Given that the pipe has varying cross-sections.
The diameter of one section is d1 = 9.1 cm and the diameter of second section is d2 = 12.6 cm.
Also, the fluid has the density,
[tex]\rho=3.43gcm^{-3}[/tex]The area of the cross-section for the first section is
[tex]\begin{gathered} A_1=\frac{\pi(d1)^2}{4} \\ =\frac{\pi(9.1)^2}{4}cm^2 \end{gathered}[/tex]The area of the cross-section for the second section is
[tex]\begin{gathered} A_2=\frac{\pi(d2)^2}{4} \\ =\frac{\pi(12.6)^2}{4}cm^2 \end{gathered}[/tex]The flow speed for the first section is v1 = 339 cm s^-1
The flow speed for the second section will be v2.
(a) The flow speed for the second section can be calculated as
[tex]\begin{gathered} A_1v1=A_2_{}v2 \\ v2=\frac{A_1v1}{A_2} \\ =\frac{\pi(9.1)^2\times339\times4}{4\times\pi\times(12.6)^2} \\ =\text{ 176.82 cm/s} \\ =1.7682\text{ m/s} \end{gathered}[/tex](b) The pressure for first section is p1 = 2.93 x 10^5 Pa
The pressure for the second section will be p2.
The pressure for the second section can be calculated by the formula,
[tex]\begin{gathered} p2=p1+\frac{1}{2}\rho\mleft\lbrace(v1)^2-(v2\mright)^2\} \\ =2.93\times10^5+\frac{1}{2}\times3.43\mleft\lbrace(339)^2-(176.82)^2\mright\rbrace \\ =4.36\text{ }\times10^5\text{ Pa} \end{gathered}[/tex]Suppose a 345-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.1 m from the ground to a branch.How much work, in joules, did the bird do on the snake? How much work, in joules, did it do to raise its own center of mass to the branch?
The work done in each case can be calculated with the change in potential energy of the body:
[tex]Work=m\cdot g\cdot h[/tex]The work done by the bird on the snake will use only the mass of the snake (in kg):
[tex]\begin{gathered} Work=0.075\cdot9.8\cdot2.1\\ \\ Work=1.5435\text{ J} \end{gathered}[/tex]The work done by the bird to raise its own center of mass will use only the bird mass (in kg):
[tex]\begin{gathered} Work=0.345\cdot9.8\cdot2.1\\ \\ Work=7.1\text{ J} \end{gathered}[/tex]50 POINTS
A velocity vs time graph is shown. What is the acceleration of the object?
Answer:
Explanation:
Given:
V₀ = 0 m/s
V = 20m/s
t = 5 s
___________
a - ?
The acceleration:
[tex]a = \frac{(V -V_{0} )}{t} \\\\[/tex]
[tex]a=\frac{(20 - 0)}{5} = 4 \frac{m}{s^{2} } \\[/tex]
what is the velocity of the boat from point x to point y
The distance between the points X and Y is,
[tex]\begin{gathered} d=9\text{ km} \\ =9000\text{ m} \end{gathered}[/tex]The time taken to move the distance is,
[tex]\begin{gathered} t=12\text{ min} \\ =12\times60\text{ s} \end{gathered}[/tex]The velocity is given by,
[tex]\begin{gathered} v=\frac{d}{t} \\ =\frac{9000\text{ m}}{12\times60\text{ s}} \\ =12.5\text{ m/s} \end{gathered}[/tex]Hence the velocity is 12.5 m/s.
A.Calculate the combined force of vector F ?B.Calculate the direction of the combined force vector F ?
Answer:
A. 282.93 N
B. 1.94 degrees
Explanation:
The combined force is found by first adding the three forces given.
We add the three forces by adding their x and y components separately and then combining the results to produce the total force,
The x component of a force is
[tex]\begin{gathered} \cos \theta=\frac{f_x}{F} \\ \Rightarrow f_x=F\cos \theta \end{gathered}[/tex]Therefore, x components of the forces is
[tex]F_x=120\cos 65+100\cos 25+200\cos (-45)[/tex]The y-component of the forces is
[tex]F_y=120\sin 120+100\sin 25+200\sin (-45)[/tex]Now evaluating the above two components gives
[tex]F_x=282.77N[/tex][tex]F_y=9.597N[/tex]Let us draw on big vector whose components are the above vectors.
