Answer:
T = 5.45 10⁻¹⁰ [tex]\sqrt{(R_e + h)^3}[/tex]
Explanation:
Let's use Newton's second law
F = ma
force is the universal force of attraction and acceleration is centripetal
G m M / r² = m v² / r
G M / r = v²
as the orbit is circular, the speed of the satellite is constant, so we can use the kinematic relations of uniform motion
v = d / T
the length of a circle is
d = 2π r
we substitute
G M / r = 4π² r² / T²
T² = [tex]\frac{4\pi ^2 }{GM} \ r^3[/tex]
the distance r is measured from the center of the Earth (Re), therefore
r = Re + h
where h is the height from the planet's surface
let's calculate
T² = [tex]\frac{4\pi ^2}{ 6.67 \ 10^{-11} \ 1.991 \ 10^{30}}[/tex] (Re + h) ³
T = [tex]\sqrt{29.72779 \ 10^{-20}} \ \sqrt[2]{R_e+h)^3}[/tex]
T = 5.45 10⁻¹⁰ [tex]\sqrt{(R_e + h)^3}[/tex]
You walk into a room and you see 4 chickens on a bed 2 cows on the floor and 2 cats in a chair. How many legs are on the ground? (I know this answer just a riddle to see who knows it) (:
Answer:
18
Explanation:
I'm pretty sure I got it right
A closely wound, circular coil with radius 2.70 cm has 800 turns. What must the current in the coil be if the magnetic field at the center of the coil is 0.0750 T
Answer:
Approximately 4.029 A
Explanation:
We can use the formula that the B field of a few loops all with current in same direction is permeability of free space (mu)* current * Number or loops divided by 2*radius. You are given B field, radius(convert into meters), number of loops and mu is 4pi * 10^-7. Solve for current and you get 4.029 Amperes.
A single-turn square loop of wire, 2.00 cm on each edge, carries a clockwise current of 0.240 A. The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30.0 turns/cm and carries a clockwise current of 15.0 A. Find:
a. the force on each side of the loop
b. the torque acting on the loop.
Answer:
Explanation:
a )
Magnetic field inside solenoid B = μ₀ NI ,
μ₀ = 4π x 10⁻⁷ ; N is no of turns per meter length in solenoid and I is current B= 4π x 10⁻⁷ x 30 x 10² x 15
= .0565 T .
Force on each side of square loop = B i L
B is external magnetic field , i is current in loop and L is length of side
Force on each side of square loop = .0565 x .24 x 2 x 10⁻²
= 2.7 x 10⁻⁴ N .
b )
Torque on the loop = F x d
F is force on one side , d is distance between two sides , that is side of the square loop
= 2.7 x 10⁻⁴ x 2 x 10⁻² N.m
= 5.4 x 10⁻⁶ N.m .
Two electrons are passing 20.0 mm apart. What is the electric repulsive force that they exert on each other
Answer:
0.5766422350752*10^-24 N
Explanation:
Couloumb's law states that states that there is an electrical force acting on 2 static charges. The magnitude is directly proportional to the product of the 2 charges.
Strength of electrons : q1 = q2 = 1.602 x 10-19. C
Substitute and solve:
F = (9*10^9)(1.602 x 10-19)(1.602 x 10-19) / (0.02)^2
Done.
The density of blood is 1055 kg/m3 . If the blood at the very top of your head exerts a minimum gauge pressure of 45 mm Hg (6000 Pa), estimate the gauge pressure at your heart in pascals.
Answer:
P = 10135.6 Pa
Explanation:
For this exercise we use that the pressure varies with the height
P = P₀ + ρ g h
where h is the height from the head to the heart, which is approximately
h = 40 cm = 0.40m and P₀ is the head pressure P₀ = 6000 Pa
P = 6000 + 1055 9.8 0.40
P = 6000 + 4135.6
P = 10135.6 Pa
a baseball is thrown vertically upward with an initial velocity of 20m/s.
A,what maximum height will it attain? B,what time will elapse before it strike the ground?
C,what is the velocity just before it strike the ground?
