Answer:
R = 7.915 10⁶ m, M = 1.04 10³⁵ kg
Explanation:
Let's start by finding the acceleration of the planet's gravity, let's use the kinematic relations
v = v₀ - g t
the velocity of the body when it falls is the same for equal height, but it is positive when it rises and negative when it falls
v = -v₀
-v₀ = v₀ - g t
g = 2v₀ / t
g = 2 15 / 2.7
g = 11.11 m / s²
I now write the law of universal gravitation and Newton's second law
F = m a
G m M / R² = m a
a = g
g = G M / R²
Now let's work with the cruiser in orbit
F = ma
acceleration is centripetal
a = v² / r
G m M / r² = m v² / r (1)
the distance from the center of the planet is
r = R + h
r = R + R = 2R
we substitute in 1
G M / 4R² = v² / 2R
G M / 2R = v²
The modulus of the velocity in a circular orbit is
v = d / T
the distance is that of the circle
d = 2π r
v = 2π 2R / T
v = 4π R / T
G M / 2R = 16pi² R² / T²
T² = 32 pi² R³ / GM
let's write the equations
g = G M / R² (2)
T² = 32 pi² R³ / GM
we have two equations and two unknowns, so it can be solved
let's clear the most on the planet and equalize
g R² / G = 32 pi² R³ / GT²
g T² = 32 pi² R
R = g T² / 32 pi²
let's reduce the period to SI units
T = 250 min (60 s / 1 min) = 1.5 104 s
let's calculate
R = 11.11 (1.5 10⁴) ² / 32 π²
R = 7.915 10⁶ m
from equation 2 we can find the mass of the planet
M = g R² / G
M = 11.11 (7.915 10⁶) ² / 6.67 10⁻¹¹
M = 1.04 10³⁵ kg
An Object, Start from rest w Confront Aiceleration 8m/s2 along a
Straight line. Find
A, the speed At the end Of 5 second
B, The average Speed for the 5second interval
Answer:
A) v = 40 m / s, B) v_average = 20 m / s
Explanation:
For this exercise we will use the kinematics relations
A) the final velocity for t = 5 s and since the body starts from rest its initial velocity is zero
v = vo + a t
v = 0 + 8 5
v = 40 m / s
B) the average velocity can be found with the relation
v_average = vf + vo / 2
v-average = 0+ 40/2
v_average = 20 m / s
Lucy moves down the hall at 3.5 m/s. When he sees Luke coming, he slows down. After 4.0 s, he is moving at 2.1. m/s. What is his acceleration?
Answer:
Acceleration, a = 0.35 m/s²
Explanation:
Given the following data;
Initial velocity, u = 3.5 m/s²
Final velocity, v = 2.1 m/s²
Time, t = 4 secs
To find the acceleration, we would use the first equation of motion;
V = u - at (the sign is negative because Lucy is slowing down).
Substituting into the formula, we have;
2.1 = 3.5 - a(4)
2.1 = 3.5 - 4a
4a = 3.5 - 2.1
4a = 1.4
a = 1.4/4
a = 0.35 m/s²
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 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
Objects A and B, of mass M and 2M respectively, are each pushed a distance d straight up an inclined plane by a force F parallel to the plane. The coefficient of kinetic friction between each mass and the plane has the same value μ.k At the highest point is:______
a. KEA > KEB
b. KEA = KEB
c. KEA < KEB
d. The work done by F on A is greater than the work done by F on B.
e. The work done by F on A is less than the work done by F on B.
Answer:
The correct answer is option (A) that is KEA > KEB .