The angle of the combined vector with respect to the x-axis is
[tex]\tan \theta=\frac{9.59}{282.77}[/tex][tex]\theta=\tan ^{-1}(\frac{9.59}{282.77})[/tex][tex]\boxed{\theta=1.94^o}[/tex]which is our answer!
The magnitude of the combined vector is
[tex]F=\sqrt[]{F^2_x+F^2_y_{}}[/tex][tex]F=\sqrt[]{(9.59)^2_{}+(282.77)^2_{}}[/tex][tex]\boxed{F=282.93N}[/tex]which is our answer!
Hence, to summerise:
A. 282.93 N
B. 1.94 degrees
Which picture below correctly identifies the effort length and lifting length of a lever?Select one:a. Ab. Bc. Cd. D
d.D
Explanation
A lever is a simple machine made of a rigid beam and a fulcrum.
where
the Load is the object which we are lifting,Fulcrum is the point at which the lever is pivoted. and Effort is he force applied to make the object move
Step 1
check for the graph that:
the lifting length goes from the fulcrum to the load
and
the effort length goes from the fulcrum to the applied force,
therefore,
the answer is
d.D
I hope this helps you
Need help 82x2682 please
ANSWER:
STEP-BY-STEP EXPLANATION:
We have the following multiplication:
[tex]undefined[/tex]A penny is dropped from a building and it takes 7.00 seconds to hit the ground.
Based on the given question, the final velocity of the penny is 68.6m/s
Calculations and ParametersBased on the definition of velocity, we can see that it has to do with the speed at which an object moves with at a particular direction.
If we observe the 1st law of motion:
v = u + gt
Where:
v = final velocity
u = initial velocity
g = acceleration due to gravity
t = time taken
We need to find the velocity as it was not given.
We already know that
u = 0 (it was at rest),
t = 7s
v = unknown
Let us put in the given values based on the equation and solve for the answer.
v = u + gt
v = 0 + ( 9.8m / s ² × 7.00s )
v = 9.8m / s ² × 7.00s
v = 68.6m/s
Therefore, the final velocity of the penny is 68.6m/s
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A penny is dropped from a building, and it takes 7 seconds to hit the ground, so the final velocity will be 68.6 m/s.
What is Velocity?The ratio of displacement to time is referred to as velocity of the object. It has SI unit meter per second or m/s and has a dimension formula LT⁻¹.
We already know that
Initial speed, u = 0
Time, t = 7 seconds
Use the equation of motion to find the final velocity,
v = u + gt
v = 0 + ( 9.8 m / s ² × 7.00s )
v = 9.8 m / s ² × 7.00s
v = 68.6 m/s
Hence, the penny hit the ground with a final velocity of 68.6 m/s.
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Analyze the benefits and consequences of sleep create (continue) their own fitness programs incorporating sleep as a fundamental pillar of fitness
Explanation:
The Intimate Relationship Between Fitness and Sleep
*“If you don’t sleep, you undermine your body,” says W. Christopher Winter, MD, the president of Charlottesville Neurology and Sleep Medicine and the author of "The Sleep Solution: Why Your Sleep Is Broken and How to Fix It."
When it comes to working out, you know that what you do in the gym is important. But what you do outside the gym — what you eat, what you drink, and especially how you sleep, is just as crucial. In fact, you must sleep in order for exercise to actually work.
“We exercise for a purpose: for cardiovascular health, to increase lean muscle mass, to improve endurance, and more. All of these 'goals' require sleep,” says W. Christopher Winter, MD, the president of Charlottesville Neurology and Sleep Medicine and the author of The Sleep Solution: Why Your Sleep Is Broken and How to Fix It.
In other words, without sleep, exercise does not deliver those benefits, Dr. Winter explains. “If you don’t sleep, you undermine your body.”
The period of a simple harmonic oscillator is the time it takes for one complete cycle of oscillation to be completed. Is this true or false?
Yes, the given statement is true.
The period of a simple harmonic oscillator is the time it takes for one complete cycle of oscillation to be completed.