Answer:
Look at explanation
Explanation:
a)Only force acting on the object is gravity, so a=-g (consider up to be positive)
use: v^2=v0^2+2a(y-y0)
plug in givens, at max height v=0
0=400-19.6(H)
Solve for H
H= 20.41m
b) Use: y=y0+v0t+1/2at^2
Plug in givens
0=0+20t-4.9t^2
solve for t
t=4.08 seconds
c) v=v0+at
v=20-39.984= -19.984m/s
3. You have a variable-voltage power supply and a capacitor in the form of two metal disks of radius 0.6 m, held a distance of 1 mm apart. What is the largest voltage you can apply to the capacitor without the air becoming highly conductive
Answer:
The breakdown of air occurs at a maximum voltage of 3kV/mm.
Explanation:
The breakdown of air occurs at a maximum voltage of 3kV/mm.
At this level of voltage the air between the plates become highly ionised and breakdown occurs. Since, the distance held between the plates is 1mm , it can withstand a maximum voltage of 3 kV.
After this voltage the air will become conductive in nature and will form ions in the air between the plates and ultimately breakdown will take place with a flash.
A 0.033-kg bullet is fired vertically at 222 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball
Answer:
The maximum height risen by the bullet-baseball system after the collision is 81.76 m.
Explanation:
Given;
mass of the bullet, m₁ = 0.033 kg
mass of the baseball, m₂ = 0.15 kg
initial velocity of the bullet, u₁ = 222 m/s
initial velocity of the baseball, u₂ = 0
let the common final velocity of the system after collision = v
Apply the principle of conservation of linear momentum to determine the common final velocity.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
0.033 x 222 + 0.15 x 0 = v(0.033 + 0.15)
7.326 = v(0.183)
v = 7.326 / 0.183
v = 40.03 m/s
Let the height risen by the system after collision = h
Initial velocity of the system after collision = Vi = 40.03 m/s
At maximum height, the final velocity, Vf = 0
acceleration due to gravity for upward motion, g = -9.8 m/s²
[tex]v_f^2 = v_i^2 +2gh\\\\0 = 40.03^2 - (2\times 9.8)h\\\\19.6h = 1602.4\\\\h = \frac{1602.4}{19.6} \\\\h = 81.76 \ m[/tex]
Therefore, the maximum height risen by the bullet-baseball system after the collision is 81.76 m.
What would the radius (in mm) of the Earth have to be in order for the escape speed of the Earth to equal (1/21) times the speed of light (300000000 m/s)? You may ignore all other gravitational interactions for the rocket and assume that the Earth-rocket system is isolated. Hint: the mass of the Earth is 5.94 x 1024kg and G=6.67×10−11Jmkg2G=6.67\times10^{-11}\frac{Jm}{kg^2}G=6.67×10−11kg2Jm
Answer:
The expected radius of the Earth is 3.883 meters.
Explanation:
The formula for the escape speed is derived from Principle of Energy Conservation and knowing that rocket is initially at rest on the surface of the Earth and final energy is entirely translational kinetic, that is:
[tex]U = K[/tex] (1)
Where:
[tex]U[/tex] - Gravitational potential energy, in joules.
[tex]K[/tex] - Translational kinetic energy, in joules.
Then, we expand the formula by definitions of potential and kinetic energy:
[tex]\frac{G\cdot M\cdot m}{r} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)
Where:
[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.
[tex]M[/tex] - Mass of the Earth. in kilograms.
[tex]m[/tex] - Mass of the rocket, in kilograms.
[tex]r[/tex] - Radius of the Earth, in meters.
[tex]v[/tex] - Escape velocity, in meters per second.
Then, we derive an expression for the escape velocity by clearing it within (2):
[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]
[tex]v = \sqrt{\frac{2\cdot G \cdot M}{r} }[/tex] (3)
If we know that [tex]v = \frac{1}{21}\cdot c[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]M = 5.94\times 10^{24}\,kg[/tex], [tex]G = 6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex] and [tex]M = 5.94\times 10^{24}\,kg[/tex], then the expected radius of the Earth is:
[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]
[tex]r = \frac{2\cdot G \cdot M}{v^{2}}[/tex]
[tex]r = \frac{2\cdot \left(6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.94\times 10^{24}\,kg)}{\left[\frac{1}{21}\cdot \left(3\times 10^{8}\,\frac{m}{s} \right) \right]^{2}}[/tex]
[tex]r = 3.883\,m[/tex]
The expected radius of the Earth is 3.883 meters.