Explanation:
Let us calculate -
If the object is straighten up and inclined plane , the work done is
[tex]W=F_d- F_f_r_id-F_gh[/tex]
[tex]W=F_d-\mu_kmgdcos\theta-mgdsin\theta[/tex]
The change in kinetic energy is ,
[tex]\Delta K=\frac{1}{2}mv^2-\frac{1}{2}m\nu_0^2[/tex]
At the top of the inclined plane , the velocity is zero
So,
[tex]\Delta K=\frac{1}{2} m(0)^2-\frac{1}{2}m\nu_0^2[/tex]
[tex]\Delta KE=-\frac{1}{2}m\nu_0^2[/tex]
From the work energy theorem , we have [tex]W=-\Delta K[/tex] in case of friction , so
[tex]\frac{1}{2}m\nu_0^2=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]
[tex]KE=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]
For object A-
[tex]KE_A=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]
For object B
[tex]KE_B= Fd -2\mu_kMgdcos\theta-2Mgdsin\theta[/tex]
[tex]KE_B= Fd -2(\mu_kMgdcos\theta-Mgdsin\theta)[/tex]
Thus , larger mass is going to mean less total work and a lower kinetic energy .
From the above results , we get
[tex]KE_A >KE_B[/tex]
Therefore , option A is correct .
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.00 s for the boat to travel from its highest point to its lowest, a total distance of 0.600 m . The fisherman sees that the wave crests are spaced a horizontal distance of 6.40 m apart.
Required:
a. How fast are the waves traveling?
b. What is the amplitude of each wave?
c. If the total vertical distance traveled by the boat were 0.30 m but the other data remained the same, how would the answers to parts (a) and (b) be affected?
Answer:
a. Speed = 1.6 m/s
b. Amplitude = 0.3 m
c. Speed = 1.6 m/s
Amplitude = 0.15 m
Explanation:
a.
The frequency of the wave must be equal to the reciprocal of the time taken by the boat to move from the highest point to the highest point again. This time will be twice the value of the time taken to travel from the highest point to the lowest point:
frequency = [tex]\frac{1}{2(2\ s)}[/tex] = 0.25 Hz
The wavelength of the wave is the distance between consecutive crests of wave. Therefore,
Wavelength = 6.4 m
Now, the speed of the wave is given as:
Speed = (Frequency)(Wavelength)
Speed = (0.25 Hz)(6.4 m)
Speed = 1.6 m/s
b.
Amplitude is the distance between the mean position of the wave and the extreme position. Hence, it will be half the distance between the highest and lowest point:
Amplitude = (0.5)(0.6 m)
Amplitude = 0.3 m
c.
frequency = [tex]\frac{1}{2(2\ s)}[/tex] = 0.25 Hz
Speed = (Frequency)(Wavelength)
Speed = (0.25 Hz)(6.4 m)
Speed = 1.6 m/s
Amplitude = (0.5)(0.3 m)
Amplitude = 0.15 m
A circular cylinder has a diameter of 3.0 cm and a mass of 25 g. It floats in water with its long axis perpendicular to the water's surface. It is pushed down into the water by a small distance and released; it then bobs up and down. Part A What is the oscillation frequency
Answer:
f = 5.3 Hz
Explanation:
To solve this problem, let's find the equation that describes the process, using Newton's second law
∑ F = ma
where the acceleration is
a = [tex]\frac{d^2 y}{dt^2 }[/tex]
B- W = m \frac{d^2 y}{dt^2 }
To solve this problem we create a change in the reference system, we place the zero at the equilibrium point
B = W
In this frame of reference, the variable y' when it is oscillating is positive and negative, therefore Newton's equation remains
B’= m [tex]\frac{d^2 y'}{dt^2 }[/tex]
the thrust is given by the Archimedes relation
B = ρ_liquid g V_liquid
the volume is
V = π r² y'
we substitute
- ρ_liquid g π r² y’ = m \frac{d^2 y'}{dt^2 }
[tex]\frac{d^2 y'}{dt^2} + \rho_liquid \ g \ \pi r^2/m ) y' \ =0[/tex]
this differential equation has a solution of type
y = A cos (wt + Ф)
where
w² = ρ_liquid g π r² /m
angular velocity and frequency are related
w = 2π f
we substitute
4π² f² = ρ_liquid g π r² / m
f = [tex]\frac{1}{2\pi } \ \sqrt{ \frac{ \rho_{liquid} \ \pi r^2 \ g}{m } }[/tex]
calculate
f = [tex]\frac{1}{2 \pi } \sqrt{ \frac{ 1000 \ \pi \ 0.03^2 \ 9.8 }{0.025} }[/tex]
f = 5.3 Hz
Scientists are constantly exploring the universe, looking for new planets that support life similar to the life on
Earth. A new planet that supports life would have all of the following characteristics except -
