Answer:
the work required to turn the crank at the given revolutions is 8,483.4 J
Explanation:
Given;
torque required to turn the crank, T = 4.50 N.m
number of revolutions, = 300 turns
The work required to turn the crank is given as;
W = 2πT
W = 2 x 3.142 x 4.5
W = 28.278 J
1 revolution = 28.278 J
300 revlotions = ?
= 300 x 28.278 J
= 8,483.4 J
Therefore, the work required to turn the crank at the given revolutions is 8,483.4 J
An object moves in a direction parallel to its length with a velocity that approaches the velocity of light. The length of this object, as measured by a stationary observer:________
a. approaches infinity.
b. approaches zero.
c. increases slightly.
d. does not change.
Answer:
b. approaches zero.
Explanation:
The phenomenon is known as length contraction.
Length contraction is a result of Einstein's special theory of relativity. This theory states that an observer in an inertial frame of reference will observe a decrease in the length of any moving object placed at another inertial frame of reference.
let the length of the train = L
Let the length observed when the train is in motion = L₀
Apply Einstein's special theory of relativity;
[tex]L_0 = L \times \sqrt{1 - \frac{v^2}{c^2} } \\\\where;\\\\v \ is \ the \ velocity \ of \ the \ train\\\\c \ is \ the \ speed \ of \ light\\\\[/tex]
from the equation above, when v = 0, the length observed is equal to the initial length of the train. (L₀ = L)
As the velocity of the train (v) approaches the speed of light (c), the length of the train observed (L₀) becomes smaller than the initial length of the train (L). (L₀ < L)
Eventually, when v equals c, we will have a square root of zero (0), and the length observed will become zero. (L₀ = 0)
Thus, the length of this object, as measured by a stationary observer approaches zero
A supertrain with a proper length of 100 m travels at a speed of 0.950c as it passes through a tunnel having a proper length of 50.0 m. As seen by a trackside observer, is the train ever completely within the tunnel? If so, by how much do the train’s ends clear the ends of the tunnel?
Answer:
19m
Explanation:
we have proper length L = 100m
the speed of the train v = 0.95
the speed of light is given as = 3x10⁸
length of the tunnel is given as = 50 meters
we can solve for the lenght contraction as
LX√1-v²/c²
= 100 * √1-(0.95*3x10⁸)²/(3x10⁸ )
= 31.22 metres
the train would be well seen at
50 - 31.22
= 18.78
= this is approximately 19 metres
we conclude tht the trains ends clears the ends of the tunnel by 19 meters.
thank you!
The Sun is a type G2 star. Type G stars (from G0 to G9) have a range of temperatures from 5200 to 5900. What is the range of log(T) for G stars? Show your work
Answer:
log T = 3.72 to 3.77
Explanation:
Temperature range is
T = 5200 to 5900
Take the log
So,
log T = log 5200 to log 5900
log T = 3.72 to 3.77
the current through a wire is measured as the potential difference is varied what is the wire resistance
Answer:
Resistance, R = 0.02 Ohms
Explanation:
Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.
Mathematically, Ohm's law is given by the formula;
V = IR
Where;
V is the voltage or potential difference.
I is the current.
R is the resistance.
From the attachment, we would pick the following values on the graph of current against voltage;
Voltage, V = 0.5 V
Current = 25 A
To find resistance;
R = V/I
R = 0.5/2.5
Resistance, R = 0.02 Ohms
Note:
Resistance (R) is the inverse of slope i.e change in current with respect to change in voltage.
A free undamped spring/mass system oscillates with a period of 4 seconds. When 10 pounds are removed from the spring, the system then has a period of 2 seconds. What was the weight of the original mass on the spring? (Round your answer to one decimal place.)
Answer:
13.3 pounds.
Explanation:
For a spring of constant K, with an attached object of mass M, the period can be written as:
T = 2*π*√(M/K)
Where π = 3.14
First, we know that the period is 4 seconds, then we have:
4s = (2*π)*√(M/K)
We know that if the mass is reduced by 10lb, the period becomes 2s.
Then the new mass of the object will be: (M - 10lb)
Then the period equation becomes:
2s = (2*π)*√((M-10lb)/K)
So we have two equations:
4s = (2*π)*√(M/K)
2s = (2*π)*√((M-10lb)/K)
We want to solve this for M.
First, we need to isolate K in one of the equations.
