Answer:
2275000 lb.ft
Explanation:
Let work done on the cable be denoted by: W_ca
Let work done on the coal be denoted by: W_co
Now, dividing the cable into segments, let x represent the length from top of the mine shaft to the segment.
Meanwhile let δx be the length of the segment.
We are told the cable weighs 8 lb/ft. Thus;
Work done on one segment = 8 × δx × x = 8x•δx
Therefore, work done on cable is;
W_ca = ∫8x•δx between the boundaries of 0 and 650
Thus;
W_ca = 4x² between the boundaries of 0 and 650
W_ca = 4(650²) - 4(0²)
W_ca = 1,690,000 lb.ft
Workdone on the 900 lb of coal will be calculated as;
W_co = 900 × 650
W_co = 585000 lb.ft
Thus,
Total work done = W_ca + W_co
Total workdone = 1690000 + 585000
Total workdone = 1690000 + 585000
Total workdone = 2275000 lb.ft
First aid for a strain or a sprain
A) the FITT formula.
B) ineffective.
C) the RICE formula. .
D) the ABC formula
Answer:
The correct answer is - C) the RICE formula.
Explanation:
The first aid for a patient with a sprain or strain mainly includes the RICE formula which stands for the rest, ice, compression, and elevation. In case of a sprain or strain, one is advised to rest, use ice on the location of the sprain, compression on the part of the sprain, and elevate the body part for less pain, less swelling or reducing chances of making it worst.
The other three formulas given are used in different conditions. The FITT formula used in the exercise planning, and the ABC formula used for the first aid of the heart attack or blockage of the airway.
There are two important isotopes of uranium: 235U and 238U. These isotopes have different atomic masses and react differently. Only 235U is very useful in nuclear reactors. One of the techniques for separating them (gas diffusion) is based on the different rms speeds of uranium hexafluoride gas, UF6.
Required:
a. The molecular masses for 235U UF6 and 238U UF6 are 349.0 g/mol and 352.0 g/mol, respectively. What is the ratio of their average velocities?
b. At what temperature would their average velocities differ by 1.00 m/s?
c. Do your answers in this problem imply that this technique may be difficult
Answer:
See explanation
Explanation:
a)Given that;
v235/v238 = √3KT/M235/√3KT/M238
v235/v238 =√M238/M235
M238 = 352.0 g/mol
M235 = 349.0 g/mol
v235/v238 = √352.0 g/mol/349.0 g/mol
v235/v238 = 1.004
b) Given that;
v235 = √3KT/M235
v238 = √3KT1/M238
√3KT/M238 - √3KT/M235 = 1
√T √3K/M238 - √3K/M235 = 1
√T = 1/√3K/M238 - √3K/M235
√T = 1/√3 * 8.31/352 * 10^-3 - √3* 8.31/349 * 10^-3
T = 768 K
c) 768 K is the same as 495 °C hence this process is achievable
what is the result of mixing 15 garm of water 80 degree celsius with 10 gram of ice -10 degree Celsius ? give specific heat capctiy of ice 0.5 calorie per gram Celsius and letent heat of fusion of ice 80 calorie per gram.
Answer:
50
Explanation:
A 25,000 kg jet is sitting still on the runway (O
m/s). If the jet goes to take off and reaches 80
m/s in 10 seconds, how much force was required
to get it there?
Answer:
The force required to get it there is 200000 N.
Explanation:
The force can be calculated by the second Newton's law:
[tex] F = ma [/tex]
Where:
m: is the mass = 25000 kg
a: is the acceleration
The acceleration is given by:
[tex] a = \frac{v}{t} [/tex]
Where:
v: is the velocity = 80 m/s
t: is the time = 10 s
[tex] a = \frac{v}{t} = \frac{80 m/s}{10 s} = 8 m/s^{2} [/tex]
Hence, the force is:
[tex] F = ma = 25000 kg*8 m/s^{2} = 200000 N [/tex]
Therefore, the force required to get it there is 200000 N.
I hope it helps you!
