Answer:
980000ft-lbsExplanation:
Step one:
given data
mass of cable= 4lb/ft
mass of coal= 1000lb
dept of mine= 700ft
Step two:
Required
the work-done to lift the coal and the rope combined
Work-done to lift coal
Wc=1000*700= 700,000 lb-ft
Work-done to lift rope
[tex]Wr=\int\limits^{700} _0 {4(700-y)} \, dx \\\\Wr=4(700y-\frac{1}{2}y^2 )\limits^{700}_0[/tex]
substitute y=700 we have, since y=0 will result to 0
[tex]Wr=4(700*700-\frac{1}{2}*700^2 )\\\\Wr=4(490000-245000)\\\\Wr=4(245000)\\\\Wr=980000ft-lbs[/tex]
A car’s brakes decelerate it at a rate of -2.40 m/s2. If the car is originally travelling at 13 m/s and comes to a stop, then how far, in meters, will the car travel during that time?
Answer:
Approximately [tex]35.2\; \rm m[/tex].
Explanation:
Given:
Initial velocity: [tex]u = 13\; \rm m \cdot s^{-1}[/tex].
Acceleration: [tex]a = -2.40\; \rm m \cdot s^{-2}[/tex] (negative because the car is slowing down.)
Implied:
Final velocity: [tex]v = 0\; \rm m \cdot s^{-1}[/tex] (because the car would come to a stop.)
Required:
Displacement, [tex]x[/tex].
Not required:
Time taken, [tex]t[/tex].
Because the time taken for this car to come to a full stop is not required, apply the SUVAT equation that does not involve time:
[tex]\begin{aligned} x &= \frac{v^2 - u^2}{2\, a} \\ &= \frac{{\left(0\; \rm m \cdot s^{-1}\right)}^2 - {\left(13\; \rm m \cdot s^{-1}\right)}^2}{2\times \left(-2.40\; \rm m\cdot s^{-2}\right)} \approx 35.2\; \rm m \end{aligned}[/tex].
In other words, this car would travel approximately [tex]35.2\; \rm m[/tex] before coming to a stop.
Can someone please answer how to convert mass into weight?
Answer:
To find the weight of something, simply multiply its mass by the value of the local gravitational field, and you get a result in newtons (N). For example, if your mass is 50 kg (about 110 pounds), then your weight is (50) (9.8). The point that must be overwhelmingly emphasized is that weight is a force.
Explanation:
A block slides down an inclined plane from rest. Initially the block is at 4.5m above the ground. Find the speed of the block when it is 1.5m above the ground. 1) 7.7m/s 2) 9.4m/s 3) 5.4m/s 4) 3.2m/s
Since, no external force is acting , so the system is in equilibrium .
Initial total energy = Final total energy
[tex]mg(4.5) = mg(1.5) + \dfrac{mv^2}{2}\\\\\dfrac{v^2}{2}=3\times g \\\\v^2=3\times 9.8\times 2\\\\v = \sqrt{58.8}\ m/s\\\\v = 7.67 \ m/s[/tex] ( Here , g = acceleration due to gravity = 9.8 m/s² )
Therefore, option 1) is correct.
Hence, this is the required solution.
A model of a helicopter rotor has four blades, each 3.4 m in length from the central shaft to the tip of the blade. The model is rotated in a wind tunnel at 550 rev/min. What is the radial acceleration of the blade tip, expressed as a multiple of the acceleration g due to gravity?
A. (5.72 × 104)g
B. (6.23 × 102)g
C. (1.15 × 103)g
D. (2.25 × 103)g
What is the energy contained in a 0.950 m3 volume near the Earth's surface due to radiant energy from the Sun?
When a potential difference of 10 V is placed across a certain solid cylindrical resistor, the current through it is 2 A. If the diameter of this resistor is now tripled, the current will be:______.A) 18 A.
B) 2/3 A.
C) 3 A.
D) 2/9 A.
E) 2 A.
Answer:
sorry I wish I could it help you
(A star if you answer this question) A school bus is traveling at 11.1 m/s and has a
momentum of 152,625 kgm/s. What is the mass of the bus?
[tex]\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}[/tex]
Actually Welcome to the Concept of the Kinematics in real world.