Let's isolate K in the first one:
4s = (2*π)*√(M/K)
(4s/2*π) = √(M/K)
(2s/π)^2 = M/K
K = M/(2s/π)^2 = M*(π/2s)^2
Now we can replace it in the other equation.
2s = (2*π)*√((M-10lb)/K)
First, let's simplify the equation:
2s/(2*π) = √((M-10lb)/K)
1s/π = √((M-10lb)/K)
(1s/π)^2 = ((M-10lb)/K
K*(1s/π)^2 = M - 10lb
Now we can use the equation: K = M*(π/2s)^2
then we get:
K*(1s/π)^2 = M - 10lb
(M*(π/2s)^2)*(1s/π)^2 = M - 10lb
M/4 = M - 10lb
10lb = M - M/4
10lb = (3/4)*M
10lb*(4/3) = M
13.3 lb = M
A boy of mass 50 kg on a motor bike is moveny coith 20m/see what is hio k.E
Use the following information to answer the next question.
Environmental Concerns
1. release of greenhouse gases
2. release of gases that cause acid rain
3. release of excess heat
4. depletion of solar energy
5. depletion of geothermal energy
6. flooding of land
Which of the above environmental concerns are associated with the production of electricity?
Select one:
O A. 2, 3, and 4
O B. 1, 2, 3, and 6
O C. 1, 2, 3, 5 and 6
O D. 1, 3, and 5
Answer:
1.Emitted primarily through the burning of fossil fuels (oil, natural gas, and coal), solid waste, and trees and wood products. Changes in land use also play a role. Deforestation and soil degradation add carbon dioxide to the atmosphere, while forest regrowth takes it out of the atmosphere.
2.Acid rain is caused by a chemical reaction that begins when compounds like sulfur dioxide and nitrogen oxides are released into the air. These substances can rise very high into the atmosphere, where they mix and react with water, oxygen, and other chemicals to form more acidic pollutants, known as acid rain.
3.Untreated, heat exhaustion can lead to heatstroke, a life-threatening condition that occurs when your core body temperature reaches 104 F (40 C) or higher. Heatstroke requires immediate medical attention to prevent permanent damage to your brain and other vital organs that can result in death.
4.The loss of solar energy in passing through the atmospheric layers is called the atmospheric deflection. ... The longer the path traversed, the greater the amount of radiant energy depleted. Various processes whereby heat energy is lost through the atmosphere are known as scattering, diffusion, absorption, and reflection.
5.Geothermal energy is renewable because the Earth has retained a huge amount of the heat energy that was generated during formation of the planet. In addition, heat is continuously produced by decay of radioactive elements within the Earth. The amount of heat within the Earth, and the amount that is lost though natural processes (e.g. volcanic activity, conduction/radiation to the atmosphere), are much, much more than the amount of heat lost through geothermal energy production. At any one geothermal field, however, the temperature of the geothermal reservoir or the fluid levels/fluid pressure in the reservoir may decrease over time as fluids are produced and energy is extracted. Produced fluids can be re-injected to maintain pressures, although this may further cool down the reservoir if care is not taken. Over time, it is commonly necessary to drill additional wells in order to maintain energy production as temperatures and/or reservoir fluid pressures decline.
6.Floods, Floodplains, and Flood-Prone Areas. ... Flooding is a result of heavy or continuous rainfall exceeding the absorptive capacity of soil and the flow capacity of rivers, streams, and coastal areas. This causes a watercourse to overflow its banks onto adjacent lands.
The environmental concerns associated with the production of electricity are the release of greenhouse gases, the release of gases that cause acid rain, the release of excess heat, the flooding of land, and the depletion of geothermal energy so, option C is correct.
What is electricity?The presence or movement of charged particles is electricity. The movement of electrons through a circuit is known as an electric current. The accumulation of electrons on an insulator causes static electricity.
Mostly released when solid trash, trees, and wood products are burned, along with fossil fuels (coal, gas, and oil).
Land use changes also have an impact. Carbon dioxide is released into the atmosphere by deforestation and soil erosion, while it is removed from it by forest regeneration.
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friction between two flat surfaces can be divided into two categories. what are the two most common kinds of friction?
Answer:
kinetic and static
Explanation:
hope it helps! ^w^
A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The ball moves so fast that the string is always taut and perpendicular to the velocity of the ball. As the ball swings from its lowest point to its highest point Group of answer choices the work done on it by gravity is -118 J and the work done on it by the tension in the string is zero. the work done on it by gravity is -118 J and the work done on it by the tension in the string is 118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to -118 J. the work done on it by gravity is 118 J and the work done on it by the tension in the string is -118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to zero.