Why is it helpful to break vectors into their horizontal and vertical components before adding vectors?
It is helpful to break vectors into their horizontal and vertical components before adding vectors because it helps to simplify calculations.
A vector at an angle will have both horizontal and vertical component in it. So resolving the vector into its horizontal and vertical components makes it easier and clearer to solve both the components separately.
Solving the components is done using trigonometry ratios and Pythagoras theorem. Some of the trigonometric ratios used are:
sin θ = Vertical component / Actual vector
cos θ = Horizontal component / Actual vector
tan θ = Vertical component / Horizontal component
Therefore, it is helpful to break vectors into their horizontal and vertical components before adding vectors because it helps to simplify calculations.
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Full working out…….2.A vibrating mass-spring system has a frequency of 0.56 Hz. How much energy ofthis vibration is carried away in a one-quantum change?
ANSWER
3.7128 x 10⁻³⁴ J
EXPLANATION
The energy carried in a one-quantum change is the product of Planck's constant, h, and the frequency of vibration, f,
[tex]E=hf[/tex]Planck's constant is 6.63 x 10⁻³⁴ J*s and, in this case, the frequency of vibration is 0.56 Hz. So, the energy carried away is,
[tex]E=0.56Hz\cdot6.63\cdot10^{-34}J\cdot s=3.7128\cdot10^{-34}J[/tex]Hence, the energy carried away in a one-quantum change is 3.7128 x 10⁻³⁴ J.
When you step off a bus moving at 2 m/s, your horizontal speed when you meet the ground isA) zero.B) less than 2 m/s but greater than zero.C) about 2 m/s.D) greater than 2 m/s.
ANSWER:
C) about 2 m/s.
STEP-BY-STEP EXPLANATION:
While step off the bus, it acquires a vertical component of velocity, but it still has the initial horizontal component of velocity due to the movement of the bus.
Which means that the velocity is either 2 m/s or about 2 m/s
A) the frictional force F newtonsB)The resultant normal reaction of the surface on the metal block
Given:
The mass of the block is.
[tex]m=10\text{ kg}[/tex]The tension on the rope is,
[tex]T=100\text{ N}[/tex]The angle with the horizontal is,
[tex]\theta=60^{\circ}[/tex]The block is moving with constant speed.
as the block is moving with constant speed, the net force on the block will be zero.
Part (A)
we can write in the horizontal direction the component of the tension will be equal to the frictional force and we write,
[tex]\begin{gathered} T\cos 60^{\circ}=F \\ F=100\cos 60^{\circ} \\ F=50\text{ N} \end{gathered}[/tex]Hence the frictional force is 50 N.
\\
Part(B)
The resultant normal reaction will be,
[tex]\begin{gathered} N=T\sin 60^{\circ}-mg \\ =100sin60^{\circ}-10\times9.8 \\ =-11.4\text{ N} \end{gathered}[/tex]hence the resultant normal reaction is -11.4 N.
The pressure is greater at the bottom of the bucket filled with water ,why?
If the mass m of the wrecking ball is 3920 kg , what is the tension TB in the cable that makes an angle of 40 ∘ with the vertical? What is the tension TA in the horizontal cable?
The tension TB in the cable makes an angle of 40 ∘ and the tension TA in the horizontal cable
TB=49380.9NTA=31741.4NThis is further explained below.
What is tension?Generally, To represent tension in a vertical direction, the term is:
[tex]T_B=\frac{m g}{\cos \theta}[/tex]
Substitute $3860kg for m, 9.8m/s^2 for g, and 40^0 for [tex]\theta[/tex].
[tex]T_B &=\frac{(3860 \mathrm{~kg})\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)}{\cos 40^{\circ}} \\[/tex]
=49380.9N
Because the cosine of the tension in the cable, which is pushing up on the item, is equal to the weight force, which is pressing down on the ground, the ball is not moving and is thus in equilibrium.