A student club is designing a trebuchet for launching a pumpkin into projectile motion. Based on an analysis of their design, they predict that the trajectory of the launched pumpkin will be parabolic and described by the equation y(x) =ax^2+bx where a=−8.0×10^−3m^−1, b=1.0(unitless), x is the horizontal position along the pumpkin trajectory and y is the vertical position along the trajectory. The students decide to continue their analysis to predict at what position the pumpkin will reach its maximum height and the value of the maximum height.
Required:
What is the derivative of the vertical position of the pumpkin trajectory with respect to its horizontal position?
Answer:
[tex]\frac{dy}{dx}=2ax+b[/tex]
Explanation:
We are given that
[tex]y(x)=ax^2+bx[/tex]
[tex]a=-8.0\times 10^{-3}m^{-1}[/tex]
[tex]b=1.0[/tex]
x=Horizontal position along the trajectory
y=Vertical position along the trajectory
We have to find the derivative of the vertical position of the pumpkin trajectory with respect to its horizontal position.
Differentiate the equation w.r.t x
[tex]\frac{dy}{dx}=2ax+b[/tex]
Hence, the derivative of the vertical position of the pumpkin trajectory with respect to its horizontal position is given by
[tex]\frac{dy}{dx}=2ax+b[/tex]
The electric field between two parallel plates is uniform, with magnitude 698 N/C. A proton is held stationary at the positive plate, and an electron is held stationary at the negative plate. The plate separation is 4.04 cm. At the same moment, both particles are released.
Required:
a. Calculate the distance (in cm) from the positive plate at which the two pass each other.
b. Repeat part (a) for a sodlum lon (Na+) and a chlorlde lon (CI-).
Answer:
x = 4.03 10⁻² m
Explanation:
Let's start by finding the acceleration for each particle due to the electric field
F = ma
the electric force is F = qE
q E = m a
a = qE / m
proton
m = 1.67 10⁻²⁷ kg
a₁ = 1.6 10⁻¹⁹ 698 /1.67 10⁻²⁷
a₁ = 6.687 10¹⁰ m / s²
directed to the right
electron
m = 9.11 10⁻³¹ kg
a₂ = 1.6 10⁻¹⁹ 698 /9.11 10⁻³¹
a₂ = 1.23 10¹⁴ m / s²
directed to the left
Taking the acceleration of the two bodies, we set a reference system with zero at the initial position of the proton on the positive plate, the point where it is located is x for the proton and x for the electron,
for the proton
x₁ = x₀₁ + v₀₁ t + ½ a₁ t²
as we start from rest vo1 = 0 and the initial position is xo = 0
x₁ = ½ a₁ t²
for the electron
x₂ = x₀₂ + v₀₂ t + ½ a₂ t²
in this case the initial velocity is zero v₀₂ = 0 and the initial position is x₀₂=d
x₂ = x₀₂ + ½ a₂ t²
at the meeting point x₁ = x₂, so we can equalize the two equations
½ a₁ t² = x₀₂ + ½ a₂ t²
½ t² (a₁ -a₂) = x₀₂
t = [tex]\sqrt{ \frac{2 x_{o2} }{ (a_1 - a_2)} }[/tex]
let's calculate
t = [tex]\sqrt{\frac{2 \ 4.04 \ 10^{-2} }{ ( 6.687^{10} + 1.23 10^{14} ) } }[/tex]
t = [tex]\sqrt{ \frac{8.08 \ 10^{-2} }{ 1.2306 \ 10^{14} } }[/tex]
t = 2.56 10⁻⁸ s
now we can calculate the position
x = ½ a₂ t²
x = ½ 1.23 10¹⁴ (2.56 10⁻⁸)²
x = 4.03 10⁻² m
Match the type of heat transfer with its description
Answer:
1. Convection
2. Radiation
3. Conduction
Hope this helps!
