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.
A wheel has a diameter of 10m and weight 360N what minimum horizontal force is necessary to pull the wheel over a brick 0.1m when a force is applied at the wheel
The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas. During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles, while during the adiabatic expansion the volume increases by a factor of 5.7. The work output of the engine is 940 J in each cycle. Compute the temperatures of the two reservoirs between which this engine
operates.
Answer:
The hot temperature is 157.5 K
The cold temperature is 48.8 K
Explanation:
Step 1: Data given
The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas.
The volume increases by a factor of 5.7
The work output of the engine is 940 J in each cycle.
During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles. This means V2 = 2*V1 (and V4 = 2*V3)
Step 2:For a carnot engine:
V2/V1 = V4/V3
Work = nR((T1)ln(V2/V1) - (T2)ln(V4/V3))
⇒with Work = the work done in the cycle = 940J
⇒with n = the number of moles = 1.50 moles
⇒with R = the gas constant = 8.314 J/mol*K
⇒with T1 = the hot temperature
⇒With T2⇒ the cold temperature
where R = 8.31 J/mol K Gas Constant
940J = 1.5moles * 8.314 J/mol*K * (T1*ln(2) - T2*ln(2)))
940 = 1.5 * 8.314 ln(2) * (T1-T2)
(T1-T2) = 940 / (1.5*8.314*ln(2))
(T1-T2) = 108.7K
For the reversible adiabatic expansion: T2 = T1*(V1/V2)^(R/Cv). Where V2/V1 = 5.7 (Because during the adiabatic expansion the volume increases by a factor of 5.7)
For a monatomic ideal gas, Cv = 3/2R
When we combine both, we'll have:
T2 = T1*(1/5.7)^(R/3/2R)
T2 = T1*(1/5.7)^(2/3)
T2= T1 * 0.31
Since we know that (T1-T2) = 108.7K
we have:
T1 - 0.31T1= 108.7K
0.69T1 = 108.7K
T1 = 157.5K
T2 = 157.5*0.31 = 48.8K
Differentiate between Scalar quantity and Vector quantity and give two examples each
The north pole of magnet A will __?____ the south pole of magnet B
Answer:
A will attract
B will repare
At which point is the kinetic energy of the pendulum the greatest?
C
A
D
B
Answer:
Point C
Explanation:
Greatest Kinetic Energy means lowest potential energy since energy is conserved. Lowest potential energy means lowest height which is at Point C.
A
Fluids in which the shear stress must reach
certain minimum value(yield stress)
before flow commences are called
Answer:
Plastic
Explanation:
Shear Modulus can be defined as the ratio of shear stress to shear strain with respect to a physical object.
This ultimately implies that, Shear Modulus arises as a result of the application of a shear force on an object or body which eventually leads to its deformation. Thus, this phenomenon is simply used by scientists to measure or determine the rigidity of an object or body.
Fluids in which the shear stress must reach certain minimum value (yield stress) before flow commences are called plastic. Thus, a plastic would only begin to flow when its shear stress attain a certain minimum value (yield stress). The unit of measurement of yield stress is usually mega pascal (MPa).
why do you like the full moon ?
Answer:
The Moon brings perspective. Observing the Moon, and I mean really looking – sitting comfortably, or lying down on a patch of grass and letting her light fill your eyes, it's easy to be reminded of how ancient and everlasting the celestial bodies are. When I do this, it always puts my life into perspective.Answer:
because it look more impressive than empty dark sky .
1. A 20.0 N force directed 20.0° above the horizontal is applied to a 6.00 kg crate that is traveling on a horizontal
surface. What is the magnitude of the normal force exerted by the surface on the crate?
N = 52.0 N
Explanation:
Given: [tex]F_a= 20.0\:\text{N}=\:\text{applied\:force}[/tex]
[tex]m=6.00\:\text{kg}[/tex]
[tex]N = \text{normal force}[/tex]
The net force [tex]F_{net}[/tex] is given by
[tex]F_{net} = N + F_a\sin 20 - mg=0[/tex]
Solving for N, we get
[tex]N = mg - F_a\sin 20[/tex]
[tex]\:\:\:\:\:\:= (6.00\:\text{kg})(9.8\:\text{m/s}^2) - (20.0\:\text{N}\sin 20)[/tex]
[tex]\:\:\:\:\:\:= 52.0\:\text{N}[/tex]
Which is the most difficult subject?
Answer:
Quantum Mechanics
Explanation:
Well, that's what I think personally.
Cold air rises because it is denser than water, is this true?
Answer:
true
Explanation:
im not sure please dont attack me
You need to calculate the volume of berm that has a starting cross-sectional area of 118 SF, and an ending cross-sectional area of 245 SF. The berm is 300 ft long and is assumed to taper evenly between the two cross-sectional areas, what is the calculated volume of the berm in cubic feet
g Consider a mass-spring system where the spring constant is 5 N/m and the mass on the spring is 0.5 kg. The mass is moved a distance of -0.9 m from its equilibrium position. How much work is done by the spring
Answer:
The work done by the spring is 2.025 J
Explanation:
Given;
mass on the spring, m = 0.5 kg
spring constant, k = 5 N/m
extension of the spring, x = 0.9 m
The work done by the spring is calculated as;
[tex]W = \frac{1}{2} kx^2\\\\W = \frac{1}{2} \times 5 \times (0.9)^2\\\\W = 2.025 \ J[/tex]
Therefore, the work done by the spring is 2.025 J
What quantity of heat is transferred when a 150.0g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of the heat flow?
Answer:
Heat is flowing into the metal.
Explanation:
From the question given above, the following data were obtained:
Mass (M) of iron = 150 g
Initial temperature (T₁) = 25.0°C
Final temperature (T₂) = 73.3°C
Direction of heat flow =?
Next, we shall determine the change in the temperature of iron. This can be obtained as follow:
Initial temperature (T₁) = 25.0 °C
Final temperature (T₂) = 73.3 °C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 73.3 – 25
ΔT = 48.3 °C
Next, we shall determine the heat transfered. This can be obtained as follow:
Mass (M) of iron = 150 g
Change in temperature (ΔT) = 48.3 °C
Specific heat capacity (C) of iron = 0.450 J/gºC
Heat (Q) transfered =?
Q = MCΔT
Q = 150 × 0.450 × 48.3
Q = 3260.25 J
Since the heat transferred is positive, it means the iron metal is absorbing the heat. Thus, heat is flowing into the metal.
You are stranded in a stationary boat. Your friend is on a dock, but the boat is just beyond his reach. There is a 5 kg anchor in the boat. You'd like to get the boat to move closer to the dock so your friend can rescue you. Select from the following list what effect each change will have on the position of the boat relative to the dock. A. The boat will move closer to the dock. B. The boat will move away from the dock. C. The position of the boat relative to the dock will not change.
Answer:
running away and launching the anchor that will give a greater speed towards the dock v₄.
Explanation:
To try to bring the boat closer to the dock, several cases can be carried out.
* move inside the ship so that the center of mass changes and since moving away you have a speed v, the ship will approach the dock at a speed v₂,
* Throw the anchor in the opposite direction to the dock so that using the conservation of the moment the boat moves towards it, it moves at a speed v₃
* A combination of the two processes running away and launching the anchor that will give a greater speed towards the dock v₄.
In all cases, the friction must be zero.
All other movements move the ship away from the dock
65. The weight of a body when totally immersed in a liquid is 4.2N if he weight of the liquid displaced is 2.5N. Find the weight of the body in air.
Answer:
Given, Apparent weight(W₂)=4.2N
Weight of liquid displaced (u)=2.5N
Let weight of body in air = W₁
Solution,
U=W₁-W₂
W₁=4.2=2.5=6.7N
∴Weight of body in air is 6.7N
a student weighs 1200N they are standing in an elevator that is moving downwards at a constant speed of
Answer:
Elevator That Is Moving Downwards At A Constant Speed Of 4.9 M/S. What Is The Magnitude Of The Net Force Acing On The Student?
This problem has been solved!
This problem has been solved!See the answer
This problem has been solved!See the answerA student weighs 1200N. They are standing in an elevator that is moving downwards at a constant speed of 4.9 m/s. What is the magnitude of the net force acing on the student?
Explanation:
use this R= m(g-a), where R = reaction = weight, m= mass, a= acceleration and g= acceleration due to gravity
A Catapult throws a payload in a circle with an arm that is 65.0 cm long. At a certain instant, the arm is rotating at 8.0 rad/s and the angular speed is increasing at 40.0 rad/s2. For this instant, find the magnitude of the acceleration of the payload.
Answer:
The acceleration of the payload is 26 m/s2.
Explanation:
length, L = 65 cm = 0.65 m
angular acceleration = 40 rad/s^2
The acceleration is given by
a = angular acceleration x length
a = 40 x 0.65
a = 26 m/s^2
No esporte coletivo, um dos principais fatores desenvolvidos é o desenvolvimento social. Qual desses não faz parte das virtudes ensinadas no esporte?
Companheirismo
Humildade
Ser justo (Fair Play)
Vencer independente do que precise ser feito
Answer:
fair palybtgshsisuehdh
The angular velocity of an object is given by the following equation: ω(t)=(5rads3)t2\omega\left(t\right)=\left(5\frac{rad}{s^3}\right)t^2ω(t)=(5s3rad)t2 What is the angular displacement of the object (in rad) between t = 2 s and t = 4 s?
Answer:
The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.
Explanation:
The angular velocity of the object ([tex]\omega[/tex]), in radians per second, is given by the following expression:
[tex]\omega(t) = 5\cdot t^{2}[/tex] (1)
Where [tex]t[/tex] is the time, measured in seconds.
The change in the angular displacement ([tex]\Delta \theta[/tex]), in radians, is found by means of the following definite integral:
[tex]\Delta \theta = \int\limits^{4}_{2} {5\cdot t^{2}} \, dt[/tex] (2)
Then we proceed to integrate on the function in time:
[tex]\Delta \theta = \frac{5}{3}\cdot (4^{2}-2^{2})[/tex]
[tex]\Delta \theta = 20\,rad[/tex]
The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.
The weight of a hydraulic barber's chair with a client is 2100 N. When the barber steps on the input piston with a force of 44 N, the output plunger of a hydraulic system begins to lift the chair. Determine the ratio of the radius of the output plunger to the radius of the input piston.
Answer:
[tex]\frac{r_1}{r_2}=6.9[/tex]
Explanation:
According to Pascal's Law, the pressure transmitted from input pedal to the output plunger must be same:
[tex]P_1 = P_2\\\\\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\\\frac{F_1}{F_2}=\frac{A_1}{A_2}\\\\\frac{F_1}{F_2}=\frac{\pi r_1^2}{\pi r_2^2}\\\\\frac{F_1}{F_2}=\frac{r_1^2}{r_2^2}[/tex]
where,
F₁ = Load lifted by output plunger = 2100 N
F₂ = Force applied on input piston = 44 N
r₁ = radius of output plunger
r₂ = radius of input piston
Therefore,
[tex]\frac{r_1^2}{r_2^2}=\frac{2100\ N}{44\ N}\\\\\frac{r_1}{r_2}=\sqrt{\frac{2100\ N}{44\ N}} \\\\\frac{r_1}{r_2}=6.9[/tex]
A uniform steel rod of length 0.9 m and mass 3.8 kg has two point masses of 2.3 kg each at the two ends. Calculate the moment of inertia of the system about an axis perpendicular to the rod, and passing through its center.
Answer: [tex]2.4705\ kg.m^2[/tex]
Explanation:
Given
length of the rod is L=0.9 m
Mass of the rod m=3.8 kg
Point masses has mass of m=2.3 kg
Moment of Inertia of the rod about the center is
[tex]\Rightarrow I_o=\dfrac{1}{12}ML^2[/tex]
Moment of inertia of combined system is the sum of rod and two point masses.
[tex]\Rightarrow I=I_o+2mr^2[/tex]
[tex]\Rightarrow I=\dfrac{1}{12}3.8\times 0.9^2+2\times 2.3\times \left(\dfrac{0.9}{2}\right)^2\\\\\Rightarrow I=1.539+0.9315\\\Rightarrow I=2.4705\ kg-m^2[/tex]
A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver immediately puts the car in neutral so that it will roll.
What will be the car’s speed as it coasts into the gas station on the other side of the valley?
Answer:
v_f = 17.4 m / s
Explanation:
For this exercise we can use conservation of energy
starting point. On the hill when running out of gas
Em₀ = K + U = ½ m v₀² + m g y₁
final point. Arriving at the gas station
Em_f = K + U = ½ m v_f ² + m g y₂
energy is conserved
Em₀ = Em_f
½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂
v_f ² = v₀² + 2g (y₁ -y₂)
we calculate
v_f ² = 20² + 2 9.8 (10 -15)
v_f = √302
v_f = 17.4 m / s
Which of the following represents the velocity time relationship for a falling apple?
Answer "a" would be correct.
Answer:
d
Explanation:
There's an acceleration from gravity, thus the velocity is becoming faster and faster as it reaches the ground. Thus its D
Brainliest please~
Hi can someon help me how to answer this?
Btw I'm from Philippines
Answer:
Test 1
1.True
2.True
3.True
4.False
5.True
6.True
7.False
8.True
9.True
10.True
yung iba nasa pic
Momentum is defined as mass ___ velocity
find the weight of a body of mass 200kg on the earth at a latitude 30°.(R=6400 km ,g=9.8m/s²,ω=7.27×10⁻⁵ rad/sec)
Answer:
................ftf6x
If you change the motor in your vehicle you need to notify the DMV within ____,
days of this change.
-20
-25
-10
-15
when you change your motor on your vehicle you need to notify the DMV within 10 days
If you change the motor in your vehicle you need to notify the DMV within 10 days of this change.
An engine or motorAn engine or motor is a machine designed to convert one or more forms of energy into mechanical energy.
Available energy sources include potential energy (e.g. energy of the Earth's gravitational field as exploited in hydroelectric power generation), heat energy (e.g. geothermal), chemical energy, electric potential, and nuclear energy (from nuclear fission or nuclear fusion). Many of these processes generate heat as an intermediate energy form, so heat engines have special importance. Some natural processes, such as atmospheric convection cells convert environmental heat into motion (e.g. in the form of rising air currents). Mechanical energy is of particular importance in transportation but also plays a role in many industrial processes such as cutting, grinding, crushing, and mixing.
Mechanical heat engines convert heat into work via various thermodynamic processes. The internal combustion engine is perhaps the most common example of a mechanical heat engine, in which heat from the combustion of fuel causes rapid pressurization of the gaseous combustion products in the combustion chamber, causing them to expand and drive a piston, which turns a crankshaft. Unlike internal combustion engines, a reaction engine (such as a jet engine) produces thrust by expelling reaction mass, by Newton's third law of motion.
Learn more about motor
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Two plastic bowling balls, 1 and 2, are rubbed with cloth until they each carry a uniformly distributed charge of magnitude 0.50 nC . Ball 1 is negatively charged, and ball 2 is positively charged. The balls are held apart by a 900-mm stick stuck through the holes so that it runs from the center of one ball to the center of the other.
Required:
What is the magnitude of the dipole moment of the arrangement?
Answer:
The right solution is "[tex]4.5\times 10^{-10} \ Cm[/tex]".
Explanation:
Given that,
q = 0.50 nC
d = 900 mm
As we know,
⇒ [tex]P=qd[/tex]
By putting the values, we get
⇒ [tex]=0.50\times 900[/tex]
⇒ [tex]=(0.50\times 10^{-9})\times 0.9[/tex]
⇒ [tex]=4.5\times 10^{-10} \ Cm[/tex]
Answer:
The dipole moment is 4.5 x 10^-10 Cm.
Explanation:
Charge on each ball, q = 0.5 nC
Length, L = 900 mm = 0.9 m
The dipole moment is defined as the product of either charge and the distance between them.
It is a vector quantity and the direction is from negative charge to the positive charge.
The dipole moment is
[tex]p = q L\\\\p = 0.5 \times 10^{-9}\times 0.9\\\\p = 4.5\times 10^{-10} Cm[/tex]
Electrical charges are of two types. True False
Answer:
Electrical charges r of 2 types its true.they are positive and negative.hope it helps.stay safe healthy and happy..Answer: Think its true
Why is the temperature constant during the melting of water?
[tex] \orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}}[/tex]
Why is the temperature constant during the melting of water?
[tex] \huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}[/tex]
[tex] \orange{\underline{\huge{\bold{\textit{\green{\bf{ REASON}}}}}}}[/tex]
THE HEAT WE R SUPPLYING TO THE WATER TO RAISE THE TEMP OF THE WATER IS USED BY THE MOLECULES TO BREAK INTERMOLECULAR BONDS WHICH HELP IN THE CHANGING OF THE LATTICE (STRUCTURE) OF THE WATER .
ICE HAS A HEXAGONAL RING LIKE STRUCTURE WHICH IS CONVERTED INTO REGULAR CRYSTALLINE STRUCTURE WHICH CAN ONLY BE FORMED WITH THE HELP OF FORMATION OF NEW BONDS AND BREAKDOWN OF OLDER ONES
THE AMOUNT OF ENERGY WHICH IS USED IN CONVERSATION OF THE STATE OF FROM SOLID TO LIQUID IS KNOWN AS LATENT HEAT OF FUSION.
SO TEMP REMAIN CONSTANT DURING CHANGE IN STATE .
[tex] \red \star{Thanks \: And \: Brainlist} \blue\star \\ \green\star If \: U \: Liked \: My \: Answer \purple \star[/tex]