So, as given here, we have to find the Mass of the bus from the given momentum, so we get as,
P = m * V
momentum = mass * velocity
here, P= 152625 kgm/s and v= 11.1 m/s
so substituting we get as,
m = 152625 ÷ 11.1 => 13,750 kg
hence,the mass of the bus is 13,750 kg.
A plane mirror is placed to the right of an object. The image formed by the mirror will be a
real image that appears to be on the right of the mirror.
real image that appears to be on the left of the mirror.
virtual image that appears to be on the right of the mirror.
virtual image that appears to be on the left of the mirror.
Hamish is studying what happens when he sends a sound wave through different mediums, and he records his data in a table.
A 2-column table with 4 rows titled Hamish's Waves. The first column labeled Wave has entries 1, 2, 3, 4. The second column labeled Information has entries liquid, solid, gas, liquid.
Which statement could made about the data collected in Hamish’s table?
Wave 1 will move the fastest.
Wave 2 will move the slowest.
Wave 3 will move the slowest.
Wave 4 will move the fastest.
What is common between transverse waves and longitudinal waves?
Both include an amplitude, crest, and rarefactions
Both move faster at higher temperatures
Both move slower through densely packed molecules
Both include a wavelength from compression to compression
An angle of refraction is the angle between the refracted ray and the
incident ray.
normal.
medium.
boundary.
Answer:
A plane mirror is placed to the right of an object. The image formed by the mirror will be a virtual image that appears to be on the left of the mirror.
Explanation:
An 7.40 kg block drops straight down from a height of 0.83 m, striking a platform spring having a force constant of 9.50 102 N/m. Find the maximum compression of the spring.
Answer:
0.25 m.
Explanation:
mass of the block = 7.40 kg, height = 0.83 m, force constant of the spring = 9.50 x [tex]10^{2}[/tex] N/m.
The maximum compression on the spring can be determined by;
Potential energy stored in the spring = [tex]\frac{1}{2}[/tex] K[tex]x^{2}[/tex]
But, potential energy = mgh
So that,
mgh = [tex]\frac{1}{2}[/tex] K[tex]x^{2}[/tex]
7.4 x 9.8 x 0.83 = 9.50 x [tex]10^{2}[/tex] x [tex]x^{2}[/tex]
60.1916 = 9.50 x [tex]10^{2}[/tex] x [tex]x^{2}[/tex]
[tex]x^{2}[/tex]= [tex]\frac{60.1916}{9.50*10^{2} }[/tex]
= 0.06336
x = 0.2517
x = 0.25 m
The maximum compression of the spring is 0.25 m.
5.
An 80 newton force and a 45 newton force act on an object
as shown below.
80 N
30°
4S N
Which of the following vectors would bets represent an
equilibrant when added to this system?
(1) 24 N to the left (3) 24 N to the right
(2) 114 N to the right (4) 45 N to the left
Tirant Showroiculations
Answer:
the answer is a time your welcome
Answer:
(1)
Explanation:
A ball is thrown 24 m/s into the air. How high does it go?
556.4 m
0 m
29.4 m
-556.4 m
Answer:
option c is correct
Explanation:
we know that
2as=vf^2-vi^2
vf=24 m/s
vi= 0 m/s
a=g= 9.8 m/s^2
s=vf^2-vi^2/2a
s=(24)²-(0)²/2*9.8
s=576/19.6
s=29.4 m
therefore option c is correct
plzzzzzzzzzzzzzzzzzzzzzzzzzz help 20 points
Answer:
1.23
Explanation:
[tex]{\underline{\pink{\textsf{\textbf{ Answer : }}}}}[/tex]
➩ 1.23 feet
[tex]{\underline{\purple{\textsf{\textbf{Explanation : }}}}}[/tex]
Given :
Simon cuts a pipe that was 4.92 feet long Then he cuts it into four equal pieces.To find :
What is the length of the each piece.Solution :
As it is told that it's divided into four equal pieces
Therefore,
We must divide it by 4 to get the length of each piece.
So,
[tex] \sf \to \: \frac{4.92}{4} \\ \sf \to \: 1.23 \: feet \: ans.[/tex]
Is Geothermal Energy renewable? Why or why not? Use in your own words.