The expression for the horizontal cable tension is,
[tex]T_A=T_B \sin \theta[/tex]
Substitute $49380.9N for T_B and $40^o for [tex]\theta[/tex]
[tex]T_A &=(49380.9 \mathrm{~N}) \sin 40^{\circ}[/tex]
=31741.4N
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A PVC pipe has a length of 45.132 centimeters.a. What are the frequencies of the first three harmonics when the pipe is open at both ends? Include units in your answers.b. What are the frequencies of the first three harmonics when the pipe is closed at one end and open at the other? Include units in your answers.
ANSWERS
a. f₁ = 380 Hz; f₂ = 760 Hz; f₃ = 1140 Hz
b. f₁ = 190 Hz; f₃ = 570 Hz; f₅ = 950 Hz
EXPLANATION
a. For a pipe of length L open at both ends, the frequencies of the first three harmonics are:
[tex]\begin{cases}f_1=\frac{v}{2L} \\ \\ f_2=2f_1=\frac{v}{L} \\ \\ f_3=3f_1=\frac{3v}{2L}\end{cases}[/tex]Assuming that the speed of the wave is the speed of sound: 343 m/s and knowing that the length of the pipe is L = 45.132 cm = 0.45132 m we can find the frequencies of the first three harmonics:
[tex]\begin{cases}f_1=\frac{343m/s}{2\cdot0.45132m}\approx380Hz \\ \\ f_2=2f_1=2\cdot380Hz\approx760Hz \\ \\ f_3=3f_1=3\cdot380Hz\approx1140Hz\end{cases}[/tex]b. For a pipe of length L closed at one end and open at the other, the frequencies of the first three harmonics are:
[tex]\begin{cases}f_1=\frac{v}{4L} \\ \\ f_2=DNE \\ \\ f_3=3f_1=\frac{3v}{4L}\end{cases}[/tex]In a closed pipe, there can only be odd harmonics (1, 3, 5...). Therefore, the second harmonic does not exist and the "third harmonic" would be the 5th,
[tex]\begin{cases}f_1=\frac{v}{4L} \\ \\ f_3=3f_1=\frac{3v}{4L} \\ \\ f_5=5f_1=\frac{5v}{4L}\end{cases}[/tex]Again, the length of the pipe is 45.132 cm = 0.45132 m, so the first three harmonics are:
[tex]\begin{cases}f_1=\frac{343m/s}{4\cdot0.45132m}\approx190Hz \\ \\ f_3=3f_1=3\cdot190Hz=570Hz \\ \\ f_5=5f_1=5\cdot190Hz=950Hz\end{cases}[/tex]How does the work needed to stretch a spring 2 cm compare to the work needed to stretch it 1 cm.A.Same amount of workB.twice the workC.4 times the work D.8 times the work
The work required to stretch a string is given by the following equation:
[tex]W=\frac{1}{2}kx^2[/tex]Where:
[tex]\begin{gathered} k=\text{ string constant} \\ x=\text{ distance the string is stretched} \end{gathered}[/tex]If the string is stretched 2 cm then we substitute the value of "x = 2" in the formula, we get:
[tex]W_2=\frac{1}{2}k(2)^2[/tex]Solving the square and simplifying:
[tex]W_2=2k[/tex]Now, if the string is stretched 1 cm we get:
[tex]W_1=\frac{1}{2}k(1)^2[/tex]Solving the operations:
[tex]W_1=\frac{1}{2}k[/tex]Now, we determine the quotient between W2 and W1:
[tex]\frac{W_2}{W_1}=\frac{2k}{\frac{1}{2}k}[/tex]Simplifying we get:
[tex]\frac{W_2}{W_1}=4[/tex]Now, we multiply both sides by W2:
[tex]W_2=4W_1[/tex]Therefore, the work required to stretch the string 2 cm is 4 times the work to stretch it 1 cm.
A person pushes a 500 kg crate with a force of 1200 N and the crate accelerates at .5 m/s^2. What is the force of friction acting on the crate?
The force of friction acting on the crate is 950 N.
What is force of friction?Force of friction is defined as the force that opposes the motion of an object when two surfaces are in contact.
The frictional force on the object is determined by applying Newton's second law of motion as shown below.
F - Ff = ma
where;
F is the applied force = 1200 NFf is the frictional forcem is the mass of the crate = 500 kga is the acceleration of the crate = 0.5 m/s²1200 - Ff = 500(0.5)
1200 - Ff = 250
Ff = 1200 - 250
Ff = 950 N
Thus, the force of friction acting on the crate preventing the motion of the crate is determined by applying Newton's second law of motion.