Explanation:
Two identical ice hockey pucks, labeled A and B, are sliding toward each other at speed v. Which one of the following statements is true concerning their momenta and kinetic energies?
a. p(A) = p(B) and KEA= KEB
b. pA= - pB and KEA= -KEB
c. pA= -pB and KEA= KEB
d. pa= pb and KEA =- KEB
Answer:
C. [tex]p_{A} = -p_{B}[/tex] and [tex]K_{A} = K_{B}[/tex].
Explanation:
The two hockey pucks travels in opposite sides with velocities of same magnitude, by definitions of linear momentum and translational kinetic energy:
Linear momentum
Hockey puck A
[tex]p_{A} = m\cdot v[/tex] (1)
Hockey puck B
[tex]p_{B} = -m\cdot v[/tex] (2)
[tex]p_{A} = -p_{B}[/tex]
Translational kinetic energy
Hockey puck A
[tex]K_{A} = \frac{1}{2}\cdot m\cdot (v)^{2}[/tex]
[tex]K_{A} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (3)
Hockey puck B
[tex]K_{B} = \frac{1}{2}\cdot m\cdot (-v)^{2}[/tex]
[tex]K_{B} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (4)
[tex]K_{A} = K_{B}[/tex]
Hence, the correct answer is C.
A crane lifts an I-beam up the side of a building. The crane's power output is 1750W for 20 seconds. After 20 seconds the I-beam was moving at 2 m/s and the mass has 200 kg. Use the work-energy process to determine the change in height of the I-beam.
Explanation:
This is one of those great problems that is harder to read than to solve!
Remember the following...
Power is the amount of work done over time. [tex]P = \frac{W}{t}[/tex]
and that work is Force times distance. [tex]W = F * d[/tex]
and that Force is just mass times acceleration. [tex]F = m * a[/tex]
So... nesting all the equations together we might get something that looks like. [tex]P = \frac{m*a*d}{t}[/tex]
We know the power, the time, the acceleration (at this given time) and the mass. We can then solve for d. [tex]d = \frac{P*t}{m*a}[/tex]
Plug and chug!
Quick Check:
12-17-20
Conservation of Energy
In the use of an energy resource like coal, do you get all
the energy back in one form(100%)? (circle one)
Yes
No
Answer:
Quick Check:
12-17-20
Conservation of Energy
In the use of an energy resource like coal, do you get all
the energy back in one form(100%)? (circle one)
Yes
No
Explanation:
Not sure
Des Linden won the Boston marathon in 2018, becoming the first American woman to win since 1985. The harsh conditions (heavy rain and a headwind) led to a winning time of 2 hrs 39 minutes and 54 seconds. If she weighed 99 pounds, her average heart rate during the race was 170 beats per minute and her heart pumped 2.5 mLs of blood per kg body mass per beat, how much blood did her heart pump during the race?
Answer:
3050.6 Litre .
Explanation:
Total time of heart beat = Total time of race = 2 hrs , 39 minutes and 54 seconds
= 2 x 60 + 39 + 54/60 min
= 120 + 39 + .9 min
= 159.9 min
rate of heart beat = 170 per min
Total no of heart beat during race = 170 x 159.9
volume of blood per kg per beat = 2.5 mL per kg of weight
body weight = 99 pounds = .4535 x 99 kg = 44.89 kg
volume of blood per beat = 2.5 mL x 44.89 mL
= 112.225 mL .
Total required volume of blood = 112.225 x 170 x 159.9 mL
= 3050612 mL
= 3050.6 L.
what causes coolant to return to the reservoir tank
Answer:
The coolant returns to the reservoir tank when the car is operated
The water level in a water tower is 45 m above the ground. What is the gauge pressure of the water at a faucet that is 20 m above the ground?
How does an image from a CT scan differ from a regular X-ray image?
A.
An X-ray image requires a computer to create it, and a CT scan does not.
B.
The CT scan uses many X-ray images in thin slices to make a 3D image.
C.
The CT scan makes a 2D image, and regular X-ray images are 3D.
D.
The CT scan uses lower amounts of radiation.