Answer:
The ball moves from lowest to highest point:
W = M g h = 3 * 9.8 * 4 = 118 J
This is work done "against" gravity so work done by gravity is -118 J
The tension of the string does no work because the tension does not
move thru any distance W = T * x = 0 because the length of the string is fixed.
A frictionless spring with a 9-kg mass can be held stretched 1.8 meters beyond its natural length by a force of 80 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 1.5 m/sec, find the position of the mass after tt seconds. meters
Answer:
the required solution is; x(t) = 0.675sin( 2.222t )
Explanation:
Given the data in the question;
Using both Newton's and Hooke's law;
m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0, [tex]x[/tex](0) = 0, [tex]x^f[/tex](0) = 1.5
given that mass m = 9 kg
[tex]x[/tex] = 1.8 m
k is F / x
hence
k = F / x
given that, F = 80 N
we substitute
k = 80 / 1.8
k = 44.44
so
m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0,
we input
9[tex]x^{ff[/tex] + 44.44[tex]x[/tex] = 0,
[tex]x^{ff[/tex] + 4.9377[tex]x[/tex] = 0
so auxiliary equation is,
r² + 4.9377 = 0
r² = -4.9377
r = √-4.9377
r = ±2.222i
hence, the solution will be;
x(t) = A×cos( 2.222t ) + B×sin( 2.222t )
⇒ [tex]x^t[/tex](t) = -2.222Asin( 2.222t ) + 2.222Bcos( 2.222t )
using initial conditions
x(0) = 0
⇒ 0 = A
[tex]x^t[/tex](t) = 1.5
1.5 = 2.222B
so
B = 1.5 / 2.222 = 0.675
Hence, the required solution is; x(t) = 0.675sin( 2.222t )
A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in the pitcher's hand is 34.0 m/s and the ball is 0.310 m from the elbow joint, what is the angular velocity (in rad/s) of the forearm
Answer:
[tex]\omega=109.67\ rad/s[/tex]
Explanation:
Given that,
The speed of the ball, u = 34 m/s
The ball is 0.310 m from the elbow joint.
We need to find the angular velocity (in rad/s) of the forearm.
We know that,
[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{34}{0.31}\\\\\omega=109.67\ rad/s[/tex]
So, the required angular velocity of the forearm is 109.67 rad/s.
A fan that is rotating at 960 rev/s is turned off. It makes 1500 revolutions before it comes to a stop. a) What was its angular acceleration(assuming it was constant)
Answer:
α = 1930.2 rad/s²
Explanation:
The angular acceleration can be found by using the third equation of motion:
[tex]2\alpha \theta=\omega_f^2-\omega_i^2[/tex]
where,
α = angular acceleration = ?
θ = angular displacement = (1500 rev)(2π rad/1 rev) = 9424.78 rad
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (960 rev/s)(2π rad/1 rev) = 6031.87 rad/s
Therefore,
[tex]2\alpha(9424.78\ rad) = (0\ rad/s)^2-(6031.87\ rad/s)^2\\\\\alpha = -\frac{(6031.87\ rad/s)^2}{(2)(9424.78\ rad)}[/tex]
α = - 1930.2 rad/s²
negative sign shows deceleration
A spinning wheel having a mass of 20 kg and a diameter of 0.5 m is positioned to rotate about its vertical axis with a constant angular acceleration, a of 6 rad/s If the initial angular velocity is 1.5 rad/s, determine The maximum angular velocity and linear velocity of the wheel after 1 complete revolution.
Answer:
ωf = 8.8 rad/s
v = 2.2 m/s
Explanation:
We will use the third equation of motion to find the maximum angular velocity of the wheel:
[tex]2\alpha \theta = \omega_f^2 -\omega_I^2[/tex]
where,
α = angular acceleration = 6 rad/s²
θ = angular displacemnt = 1 rev = 2π rad
ωf = max. final angular velocity = ?
ωi = initial angular velocity = 1.5 rad/s
Therefore,
[tex]2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}[/tex]
ωf = 8.8 rad/s
Now, for linear velocity:
v = rω = (0.25 m)(8.8 rad/s)
v = 2.2 m/s
Find the ratio of the Coulomb electric force Fe to the gravitational force Fo between two
electrons in vacuum.