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A runner runs around a circular track. He completes one lap at a time of t = 314 s at a constant speed of v = 3.1 m/s. t = 314 sv = 3.1 m/sWhat is the radius, r in meters, of the track? What was the runners centripetal acceleration, ac in m/s2, during the run?
Since the runner completes 1 lap in 314 seconds, and its velocity is 3.1m/s, then the total distance covered in 1 lap is:
[tex]\begin{gathered} d=vt \\ =(3.1\frac{m}{s})(314s) \\ =973.4m \end{gathered}[/tex]That distance corresponds to the perimeter of the circumference. The perimeter of a circumference with radius r is 2πr. Then:
[tex]\begin{gathered} 2\pi r=d \\ \\ \Rightarrow r=\frac{d}{2\pi} \\ =\frac{973.4m}{2(3.14...)} \\ =154.9...m \end{gathered}[/tex]The centripetal acceleration of an object in a circular trajectory with radius r and speed v is:
[tex]a_c=\frac{v^2}{r}[/tex]Replace v=3.1m/s and r=154.9m to find the centripetal acceleration:
[tex]a_c=\frac{(3.1\frac{m}{s})^2}{(154.9m)}=0.062\frac{m}{s^2}[/tex]Therefore, the radius of the track is approximately 155m and the centripetal acceleration of the runner is approximately 0.062 m/s^2.
55. Write the shortened notation for the ion symbol next to each element given below. A.) Calcium loses 2 negative electronsB.) Chlorine gains 1 negative electronsC.) Hydrogen gains 1 negative electron
We are asked to write shortened notation for the ion symbol next to each element given below.
Convention:
First of all, write the shortened notation of the element.
The superscript (at the top) is the number of charges on the ion and also a + sign for a positive ion and - sign for a negative ion.
When an atom loses an electron then the charge is positive (+).
When an atom gains an electron then the charge is negative (-)
A.) Calcium loses 2 negative electrons
Since it loses 2 electrons it will have 2+ superscript.
Calcium has an atomic number of 20 (the number of protons)
[tex]_{20}Ca^{2+}[/tex]B.) Chlorine gains 1 negative electron
Since it gains 1 electron it will have - superscript (we don't write 1 when the charge is 1)
Chlorine has an atomic number of 17 (the number of protons)
[tex]_{17}Cl^-[/tex]C.) Hydrogen gains 1 negative electron
Since it gains 1 electron it will have - superscript
Hydrogen has an atomic number of 1 (the number of protons)
[tex]_1H^-[/tex]An object is dropped from rest out of the window of a building, and the time to hit the ground is found to be 5 seconds. The same object is then dropped from rest out of a window twice as high above the ground as the original window. The time it takes the object to hit the ground is closest to:
ANSWER:
7 s
STEP-BY-STEP EXPLANATION:
Given:
u = 0m/s
t = 5 sec
g = 9.8 m/s^2
The first thing is to calculate the height of the building, using the following formula:
[tex]\begin{gathered} s=ut+\frac{1}{2}gt^2 \\ \text{ Replacing} \\ s=0\cdot5+\frac{1}{2}\cdot9.8\cdot5^2 \\ s=122.5\text{ m} \end{gathered}[/tex]Now, we apply the same formula, but we substitute the double value of the distance and solve for t, just like this:
[tex]\begin{gathered} 2\cdot122.5=\frac{1}{2}\cdot9.8\cdot\: t^2 \\ 9.8\cdot t^2=245\cdot2 \\ t^2=\frac{490}{9.8} \\ t=\sqrt[]{50} \\ t=7.07\text{ sec} \\ t\approx7\text{ sec} \end{gathered}[/tex]The time it takes for the object to fall is 7 seconds.
Two drops of mercury each has a charge on 2.42 nC and a voltage of 293.97 V. If the two drops are merged into one drop, what is the voltage on this drop?