A. a gaseous atmosphere.
B. an orbiting moon.
C. liquid water.
D. protection from radiation.
A new planet that supports life would have all the following characteristics except an orbiting moon. Hence, option B is correct.
What is a Planet?An enormous, spherical celestial object known as a planet is neither a star nor its remains. The nebular hypothesis, which states how an interstellar cloud falls out of a nebula to produce a young protostar encircled by a protoplanetary disk, is now the best explanation for planet formation.
By gradually accumulating material under the influence of gravity, or accretion, planets develop in this disk.
The rocky planets Mercury, Venus, Earth, and Mars, as well as the giant planets Jupiter, Saturn, Uranus, and Neptune, make up the Solar System's minimum number of eight planets. These planets all revolve around axes that are inclined relative to their respective polar axes.
To know more about Planet:
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The model shows the star Wolf 359, the sun, and Star X. It takes 7.8 years for light produced on Wolf 359 to reach the sun.
If Star X is 50 times as far from the sun as Wolf 359 is, how far is Star X from the sun, to the nearest light-year?
Answer:
390 light-years
Explanation:
50 x 7.8 =390
Do anyone answer this question
Answer:
B) 10^-2 cm/s
in term of meter. it is 10^-4 m/s
Explanation:
How does the presence of a nucleus provide a method of basic cell
classification? *
Answer:
The nucleus-containing cells are called eukaryotic cells. Eukaryote means having membrane-bound organelles.
Explanation:
I hope this is what you were looking for?!
Hope this helps!
Have a great day!
-Hailey!
The cells which have a nucleus are called Eukaryotic cells. The cells which do not have a nucleus are called prokaryotic cells.
What is a nucleus?
The nucleus is the controlling center of the body. It performs all the major activities in the cell. The cell which possesses a nucleus is called a eukaryote. The cell which does not have a nucleus is called prokaryote.
A prokaryotic cell is a straightforward, one-celled (unicellular) organism that is devoid of a nucleus or any other organelle that is membrane-bound. The nucleoid, a dark area in the center of the cell, is where prokaryotic DNA is located.
Therefore, The cells which have a nucleus are called Eukaryotic cells. The cells which do not have a nucleus are called prokaryotic cells.
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A 2:2 kg toy train is con ned to roll along a straight, frictionless track parallel to the x-axis. The train starts at the origin moving at a speed of 1:6m=s in the +x direction, and continues until it reaches a position 7:5m down the track from where it started. During its journey, it experiences a force pointing in the same direction as the vector 0:6 +0:8 , with magnitude initially 2:8N and decreasing linearly with its x-position to 0N when the train has finished its journey.
Required:
a. Calculate the work done by this force over the entire journey of the train.
b. Find the speed of the train at the end of its journey.