Answer:
thus the coulomb force is F – 8.19x10-8N. this is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. the ratio of the magnitude of the electrostatic force to gravitational force in this case is,thus,FFG – 2.27x1039 F F G – 2.27x 10 39.
What is the approximate radius of an equipotential spherical surface of 30 V about a point charge of +15 nC if the potential at an infinite distance from the surface is zero?
Answer:
V = k Q / R potential at distance R for a charge Q
R = k Q / V
R = 9 * 10E9 * 15 * 10E-9 / 30 = 9 * 15 / 30 = 4.5 m
Note: Our equation says that if R if infinite then V must be zero.
A falcon is hovering above the ground, then suddenly pulls in its wings and begins to fall toward the ground. Air resistance is not negligible.
Identify the forces on the falcon.
a. Kinetic friction
b. Weight w
c. Static friction
d. Drag D
e. Normal force n
f. Thrust
g. Tension T
Answer:
Explanation:
When a falcon is hovering, the force of up thrust is balanced by the weight.
When it begins to fall towards the ground, the weight acts downwards, kinetic friction is upwards, drag is upwards, normal force is upwards, thrust is upwards.
B.F.Skinner emphesized the importance of-----?
Answer:
BFSkinner enfatizó la importancia de creía en la importancia de desarrollar la psicología experimental y dejar atrás el psicoanálisis y las teorías acerca de la mente basadas en el simple sentido común.
Explanation:
Question 1 of 10
Which nucleus completes the following equation?
239UHe+?
A. 228 Th
B. 2220
c. 23. Pu
D. 78Th
SUBMIT
Answer:
Option D. ²²²₉₀Th
Explanation:
Let the unknown be ⁿₘZ. Thus, the equation becomes:
²²⁶₉₂U —> ⁴₂He + ⁿₘZ
Next, we shall determine n, m and Z. This can be obtained as follow:
For n:
226 = 4 + n
Collect like terms
226 – 4 = n
222 = n
n = 222
For m:
92 = 2 + m
Collect like terms
92 – 2 = m
90 = m
m = 90
For Z:
ⁿₘZ => ²²²₉₀Z => ²²²₉₀Th
Therefore, the complete equation becomes:
²²⁶₉₂U —> ⁴₂He + ⁿₘZ
²²⁶₉₂U —> ⁴₂He + ²²²₉₀Th
Thus, the unknown is ²²²₉₀Th
What bet force is required to stop a 2250 kg car if the decelerates at a rate of -4.3 m/s^2 please answer fast
Answer:
Force = Mass × Acceleration
[tex]{ \tt{force = 2250 \times 4.3}} \\ = { \tt{9675 \: newtons}}[/tex]
A magnetic force acting on an electric charge in a uniform magnetic field what happend
Answer:
hgff
Explanation:
Answer:
The charge moves to equilibrium.
E.e = B.e.V
E is electric field force.
e is the charge.
B is magnetic field force.
V is acceleration voltage.
you decide to work part time at a local supermarket. The job pays eight dollars and 60 per hour and you work 20 hours per week. Your employer withhold 10% of your gross pay federal taxes, 7.65% for FICA taxes, and 5% for state taxes
I guess that we want to find how much money you get each week.
We know that the job pays $8.60 per hour.
We know that you work 20 hours per week.
Then the gross pay (the total money that you earn) in a week is 20 times $8.60, or:
20*$8.60 = $172.
Now we know that your employer witholds:
10% + 7.65% + 5% = 22.65%
Then your employer withholds 22.65% of your gross pay.
if the 100% of your gross pay is $172
Then the 22.65% will be:
(22.65%/100%)*$172 = 0.2265*$172 = $38.96
This means that your employer withholds $38.96 of your weekly gross pay.
Then each week you get:
$172 - $38.96 = $133.04
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In what kind of reaction is water (H20) broken down into hydrogen gas (H2) and oxygen gas (O2)?
A. Combination
B. Decomposition
C. Displacement
D. Combustion
Answer:
Answer is B (Decomposition)
Sorry I really see ur questions but I don't know the answer but next time I will try to answer sorry:(
(a) What is the efficiency of an out-of-condition professor who does 1.90 ✕ 105 J of useful work while metabolizing 500 kcal of food energy? % (b) How many food calories would a well-conditioned athlete metabolize in doing the same work with an efficiency of 25%? kcal
Answer:
a) The energy efficiency of the out-of-condition professor is 9.082 %.
b) The food calories needed by the well-conditioned athlete is 181.644 kilocalories.
