3 \ m[/tex] Distance, [tex]d = 2.8 \ m[/tex]The angle of incidence will be:
→ [tex]tan \Theta_1= \frac{l_1}{h_1}[/tex]
[tex]= \frac{2.1}{1.3}[/tex]
[tex]= 2.076[/tex]
then,
[tex]\Theta_1 = 64.3^{\circ}[/tex]
From air into water, we get
→ [tex]n_{air} Sin \Theta_1 = n_{water} Sin \Theta_2[/tex]
[tex](1.00) Sin 64.3^{\circ} = (1.3) Sin \Theta_2[/tex]
[tex]\Theta_2 = 42.6^{\circ}[/tex]
hence,
The horizontal distance,
→ [tex]l_2 = l_1+h_2 tan \Theta[/tex]
By substituting the values, we get
[tex]= 2.1+(2.1)tan 42.6[/tex]
[tex]= 4.6 \ m[/tex]
Thus the above approach is right.
Learn more about distance here:
https://brainly.com/question/22371581
At takeoff, a commercial jet has a 60.0 m/s speed. Its tires have a diameter of 0.850 m.(a) At how many rev/min are the tires rotating?(b) What is the centripetal acceleration at the edge of the tire?(c) With what force must a determined 1.00×10−15 kg bacterium cling to the rim?(d) Take the ratio of this force to the bacterium's weight.
Answer:
(a) Angular speed = 1344.8rev/min
(b) Centripetal acceleration = 8473.4m/s^2
(c) Force = 8.4734×10^-12N
(d) Ratio of force to bacterium's weight is 864.6
Explanation:
(a) Angular speed (w) = v/r
v = 60m/s, d = 0.850m, r = 0.850m/2 = 0.425m
w = 60/0.425 = 141.2rad/s = 141.2×9.524rev/min = 1344.8rev/min
(b) Centripetal acceleration = w^2r = 141.2^2 × 0.425 = 8473.4m/s^2
(c) Force = mass × centripetal acceleration = 1×10^-15 × 8473.4 = 8.4734×10^-12N
(d) Bacterium's weight = mass × acceleration due to gravity = 1×10^-15 × 9.8 = 9.8×10^-15N
Ratio of force to bacterium's weight = 8.4734×10^-12N/9.8×10^-15N = 864.6
The commercial jet's tires' rotation speed is calculated using their speed and diameter. The centripetal acceleration at the tire's edge is calculated using the speed and radius. The force with which a tiny bacterium clings to the rim is found using its mass and the centripetal acceleration.
Explanation:We start by calculating the rotation speed of the tires. To do this, we use the known values of tire speed (60 m/s) and diameter (0.85 m). The first step is to convert speed into distance travelled per revolution, and diameter into circumference, then divide speed by circumference to get revolutions per second. We convert this into revolutions per minute for the answer to part (a). For part (b), we use the formula for centripetal acceleration, which involves the square of speed and the radius (half of the tire's diameter) to find acceleration at the rim. For part (c), we find the force exerted by the bacterium by multiplying its mass by the centripetal acceleration. Part (d) compares this force to the bacterium's weight, which is found by multiplying its mass by the acceleration due to gravity.
Learn more about Physics of Rotation here:https://brainly.com/question/34647357
#SPJ3
If a car is moving with a velocity v, the distance the car takes to come to a halt is ___________ the distance that same car would take if it was starting with a velocity 2v.
Answer:
one forth.
Explanation:
car is moving at the speed of v
car stops, final speed = s
distance to stop the car = ?
using equation of motion
u_i² = u_f² + 2 a s
0² = v² - 2 a s
-ve sign is used because the car is decelerating.
[tex]s = \dfrac{v^2}{2a}[/tex]
now, if the velocity of the car is 2v distance to stop
[tex]s' = \dfrac{(2v)^2}{2a}[/tex]
[tex]s' = 4\dfrac{v^2}{2a}[/tex]
[tex]s' = 4 s[/tex]
[tex]s = \dfrac{s'}{4}[/tex]
now, the distance is one forth.
so, car with speed v has to cover one forth of the distance cover by car with speed 2 v.