The electric potential is given by:
[tex]\begin{gathered} V=\frac{Kq}{r} \\ \end{gathered}[/tex]Let's find r first:
[tex]\begin{gathered} r=\frac{Kq}{V}=\frac{8.988\times10^9\cdot2.42\times10^{-9}}{293.97} \\ r\approx0.074m \end{gathered}[/tex]Now we can find the radius of the new drop:
[tex]r_t=2(r)=2(0.074)=0.148[/tex]So:
[tex]\begin{gathered} V=\frac{K(2q)}{r_t}=\frac{8.988\times10^9\cdot2(2.42\times10^{-9})}{(0.148)} \\ V=293.93V \end{gathered}[/tex]Write a 5 page research on the topic 'How do migrating birds / animals find their direction? (Two birds / animals)'. Need this ASAP.
Answer: Birds migrate to move from areas of low or decreasing resources to areas of high or increasing resources. The two primary resources being sought are food and nesting locations. Here’s more about how migration evolved.
Birds that nest in the Northern Hemisphere tend to migrate northward in the spring to take advantage of burgeoning insect populations, budding plants and an abundance of nesting locations. As winter approaches and the availability of insects and other food drops, the birds move south again. Escaping the cold is a motivating factor but many species, including hummingbirds, can withstand freezing temperatures as long as an adequate supply of food is available.
Explanation: sorry about the anther
25. A student cycles along a level road at a speed of 5.0 m / s. The total mass of the student and bicycle is 120 kg. The student applies the brakes and stops. The braking distance is 10 m. What is the average braking force?
If a student bikes at a pace of 5.0 m/s down a straight route. The student's bicycle weighs 120 kg in total. The pupil puts on the brakes and comes to a stop. The average braking force of the automobile would be 150 newtons if the stopping distance were 10 meters.
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
Work done by the braking force of the cycle = change in kinetic energy of the student
Force × distance = 1/2 × mass × velocity²
F = 0.5 x 120 x 5² / 10
F = 150 Newtons
Thus, the average braking force of the cycle would be 150 Newtons.
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a 298 kg boat is being propelled forward with a force of 2,365 N. What is the acceleration of the boat if it has a resistance force (rewarded) due to wind and water of 878 N? (Write answer as a 2 digit number)
The acceleration of the boat is 4.9m/s²
Mass of boat= 298 kg
Forward force= 2365 N
Resistance force= 878N
We need to apply the concept of laws of motion
Net force= Forward force- Resistance force
Net force= 2365-878 N
= 1487 N
Net force= mass x acceleration
2365= acceleration x 298
acceleration = 4.9 m/s²
Therefore, the acceleration of the boat is 4.9 m/s²
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If the planet Mercury has a mass of planet 3.3×10²³ kg and a radius of 2400 km - calculate the magnitude of the gravitational field on its surface?
The magnitude of the gravitational field on the surface of the planet mercury is 3.82 m/s²
Explanation:The mass of mercury, m = 3.3×10²³ kg
The radius, r = 2400 km
r = 2400 x 1000m
r = 2.4 x 10⁶m
Note that the magnitude of the gravitational field on the surface of the planet is the acceleration due to gravity on that planet
It is given by the formula:
[tex]g=\frac{Gm}{r^2}[/tex]Substitute the values of G, m, and r into the formula above
[tex]\begin{gathered} g=\frac{6.67\times10^{-11}\times3.3\times10^{23}}{(2.4\times10^6)^2} \\ g=\frac{6.67\times10^{-11}\times3.3\times10^{23}}{(2.4\times10^6)^2} \\ g=\frac{2.2\times10^{13}}{5.76\times10^{12}} \\ g=3.82m/s^2 \end{gathered}[/tex]The magnitude of the gravitational field on the surface of the planet mercury is 3.82 m/s²
Describe the mathematical relationship between the distance (d) and the attractive force (F) between protons and electrons.
The attractive force and the distance are inversely proportional.
[tex]F\propto\frac{1}{r}[/tex]This relation means that the attractive force decreases as the distance increases, and the attractive force increases as the distance decrease.
Toaster uses a nichrome heating coil and operates at 120 V. When the toaster is turned on at 20°C, the current in the cold coil is 1.5 A. When the coil warms up, the current has a value of 1.3 A. If the thermal coefficient of resistivity for nichrome is 4.5x10-4 1/Co, what is the temperature of the coil?Group of answer choices68oC490oC160oC360oC260oC
Given that the operating voltage is V = 120 V.