Explanation:
a) The energy efficiency of the food metabolization ([tex]\eta[/tex]), no unit, is defined by following formula:
[tex]\eta = \frac{W}{E}\times 100\,\%[/tex] (1)
Where:
[tex]W[/tex] - Useful work, in joules.
[tex]E[/tex] - Food energy, in joules.
If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]E = 2.092\times 10^{6}\,J[/tex], the energy efficiency of the food metabolization is:
[tex]\eta = \frac{1.90\times 10^{5}\,J}{2.092\times 10^{6}\,J} \times 100\,\%[/tex]
[tex]\eta = 9.082\,\%[/tex]
The energy efficiency of the out-of-condition professor is 9.082 %.
b) If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]\eta = 25\,\%[/tex], then the quantity of food energy is:
[tex]E = \frac{W}{\eta}\times 100\,\%[/tex]
[tex]E = 1.90\times 10^{5}\,J\times \frac{100\,\%}{25\,\%}[/tex]
[tex]E = 7.60\times 10^{5}\,J[/tex]
[tex]E = 181.644\,kcal[/tex]
The food calories needed by the well-conditioned athlete is 181.644 kilocalories.
plz help me with hw A bus of mass 1000 kg moving with a speed of 90km/hr stops after 6 sec by applying brakes then calculate the distance travelled and amount of force applied.
Answer:
Mass, M = 1000 kg
Speed, v = 90 km/h = 25 m/s
time, t = 6 sec.
Distance:
[tex]{ \tt{distance = speed \times time }} \\ { \tt{distance = 25 \times 6}} \\ { \tt{distance = 150 \: m}}[/tex]
Force:
[tex]{ \tt{force = mass \times acceleration}} \\ { \bf{but \: for \: acceleration : }} \\ from \: second \: equation \: of \: motion : \\ { \bf{s = ut + \frac{1}{2} {at}^{2} }} \\ \\ { \tt{150 = (0 \times 6) + ( \frac{1}{2} \times a \times {6}^{2} ) }} \\ \\ { \tt{acceleration = 8.33 \: {ms}^{ - 2} }} \\ \\ { \tt{force = 1000 \times 8.33}} \\ { \tt{force = 8333.3 \: newtons}}[/tex]
True or False: The forces applied by our muscles on our bones are usually several times larger than the forces we exert on the outside world with our limbs.
Answer:
True
Explanation:
This is because of the point where the forces are applied by our muscles and
the angle they have about the bones. Take for example the diagram I uploaded.
If we do a free body diagram and a sum of torques, we would get that:
[tex]F_{muscle}sin \theta r1 - mg r2 = 0[/tex]
In this case, mg is the same in magnitude as the force made by the hand to hold the ball, so:
[tex]F_{muscle}sin \theta r_{1} - F_{hand} r_{2} = 0[/tex]
If we solve the equation for the force of the muscle we would get that:
[tex]F_{muscle}=\frac{F_{hand}r_{2}}{r_{1}sin \theta}[/tex]
Since r2 is greater than r1 and the sin function can only return values that are less than 1, this means that the force of the muscle is much greater than the force used by the hand to hold the weight.
Let's use some standard values to prove this, let's say that r1=10cm, r2=35cm and theta=60 degrees. When inputing the values into the equation we get:
[tex]F_{muscle}=\frac{F_{hand}(35cm)}{(10cm)sin (60^{o})}[/tex]
which yields:
[tex]F_{muscle}=4.04 F_{hand}[/tex]
so in this example, the force made by the muscle is 4 times as big as the force exerted by the hand.
A 77 turn, 10.0 cm diameter coil rotates at an angular velocity of 8.00 rad/s in a 1.18 T field, starting with the normal of the plane of the coil perpendicular to the field. Assume that the positive max emf is reached first.
a. What is the peak emf?
b. At what time is the peak emf first reached?
c. At what time is the emf first at its most negative?
d. What is the period of the AC voltage output?