What is the distance between the motion sensor and the object?
The motion sensor acts as a transmitter, when it sends out pulses, and as a receiver, when it listens for echoes. The delay t between sending a pulse and receiving the echo is equal to 8.8 ms (=0.0088 s). The air temperature id 20 degrees Celsius.
Answer:
1.5092 m
Explanation:
Time taken by the wave for the round trip is 8.8 ms
Velocity of sound at 20 degrees Celsius is 343 m/s
The distance to the object will be given by the one way trip
Time taken for one way trip is
[tex]\dfrac{8.8}{2}=4.4\ ms[/tex]
Distance is given by
[tex]Distance=Speed\times time\\\Rightarrow Distance=343\times 4.4\times 10^{-3}\\\Rightarrow Distance=1.5092\ m[/tex]
The distance between the motion sensor and the object is 1.5092 m
The distance between the motion sensor and the object is 1.5092 m
What is a motion sensor?The motion sensor serves as a transmitter for pulses when it sends out pulses, and as a receiver, when it listens for echoes.
From the parameters given:
The time delay = 0.0088 sThe speed of sound (s) = 343 m/sThe temperature of air = 20° CThe distance traveled by the signal is:
d = s × Δ t
d = 343 m/s × 0.0088s
d = 3.0184 m
Provided that the distance is the circular trip between the motion sensor and the object,
Therefore, we can conclude that the distance between the motion sensor and the object will be equal to half the distance traveled by the signal.
i.e.
= 3.0184 m/2
= 1.5092 m
Learn more about motion sensors here:
https://brainly.com/question/13478226
. The mass flow rate through the nozzle of a rocket engine is 200 kg/s. The areas of the nozzle inlet and exit planes are 0.7 and 2.4 m2, respectively. On the nozzle inlet plane, the pressure and velocity are 1600 kPa and 150 m/s, respectively, whereas on the nozzle exit plane, the pressure and velocity are 80 kPa and 2300 m/s, respectively. Find the thrust force acting on the nozzle.
To solve this problem we will define the data obtained in each of the sections. We know that the Net Force is equivalent to the Force in section 1, which can be found through mass flow and velocity, plus the force in section two, which can be found as the product between pressure and Area, so therefore we have
State 1:
[tex]\dot{m} = 200kg/s[/tex]
[tex]v_1 = 150m/s[/tex]
[tex]P_1 = 1600kPa[/tex]
State 2
[tex]v_2 = 2300m/s[/tex]
[tex]A_2 = 2.4m^2[/tex]
[tex]P_2 = 80kPa[/tex]
We have that net force is equal to
[tex]F_{net} = F_1 + F_2[/tex]
[tex]F_1 = \dot{m} (v_2-v_1)[/tex]
[tex]F_2 = (P_1-p_2)A_2[/tex]
Replacing,
[tex]F = \dot{m} (v_2-v_1) + (P_1-p_2)A_2[/tex]
[tex]F = 200(2300-150)+(1600-80)*10^3(2.4)[/tex]
[tex]F = 4078kN[/tex]
Therefore the thrust force acting on the nozzle is 4078kN
If three uncharged styrofoam balls are placed together and agitated so that one gains 3 CC of charge and another gains 4 CC of charge, how much charge must there be on the third one
Answer:
-7 C
Explanation:
Assuming that object other than the styrofoam balls was part of the charge transfer. In order to maintain charge balance, the initial charge of the system must equal the ending charge. If all balls were uncharged initially, the ending charge on the third ball must be:
[tex]C_1+C_2+C_3 = 0\\3+4+C_3=0\\C_3=-7[/tex]
There must be -7 C of charge on the third ball.
Determine the magnitude of force F so that the resultant FR of the three forces is as small as possible.