Answer:
The Answer is B
Explanation:
I just took the test
An image from a CT scan differ from a regular X-ray image in such a way that B)The CT scan uses many X-ray images in thin slices to make a 3D image.
What is a CT scan ?A CT scan also stands for Computed tomography scan is a medical technique to obtain detail of internal images of the body .
x ray image are in 2D while CT Scan image are in 3D (hence option c) is wrong ). Also CT scans have a doughnut shaped tube that rotated the x ray 360° degree around the body . The data provides a detailed 3D view of the inside of the body
hence , option B
The CT scan uses many X-ray images in thin slices to make a 3D image.
is correct
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A sound wave travels in air toward the surface of a freshwater lake and enters into the water. The frequency of the sound does npt change when the sound enters the water. The wavelength of the sound is 5.67 m in the air, and the temperature of both the air and the water is 20 oC. What is the wavelength in the water
Answer:
wavelength in the water = 1.311 m
Explanation:
givendata
wavelength of the sound in air = 5.67 m
temperature = 20°C
solution
as we know velocity c in air = 343.3 m/s and velocityˇc in water= 1484m/s
and we take wavelength = w
and
wavelength=velocity ÷ frequency .................1a
as we know here that In air and in water if the frequency not change then there wavelength depend on speed of sound in air and water
so
wavelength of sound in water = w in air × c in water ÷ c in air
wavelength in the water in water = 5.67 × 343.2 ÷ 1484
wavelength in the water = 1.311 m
i will give the brainliest answer to whoever answers this. how did Sir William Gilbert support his theory of magnetic fields
Answer:
whan he observed that magnetic forces often produced circular motions,he began to connect the phenomenon of magnetism with the rotation of the earth.This led to his discover of the earth's own magnesium,and provided the theoretical foundation for the science of geomagnetism.
hope it's helpful ❤❤❤
THANK YOU.
A hollow ball weighs 40 newtons. In a water tank, it displaces 15 newtons of water. What is the buoyant force on the ball? Will the ball float or sink? Explain your reasoning.
Answer:
25N
it will float since the water displaced is less than the weight. buoyancy = 25n down
Explanation:
40 newtons - 25 newtons = 25 newtons
The buoyant force on the ball is 25 Newton and the ball will float.
From the information given, we're informed that a hollow ball weighs 40 newtons and a water tank displaces 15 newtons of water.
Therefore, the buoyant force on the ball will be:
= 40N - 15N = 25N
Since we got a positive value, it means that the ball will float on water.
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Which of the the following quantities has the same unit as kilowatt - hour?
(a) Force × time
(b) Force × distance
(c) Force × acceleration
(d) Force × velocity
Answer:
I think (d) is your answer...
Force × velocity is the quantities has the same unit as kilowatt - hour. hence option D is correct.
What is Force ?A force in physics is an effect that changes the velocity of a mass-moving object, causing it to accelerate from rest, for example. It is a vector quantity since it can be a push or a pull and always has magnitude and direction. It is denoted by the letter F (formerly P) and is measured in newtons (N), the SI unit of force.
The net force exerted on an object is equal to the rate at which its momentum varies over time, according to Newton's second law in its original formulation. According to this equation, an object's acceleration is precisely proportional to the net force applied on it if its mass remains constant.
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A river flows due east at 1.70 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant velocity of 14.0 m/s due north relative to the water.
(a) What is the velocity of the boat relative to shore?
m/s
° (north of east)
(b) If the river is 340 m wide, how far downstream has the boat moved by the time it reaches the north shore in meters?