Answer:
a) 10.51 J
b) 3.48 m/s
Explanation:
Given data :
mass of train ( M ) = 2.2 kg
Given initial velocity ( u ) = 1.6 m/s
a) calculating work done by the force over the journey of the train
F = mx + b ------ ( 1 )
m = slope = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m
x = distance travelled on the x axis by the train = 7.5 m
F = force experienced by the train = 2.8 N
x = 0
∴ b = 2.8
hence equation 1 can be written as
F = ( -0.373) x + 2.8 ----- ( 2 )
hence to determine the work done by the force
W = [tex]\int\limits^7_0 { ( -0.373) x + 2.8 )} \, dx[/tex] Note: the limits are actually 7.5 and 0
∴ W ( work done ) = -10.49 + 21 = 10.51 J
b) calculate the speed of the train at the end of its journey
we will apply the work energy theorem
W = 1/2 m*v^2 - 1/2 m*u^2
∴ V^2 = 2 / M ( W + 1/2 M*u^2 ) ( input values into equation )
V^2 = 12.11
hence V = 3.48 m/s
In an elastic collision between a moving 10-kg mass and a stationary 10-kg mass half the momentum is transferred to the stationary mass. In this situation the total kinetic energy after the collision is less than it was before the collision. Where did the kinetic energy go?
A) The kinetic energy was destroyed during the collision.
B) Some of the kinetic energy was turned into momentum during the collision.
C) Some of the kinetic energy was turned into heat or used to deform the masses.
D) Some of the kinetic energy was turned into potential energy during the collision.
Answer: C
Explanation:
USAtestprep
describe briefly how you can a body
Answer:
what
you need to elaborate
Answer: Can you please write question clearly.
Explanation:
NASA’s Tracking and Data Relay Satellite (TDRS) System constellation resides at geosynchronous orbit (35,000km) altitude. If a technician at the Goddard Spaceflight Center in Maryland initiates a transmission to the Johnson Spaceflight Center in Houston over TDRS, how long will it be until JSC detects the transmission (one-way latency)? You may assume there is negligible processing delay on the satellite, and that c = 3x108 m/sec.
Answer:
35,000 km = 35,000,000 m = 3.5 E107 m
t = S / v = 3.5 * 10E7 / 3.0 E10E8 = .117 sec
A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 s apart. Part A How far away did the impact occur? (Use vair=343m/s , vconcrete=3000m/s )
Answer:
The impact occured at a distance of 2478.585 meters from the person.
Explanation:
(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:
A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)
Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete ([tex]\Delta t[/tex]), in seconds:
[tex]\Delta t = t_{A}-t_{C}[/tex] (1)
Where:
[tex]t_{C}[/tex] - Time spent by the sound in concrete, in seconds.
[tex]t_{A}[/tex] - Time spent by the sound in the air, in seconds.
By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:
[tex]\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}[/tex]
[tex]\Delta t = x\cdot \left(\frac{1}{v_{A}}-\frac{1}{v_{C}} \right)[/tex]
[tex]x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }[/tex] (2)
Where:
[tex]v_{C}[/tex] - Speed of the sound in concrete, in meters per second.
[tex]v_{A}[/tex] - Speed of the sound in the air, in meters per second.
[tex]x[/tex] - Distance traveled by the sound, in meters.
If we know that [tex]\Delta t = 6.4\,s[/tex], [tex]v_{C} = 3000\,\frac{m}{s}[/tex] and [tex]v_{A} = 343\,\frac{m}{s}[/tex], then the distance travelled by the sound is:
[tex]x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }[/tex]
[tex]x = 2478.585\,m[/tex]
The impact occured at a distance of 2478.585 meters from the person.
Q10. Refer to the Condon table to answer question
Second letter
UUU
UCU
UGU
OUC
UCO
UAC
Leu
UAA Btop UOA Stop
UCG UAG Stop UGOT
CU
CCU
CAU CGU
CUC
ССС
САС
Leu
CUA
CGC
Pro
CCA
CAA COA
CUG
CCG
AD
RoecoDoo Do
Asn
AUU
ACU
AUC File
ACC
AUA АСА
AUG Met ACO
AAU
ААС
ΑΛΛΑ
LANG
AGU
Ser
AGC
Thr
Jue AGA
Jara
AGG
sp
GU
QUC
GUA
GUG
GCU GAU
GCC
Ala
GAC
GCA GAA
OCG GAG
GOC
GGA
Jolu 900
Write the mRNA copy of this DNA CCG GA
GCT (original) | Imk]
Use the Condon table above to list all the amin
Answer:
so you have a question
Explanation:
either way, have a nice day
I am b o r e d, I am very very b o r e d!
