Imagine you are on a space mission and you are 6 AU's from the Sun and you use a light sensor to measure the brightness of the Sun. The amount of sunlight received per square centimeter would be different by what factor compared to the same measurement on Earth at AU
Answer:
36 times less.
Explanation:
The distance from you to the sun is 6AU's, and from the sun to the earth is 1 AU.
Therefore,
At Earth sunlight received per unit cm² is:
[tex]I_{earth} = \dfrac{I_o}{4 \pi \times (1)^2}[/tex]
[tex]I_{earth} = \dfrac{I_o}{4 \pi}[/tex]
[tex]I_{me} =\dfrac{I}{4 \pi (6)^2}[/tex]
[tex]I_{me} = \dfrac{I_o}{36(4 \pi)}[/tex]
Thus, [tex]I_{earth} = 36 \times I_{me}[/tex]
Thus, the right answer is 36 times less.
Why does Marx’s workers’ paradise resolve the problems of capitalism?
A. Everything is free, and no one has to work.
B. Workers are divided into three classes, much like in Plato’s ideas.
C. There is no currency in the paradise so no economic problems.
D. People work for their own good instead of a factory owner’s.
A heating coil operates on 220 V if it draws 15.0 A. Find it's resistance
Answer:
R ≈ 15 ohms
Explanation:
Using ohm's law equation,
I = V/R, to solve for the resistance of the heating coil.
R = V/I
Known:
V = 220 v = 220 kgm^2s^-3A^-1
I = 15 A
Unknown:
R =?
Solution:
R = (220 kgm^2s^-3A^-1)/ 15.0 A
R = 14.6 kgm^2s^-3A^-2
R ≈ 15 kgm^2s^-3A^-2
R ≈ 15 ohms
A bucket of water of mass 14.6 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.320 m with mass 11.1 kg . The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 11.0 m to the water. You can ignore the weight of the rope.
Required:
a. What is the tension in the rope while the bucket is falling?
b. What is the time of fall?
c. While the bucket is falling, what is the force exerted on the cylinder by the axle?
Answer:
a. T = 39.41 N
b. t = 1.76s
c. 150.78 N
Explanation:
Given:
Mass of bucket of water, Mb = 14.6 kg
Mass of cylinder, Mc = 11.1 kg
Diameter of cylinder, D = 0.320 m, or radius, r = D/2 = 0.16m
Displacement of the bucket from the top, that is the vertical displacement , y = 11.0 m
a. The tension in the rope while the bucket is falling is:
F = mg - T = ma
Where F= The force
m= mass
g= Acceleration due to gravity
T = tension in the rope
a = acceleration
T= m(g - a)
Then, calculating the angular acceleration of the pulley system in relation to its radial acceleration
T= 1/2Ma
Merging the two final equation so as to solve for a
M(g - a) = 1/2Ma
Make a the subject of the formula
Mg - Ma = 1/2Ma
1/2Ma + Ma = Mg
a (1/2 M + M) = Mg
Divide both side by (1/2 M + M)
a = Mg ÷ (1/2 M + M)
Inputing the given value in the formula above
g= 9.8m/s2
a = (14.6 kg) (9.8m/s2) ÷ 1/2 (11.1 kg) + 14.6 kg
a = 7.1007m/s2
Now it is easy to input the value into T= 1/2Ma
T = 1/2 (11.1 kg) (7.1007m/s2) = 39.41 N
B. Time of fall is:
Using one of the equation of motion
s = ut + 1/2 at^2
U = Initial velocity
t = time
a = acceleration
s= distance in this case displacement y
making t the subject of the formula
t = √(2s ÷ a)
u is 0 since the bucket starts from rest
so, t = √((2)(11.0 m) ÷ 7.1007m/s2)
t = 1.76s
c. the force exerted on the cylinder by the axle = T + Mg
= 42 N + (11.1 kg) (9.8m/s2)
= 150.78 N
What does it mean if the reflected beam is above the incident beam? What does it mean if reflected beam is below the incident beam?
Answer:
aim at prisma and will have all colors
Explanation:
Answer:
If the vision position is above the actual image location then the light travel from the object in such a way that the angle of incidence is less than the angle of reflected ray which means that the reflected beam is above the incident beam.
Explanation:
g 1. Water flows through a 30.0 cm diameter water pipe at a speed of 3.00 m/s. All of the water in the pipe flows into a smaller pipe that is 10.0 cm in diameter. Determine: a) The speed of the water flowing through the 10.0 cm diameter pipe. b) The mass of water that flows through the larger pipe in 1.00 minute. c) The mass of water that flows through the smaller pipe in 1.00 minute.