The sound level measured in a room by a person watching a movie on a home theater system varies from 40 dB during a quiet part to 80 dB during a loud part. Approximately how many times louder is the latter sound
Answer:
[tex]\alpha=-3.01dB[/tex]
Explanation:
From the question we are told that:
Sound level intensity
[tex]\triangle I=40dB-80dB[/tex]
Generally the equation for intensity level is mathematically given by
[tex]\alpha=10log_{10}(I/I_x)dB[/tex]
Where
I= Intensity measured
[tex]I_x=Threshold\ of\ audibility[/tex]
[tex]I_x= 10-12 W / m2[/tex]
[tex]\alpha= 10 log10 \frac{I_1}{I_x} - 10 log10 \frac{}I_2{I_x}[/tex]
[tex]\alpha= 10 log10 \frac{I_1}{I_2}[/tex]
[tex]\alpha=10 log10\frac{40}{80}[/tex]
[tex]\alpha=-3.01dB[/tex]
The period of a pendulum is the time it takes the pendulum to swing back and forth once. If the only dimensional quantities that the period depends on are the acceleration of gravity, g, and the length of the pendulum, l, what combination of g and l must the period be proportional to
Explanation:
Let T is the period of a pendulum. The SI unit of time is seconds (s).
It depends on the acceleration of gravity, g, and the length of the pendulum, l.
The SI unit of acceleration of gravity, g and the length of the pendulum, l are m/s² and m respectively.
If we divide m and m/s², we left with s². If the square root of s² is taken, we get s only i.e. the SI unit of period of a pendulum.
So,
[tex]T\propto \sqrt{\dfrac{l}{g}}[/tex]
Hence, this is the required solution.
A flag pole 18m high casts a shadow 9.6m long . Find the distance of top of pole from the far of end of Shadow.
Answer:
[tex]{ \bf{pythogras \: theorem :}} \\ \\ { \tt{ = \sqrt{ {9.6}^{2} + {18}^{2} } }} \\ = 20.4 \: cm[/tex]
If the sum of the external forces on an object is zero, then the sum of the external torques on it
must also be zero.
A) True
B) False
Answer:
True.
Explanation:
If the sum of the external forces on an object is zero, then the sum of the external torques on it must also be zero.
The net external force and the net external torque acting on the object have to be zero for an object to be in mechanical equilibrium.
Hence, the given statement is true.
You are helping your friend move a new refrigerator into his kitchen. You apply a horizontal force of 275 N in the positive x direction to try and move the 61 kg refrigerator. The coefficient of static friction is 0.58. (a) How much static frictional force does the floor exert on the refrigerator
Answer:
f = 347.08 N
Explanation:
The frictional force exerted by the floor on the refrigerator is given as follows:
[tex]f = \mu R = \mu W[/tex]
where,
f = frictional force = ?
μ = coefficient of static friction = 0.58
W = Weight of refrigerator = mg
m = mass of refrigerator = 61 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]f = \mu mg\\f = (0.58)(61\ kg)(9.81\ m/s^2)\\[/tex]
f = 347.08 N
A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels, from left to right along a long, horizontal stretched string with a speed of 36.0 m s. I Take the origin at the left end of the undisturbed string. At time t = 0 the left end of the string has its maximum upward displacement,
(a) What is the frequency of the wave?
(b) What is the angular frequency of the wave?
(c) What is the wave number of the wave?
(d) What is the function y(x,t) that describes the wave?
(e) What is y(t) for a particle at the left end of the string?
(f) What is y(t) for a particle 1.35 m to the right of the origin?
(g) What is the maximum magnitude of transverse velocity of any particle of the string?
(h) Find the transverse displacement of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
(i) Find the transverse velocity of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
Explanation:
Given that,
Amplitude, A = 2.5 nm
Wavelength,[tex]\lambda=1.8\ m[/tex]
The speed of the wave, v = 36 m/s
At time t = 0 the left end of the string has its maximum upward displacement.