Answer: FR=2.330kN
Explanation:
Write down x and y components.
Fx= FSin30°
Fy= FCos30°
Choose the forces acting up and right as positive.
∑(FR) =∑(Fx )
(FR) x= 5-Fsin30°= 5-0.5F
(FR) y= Fcos30°-4= 0.8660-F
Use Pythagoras theorem
F2R= √F2-11.93F+41
Differentiate both sides
2FRdFR/dF= 2F- 11.93
Set dFR/dF to 0
2F= 11.93
F= 5.964kN
Substitute value back into FR
FR= √F2(F square) - 11.93F + 41
FR=√(5.964)(5.964)-11.93(5.964)+41
FR= 2.330kN
The minimum force is 2.330kN
To determine the magnitude of force F to minimize the resultant FR of three forces, we can use the rule for finding the magnitude of a vector. By taking the square root of the sum of the squares of the components, we can find the magnitude of F.
The question is asking to determine the magnitude of force F so that the resultant FR of the three forces is as small as possible. To find the magnitude of the resultant force, we need to use the rule for finding the magnitude of a vector. We take the square root of the sum of the squares of the components. In this case, we have F = F₁ + F₂, and we can plug in the values given to find the magnitude of F.
For more such questions on magnitude of force, click on:
https://brainly.com/question/33958444
#SPJ3
Archimedes' principle can be used not only to determine the specific gravity of a solid using a known liquid; the reverse can be done as well. As an example, a 4.00-kg aluminum ball has an apparent mass of 2.10 kg when submerged in a particular liquid: calculate the density of the liquid in kg/m^3
Derive a formula for determining the density of a liquid using this procedure.
Express your answer in terms of the variables mobject, mapparent, and rhoobject.
The density of a fluid in which an object is submerged can be determined using the original mass of the object, the apparent submerged mass of the object, and the density of the object, using the formula provided above. Applying this formula to a 4.00-kg aluminum ball submersed in a liquid and appearing to be 2.10 kg yields a liquid density of 1247 kg/m^3.
Explanation:According to Archimedes' Principle, the buoyant force on a submerged object is equal to the weight of the fluid displaced by the object. From this principle, if an object is fully submerged in a fluid then the volume of the fluid displaced is equal to the volume of the object.
To calculate the density of a liquid, using the mass of the object, the apparent mass, and the density of the object, the formula commonly used is:
ρliquid = (mobject - mapparent) / (mobject / ρobject - mapparent / ρobject).
Let's apply this formula to the situation you described. An aluminum ball of mass 4.00 kg is submerged in a liquid and has an apparent mass of 2.10 kg. The density of aluminum is approximately 2700 kg/m3. Therefore, the density of the liquid is (4.00kg - 2.10kg) / (4.00kg / 2700kg/m3 - 2.10kg / 2700kg/m3) = 1247 kg/m3.
Learn more about Density here:https://brainly.com/question/34199020
#SPJ3
Archimedes' principle states that when a body is submerged in a liquid, it experiences an upward force equal to the weight of the liquid displaced by the object. The magnitude of the buoyant force is equal to the weight of the displaced liquid and is directed upward. The buoyant force can be used to determine the density of an object as well as the density of a liquid. The formula for Archimedes' principle is as follows:
Buoyant force = Weight of displaced fluid = density of fluid x volume of fluid displaced x g
where g is the acceleration due to gravity.
In this case, we can use the formula to determine the density of the liquid as follows:
Buoyant force = Weight of the object - Apparent weight of the object
Density of fluid x Volume of fluid displaced x g = m_object x g - m_apparent x g
Density of fluid = (m_object - m_apparent) / volume of fluid displaced
Now, substituting the given values, we get:
Density of fluid = (4.00 kg - 2.10 kg) / [(4.00 kg - 2.10 kg)/1000 kg/m³]
Density of fluid = 1900 kg/m³
Therefore, the density of the liquid is 1900 kg/m³.