Answer:
a) v₂ = 30 m/s
b) m₁ = 12600 kg
c) m₂ = 12600 kg
Explanation:
a)
Using the continuity equation:
[tex]A_1v_1 = A_2v_2[/tex]
where,
A₁ = Area of inlet = π(0.15 m)² = 0.07 m²
A₂ = Area of outlet = π(0.05 m)² = 0.007 m²
v₁ = speed at inlet = 3 m/s
v₂ = speed at outlet = ?
Therefore,
[tex](0.07\ m^2)(3\ m/s)=(0.007\ m^2)v_2\\\\v_2 = \frac{0.21\ m^3/s}{0.007\ m^2}[/tex]
v₂ = 30 m/s
b)
[tex]m_1 = \rho A_1v_1t[/tex]
where,
m₁ = mass of water flowing in = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,
[tex]m_1 = (1000\ kg/m^3)(0.07\ m^2)(3\ m/s)(60\ s)\\[/tex]
m₁ = 12600 kg
c)
[tex]m_1 = \rho A_1v_1t[/tex]
where,
m₂ = mass of water flowing out = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,
[tex]m_2 = (1000\ kg/m^3)(0.007\ m^2)(30\ m/s)(60\ s)\\[/tex]
m₂ = 12600 kg
If a runner is running at 100 meters per minute at the exact moment they cross the finish line of a race, it is the
instantaneous speed.
O True
O False
Answer:
Hello! Your answer would be, O False
Explanation:
Hope I helped! Brainiest plz!♥ Have a nice morning! Hope you make a 100%! -Abby
Answer:
False is the correct answer.
Explanation:
plz mark me as brainliest.
PLZ help 10 points!!! space question!
Answer:
B. They are smaller and made of rocky material
Explanation:
i think it's right??
If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 15.0 km ? Take the mass of the sun to be ms = 1.99×1030 kg , the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 , and the free-fall acceleration at the earth's surface to be g = 9.8 m/s2 . Express your weight wstar in newtons.
Answer:
W' = 1.66 x 10¹⁴ N
Explanation:
First, we will calculate the mass:
[tex]W = mg[/tex]
where,
W = weight on earth = 690 N
m = mass = ?
g = acceleration due to gravity on earth = 9.8 m/s²
Therefore,
[tex]m = \frac{W}{g} = \frac{690\ N}{9.8\ m/s^2}\\\\m = 70.4\ kg[/tex]
Now, we will calculate the value of g on the neutron star:
[tex]g' = \frac{GM}{R^2}[/tex]
where,
g' = acceleration due to gravity on the surface of the neutron star = ?
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of the Neutron Star = 1.99 x 10³⁰ kg
R = Radius of the Neutron Star = 15 km/2 = 7.5 km = 7500 m
Therefore,
[tex]g' = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1.99\ x\ 10^{30}\ kg)}{(7500\ m)^2}\\\\g' = 2.36\ x\ 10^{12}\ m/s^2[/tex]
Therefore, the weight on the surface of the neutron star will be:
[tex]W' = mg'\\W' = (70.4\ kg)(2.36\ x\ 10^{12}\ m/s^2)[/tex]
W' = 1.66 x 10¹⁴ N
A right triangle has side lengths of 4 centimeters and 5 centimeters. What is the length of the hypotenuse?
O A. 3 cm
B. 4 cm
C. 5 cm
D. 41 cm
9. Mr. Smith went skiing in Maine last weekend. He traveled 523 kilometers to Sugarloaf from
Leominster. His average speed was 109 km/hr. How long did it take Mr. Smith to hit the slopes?
Answer:
Time taken by Mr. smith = 4.80 hour (Approx.)
Explanation:
Given:
Distance travel by Mr. smith = 523 kilometer
Average speed of Mr. smith = 109 km/hr
Find;
Time taken by Mr. smith
Computation:
Time taken = Distance cover / Speed
Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.
smith
Time taken by Mr. smith = 523 / 109
Time taken by Mr. smith = 4.798 hr
Time taken by Mr. smith = 4.80 hour (Approx.)
fule cells have been developed that can generate a large amount of energy. for example, a hydrogen fuel cell works by combining hydrogen and oxogen gas to produce water and electrical energy. if a fuel cell can generate 10.0 kilowatts of power and the current is 15.8 amps, what is the voltage of the electricity?
Answer:
Where are fuel cells used?
Fuel cells are used for primary and backup power for commercial, industrial and residential buildings and in remote or inaccessible areas. They are also used to power fuel cell vehicles, including forklifts, automobiles, buses, boats, motorcycles and submarines
Explanation:
Why would the kinetic energy at the bottom of the track be less than the potential energy at the top of the track?
Answer:
This is because of friction and heat lost.