Answer:
a)
v = 14.1028 m/s
∅ = 83.0765° north of east
b)
the required distance is 40.98 m
Explanation:
Given that;
velocity of the river u = 1.70 m/s
velocity of boat v = 14.0 m/s
Now to get the velocity of the boat relative to shore;
( north of east), we say
a² + b² = c²
(1.70)² + (14.0)² = c²
2.89 + 196 = c²
198.89 = c²
c = √198.89
c = 14.1028 m/s
tan∅ = v/u = 14 / 1.7 = 8.23529
∅ = tan⁻¹ ( 8.23529 ) = 83.0765° north of east
Therefore, the velocity of the boat relative to shore is;
v = 14.1028 m/s
∅ = 83.0765° north of east
b)
width of river = 340 m,
ow far downstream has the boat moved by the time it reaches the north shore in meters = ?
we say;
340sin( 90° - 83.0765°)
⇒ 340sin( 6.9235°)
= 40.98 m
Therefore, the required distance is 40.98 m
what is surface tension
Answer:
the tension of the surface film of a liquid caused by the attraction of the particles in the surface layer by the bulk of the liquid, which tends to minimize surface area.
Explanation:
A glideron an air trackmoves in the +x direction with a constant acceleration. It has two flags, each exactly 0.02 mlong, with the midpoints of the flags separated by 0.150m. The first flag interrupts the photogate timer for a time 0.058s, and the second flag interrupts the photogate timer for a time 0.047s.1.What was the average velocity of the glider during the interval when the first flag was interrupted?2. What was the average velocity of the glider during the interval when the second flag was interrupted?3. Approximately how much time elapsed between the passage of the first flag and the passage of the second flag?4. Calculatethe approximate accelerationof the glider.
Answer:
1) Average Velocity V1 = 0.3448 m/s
2) Average Velocity V2 = 0.4255 m/s
3) time elapsed t = 0.39 seconds
4) Approximate acceleration of the glider = 0.207 m/s²
Explanation:
Given the data in the question;
1) What was the average velocity of the glider during the interval when the first flag was interrupted
V1 ( first interruption) = distance travelled / elapsed time
= 0.02 m / 0.058 s = 0.3448 m/s
2) What was the average velocity of the glider during the interval when the second flag was interrupted?
V2( second interruption) = distance travelled / elapsed time
= 0.02 m / 0.047 s = 0.4255 m/s
3) Approximately how much time elapsed between the passage of the first flag and the passage of the second flag?
Average Speed = ( v1 + v2 ) / 2 = ( 0.3448 m/s + 0.4255 m/s) ) / 2
= 0.7703 / 2 = 0.3851 m/s
so time elapsed will be
t = 0.150 m / 0.3851 m/s
t = 0.3895 ≈ 0.39 seconds
4) Calculate the approximate acceleration of the glider
acceleration = ( v2 - v1 ) / t
a = ( 0.4255 m/s - 0.3448 m/s ) / 0.3895
a = 0.0807 m/s / 0.3895 s
a = 0.207 m/s²
Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 290 m/s2 for 20 ms, then travels at constant speed for another 30 ms. During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?
Answer:
0.232m
Explanation:
We are given that
Acceleration,[tex]a=290m/s^2[/tex]
[tex]t_1=20ms=20\times 10^{-3}s=0.02s[/tex]
[tex]1ms=10^{-3}s[/tex]
[tex]t_2=30ms=30\times 10^{-3}s=0.03s[/tex]
We have to find the distance reached by tongue.
We know that
[tex]s=ut+\frac{1}{2}at^2[/tex]
Initial speed, u=0
Using the formula
[tex]x_1=0+\frac{1}{2}(290)(0.02)^2[/tex]
[tex]x_1=0.058m[/tex]
[tex]v=u+at[/tex]
Using the formula
Speed after 0.02s
[tex]v=0+(290)(0.02)[/tex]
[tex]v=5.8m/s[/tex]
Distance=vt
By using the formula
Distance traveled in 0.03 s
[tex]x_2=5.8(0.03)=0.174m[/tex]
Total distance traveled in 50 ms
[tex]\Delta x=x_1+x_2=0.058+0.174[/tex]
=0.232m
Bart runs up a 2.5 meter high flight of stairs at a
constant speed in 4.25 seconds chasing his
adolescent gazelle. If Bart's mass is 60 kg,
determine the work he did.
Answer:
1470J
Explanation:
Given parameters:
Height of stairs = 2.5m
Time taken = 4.25s
Mass of Bart = 60kg
Unknown:
Work done by Bart = ?