A world-class shotputter can put a 7.26 kg shot a distance of 22 m. Assume that the shot is constantly accelerated over a distance of 2 m at an angle of 45 degrees and is released from a height of 2 m above the ground. Estimate the weight that this athlete can lift with one hand.
To solve this problem we will make a free body diagram to better understand the displacement measurements made by the body. From there we will apply the linear motion kinematic equations that describe the position of the body in reference to its vertical displacement, acceleration and speed. With this speed found we will apply the energy conservation theorem that will allow us to find the Force.
Equation of trajectory of a projectile is
[tex]y = xtan\theta - x^2 \frac{g}{2u^2cos^2\theta}[/tex]
Here
u = Initial velocity
x = Horizontal displacement
g = Acceleration due to gravity
y = Vertical displacement
We have that
[tex]x = 22m[/tex]
[tex]y = -2\sqrt{2m}[/tex]
Replacing we have that,
[tex]-2\sqrt{2} = 22tan45\° -\frac{9.8}{2u^2cos^2 45}(22)^2[/tex]
[tex]-2\sqrt{2} =22 -\frac{9.8*484}{2u^2(1/2)}[/tex]
[tex]-2\sqrt{2} =22 -\frac{4743.2}{u^2}[/tex]
[tex]u^2 = \frac{4743.2}{22+2\sqrt{2}}[/tex]
[tex]u = 191.03m/s[/tex]
From the work energy theorem
[tex]W_{net} = K_f +K_i[/tex]
Here,
[tex]K_f = \frac{1}{2} mu^2[/tex]
[tex]K_i = \frac{1}{2} m(0)^2 = 0[/tex]
[tex]W_{net} = W_m+W_g[/tex]
Where,
[tex]W_m =\text{Work by man} = F_s[/tex]
[tex]W_{g} = \text{Work by gravity} = -mgh[/tex]
Therefore
[tex]F_s -mgh = \frac{1}{2} mu^2[/tex]
[tex]F_s = \frac{m}{s} (\frac{u^2}{2}+gh)[/tex]
[tex]F_s = \frac{7.26}{2\sqrt{2}}(\frac{191.03}{2}+9.8*2)[/tex]
[tex]F_s = 295.477N[/tex]
A rocket has a mass 340(103) slugs on earth. Specify its mass in SI units, and its weight in SI units.
a) Specify mass of the rocket in SI units.
b) Specify weight of the rocket in SI units.
c) If the rocket is on the moon, where the acceleration due to gravity is gm= 5.30ft/s2, determine to three significant figures its weight in SI units.
d) Determine to three significant figures its mass on the moon in SI units.
Answer:
[tex]49.6\times 10^5\ kg[/tex]
[tex]48.6\times 10^6\ N[/tex]
[tex]80.1\times 10^5\ N[/tex]
[tex]49.6\times 10^5\ kg[/tex]
Explanation:
[tex]1\ slug=14.5939\ kg[/tex]
[tex]340\times 10^3\ slug=340\times 10^3\times 14.5939=4961926\ kg[/tex]
The mass in SI unit is [tex]49.6\times 10^5\ kg[/tex]
Weight would be
[tex]W=mg\\\Rightarrow W=4961926\times 9.81\\\Rightarrow W=48676494.06\ N[/tex]
The weight in SI unit is [tex]48.6\times 10^6\ N[/tex]
[tex]1\ ft/s^2=0.3048\ m/s^2[/tex]
[tex]5.30\ ft/s^2=5.30\times 0.3048=1.61544\ m/s^2[/tex]
[tex]W=mg\\\Rightarrow W=4961926\times 1.61544\\\Rightarrow W=8015693.73744\ N[/tex]
The weight on the moon is [tex]80.1\times 10^5\ N[/tex]
The mass of an object is same anywhere in the universe.
So, the mass of the rocket on Moon is [tex]49.6\times 10^5\ kg[/tex]