Explanation:
A cold block of metal feels colder than a block of wood at the same temperature. Why? A hot block of metal feels hotter than a block of wood at the same temperature. Again, why? Is there any temperature at which the two blocks feel equally hot or cold? What temperature is this?
: In general, metals feel colder or hotter to the touch than other materials at the same temperature because they're good thermal conductors. This means they easily transfer heat to colder objects or absorb heat from warmer objects
It is because the metal conducted heat faster than it feels colder than the wood, which conducted heat slower. Even tho they are similar temperatures. The metal will feel colder than the wood because of the thermal conductivity of the metal, compared to the wood.
The fictional rocket ship Adventure is measured to be 65 m long by the ship's captain inside the rocket.When the rocket moves past a space dock at 0.5c. As rocket ship Adventure passes by the space dock, the ship's captain flashes a flashlight at 1.20-s intervals as measured by space-dock personnel.
Required:
How often does the flashlight flash relative to the captain?
Answer:
1.04 s
Explanation:
The computation is shown below:
As we know that
t = t' × 1 ÷ (√(1 - (v/c)^2)
here
v = 0.5c
t = 1.20 -s
So,
1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)
1.20 = t' × 1 ÷ (√(1 - (0.5)^2)
1.20 = t' ÷ √0.75
1.20 = t' ÷ 0.866
t' = 0.866 × 1.20
= 1.04 s
The above formula should be applied
What is the period of a wave with a speed of 20.0 m/s and a frequency of 10.0 Hz?
im confused hold on imma send you a link to the answerExplanation:
Model the Earth's atmosphere as 79% N2, 19% O2, and 2% Argon, all of which are in thermal equilibrium at 280 K. At what height is the density of O half its value at sea level
Answer:
[tex]9.495 \times 10^3\ m[/tex]
Explanation:
From the given information:
Using the equation of Barometric formula as related to density, we have:
[tex]\rho (z) = \rho (0) e^{(-\dfrac{z}{H})} \ \ \ \ --- (1)[/tex]
Here;
[tex]p(z) =[/tex] the gas density at altitude z
[tex]\rho(0) =[/tex] the gas density at sea level
H = height of the scale
[tex]H = \dfrac{RT}{M_ag } \ \ \ --- (2)[/tex]
Also;
R represent the gas constant
temperature (T) a= 280 K
g = gravity
[tex]M_a =[/tex] molaar mass of gas; here, the gas is Oxygen:
∴
[tex]M_a =[/tex] 15.99 g/mol
= 15.99 × 10⁻³ kg/mol
[tex]H = \dfrac{8.3144 \times 280}{15.99 \times 10^{-3} \times 9.8 }[/tex]
[tex]H =14856.43 \ m[/tex]
Now we need to figure out how far above sea level the density of oxygen drops to half of what it is at sea level.
This implies that we have to calculate z;
i.e. [tex]\rho(z) =\dfrac{\rho(0) }{(2)}[/tex]
By using the value of H and [tex]\rho(z)[/tex] from (1), we have:
[tex]\dfrac{\rho(0) }{(2)} = \rho (0) e^{(-\dfrac{z}{14856.43})}[/tex]
∴
[tex]\dfrac{1}{2} = e^{(-\dfrac{z}{14856.43})} \\ \\ e^{(-\dfrac{z}{14856.43})} =\dfrac{1}{2}[/tex]
By rearrangement and taking the logarithm of the above equation; we have:
[tex]- z = 14856.43 \times \mathtt{In}\dfrac{1}{2} \\ \\ -z = 14856.43 \times (-0.6391) \\ \\ z = 9495 \ m \\ \\ z = 9.495 \times 10^3\ m[/tex]
As a result, the oxygen density at [tex]9.495 \times 10^3\ m[/tex] is half of what it is at sea level.
A negative charge of 4.0 x 10 C and a positive charge of 7.0 x 10 C are separated by 0.15 m. What is the force between the two charges?
Can you please help me ?
Answer:
Distance is 300 and displacement is 100
Explanation:
Distance= 100+100+100=300
Displacement=100
plz mark me as brainliest.
The first extra-solar planet around a main sequence star (i.e., not a neutron star or white dwarf) was found around the star 51 Pegasi in 1995. The large planet causes a measureable motion of the star around the center of mass of the system. Pegasi 51's orbital motion had a period of 4 days indicating a very large planet very close to the star. A large planet of mass M around a small star of mass 4M. The distance between planet and star is L. Both the planet and star will orbit around the center of mass of the system (marked by the red X).
Required:
Where is the center of mass of the star-planet system?