(a) Let f is the frequency. So,
[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{36}{1.8}\\\\f=20\ Hz[/tex]
(b) Angular frequency of the wave,
[tex]\omega=2\pi f\\\\=2\pi \times 20\\\\=125.7\ rad/s[/tex]
(c) The wave number of the wave[tex]=\dfrac{1}{\lambda}[/tex]
[tex]=\dfrac{1}{1.8}\\\\=0.56\ m^{-1}[/tex]
Question 9 of 10
According to the law of conservation of momentum, the total initial
momentum equals the total final momentum in a(n)
A. Interacting system
B. System interacting with one other system
C. Isolated system
D. System of balanced forces
Answer:
The answer is C. Isolated System
Answer:
C. Isolated system
Explanation :
∵According to law of conservation of momentum ,In an isolated system ,the total momentum remains conserved.
A scientist who studies fossils of ancient life forms .O ornithologist O Paleontologist O Ichthyologist O Marine Biologist .
Hurry !! First one to answer gets points !
Answer:
paleontologist
Explanation:
Paleontologists are scientists that investigate the fossils of extinct life forms. Thus, the correct option is B.
What is Fossil?A fossil is defined as the preserved trace, imprint, or proof of a once-living entity from a past geological era. Exoskeletons, bones, shells, impressions of animals or microbes in stone, objects preserved in amber, hair, petrified wood, and genetic traces are a few examples. The collection of all the fossils is called as the fossil record.
An organism from a past geologic era that has been preserved in the Earth's crust is referred to as a fossil. Paleontologists are scientists that investigate the fossils of extinct life forms. The intricate system of fossil records is the main source of information about the evolution of life on Earth.
Therefore, the correct option is B.
Learn more about Fossils, here:
https://brainly.com/question/5431129
#SPJ7
A resident of a lunar colony needs to have her blood pressure checked in one of her legs. Assume that we express the systemic blood pressure as we do on earth and that the density of blood does not change. Suppose also that normal blood pressure on the moon is still 120/80 (which may not actually be true).
Required:
If a lunar colonizer has her blood pressure taken at a point on her ankle that is 1.5 m below her heart, what will be her systemic blood-pressure reading, expressed in the standard way, if she has normal blood pressure? The acceleration due to gravity on the moon is 1.67 m/s^2
Answer:
The pressure is 2505 Pa.
Explanation:
Height, h = 1.5 m
density of blood, d = 1000 kg/cubic meter
Gravity, g = 1.67 m/s^2
let the pressure is P.
The pressure due to the fluid is given by
P = h d g
P = 1.5 x 1000 x 1.67
P = 2505 Pa
g A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 3.00 m at constant speed. If the coefficient of kinetic friction between the desk and the floor is 0.400, how much work did she do
The desk is in equilbrium, so Newton's second law gives
∑ F (horizontal) = p - f = 0
∑ F (vertical) = n - mg = 0
==> n = mg
==> p = f = µn = µmg = 0.400 (80.0 kg) g = 313.6 N
The student pushes the desk 3.00 m, so she performs
W = (313.6 N) (3.00 m) = 940.8 Nm ≈ 941 J
of work.
any one tell me about the earth rotation it rotatining or not with any proof?
1.- Que distancia recorrió una carga de 2,5x10-6 coul, generando así un campo eléctrico de 55new/coul.
Answer:
r = 20.22 m
Explanation:
Given that,
Charge,[tex]q=2.5\times 10^{-6}\ C[/tex]
Electric field, [tex]E=55\ N/C[/tex]
We need to find the distance. We know that, the electric field a distance r is as follows :
[tex]E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m[/tex]
So, the required distance is 20.22 m.
The gravitational force Asteroid A experiences is the gravitational force Asteroid C experiences
Answer:
The gravitational force Asteroid A experiences is greater than the gravitational force Asteroid C experiences
The atoms in your body are mostly empty space . And so are the atoms in any wall. Why then is your body unable to pass through walls ?
First of all, both are not a single sheet of atom. There are many layers of atoms, so the empty part gets beside each other, so there are less empty part. Secondly, there are so many atoms that the probability that they will have empty space at the same place necessary, is negligible.
This was something from logic.
The reason I was taught in my class was that only a limited number of electrons can be in a given orbit, so atoms cannot overlap each other.