I'm b o r e d with Lazarbeam Quarantine edition
episode 2352 because apparently the quarantining never ends :(
2. A 4kg object possesses 18J of Kinetic energy. What is the velocity?
Plz help I’ll give you points!
Which statement best describes work in the scientific sense?
O A. Work is the sum of the distances an object moves due to the
forces applied to it.
O B. Work is the number of tasks done in the amount of time needed to
complete them.
O C. Work is the ratio of the force acting on an object and the distance
the object travels.
O D. Work is the product of a force and the distance an object moves
because of the force.
Answer:
the answer is D I tought
HURRY IM TIMED!
The age and gender of an audience are important to consider when deciding on a subject.
True
False
Answer:
True.
Explanation:
true
knowing your audience there General age gender education level religion language culture and gave group members is the single most important aspect of developing your speech this means the speaker dock smart the audience wasn't often without asking question or spending with any feedback
Which two options describe physical properties of matter
stored energy is _________ ___________
kinetic energy
energy in motion
potential energy
Answer:
Potential energy
Explanation:
Potential energy is stored energy
Plutonium-238 has a half life of 87.7 years. What percentage of a 5 kilogram (kg) sample remains after 50 years?
Answer:
i dont know but i should know try g o o g l e
Explanation:
the minimum speed on the interstate highway is
1. 40 mph
2. 60 mph
3. 55 mph
4. 50 mph
I'm in Nebraska btw
A centroid is an object's geometric center. For an object of uniform composition, its centroid is also its center of mass. Often the centroid of a complex composite body is found by, first, cutting the body into regular shaped segments, and then by calculating the weighted average of the segments' centroids.An object is made from a uniform piece of sheet metal. The object has dimensions of a
This question is not complete, the complete question is;
A centroid is an object's geometric center. For an object of uniform composition, its centroid is also its center of mass. Often the centroid of a complex composite body is found by, first, cutting the body into regular shaped segments, and then by calculating the weighted average of the segments' centroids.
An object is made from a uniform piece of sheet metal. The object has dimensions of α = 1.50 ft, where α is the diameter the semi-circle, b= 3.51 ft, and c = 2.20 ft. A hole with diameter d = 0.500 ft is centered at ( 1.21, 0.750 ).
Find x", y", the coordinates of the body's centroid.
Answer:
x" = 1.4857 ft
y" = 0.668 ft
Explanation:
Given the data in the question and as illustrated in the second image below;
from the image;
BC² = DC² - BD²
BC² = 2.2² - 1.5² = 4.84 - 2.25 = 2.59
BC = √2.59 = 1.61 ft
AB = 3.51 ft - 0.75 ft - 1.61 ft = 1.15 ft
so;
A₁ = [tex]\frac{1}{2}[/tex] × 1.51 ft × 1.61 ft = 1.2075 ft²
x₁ = 0.75 + 1.15 + [tex]\frac{1}{3}[/tex](1.61 ft) = 2.44 ft
