Answer:
θ₂/ θ₁= 2.58
Explanation:
In this exercise you are asked to compare the angular diameters of Mars and Jupiter. The angular diameter or angle in radians is
θ = D / R
where D is the diameter of the body, the distance from Earth to the body of interest and θ is angle in radians
The different distances are tabulated with respect to the Sun
Sun -Earth 1,496 10¹¹ m
Sun- Mars 2.28 10¹¹ m
Sun - Jupiter 7.78 10 m
The Radii of the planets are
Mars 3.37 10⁶ m
Jupiter 6.99 10⁷ m
let's calculate the angles for each body
a) Mars
θ₁ = 2r / R'
the distance from the ground is
R ’= D_planet - D_earth
R ’= 2.28 10¹¹ - 1.496 10¹¹
R ’= 0.784 10¹¹ m
let's calculate
θ₁ = [tex]\frac{2 \ 3.37 \ 10^6 }{0.784 \ 10^{11}}[/tex]
θ₁ = 8.6 10⁻⁵ radians
b) Jupiter
R ’= 7.78 10¹¹ - 1.496 10¹¹
R ’= 6.284 10¹¹ m
let's calculate
θ₂ = [tex]\frac{2 \ 6.99 \ 10^7}{6.284 \ 10^{11}}[/tex]
θ₂ = 2.22 10⁻⁴ radians
the ratio of the angular diameters is
θ₂/ θ₁ = [tex]\frac{2.22 \ 10^{-4}}{8.6 \ 10^{-5}}[/tex]
θ₂/ θ₁= 2.58
A circuit has 12 Amps and 220 Volts. What is the Resistance of the circuit?
Answer:
:To find the Voltage, ( V ) [ V = I x R ] V (volts) = I (amps) x R (Ω)
To find the Current, ( I ) [ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)
To find the Resistance, ( R ) [ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)
To find the Power (P) [ P = V x I ] P (watts) = V (volts) x I (amps)
why no tempature can be lower than 0 kelvin
Answer:
At zero kelvin (minus 273 degrees Celsius) the particles stop moving and all disorder disappears. Thus, nothing can be colder than absolute zero on the Kelvin scale. Physicists have now created an atomic gas in the laboratory that nonetheless has negative Kelvin values.
Explanation:
Two small charged objects repel each other with a force F when separated by a distance d. If the charge on each object is reduced to 1/ 14 of its original value and the distance between them is reduced to d/ 24 the force becomes Group of answer choices F F * 24 / 196 F * 576 / 14 F * 576 / 196 F * 196 / 576
Answer:
Ff = F₀ *(576/196)
Explanation:
Assuming that both charges are equal each other, we can express the repulsion force between the charges (assuming that we can treat them as point charges) using Coulomb's Law, as follows:[tex]F_{o} = \frac{k*q^{2} }{d^{2}} (1)[/tex]
Now, if q reduces to q/14, and d is reduced to d/24, the new value of the force will be:[tex]F_{f} = \frac{k*(q/14)^{2} }{(d/24)^{2}} = \frac{k*q^{2}}{d^{2}} * \frac{(24)^{2}}{(14)^{2}} (2)[/tex]
⇒[tex]F_{f} =\frac{k*q^{2}}{d^{2}} * \frac{(24)^{2}}{(14)^{2}} = F_{o} * \frac{576}{196} (3)[/tex]
A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antinode is 2.20 mm and the speed of propagation of transverse waves on the string is 260 m/s. The string extends along the x-axis, with one of the fixed ends at x= 0, so that there is a node at x =0. The smallest value of x where there is an antinode is x= 0.150m.
Required:
a. What is the maximum transverse speed of a point on the string at an antinode?
b. What is the maximum transverse speed of a point on the string at x = 0.075 m?
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so [tex]V_{max1}[/tex] = A × ω
[tex]V_{max1}[/tex] = 2.2×10⁻³ × 2722.69 m/s
[tex]V_{max1}[/tex] = 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
[tex]V_{max2}[/tex] = A' × ω
[tex]V_{max2}[/tex] = 1.555×10⁻³ × 2722.69
[tex]V_{max2}[/tex] = 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
two objects are moving in the xy plane. no external forces are acting on the objects. object a has a mass of 3.2 kg and has a velocity of v= (2.3m/s)i + (4.2m/s)j and object b has a mass of 2.9kg and has a velocity of v=(-1.8m/s)i = (2.7 m/s)j. sometime later object a is seen to have a velocity va=(1.7m/s)i+(3.5m/s)j what is the velocity of object b at that instant
Answer:
58.469 kg.m/s are moving in the xy plane
PHYSICS QUESTION PLS HELP
The coaster starts at rest, so the kinetic energy (KE) at point A is 0. It is situated 33 m above ground, so its potential energy (PE) at A is
mgh = (3000 kg) (9.80 m/s²) (33 m) = 970,200 J
The total energy is the same, 970,200 J.
Assuming no energy is lost to friction or sound etc, energy is conserved throughout the coaster's motion, so the total energy should be the same at each point.