Answer:
[tex]r_{cm}[/tex]= 1/5 L
Explanation:
To find the center of mass of the system let's use
[tex]r_{cm} = \frac{1}{M}[/tex] ∑ r_i x_i
where m is the total mass of the system
let's apply this expression to our case
Let's set the reference frame on the star
[tex]r_{cm} = \frac{1}{M +4M} ( 4M 0 + M L)[/tex]
r_{cm} = [tex]\frac{1}{5}[/tex] L
[tex]r_{cm}[/tex]= 1/5 L
Hey, the breast center is 1/5 of the distance between the star and the planet.
0)
A-B-C-D-E
Describe the mass of the reactants and products given the balanced chemical reaction
A)
The mass of the products is 2/3 that of the reactants.
B)
The mass of the reactants equals the mass of the products.
C)
The mass of the reactants is less than the mass of the products.
D)
The mass of the reactants is greater than the mass of the products.
Answer:
D)
The mass of the reactants is greater than the mass of the products.
How much heat is needed to warm 365 mL of water in a baby bottle from 240C to 38C
Answer:98.6 degrees Fahrenheit
Explanation:
The density of 1 kilogram of gold is
Answer:
0.02 kg/cm³
Explanation:
Suppose a 65 kg person stands at the edge of a 6.5 m diameter merry-go-round turntable that is mounted on frictionless bearings and has a moment of inertia of 1850 kgm2. The turntable is at rest initially, but when the person begins running at a speed of 3.8 m/s (with respect to the turntable) around its edge, the turntable begins to rotate in the opposite direction. Calculate the angular velocity of the turntable. (Hint: use what you know about relative velocity to help solve the problem
Answer:
[tex]0.3165\ \text{rad/s}[/tex]
Explanation:
m = Mass of person = 65 kg
d = Diameter of round table = 6.5 m
r = Radius = [tex]\dfrac{d}{2}=3.25\ \text{m}[/tex]
v = Velocity of person running = 3.8 m/s
[tex]I_t[/tex] = Moment of inertia of turntable = [tex]1850\ \text{kg m}^2[/tex]
Moment of inertia of the system is
[tex]I=I_t+mr^2\\\Rightarrow I=1850+65\times 3.25^2\\\Rightarrow I=2536.5625\ \text{kg m}^2[/tex]
As the angular momentum of the system is conserved we have
[tex]L_i=L_f\\\Rightarrow mvr=I\omega_f\\\Rightarrow \omega_f=\dfrac{mvr}{I}\\\Rightarrow \omega_f=\dfrac{65\times 3.8\times 3.25}{2536.5625}\\\Rightarrow \omega_f=0.3165\ \text{rad/s}[/tex]
The angular velocity of the turntable is [tex]0.3165\ \text{rad/s}[/tex].
two bodies of masses 10 kg and 4 kg are moving in opposite direction with the same velocity of 5 m per second if they Collide and stick together after the impact calculate their velocities
Answer:
dasd
Explanation:
dsdas
What type of force is F-?
Answer:
Frictional Force
Explanation:
i think so
Two risks of exposure to High levels of UV radiation
Answer:
uv radiation cause cancer
uv radiation effect our eyes
Answer:
you can get
1:skin cancer
2:eye damage
3:skin damage
4:immune system suppression
choose which two u want
hope this helped
:)
Explanation:
what is force,momentum,and velocity.
Answer:
A force is a push or pull upon an object resulting from the object's interaction with another object.
Momentum is force or speed of movement.
Velocity defines the path of the motion of the frame or the object
An object’s
✔ mass
will remain constant throughout the universe, but its
can change from planet to planet.
If you increase the mass of a planet, what happens to its gravity?
If the gravity on a planet decreases, what happens to the weight of an object on that planet?
Answer:
mass, weight, strength of gravity increases, weight decreases
Explanation:
got it on edge
Answer:
An object’s
✔ mass
will remain constant throughout the universe, but its
✔ weight
can change from planet to planet.
If you increase the mass of a planet, what happens to its gravity?
✔ strength of gravity increases
If the gravity on a planet decreases, what happens to the weight of an object on that planet?
✔ weight decreases
Explanation:
right on edge 22
An ultrasonic tape measure uses frequencies above 20 MHz todetermine dimensions of structures such as buildings. It does so byemitting a pulse of ultrasound into air and then measuring the timeinterval for an echo to return from a reflecting surface whosedistance away is to be measured. The distance is displayed as adigital read-out. A tape measure emits a pulse of ultrasound with afrequency of 25.0 MHz.
(a) What is the distance to an object fromwhich the echo pulse returns after 24ms when the air temperature is 26°C?
(b) What should be the duration of the emitted pulse if it is toinclude 10 cycles of the ultrasonic wave?
(c) What is the spatial length of such a pulse?
Answer:
a) 1m
b) 2μs
c) 3mm
Explanation:
