Answer:
The tension in [tex]T_1[/tex] is 0.1024 Newtons.
The tension in [tex]T_2[/tex] is 0.1157 Newtons.
Explanation:
Lets create a free body diagram showing all of the forces; we need to show the vertical and horizontal components of the tension.
I will attach a picture of my free body diagram. Notice I created 2 new triangles with the adjacent angles of angle A and C from the original picture.
Lets make a list of all the variables we have now. Also lets write down the information we are given.
[tex]A=27\\C=38\\m=12\\g=9.81\\W\\T_{1} \\T_{1x} \\T_{1y}\\T_{2} \\T_{2x} \\T_{2y} \\[/tex]
In this situation the sum of of the vertical tension components must support the weight. To find the vertical components we can use the SIN function.
[tex]sin(x)=\frac{O}{H} \\O=Hsin(x)[/tex]
Therefore we can write that the sum of the forces in the y direction is
[tex]\sum F_y=T_1sin(A)+T_2sin(C)=W[/tex]
This system is in equilibrium; the object should not move along the x-axis. Therefore, the horizontal components of [tex]T_1[/tex] and [tex]T_2[/tex] must then equal each other. To find the horizontal components we can use the COS function.
[tex]cos(x)=\frac{A}{H} \\A=Hcos(x)[/tex]
Therefore we can write that the sum of the forces in the x direction is
[tex]\sum F_x=T_1cos(A)=T_2cos(C)[/tex]
Now we have to equations to help us solve the problem.
[tex]T_1cos(A)=T_2cos(C)[/tex]
[tex]T_1sin(A)+T_2sin(C)=W[/tex]
We do not know the numerical values of [tex]T_1[/tex] and [tex]T_2[/tex] so we will have to manipulate algebraically to solve them.
In the first equation lets solve for [tex]T_1[/tex].
[tex]T_1cos(A)=T_2cos(C)[/tex]
Divide both sides by [tex]cos(A)[/tex].
[tex]T_1=\frac{T_2cos(C)}{cos(A)}[/tex]
Separate the right side into two fractions.
[tex]T_1=\frac{T_2}{1} *\frac{cos(C)}{cos(A)}[/tex]
Use the reciprocal trig identity for cosine.
[tex]sec(x)=\frac{1}{cos(x)}[/tex]
[tex]T_1=\frac{T_2}{1} *cos(C)}*sec(A)[/tex]
[tex]T_1=T_2*cos(C)}*sec(A)[/tex]
Now insert our answer for [tex]T_1[/tex] into the second equation.
[tex]T_2*cos(C)}*sec(A)*sin(A)+T_2*sin(C)=W[/tex]
Solve for [tex]T_2[/tex]. Lets replace each trig function with its own variable to make this easier.
[tex]T_2*cos(C)}*sec(A)*sin(A)+T_2*sin(C)=W[/tex]
[tex]cos(C)=x\\sec(A)=y\\sin(A)=z\\sin(C)=u[/tex]
[tex]T_2*x*y*z+T_2*u=W\\T_2xyz+T_2u=W[/tex]
Now lets solve for [tex]T_2[/tex].
Factor [tex]T_2[/tex] out of each term.
[tex]T_2(xyz)+T_2(u)=W[/tex]
Factor [tex]T_2[/tex] out of each term.
[tex]T_2(xyz+u)=W[/tex]
Divide each side by [tex]xyz+u[/tex].
[tex]T_2=\frac{W}{xyz+u}[/tex]
Lets substitute the trig functions back in for the variables
[tex]T_2=\frac{W}{cos(C)*sec(A)*sin(A)+sin(C)}[/tex]
[tex]W[/tex] is the weight.
The formula for weight is [tex]W=mg[/tex]. Where [tex]m[/tex] is the mass in kilograms.
12 grams is 0.012 kilograms.
[tex]W=0.012*9.81[/tex]
[tex]W=0.11772[/tex]
Numerical Evaluation
Lets evaluate [tex]T_2[/tex].
[tex]T_2=\frac{0.11772}{cos(38)*sec(27)*sin(27)+sin(38)}[/tex]
[tex]T_2=0.11573[/tex]
Lets evaluate [tex]T_1[/tex].
[tex]T_1cos(A)=T_2cos(C)[/tex]
[tex]T_1*cos(27)=0.11573*cos(38)\\T_1=0.10235443[/tex]
A rating/review would be much appreciated.
Answer:
T₁ = 102.25 N (2 d.p.)
T₂ = 115.61 N (2 d.p.)
Explanation:
The diagram shows a body of mass 12 kg (weight = 12g N) held in equilibrium by two light, inextensible strings. One string makes an angle of 27° with the positive horizontal and the other string makes an angle of 38° with the negative horizontal.
The body is in equilibrium, so both the horizontal and vertical components of the forces must sum to zero.
Resolving horizontally, taking (→) as positive:
[tex]\implies -T_1 \cos (27^{\circ})+T_2 \cos (38^{\circ})=0[/tex]
[tex]\implies T_1 \cos (27^{\circ})=T_2 \cos (38^{\circ})[/tex]
[tex]\implies T_1=\dfrac{T_2 \cos (38^{\circ})}{ \cos (27^{\circ})}[/tex]
Resolving vertically, taking (↑) as positive:
[tex]\implies T_1 \sin(27^{\circ})+T_2 \sin(38^{\circ})-12\text{g}=0[/tex]
[tex]\implies T_1 \sin(27^{\circ})+T_2 \sin(38^{\circ})=12\text{g}[/tex]
Substitute the found expression for T₁ into the second equation and take g = 9.8 ms⁻²:
[tex]\implies \left(\dfrac{T_2 \cos (38^{\circ})}{ \cos (27^{\circ})}\right) \sin(27^{\circ})+T_2 \sin(38^{\circ})=12\text{g}[/tex]
[tex]\implies T_2 \cos (38^{\circ})\tan(27^{\circ})+T_2 \sin(38^{\circ})=12\text{g}[/tex]
[tex]\implies T_2 \left(\cos (38^{\circ})\tan(27^{\circ})+ \sin(38^{\circ})\right)=12\text{g}[/tex]
[tex]\implies T_2 =\dfrac{12\text{g}}{\cos (38^{\circ})\tan(27^{\circ})+ \sin(38^{\circ})}[/tex]
[tex]\implies T_2=115.614550...\:\text{N}[/tex]
Substitute the found value of T₂ into the equation for T₁ and take g = 9.8 ms⁻²:
[tex]\implies T_1=\dfrac{\left(\dfrac{12\text{g}}{\cos (38^{\circ})\tan(27^{\circ})+ \sin(38^{\circ})}\right) \cos (38^{\circ})}{ \cos (27^{\circ})}[/tex]
[tex]\implies T_1=102.250103...\text{N}[/tex]
Therefore, the value of T₁ and T₂ in the given diagram is:
T₁ = 102.25 N (2 d.p.)T₂ = 115.61 N (2 d.p.)Burning fossil fuels generates heat energy but also emits air pollution and generates greenhouse gases such as CO2 and N2O. The combustion of gigatons of fossil fuels over the last century has lead to the emission of so much greenhouse gas that we have changed the chemical composition of Earth's atmosphere. This chemical change in the atmosphere has enhanced the greenhouse effect leading to global warming and climate change today.
Over 90% of the warming that has occurred on Earth in the last 60 years has occurred in the world's oceans. Most of the excess heat energy acquired by the oceans remains in the uppermost 700 m near the surface of the water, but almost 30% of the heat energy has conducted into the ocean depths. Measurements indicate that in the last 60 years of global warming the oceans have acquired an extra 228*1021J of heat energy.
The ocean covers approximately 71% of Earth's surface, an area of 361*106 km2 and has an average depth of the ocean is 3.682 km. The density of sea water is 1023.6 kg/m3 , the specific heat of sea water is approximately 3850 J/(kgK) ,and the coefficient of volume expansion can be taken to be .21*10-3 C-1.
a) What is the temperature increase of ocean water on average over the last 60 years.
b) Find the volumetric expansion of the ocean as a result of this temperature increase.
c) If the area of the ocean is considered constant by how much has the average height of the ocean surface increased due to this heat influx?
The temperature increase over the last 60 years is 4.35 * 10¹⁰ K, the volumetric expansion over this temperature increase is 1.24 * 10¹⁶ km³ and the height of the ocean is increased by 3.44*10¹⁰ km
What is Heat TransferHeat transfer refers to the movement of thermal energy from a high-temperature body to a low-temperature body. There are three main types of heat transfer: conduction, convection, and radiation.
a) To find the temperature increase of ocean water on average over the last 60 years, we can use the formula for heat energy:
q = mcΔT
Where:
q = heat energy acquired by the oceans (228*1021J)
m = mass of the ocean water (density x volume)
c = specific heat of sea water (approximately 3850 J/(kg K))
ΔT = change in temperature
Given that the area of the ocean is 361106 km² and the average depth of the ocean is 3.682 km, the volume of the ocean can be calculated as:
V = A x D = 361106 km² x 3.682 km = 1.329*10⁶ km³
Given that the density of sea water is 1023.6 kg/m³, the mass of the ocean water can be calculated as:
m = V x ρ = 1.329 * 10⁶ km³ x 1023.6 kg/m³ = 1.36 * 10⁹ kg
Now, we can substitute these values into the equation for heat energy to find the change in temperature:
q = mcΔT
ΔT = q / (m x c)
Substituting the values we get:
ΔT = (228*10²¹J) / (1.36 *10⁹ kg x 3850 J/(kg K)) = 4.35 * 10¹⁰ K
Therefore, the temperature increase of ocean water on average over the last 60 years is 4.35 * 10¹⁰ K.
b) To find the volumetric expansion of the ocean as a result of this temperature increase, we can use the formula for volume expansion due to temperature change:
ΔV = βVΔT
Where:
ΔV = change in volume
V = initial volume of the ocean
β = coefficient of volume expansion
ΔT = change in temperature
Given that the coefficient of volume expansion can be taken to be 0.21*10⁻³ C⁻¹, we can substitute the values we have into the equation:
ΔV = 0.21*10⁻³ C⁻¹ * 1.36*10⁹ km³ * 4.35 * 10¹⁰ K = 1.24 *10¹⁶ km³
Therefore, the volumetric expansion of the ocean as a result of this temperature increase is 1.24 * 10¹⁶ km³
c) If the area of the ocean is considered constant, we can use the volumetric expansion to find the change in height of the ocean surface. The change in height can be calculated by:
ΔH = ΔV / A
Where:
ΔH = change in height
ΔV = change in volume
A = area of the ocean
Substituting the values we get:
ΔH = (1.24 * 10¹⁶ km³) / (361106 km2) = 3.44 *10¹⁰ km
Therefore, the average height of the ocean surface has increased by 3.44*10¹⁰ km due to this heat influx.
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Samuel placed a plastic container upside down over a paperclip. He held a strong magnet up to the side of the bowl and the paperclip moved toward the magnet. Which best explains the result of Samuel's investigation. A. The magnet caused a balanced force on the paperclip. B. The magnet caused a non-contact force on the paperclip. C. The magnet increased the amount of energy in the paperclip. D. The magnet reduced the net force acting on the paperclip.
Answer:
The magnet caused a non-contact force on the paperclip.
Oxygen has a molar mass of 32 g/mol. If 12 moles of oxygen are in a 0.1-m' container with an rms speed of 480 m/s, what is the pressure of the gas?
The pressure of the oxygen gas is 2.95 x 10⁵ Pa.
What is the pressure of the oxygen gas?
The pressure of the oxygen gas is calculated by applying the following formula as shown below.
Mathematically, the formula for the pressure of a gas at given root mean square pressure is given as
P ( O₂ ) = [ ( n R ( v² M / 3R ) / ( VO₂ ) ]
where;
n is the number of moles of the gasv is the root mean square speed of the gasR is the ideal gas constantM is the molar mass of the oxygen gasVO₂ is the volume of the oxygen gasP ( O₂ ) = [ ( n x v² M ) / (3 x VO₂ ) ]
The pressure of the oxygen gas is calculated as follows;
P ( O₂ ) = [ ( 12 x 480² x 0.032 ) / ( 3 x 0.1 ) ]
P ( O₂ ) = 2.95 x 10⁵ Pa
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Frici the cat has excellent hearing. He hears the tapping of his favorite treat from afar
in his plate in front of the door.
His owner dropped the treat into his plate from a height of 125 cm, the door is 30 meters from the garden gate, Frici
and he was wandering at the edge of the forest 300 meters from the garden entrance. Its reaction time is 0.1 seconds and 6.6
is speeding home at a speed of m/s.
How long after dropping the reward snack does Frici arrive if the speed of sound is 330 m/s?
Help:
Sub-question 1: How long will the reward wall fall if the difficulty acceleration is set to 10 m/s²
shall we round?
Sub-question 2: How long does it take for the voice to reach Frici?
Sub-question 3: How long does it take Frici The cat to reach her plate?
The time spent after dropping the reward snack is
0.5 seconds1 second4.55 secondsWhat are Frici's resolves?Generally, To determine how long it takes Frici to arrive at his plate after the treat is dropped, we need to calculate the time it takes for the sound of the treat hitting the plate to reach Frici, and then add the time it takes for Frici to run back home.
Sub-question 1: The time it takes for the treat to fall from a height of 125 cm can be calculated using the formula:
t = √(2h/g)
where
h is the height (125 cm) and g is the acceleration due to gravity (10 m/s²). The height must be converted to meters first. So,t = √(2*1.25/10)
= 0.5 seconds
Sub-question 2: The time it takes for the sound to reach Frici can be calculated using the formula:
t = d/s where d is the distance
30 meters + 300 meters = 330 meters
and s is the speed of sound (330 m/s).
So, t = 330/330
= 1 second
Sub-question 3: Frici's reaction time is 0.1 seconds and he runs at a speed of 6.6 m/s, so the time it takes for him to reach his plate can be calculated using the formula:
t = d/s
where
d is the distance (30 meters) and s is his running speed (6.6 m/s).
So,
t = 30/6.6
= 4.55 seconds
Therefore, the total time it takes for Frici to arrive at his plate is: 0.11 seconds (time for treat to fall) + 1 second (time for sound to reach Frici) + 4.55 seconds (time for Frici to run back) = 5.66 seconds
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Place the following in order according to how fast sound travels through them: Gas, Liquid, Solid, Vacuum.
Answer:
1Vacuum, 2Gas, 3Liquid, 4Solid.
Explanation:
300,000 273,400 205,600 116,878
Answer:
Solids - fastest
Liquids - faster
Gases - slowest
Vacuum - Not at all
Explanation:
Sound waves have different speed in different mediums because:
different mediums have different densities so that's why speed is different (because speed of sound depends on density of mediums).
in a resonant vibration a body oscillates under the action of a force along a line such that maximum acceleration is equal to maximum velocity. Then frequency of applied force is
If the maximum acceleration is equal to maximum velocity, then frequency of applied force is equal to natural frequency of the vibrating body.
What is simple harmonic motion?
A simple harmonic motion is characterized by this changing acceleration that always is directed toward the equilibrium position and is proportional to the displacement from the equilibrium position.
Mathematically, the formula for the maximum acceleration of a simple harmonic motion is given as;
a ( max ) = ω²A
where;
ω is the angular speed of the oscillatorA is the amplitude of the oscillationResonance vibration is a type of forced vibration in which the natural frequency of vibration of the body is the same as the impressed frequency of external applied force and the amplitude of forced vibration is maximum.
Thus, we can conclude that, the natural frequency of the vibrating body or object is equal to the frequency of the applied force, hence the term resonant vibration.
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2. A football of mass 2.5kg is lifted up to the top of a cliff that is 180 m high. How much potential energydoes the foot ball gain?
Answer: 4.41 kN
Explanation:
Potential energy formula = mass x acceleration due to gravity x height
= 2.5 x 180 x. 9.8 = 4410 N or 4.41 kN
A =1.25
kg object hangs in equilibrium at the end of a rope (taken as massless) while a wind pushes the object with a W=18.3
N horizontal force. Find the magnitude of the tension in the rope and the rope's angle from the vertical. The acceleration due to gravity is =9.81
m/s2.
The magnitude of the tension in the rope is 12.2625 N, and the rope is being pulled horizontally to the left by the wind force..
What is tension?
Tension is the force transmitted through a string, rope, cable or wire when it is pulled tight by forces acting on it from opposite ends. It is a force that is directed along the length of the string, rope, cable or wire and is always directed away from the object that is creating the tension.
When an object is suspended from a rope or cable, the tension in the rope or cable must be equal to the weight of the object in order for the object to remain stationary or move at a constant velocity. If the tension is less than the weight of the object, the object will fall, and if the tension is greater than the weight of the object, the object will rise.
Tension plays an important role in many areas of physics and engineering, such as in the design of suspension bridges, elevator systems, and the motion of satellites in orbit.
To find the tension in the rope and the rope's angle from the vertical, we can start by drawing a free body diagram of the object:
Tension (upwards)
|
|
|
|
Weight (downwards)
|
|
|
|
Wind force (to the right)
where Tension is the tension in the rope, Weight is the weight of the object, and Wind force is the horizontal force due to the wind.
The object is in equilibrium, so the net force on it is zero. We can write two equations based on the forces in the vertical and horizontal directions:
Vertical equilibrium:
Tension - Weight = 0
Tension = Weight
Horizontal equilibrium:
Wind force = Tension * sin(theta)
where theta is the angle between the rope and the vertical.
We can solve these equations for the tension and theta:
Weight = m * g = 1.25 kg * 9.81 m/s^2 = 12.2625 N
Tension = Weight = 12.2625 N
sin(theta) = Wind force / Tension = 18.3 N / 12.2625 N = 1.492
Since sin(theta) is greater than 1, there is no solution for theta. This means that the rope is not making an angle with the vertical, and is instead being pulled directly to the left by the wind force.
Therefore, the magnitude of the tension in the rope is 12.2625 N, and the rope is being pulled horizontally to the left by the wind force.
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Which of the following correctly displays the four fundamental forces in terms of increasing magnitude?
A. Electromagnetic force, gravity, weak nuclear force, strong nuclear force
B. Gravity, electromagnetic force, strong nuclear force, weak nuclear force
C. Weak nuclear force, gravity, strong nuclear force, electromagnetic force
D. Gravity, weak nuclear force, electromagnetic force, strong nuclear force
D. Gravity, weak nuclear force, electromagnetic force, strong nuclear force
The four fundamental forces in the universe, in order of increasing magnitude, are:
Gravity: This is the weakest of the four fundamental forces. It is the force that holds everything together and is responsible for the attraction between masses.
Weak Nuclear Force: This force is responsible for certain types of radioactive decay and the fusion of nuclei in stars. It is about 10^25 times weaker than the strong nuclear force.
Electromagnetic Force: This force is responsible for electric and magnetic interactions, such as the forces between charges, the forces between magnets, and the forces between electric currents.
Strong Nuclear Force: This is the strongest of the four fundamental forces. It holds the protons and neutrons in an atom's nucleus together, and it is about 10^38 times stronger than the electromagnetic force.
Therefore, the correct order of the four fundamental forces in terms of increasing magnitude is D: Gravity, weak nuclear force, electromagnetic force, strong nuclear force
A fish tank is 20 inches by 12 inches by 12 inches. What is its volume in m * m ^ 3
As the fish tank is 20 inches by 12 inches by 12 inches, its volume is 47194744.32 mm³.
What is volume?The space that any three-dimensional solid occupies is known as its volume. These solids can take the form of a cube, cuboid, cone, cylinder, or sphere.
1 inch = 0.0254 meters.
20 inches = 20 × 0.0254 meters = 0.508 meters = 508 mm.
12 inches = 12 × 0.0254 meters = 0.3048 meters = 304.8 mm.
Hence, The volume of the fish tank = length × width × height
= 508 mm × 304.8 mm × 304.8 mm.
= 47194744.32 mm³
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Who was the first person to discover and name gas laws 
Answer:
Charles law and Boyle's law, Dalton law of partial pressure, Graham law of diffusion of gases.
Explanation:
let's hangout sometime
A potter’s wheel moves from rest to an angular speed of 0.20 rev/s in 32.4 s.
Assuming constant angular acceleration,
what is its angular acceleration in rad/s2?
Answer in units of rad/s^2
A potter’s wheel moves from rest to an angular speed of the angular acceleration is 0.006 rad/s^2.
What is acceleration ?
Acceleration is the rate of change of velocity over time. It is the rate at which an object's speed, or the magnitude of its velocity, changes over time. Acceleration can be positive, negative, or zero depending on the direction of the velocity. For example, if an object is moving in a straight line and its speed increases,
it is said to have a positive acceleration; if its speed decreases, it has a negative acceleration. If the object's speed does not change, then the object is said to have a zero acceleration. Acceleration is a vector quantity, meaning it has both magnitude and direction.
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what are some factors that are involved when an object re-enters earth's atmosphere? (ex. red bull free fall)
Answer:
Explanation:
1. Heat: Re-entry into Earth’s atmosphere generates a huge amount of heat due to friction between the object and the atmosphere. This is especially true for objects traveling at high speeds, like a spacecraft. The heat generated from re-entry can cause the object to break apart due to the extreme temperatures.
2. Air Resistance: Re-entry into Earth’s atmosphere also generates a great deal of air resistance. This air resistance can cause the object to slow down or even change direction. This can be dangerous if the object is traveling too fast or if the air resistance is too great.
3. Drag Force: Drag force is the force generated by the air molecules around the object. This force is responsible for slowing down the object as it re-enters the atmosphere. The amount of drag force depends on the size and shape of the object, as well as the speed and direction of the object.
4. Angle of Attack: The angle of attack is important in determining how an object enters the atmosphere. If the angle of attack is too high, the object can be destroyed due to the extreme heat and air resistance. If the angle of attack is too low, the object can skip off the atmosphere and continue in its original trajectory.
5. Ablative Heat Shield: Ablative heat shields are commonly used to protect objects from the extreme heat and air resistance encountered during re-entry. These shields are made of specialized materials that can withstand the extreme temperatures and ablate (or burn away) as the object re-enters the atmosphere.
The figure shows blocks with different masses hanging from a spring or combination of two
springs. All the springs are ideal and have spring constants of k. Which of the following
correctly ranks the systems in order of most stored elastic potential energy to least stored
Without the figure, it's difficult to determine the correct ranking of the systems in order of most stored elastic potential energy to least stored. The ranking would depend on the specific masses of the blocks and the displacement of the spring(s) in each system. The spring constant (k) is the same in all systems, so it does not affect the ranking.
In general, the system with the greatest stored elastic potential energy will be the one in which the spring(s) is most stretched or compressed. The more the spring(s) is stretched or compressed, the more potential energy it has stored. The mass of the block will also affect the amount of stored energy, as a heavier block will stretch or compress the spring(s) more than a lighter block.
Without the exact information about the masses and the displacement of the springs, It's not possible to determine the ranking of the systems.
Calculate the kinetic energy of electron emitted from a surface of potassium metal of wavelength 5.5x10-⁸cm. If the work function of the metal is 3.62x10-¹² joules
Answer:
do it your self like who does that
What happens to Compton after
he jumps from the plane?
Answer:
In January 2012, Compton suffered a heart attack. On 25 February 2012, he died at one of his daughters' home in Washington. His wife Donna preceded him in death in 1994. Buck Compton was survived by two daughters and four grandchildren.
Explanation:
In January 2012, Compton suffered a heart attack. On 25 February 2012, he died at one of his daughters' home in Washington. His wife Donna preceded him in death in 1994. Buck Compton was survived by two daughters and four grandchildren.
. A tire typically has a coefficient of friction around 0.7 in dry conditions. In wet
conditions, the coefficient of friction drops to around 0.4. What is the difference
between the frictional force in dry conditions compared to wet conditions on a car
that has a weight of 14000 N.
Answer:
4300N
Explanation:
FrictionBased on the free body diagram, the friction (fFr in the attached diagram) is dependent on the reaction force (Rn in the attached diagram) acting on the object, in this case a car. The frictional force will always act against the direction of movement (DOM in the attached diagram.)
SolutionGiven the condition is dry,
Summation of forces in the vertical direction (Positive for upward forces) = 0
Reaction Force - Weight of car = 0
Reaction Force = Weight of car = 14000N
Since the friction is dependent on the reaction force with the formula: Coefficient of friction X Reaction Force,
Friction (Dry Condition) = 0.7 * 14000N = 9800N
Now given the condition is wet,
We will use the same formula but with a change to the coefficient of friction.
Friction (Wet Condition) = 0.4 * 14000N = 5600N
Difference between them = 9800N - 5600N = 4300N.
At its closest distance from Earth, the moon experiences a gravitational force of 2.22 E 20 N. The mass of the Earth is 5.97 E24 kg. The mass of the moon is 7.35 E22 kg. What is the distance between the Earth and the moon at that point if G = 6.67 E−11 N*m2/kg2?
A. 1.32 E 4 km
B. 3.63 E 6 km
C. 3.63 E 8 km
D. 1.32 E 17 km
Answer:
Explanation:
Answer: B. 3.63 E 6 km
The equation for gravitational force is F = G*m1*m2/r2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them. We can rearrange the equation to solve for r: r = (G*m1*m2/F)1/2.
Using the given values, we can calculate the distance between the Earth and the moon at the closest point: r = (6.67 E−11 N*m2/kg2 * 5.97 E24 kg * 7.35 E22 kg / 2.22 E 20 N)1/2 = 3.63 E 6 km.
Therefore, the answer is B. 3.63 E 6 km.
Two forces that are F1 =15N (east) and F2 = 7N (west) act on a body in a frictionless floor. The displacement of the body is 6 m, what is the work done by forces on the body?
Answer:
The net force acting on the body is F1 - F2 = 15 N - 7 N = 8 N (east). Since the displacement of the body is 6 m in the same direction as the net force, the work done by the forces on the body is W = Fnet x d = 8 N x 6 m = 48 J (Joules).
Explanation:
above
You weigh 650 N.
What would you weigh if the Earth were
five times as massive as it is and its radius
were five times its present value?
Answer in units of N.
If the earth is 5 times as massive and the radius 5 times its present value, your weight will be 130 N.
How to find the weight of the body?Weight is the amount of gravitational force acting on an object as a result of the gravitational attraction of the Earth.
Assuming gravitational force = [tex]F = \frac{Gm_{1} m_{2} }{r^{2} }[/tex]
where G = 6.67 x 10⁻¹¹m³/kgs², a universal constant.
m = masses of two objects and
r = distance of two objects.
Given that W = 650 N
m₁ = 5Me, new mass of earth
r = 5re, new radius of earth
Then [tex]650 N = \frac{GM_{E} m}{r^{2}_{E} }[/tex]
Let the new weight be [tex]F = \frac{GM m}{r^{2} }[/tex]
Substituting the new mass and radius;
[tex]F = \frac{G(5M_{E}) m}{(5r_{E})^{2}}[/tex]
[tex]F = \frac{5}{25} \frac{GM_{E} m}{r^{2}_{E}}[/tex]
Knowing that [tex]650 N = \frac{GM_{E} m}{r^{2}_{E} }[/tex]
[tex]F = \frac{5}{25}(650 N)[/tex] = 0.2 x 650N
F = 130 N
When the earth is 5 times its radius and 5 times its mass a weight of 650N will become 130 N.
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Erosion includes what processes?
A) building up sediment to make new rocks
B) cleaning rocks to make them shiner
C) wearing away and removal of surface materials
D) combing earth materials to form new minerals
Hannah sees a caterpillar on the leaves of the plants.what components of the ecosystem is a caterpillar
Answer:
Biotic
Explanation:
In a particular ecosystem, caterpillar and parts of the soil are biotic components
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Two bodies separated from
each other at a certain distance
started moving simultaneously
to meet each other - one with ar
acceleration of 2.4 m/s, and the
other with an acceleration of 4.8
m/s2. Determine the ratio of the
displacement module of the first
body to the displacement
module of the second body at
the moment of their meeting.
The result of the ratio of the displacement module of the second body at the point of meeting is 0.5.
How to find displacement ratio?To determine the ratio of the displacement of the first body to the displacement of the second body at the moment of their meeting, use the equation of motion:
d = vt + 1/2at²
where d is the displacement, v is the initial velocity, t is the time, and a is the acceleration.
Since the bodies are moving simultaneously towards each other, then assume that their initial velocities are zero. Also, at the moment of their meeting, their displacement will be the same, d₁ = d₂.
Assume that the time at which they meet is t, then:
d₁ = 1/2 * 2.4t²
And the equation for the displacement of the second body:
d₂ = 1/2 * 4.8t²
If d₁ = d₂
then, 1/2 * 2.4t² = 1/2 * 4.8t²
Solving this equation for t and substituting it into the equation for d₁ or d₂, the ratio of the displacement of the first body to the displacement of the second body: d₁/d₂ = 2.4/4.8 = 0.5 or 1/2
So, at the moment of their meeting, the displacement of the first body is half of the displacement of the second body.
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Reginald Esuke from Cameroon ran down a mountain slope in just 62.25 min. How much work was done if the power developed during Esuke’s descent was 585.0 W?
The amount of work which was done if the power developed during Esuke’s descent was 585.0 W is equal to 2,184,975 Joules.
What is work done?The work done (W) by an object can be calculated by multiplying the force acting on it by the perpendicular distance covered by the physical object over a specific period of time. This ultimately implies that, work done (W) and energy (E) have the same unit i.e Joules (J).
Mathematically, the work done (W) by an object can be calculated by using the following formula:
Work done = power × time
Additionally, we would convert the unit of time in minutes to seconds by multiplying by 60 seconds.
Substituting the given parameters into the work done (W) formula, we have the following;
Work done = 585.0 × 62.25 × 60
Work done = 2,184,975 Joules.
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If wind blows 40m/s over house, what will be net force on the roof it has surface area 250cm²? (density of air = I, 29kg/m²)
According to the question: the net force on the roof is 8.45 kN.
What is net force?The vector sum of the forces operating on a particle of object is known as the net force in mechanics. The original forces' impact on the motion of the particle is replaced by the net force, which is a single force. In accordance with Newton's second rule of motion, it causes the particle to accelerate at a rate equal to the sum of all those actual forces.
The net force on the roof is equal to the air pressure multiplied by the surface area of the roof. The air pressure is equal to the density of the air multiplied by the wind speed squared. Therefore, the net force on the roof can be calculated as follows:
Force = (density x wind speed²) x surface area
Force = (1.29 kg/m² x (40 m/s)²) x (250 cm²)
Force = 8.45 kN
Therefore, the net force on the roof is 8.45 kN.
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If you wanted to determine how mass affects acceleration, lost the independent and dependent variables and the constant.
This is the equations for a hyperbola if we plot mass as both the independent variable & acceleration as such dependent variable, again understanding Newton's second law.
What was the experiment's independent variable?The measured reaction is the dependent variable. The dependent variable is dependent on the variation in the independent variable, to put it another way.
What distinguishes dependent from independent behavior?In a scientific experiment, variables are the two most crucial variables. The independent variable is the one that is under the experimenter's control. The variable that adapts to the explanatory variables is known as the dependent variable.
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In the human femur, bone tissue is strongest in resisting compressive force,
approximately half as strong in resisting tensile force, and only about one-
fifth
as strong in resisting shear force. If a tensile force of 8000 N is sufficient to
produce a fracture, how much compressive force will produce a fracture?
How much shear force will produce a fracture
The compressive force that would be enough to produce a fracture is 4000 N.
What is the force?We know that the femur is one of the most important bones that we have in the human body. In this case, we have been told that In the human femur, bone tissue is strongest in resisting compressive force, approximately half as strong in resisting tensile force, and only about one- fifth as strong in resisting shear force.
Then we know that;
Tensile force = 8000
The compressive force would be half of this magnitude as such;
Compressive force = 4000 N
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Using the density- temperature graph, compare the behaviour of water density and the density of any other liquid as the temperation increases
Four research assistants were asked to observe a gamer and count the number of times the gamer exhibited aggressive behaviors during a one hour gaming session. The first research assistant counted five aggressive behaviors, the second recorded eight, the third observed only two, and the fourth observed ten. If we assume that the research assistants are equally accurate in what they observed, what is the mostly explanation for these discrepancies?
There could be several explanations for the discrepancies in the number of aggressive behaviors observed by the research assistants, but the most likely explanation is that the observations were subjective and prone to bias.
1. Observer bias: The research assistants may have different definitions of what constitutes aggressive behavior, or they may have different levels of sensitivity to detecting aggressive behaviors. This can lead to different interpretations of the same behavior and result in different counts.
2. Different focus: The research assistants may have focused on different aspects of the gaming session and paid attention to different details, resulting in different counts.
3. Fatigue: The research assistants may have become less attentive or less accurate in their counting as the gaming session progressed, which could explain why some of the counts are higher than others.
4. Confirmation bias: Research assistants may have been more likely to notice behaviors that confirm their preconceptions about the gamer, leading to counts that are higher or lower than what was actually observed.
5. Different cultural background: Research assistants from different cultural background may have different understanding of what is aggressive behavior.
The most powerful ice breaker in the world was built in the former Soviet Union. The ship is almost 150 m long, and its nuclear engine generates 56 MegaWatts (56 E^6 W) of power. How much work can this engine do in 1.0 hour? (1 hour = 3600 seconds)
The work done by the engine is 2.016×10¹¹ J.
What is work done?Work is said to be done whenever a force moves an object through a distance.
To calculate the work done by the engine, we use the formula below.
Formula:
W = Pt..................... Equation 1Where:
W = Work doneP = Powert = TimeFrom the question,
Given:
P = 56 mega Watts = 56×10⁶ Wt = 1 hours = 3600 secondSubstitute these values into equation 1
W = 56×10⁶×3600W = 2.016×10¹¹ JHence, the work done is 2.016×10¹¹ J.
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