Solution:
Work done is a function of the force acting to move a body through a distance.
Work done = Weight x height
Work done = mass x acceleration due to gravity x height
Work done = 60 x 9.8 x 2.5 = 1470J
a 70 kg student stands on top of a 5.0 m platform diving board . how much gravitational potential energy does the student have?
how much work did it take for the student to travel from the ground to the top of the platform diving board?
Answer:
a. P.E = 3430Joules.
b. Workdone = 3430Nm
Explanation:
Given the following data;
Mass = 70kg
Distance = 5m
We know that acceleration due to gravity is equal to 9.8m/s²
To find the potential energy;
Potential energy = mgh
P.E = 70*9.8*5
P.E = 3430J
b. To find the workdone;
Workdone = force * distance
But force = mass * acceleration
Force = 70*9.8
Force = 686 Newton.
Workdone = 686 * 5
Workdone = 3430Nm
A large truck traveling at +15 m/s collides head-on with a small car traveling at -15 m/s, which of the following statements is correct?
a) the large truck will experience the greater impulse
b) the small car will experience the greater impulse
c) the magnitude of the impulses for both vehicles is the same
Dangers of physical in heat and
humidity include:
A)hypothermia
B)frostbite
C)muscle cramps
D)wind chill index
A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate
the following using the energy/work formulae only:
1.The kinetic energy at the top of the cliff
2.The total mechanical energy at the top of the cliff
3.The kinetic energy of the Rock half way down 4.the speed of the Rock half way down
5.The speed of the Rock as it hits the ground
Answer:
Answers and steps in the pic..hope you can u understand the calculations....
Which of the following is the ability of a physical system to do work?
O force
O energy
Ospring constant
Odisplacement
1. Two spherical objects have equal masses and
experience a gravitational force of 85 N towards one
another. Their centers are 36mm apart
Answer:
4.06 x 10⁴kg
Explanation:
Given parameters:
Gravitational force = 85N
Distance or separation = 36mm = 0.036m
Unknown:
Their masses = ?
Solution:
Both of the spherical objects have the same mass as indicated by the object.
From the newton's law of universal gravitation:
F = [tex]\frac{G m1 m2}{r^{2} }[/tex]
F is the gravitational force
G is the universal gravitation constant = 6.67 x 10⁻¹¹
m1 is the mass of object 1
m2 is the mass of object 2
r is the distance between them
Since m1 = m2 = m
So;
85 = [tex]\frac{6.67 x 10^{-11} x m^{2} }{0.036^{2} }[/tex]
Therefore;
m = 4.06 x 10⁴kg
All object in the universe attract each other on the basis of the masses and distance between them. The mass of the given particle is [tex]\bold {4.06 x 10^4kg}[/tex].
Universal law of gravitation:
It states that the all object in the universe attract each other on the basis of the masses and distance between them.
[tex]\bold{F=G{\frac{m_1m_2}{r^2}}}[/tex]
Where,
F = force = 85 N
G = gravitational constant = [tex]\bold{6.67 x 10\\^-^1^1}[/tex]
m1 = mass of object 1 = ?
m2 = mass of object 2 = ?
r = distance between centers of the masses = 36 mm
As stated in the question,
m1 = m2 = m
Now, put the values in the formula,
[tex]\bold { 85 = \dfrac {6.67\times 10^-^1^1 \times m^2}{0.036^2}}\\\\\bold { m = 4.06 x 10^4kg}[/tex]
Therefore, the mass of the given particle is [tex]\bold {4.06 x 10^4kg}[/tex].
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A 10 kg remote control plane is flying at a height of 111 m. How much
potential energy does it have?
Answer:
10.88kJ
Explanation:
Given data
mass= 10kg
heigth= 111m
Applying
PE= mgh
assume g= 9.81m/s^2
substitute
PE= 10*9.81*111
PE=10889.1 Joules
PE=10.881kJ
Hence the potential energy is 10.88kJ