The value found for the universal gravitational constant, G, will vary depending on the materials used for the balls of a Cavendish balance. Question 11 options: True False
Answer:
false
Explanation:
took the test
what is the time taken by moving body with acceleration 0.1m/s2 if the initial or finak velocities are 20m/s and 30m/s respectively?
Answer:
t= 100s
Explanation:
use v=v0+at
plug in givens and solve for t
30=20+0.1*t
t= 100s
Question 4(Multiple Choice Worth 4 points)
(02.04 MC)
Which explanation justifies why the theory of evolution is a theory and not a law?
Predicts an organism's ability to adapt to its environment
It can be expressed as a simple mathematical statement
Explains the existence of diverse forms of life on Earth
O Additional evidence will change the theory into a law
Answer:
A(predicts an organisms ability to adapt to its enviroment, it is not a fact that each organization can adapt)
Explanation:
what is conservation energy?
Explanation:
Conservation of energy, principle of physics according to which the energy of interacting bodies or particles in a closed system remains constant
hope it is helpful to you
In Trial II, the same spring is used as in Trial I. Let us use this information to find the suspended mass in Trial II. Use 0.517 ss for the value of the period.
Trial 1 Spring constant is 117N/m, period of oscillations .37s, mass of the block is .400kg .
Trial 2 oscillation period is .52s
Answer:
[tex]M_2=0.79kg[/tex]
Explanation:
From the question we are told that:
Period [tex]T=0.517s[/tex]
Trial 1
Spring constant [tex]\mu=117N/m[/tex]
Period [tex]T_1=0.37[/tex]
Mass [tex]m=0.400kg[/tex]
Trial 2
Period [tex]T_2=0.52[/tex]
Generally the equation for Spring Constant is mathematically given by
\mu=\frac{4 \pi^2 M}{T^2}
Since
[tex]\mu _1=\mu_2[/tex]
Therefore
[tex]\frac{4 \pi^2 M_1}{T_1^2}=\frac{4 \pi^2 M_2}{T_2^2}[/tex]
[tex]M_2=M_1*(\frac{T_2}{T_1})^2[/tex]
[tex]M_2=0.400*(\frac{0.52}{0.37}})^2[/tex]
[tex]M_2=0.79kg[/tex]
When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C where C is a constant. Suppose that at a certain instant the volume is 420 cubic centimeters and the pressure is 99 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?
Answer:
[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]
Explanation:
We are given that
[tex]PV^{1.4}=C[/tex]
Where C=Constant
[tex]\frac{dP}{dt}=-7KPa/minute[/tex]
V=420 cubic cm and P=99KPa
We have to find the rate at which the volume increasing at this instant.
Differentiate w.r.t t
[tex]V^{1.4}\frac{dP}{dt}+1.4V^{0.4}P\frac{dV}{dt}=0[/tex]
Substitute the values
[tex](420)^{1.4}\times (-7)+1.4(420)^{0.4}(99)\frac{dV}{dt}=0[/tex]
[tex]1.4(420)^{0.4}(99)\frac{dV}{dt}=(420)^{1.4}\times (7)[/tex]
[tex]\frac{dV}{dt}=\frac{(420)^{1.4}\times (7)}{1.4(420)^{0.4}(99)}[/tex]
[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]
Answer:
[tex]\dot V=2786.52~cm^3/min[/tex]
Explanation:
Given:
initial pressure during adiabatic expansion of air, [tex]P_1=99~kPa[/tex]
initial volume during the process, [tex]V_1=420~cm^3[/tex]
The adiabatic process is governed by the relation [tex]PV^{1.4}=C[/tex] ; where C is a constant.
Rate of decrease in pressure, [tex]\dot P=7~kPa/min[/tex]
Then the rate of change in volume, [tex]\dot V[/tex] can be determined as:
[tex]P_1.V_1^{1.4}=\dot P.\dot V^{1.4}[/tex]
[tex]99\times 420^{1.4}=7\times V^{1.4}[/tex]
[tex]\dot V=2786.52~cm^3/min[/tex]
[tex]\because P\propto\frac{1}{V}[/tex]
[tex]\therefore[/tex] The rate of change in volume will be increasing.
Three forces of magnitude 10N, 5N and 4N act on an object in the directions North, West and East respectively. Find the magnitude and directions of their resultant
Answer:
19N to the south
Explanation:
F =10N + 5N + 4N