The initial temperature of the toaster is T1 = 20 degrees Celsius
The initial current in the coil is I1 = 1.5 A
The final current in the coil is I2 = 1.3 A
The thermal coefficient of resistivity for nichrome is
[tex]\alpha=4.5\times10^{-4}^{}\text{ }^{\circ}C^{-1}[/tex]We have to find the final temperature of the coil, T2.
The initial resistance of the coil is
[tex]\begin{gathered} R1=\frac{V}{I1} \\ =\frac{120}{1.5} \\ =80\Omega \end{gathered}[/tex]The final resistance of the coil is
[tex]\begin{gathered} R2\text{ =}\frac{V}{I2} \\ =\frac{120}{1.3} \\ =92.307\Omega \end{gathered}[/tex]The formula to calculate the final temperature of the coil is
[tex]\begin{gathered} \alpha=\frac{(R2-R1)}{R1(T2-T1)} \\ T2-T1=\frac{(R2-R1)}{\alpha\times R1} \\ T2=\frac{(R2-R1)}{\alpha\times R1}+T1 \end{gathered}[/tex]Substituting the values, the final temperature will be
[tex]\begin{gathered} T2=\text{ }\frac{92.307-80}{4.5\times10^{-4}\times80}+20 \\ \approx360^{\circ}\text{ C} \end{gathered}[/tex]Thus, the final temperature is 360 degrees Celsius.
An intrepid hiker reaches a large crevasse in his hiking route. He sees a nice landing ledge 60.0 cm below his position but it is across a 2.3 m gap. He spends 1.2 s accelerating horizontally at 5.92 m/s2 [right] in an attempt to launch himself to the safe landing on the far side of the gap. Does he make it?
The hiker made it to a safe landing on the other side of the gap after travelling horizontally at 2.49 m.
What is the time motion from the vertical height?
The time taken for the hiker to fall from the given height is calculated as follows;
h = vt + ¹/₂gt²
where;
v is the vertical velocity = 0t is the time of motiong is acceleration due to gravityh is the height of fallh = ¹/₂gt²
t = √(2h/g)
t = √[(2 x 0.6) / (9.8)]
t = 0.35 seconds
The horizontal velocity of the hiker during the period of acceleration is calculated as follows;
Vₓ = at
Vₓ = (5.92 m/s²) x (1.2 s)
Vₓ = 7.104 m/s
The horizontal distance travelled during the time period of 0.35 seconds;
X = Vₓt
X = 7.104 x 0.35
X = 2.49 m
Thus, the hiker made it to a safe landing on the other side of the gap which is 2.3 m wide and smaller to his horizontal displacement of 2.49 m.
Learn more about horizontal distance here: https://brainly.com/question/24784992
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If John Glenn weighed 640 N on Earth's surface, a) how much would he haveweighed if his Mercury spacecraft had (hypothetically) remained at twice thedistance from the center of Earth? b) Why is it said that an astronaut is nevertruly "weightless?"
Given:
The weight of John Glenn, w=640 N
To find:
a) The weight if the distance was twice that of the initial value.
b) Why is an astronaut never weightless.
Explanation:
a)
Let the distance between the spacecraft and the earth be r.
If it becomes twice, then the distance is 2r.
The initial gravitational force on John Glenn is,
[tex]F=w=\frac{GMm}{r^2}[/tex]Where G is the gravitational constant, M is the mass of the earth and m is the mass of John Glenn.
The force when the distance is twice,
[tex]\begin{gathered} w_n=\frac{GMm}{(2r)^2} \\ =\frac{GMm}{4r^2} \\ =\frac{w}{4} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} w_n=\frac{640}{4} \\ =160\text{ N} \end{gathered}[/tex]b)
Even when the astronaut is in space they still have the mass and so does the earth. Thus there will always be a gravitational force of attraction between the earth and the astronaut. The astronaut does not feel the weight because there will be nothing in space that pushes them back. That is why an astronaut is never truly weightless.
Final answer:
a) Thus the weight of John Glenn will be 160 N