y₁ = [tex]\frac{1}{3}[/tex](1.5 ft) = 0.5 ft
A₂ = 1.15 ft × 1.5 ft = 1.725 ft²
x₂ = 0.75 ft + ( 1.15/2 )ft = 1.325 ft
y₂ = ( 1.5/2 ) ft = 0.75 ft
A₃ = [tex]\frac{\pi }{2}[/tex](0.75 ft)² = 0.88 ft²
x₃ = 0.75 - ([tex]\frac{4 }{3\pi }[/tex](0.75 ft)) = 0.43 ft
y₃ = 0.75 ft
diameter d = 0.5 ft and centered at ( 1.21, 0.750 )
A₄ = [tex]\frac{\pi }{4}[/tex]( d )² =
x₄ = 1.21 ft
y₄ = 0.75 ft
Thus;
x" = [tex]\frac{A_1 x_1 + A_2 x_2 + A_3 x_3 - A_4x_4 }{A_1+A_2+A_3-A_4}[/tex]
so we substitute
x" = [tex]\frac{(1.2075X2.44) + (1.725 X 1.325) + (0.88X0.43) - (0.196 X 1.21 )}{ ( 1.2075 + 1.725 + 0.88 - 0.196 )}[/tex]
x" = [tex]\frac{ (2.9463 + 2.285625 + 0.3784 - 0.23716)}{ 3.6165 }[/tex]
x" = 5.373165 / 3.6165
x" = 1.4857 ft
y" = [tex]\frac{A_1 y_1 + A_2 y_2 + A_3 y_3 - A_4y_4 }{A_1+A_2+A_3-A_4}[/tex]
so we substitute
y" = [tex]\frac{(1.2075X0.5) + (1.725 X 0.75) + (0.88X0.75) - (0.196 X 0.75 )}{ ( 1.2075 + 1.725 + 0.88 - 0.196 )}[/tex]
y" = [tex]\frac{ (0.60375 + 1.29375 + 0.66 - 0.14112)}{ 3.6165 }[/tex]
y" = 2.41638 / 3.6165
y" = 0.668 ft
Therefore,
x" = 1.4857 ft
y" = 0.668 ft
Two point charges, initially 3 cm apart, are moved to a distance of 1 cm apart. By what factor does the resulting electric force between them change?
A. 3
B. 1/9
C. 1/3
D. 9
A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (The moment of inertia of the rod about this axis is ML2/3.) The rod is released when it makes an angle of 37° with the horizontal. What is the angular acceleration of the rod at the instant it is released (in rad/s^2)?
Answer:
A: 9.8 rad/s2
B: 7.4 rad/s2
C: 8.4 rad/s2
D: 5.9 rad/s2
E: 6.5 rad/s2
I think the answer is A 9.8rad/s2
Find the speed of a wave with a frequency of 18 Hz and a wavelength of 6 meters. Show work. WILL MARK BRAINLIEST IF CORRECT
Answer:
so i would say 11.4 i dont have work only this link
Explanation:
https://flexbooks.ck12.org/cbook/ck-12-physics-flexbook-2.0/section/11.4/primary/lesson/wave-speed-ms-ps
A community plans to build a facility to convert solar radiation to electrical power. The community requires 2.20 MW of power, and the system to be installed has an efficiency of 30.0% (that is, 30.0% of the solar energy incident on the surface is converted to useful energy that can power the community). Assuming sunlight has a constant intensity of 1 020 W/m2, what must be the effective area of a perfectly absorbing surface used in such an installation
Answer:
The answer is "[tex]\bold{7.18 \times 10^3 \ m^2}[/tex]".
Explanation:
The efficiency system:
[tex]\eta =\frac{P_{req}}{P} \times 10\\\\P =\frac{P_{req}}{\eta} \times 10\\\\[/tex]
[tex]=(\frac{2.20 \times 10^6 \ W}{30})\times 100\\\\=(\frac{220 \times 10^6 \ W}{30})\\\\=(\frac{22 \times 10^6 \ W}{3})\\\\=7.33 \times 10^6 \ W[/tex]
Using formula:
[tex]A=\frac{P}{I}[/tex]
Effective area:
[tex]A= \frac{7.33 \times 10^6 \ W}{1020\ \frac{W}{m^2}}\\\\[/tex]
[tex]=\frac{7.33 \times 10^6 }{1020}\ m^2 \\\\ =0.0071862 \times 10^6 \ m^2 \\\\=7.1862 \times 10^3 \ m^2 \\\\[/tex]