At point B, the coaster has dropped to a height of 10 m, so it has PE
mgh = (3000 kg) (9.80 m/s²) (10 m) = 294,000 J
which means it must have KE
970,200 J = KE + 294,000 J → KE = 676,200 J
which gives the coast a speed v at point B of
1/2 mv ² = 1/2 (3000 kg) v ² = 676,200 J → v ≈ 21.2 m/s
At point C, the coaster has a speed of 16.0 m/s, so it has KE
1/2 mv ² = 1/2 (3000 kg) (16.0 m/s)² = 384,000 J
and hence PE
970,200 J = 384,000 J + PE → PE = 586,200 J
This lets us determine the height h at C:
mgh = (3000 kg) (9.80 m/s²) h = 586,200 J → h ≈ 19.939 m
which means the loop has diameter h - 10 m ≈ 9.94 m.
At point D, the coaster is 15 m above the ground so its PE at D is
mgh = (3000 kg) (9.80 m/s²) (15 m) = 441,000 J
and so its KE is
970,200 J = KE + 441,000 J → KE = 529,200 J
and hence has speed v at D
1/2 mv ² = 1/2 (3000 kg) v ² = 529,200 J → v ≈ 18.9 m/s
If the mass of the book is 50 sliding with acceleration 1.2 m/s ^ 2 then the friction force is
364N
185N
173N
73N
Answer and I will give you brainiliest
Define the following soil conservation technique. Make sure to include pro’s and con’s of this method.
(Terraces)
Answer: Terraces on moderate to steep irregular slopes pro- ... sure of infertile or toxic soils. ... Following are terms used to define distances mea- ... the soil in the entire field will be disturbed to con-.
Explanation:
A student pulls on a cart and applies a 20N force at an angle of 25 degrees above the horizontal to
move a crate a distance of 10m across the floor. How much work does the student do?
Answer:
W = 181.26 J
Explanation:
Given that,
The force acting on the cart, F = 20 N
It is at an angle of 25 degrees above the horizontal to move a crate a distance of 10m across the floor.
We need to find work done by the student. The work done by the student is given by :
[tex]W=Fd\cos\theta\\\\W=20\times 10\times \cos25\\W=181.26\ J[/tex]
So, the required work done is 181.26 J.
A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M and positive charge Q hovers motionless above the center of the disk, suspended by the Coulomb repulsion due to the charged disk.
Required:
a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?
b. At what height h does the sphere hover?
Answer:
a. F = Qs/2ε₀[1 - z/√(z² + R²)] b. h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]
Explanation:
a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?
The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by
E = s/2ε₀[1 - z/√(z² + R²)]
So, the net force on the small plastic sphere of mass M and charge Q is
F = QE
F = Qs/2ε₀[1 - z/√(z² + R²)]
b. At what height h does the sphere hover?
The sphere hovers at height z = h when the electric force equals the weight of the sphere.
So, F = mg
Qs/2ε₀[1 - z/√(z² + R²)] = mg
when z = h, we have
Qs/2ε₀[1 - h/√(h² + R²)] = mg
[1 - h/√(h² + R²)] = 2mgε₀/Qs
h/√(h² + R²) = 1 - 2mgε₀/Qs
squaring both sides, we have
[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²
h²/(h² + R²) = (1 - 2mgε₀/Qs)²
cross-multiplying, we have
h² = (1 - 2mgε₀/Qs)²(h² + R²)
expanding the bracket, we have
h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²
collecting like terms, we have
h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²
Factorizing, we have
[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²
So, h² = (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]
taking square-root of both sides, we have
√h² = √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]
h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]
Cara is building a model of the solar system, which includes the Sun. She plans to include a written description to provide details about each piece in her model. In order for her model to be realistic, which of the following should she include in her representation of the Sun?
She should show that sunspots can ve seen as white areas on the Sun's surface.
She should explain that the Sun is made up of gaseous layers that surround an iron core.
She should show that the Sun revolves around the planets, determining the length of the year.
She should explain that the Sun rotates, even though different parts rotate at different rates.
Answer:
She should explain that the Sun is made up of gaseous layers that surround an iron core.
A material through which electricity cannot flow is called:
a conductor
an insulator
an electrode
a wet cell
Answer:
el conductor
Explanation:
gracias por los puntitoss
Answer:
conductor
Explanation:
HELP ! ILL MARK BRAINLIEST HELP ASAP
Answer:
A
Explanation:
In 5 minutes, they went 10 miles at both 2, 3, and 4 checkpoints. The bus then starts to speed up.
Hope this helps!
3. A 10-centimeter diameter solid sphere made of a conducting material has 10 micro-Coulombs of charge placed upon it. What is the potential difference between a point on one side of the sphere to a point on the exact opposite side of the sphere
Answer:
zero
Explanation:
For a solid conducting sphere, charges are present on the surface of the sphere due to a phenomenon known as electrostatic sheilding. This affects the charge present in the body and makes it zero. However, the electrostatic potential appears to be equal to the whole present point that shows on the surface. The surface of a spherical conducting solid sphere is known as an equipotential surface. Thus, the potential difference between the two opposite points on the surface of the sphere will also be zero.