Answer:
a) fem = 5.709 V, b) t = 0.196 s, c) t = 0.589 s, d) T = 0.785 s
Explanation:
This is an exercise in Faraday's law
fem= - N [tex]\frac{d \Phi _B}{dt}[/tex]
fem = - N [tex]\frac{d \ (B A cos \theta)}{dt}[/tex]
The magnetic field and the area are constant
fem = - N B A [tex]\frac{d \ cos \ \theta}{dt}[/tex]
fem = - N B A (-sin θ) [tex]\frac{d \theta}{dt}[/tex]
fem = N B (π d² / 4) sin θ w
fem= [tex]\frac{\pi }{4}[/tex] N B d² w sin θ
with this expression we can correspond the questions
a) the peak of the electromotive force
this hen the sine of the angle is 1
sin θ = 1
fem = [tex]\frac{\pi }{4}[/tex] 77 1.18 0.10² 8.0
fem = 5.709 V
b) as the system has a constant angular velocity, we can use the angular kinematics relations
θ = w₀ t
t = θ/w₀
Recall that the angles are in radians, so the angle for the maximum of the sine is
θ= π/2
t = [tex]\frac{\pi }{2} \ \frac{1}{8}[/tex]
t = 0.196 s
c) for the electromotive force to be negative, the sine function of being
sin θ= -1
whereby
θ = 3π/ 2
t = [tex]\frac{3\pi }{2} \ \frac{1}{8}[/tex]
t = 0.589 s
d) This electromotive force has values that change sinusoidally with an angular velocity of
w = 8 rad / s
angular velocity and period are related
w = 2π / T
T = 2π / w
T = 2π / 8
T = 0.785 s
A horizontail rod (oriented in the east -west direction) is moved northward at a constant velocity through a magnetic field that points straight down. Make a statement concerning the potential induced across the rod.
A) The east end of the rod is at higher potential than the west end.
B) The bottom surface of the rod is at higher potential than the top surface.
C) The top surface of the rod is at higher potential than the bottom surface.
D) The west end of the rod is at higher potential than the east end.
E) The potential is uniform.
Answer:
a
Explanation:
3. Define 1 standard kilogram?
Answer:
standard kilogram is the SI unit of mass
Answer:
The total mass of platinum-irridum cylinder whose diameter is equal to its height and stored at 0°C in the bureau of weight and measure in France is called 1 standard kilogram
A plane flying horizontally at an altitude of 1 mi and a speed of 480 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station.
Answer:
First remember that the distance between two points (a, b) and (c, d) is given by the equation:
[tex]d = \sqrt{(a - c)^2 + (b - d)^2}[/tex]
Now let's define the position of the radar as:
(0mi, 0mi)
Then we can write the position of the plane as:
(480mi/h*t, 1mi)
where t is time in hours.
Then we can write the distance equation as:
[tex]d(t) = \sqrt{(480\frac{mi}{h}*t - 0mi)^2 + (1mi -0mi)^2 } \\\\d(t) = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }[/tex]
Now we want to get:
the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station.
So first we want to find the value of t such that:
d(3) = 3mi
We will look at the positive value of t, because at this point the plane is increasing its distance to the station.
[tex]3mi = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }\\\\(3mi)^2 = (480\frac{mi}{h}*t )^2 + (1mi)^2\\\\9mi^2 - 1mi^2 = (480\frac{mi}{h}*t )^2\\\\8mi^2 = (230,400 mi^2/h^2)*t^2\\\\\\\sqrt{\frac{8mi^2}{230,400 mi^2/h^2} } = t = 0.0059 h[/tex]
The rate of change when the plane is 3 mi away from the station is:
d'(0.0059h)
remember that:
d'(t) = dd(t)/dt
We can write:
d(t) = h( g(t) )
such that:
h(x) = √x
g(t) = (480mi/h*t)^2 + (1mi)^2
then:
d'(t) = h'(g(t))*g'(t)
This is:
[tex]d'(t) = \frac{dd(t)}{dt} = \frac{1}{2}*\frac{2*t*480mi/h}{\sqrt{(480mi/h*t)^2 + (1mi)^2} }[/tex]
The rate of change at t = 0.0059h is then:
[tex]d'(0.0059h) = \frac{1}{2}*\frac{2*0.0059h*(480mi/h)^2}{\sqrt{(480mi/h*0.0059h)^2 + (1mi)^2} } =452.6 mi/h^2[/tex]
A Man has 5o kg mass man in the earth and find his weight
Answer:
49 N
Explanation:
Given,
Mass ( m ) = 50 kg
To find : Weight ( W ) = ?
Take the value of acceleration due to gravity as 9.8 m/s^2
Formula : -
W = mg
W = 50 x 9.8
W = 49 N