Heeeeeeeeelp please
OK please your picture not perfect please try again
Which hormone do ovaries release?
A. estrogen
B. glucagon
C. insulin
D. testosterone
Answer:
A. estrogen
Explanation:
This is released in the female reproductive organ.
Connecting math to physics
Answer:
wat
Explanation:
Please Help!!!!
When energy is transferred in a system, the total amount of energy before the transfer is _____________ after the transformation is complete, just in different forms.
Group of answer choices
different
lost
transformed into light
the same
I need help will mark brainliest
Answer: ITS 1 TRUST ME MAN BYE K
Explanation: OK BYE TRUST YEAH
Do you believe you can create a Controlled experiment without an Observational Study? Why or Why not. Include scientific evidence to support your response. PLEASE HELP I BEG YOU.
Answer:
No, it is very unlikely to perform a controlled experiment, because you need to observe the amount or anything from something. Consider someone on the busy street of a New York neighborhood asking random people that pass by how many pets they have, then taking this data and using it to decide if there should be more pet food stores in that area.
QUICK SOMEONE PLEASE HELP!!!! I’LL MARK BRAINLIEST!!!
Certain neutron stars (extremely dense stars) are believed to be rotating at about 500 rev/s. If such a star has a radius of 17 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation
Answer:
7.22 × 10²⁹ kg
Explanation:
For the material to be in place, the gravitational force on the material must equal the centripetal force on the material.
So, F = gravitational force = GMm/R² where M = mass of neutron star, m = mass of object and R = radius of neutron star = 17 km
The centripetal force F' = mRω² where R = radius of neutron star and ω = angular speed of neutron star
So, since F = F'
GMm/R² = mRω²
GM = R³ω²
M = R³ω²/G
Since ω = 500 rev/s = 500 × 2π rad/s = 1000π rad/s = 3141.6 rad/s = 3.142 × 10³ rad/s and r = 17 km = 17 × 10³ m and G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²
Substituting the values of the variables into M, we have
M = R³ω²/G
M = (17 × 10³ m)³(3.142 × 10³ rad/s)²/6.67 × 10⁻¹¹ Nm²/kg²
M = 4913 × 10⁹ m³ × 9.872 × 10⁶ rad²/s²/6.67 × 10⁻¹¹ Nm²/kg²
M = 48,501.942 × 10¹⁵ m³rad²/s² ÷ 6.67 × 10⁻¹¹ Nm²/kg²
M = 7217.66 × 10²⁶ kg
M = 7.21766 × 10²⁹ kg
M ≅ 7.22 × 10²⁹ kg
Help me !!!
What is the velocity of a ball with
a momentum of -4.5 kg*m/s and a
mass of 0.45 kg?
Answer:−4.05
Explanation:
what is the difference between mass and weight
Answer:
The mass of an object is a measure of the object's inertial property, or the amount of matter it contains. The weight of an object is a measure of the force exerted on the object by gravity, or the force needed to support it. The pull of gravity on the earth gives an object a downward acceleration of about 9.8 m/s2.
Answer:
Explanation:
The mass is essentially "how much stuff" is in an object. ... Weight: There is a gravitational interaction between objects that have mass. If you consider an object interacting with the Earth, this force is called the weight. The unit for weight is the Newton (same as for any other force).
1.0 kg clay ball traveling straight down at -10 m/s hits the floor and and sticks on it
Answer:
What am I suppose to solve
Explanation:
Describe Kinetic Energy and Potential Energy (in your own words please!!)
Answer:
Energy stored in an object due to its position is Potential Energy. · Energy that a moving object has due to its motion is Kinetic Energy.
Explanation:
what is the angle between 3i-2j-3k and the negative x axis
Answer:
Um its the vbuck card on the 3 thrid level
Explanation:
Bc its a vbuck card you know sihdg;aig
A monk is sitting atop a mountain in complete rest in meditation. What is the kinetic Energy of the monk? (assume mass of 65 kg and the mountain's height was 1000 m)
Answer:
no kinetic energy
hope this helps! :-D
Explanation:
the monk is not moving
Collisions question plzz help mee
A girl jogs around a horizontal circle with a constant speed. She travels one fourth of a revolution, a distance of 25 m along the circumference of the circle, in 5.0 s. The magnitude of her acceleration is
Answer:
The centripetal acceleration of the girl is 2.468 m/s²
Explanation:
Given;
number of turns, = ¹/₄ Revolution
distance traveled by the girl, d = 25 m
time of motion, t = 5.0 s
The linear speed of the of the girl is calculated as;
[tex]v = \omega \ r\\\\v =(\frac{1}{4}rev \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1}{5 \ s} ) (25 \ m)\\\\v = (0.3142 \ \frac{rad}{s} )(25 \ m)\\\\v = 7.855 \ m/s[/tex]
The centripetal acceleration of the girl is calculated as;
[tex]a_c = \frac{v^2}{r} \\\\a_c = \frac{(7.855)^2}{25} \\\\a_c = 2.468 \ m/s^2[/tex]
Therefore, the centripetal acceleration of the girl is 2.468 m/s²