Answer:
Yes, geothermal energy is a renewable energy resource because the water can be heated by pumping it through the rocks.
Three moles of a monatomic ideal gas are heated at a constant volume of 1.20 m3. The amount of heat added is 5.22x10^3 J.(a) What is the change in the temperature of the gas?________ ? K(b) Find the change in its internal energy.________ ? J(c) Determine the change in pressure.________ ? Pa
Answer:
A) 140 k
b ) 5.22 *10^3 J
c) 2910 Pa
Explanation:
Volume of Monatomic ideal gas = 1.20 m^3
heat added ( Q ) = 5.22*10^3 J
number of moles (n) = 3
A ) calculate the change in temp of the gas
since the volume of gas is constant no work is said to be done
heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT
make ΔT subject of the equation
ΔT = Q / n.(3/2).R
= (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )
= 140 K
B) Calculate the change in its internal energy
ΔU = Q this is because no work is done
therefore the change in internal energy = 5.22 * 10^3 J
C ) calculate the change in pressure
applying ideal gas equation
P = nRT/V
therefore ; Δ P = ( n*R*ΔT/V )
= ( 3 * 8.3144 * 140 ) / 1.20
= 2910 Pa
A) The change in the temperature of the gas is; ΔT = 139.5 K
B) The change in internal energy of the gas is; ΔU = 5.22 × 10³ J
C) The change in pressure of the gas is; ΔP = 2899.5 Pa
We are given;
Volume of Monatomic ideal gas; V = 1.2 m³
Amount of heat added; Q = 5.22 × 10³ J
number of moles; n = 3
A) To calculate the change in temperature of the monatomic idea gas, we will use formula;
Q = ³/₂nRΔT
Where R is a constant = 8.314 J/mol.K
ΔT is the change in temperature
Making ΔT the subject of the formula;
ΔT = ²/₃(Q/(nR))
ΔT = ²/₃(5.22 × 10³)/(3 × 8.314)
ΔT = 139.5 K
B) Due to the fact that no work was done, then from first law of thermodynamics, we can say that;
ΔU = Q
Thus;
change in internal energy; ΔU = 5.22 × 10³ J
C) The change in pressure will be calculated from the formula;
ΔP = (n*R*ΔT)/V
ΔP = (3 * 8.314 * 139.5)/1.2
ΔP = 2899.5 Pa
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A box of books weighing 319 N is shoved across the floor by a force of 485 N exerted downward at an angle of 35 degres below the horizontal.a) If the coeficent of friction between the box and the floor is 0.57, how long does it take to move the box 4 meters, starting from rest?b) If If the coeficent of friction between the box and the floor is 0.75, how long does it take to move the box 4 meters, starting from rest?
Let w, n, p, and f denote the magnitudes of the 4 forces acting on the box.
• w = weight = 319 N
• n = normal force
• p = pushing force = 485 N
• f = friction = µ n, where µ is the coefficient of static friction
The net force on the box points in the direction that the box moves, which is to the right. In particular, this means the box is vertically in equilibrium. Split up the vectors into their vertical and horizontal components, and apply Newton's second law. (I take up and right to be the positive vertical and horizontal directions, respectively.)
• vertical:
p sin(-35°) + n - w = 0
and solving for n,
- (485 N) sin(35°) + n - 319 N = 0
n ≈ 597 N
• horizontal:
p cos(-35°) - f = m a
where a is the magnitude of the net acceleration on the box. Solve for a. Since f = µ n and m = w / g (where g = 9.80 m/s² is the mag. of the acc. due to gravity) we get
p cos(35°) - µ n = (w / g) a
(485 N) cos(35°) - µ (597 N) = (319 N) / (9.80 m/s²) a
a ≈ (12.2 - 18.3 µ) m/s²
(a) If µ = 0.57, then the net acceleration on the box is
a ≈ (12.2 - 18.3 • 0.57) m/s² ≈ 1.75 m/s²
so that the time t required to move the box 4 m is
4 m = 1/2 a t ²
t ≈ √((8 m) / (1.75 m/s²))
t ≈ 2.14 s
(b) The box does not move.
If µ = 0.75, then
a = (12.2 - 18.3 • 0.75) m/s² ≈ -1.55 m/s²
but a negative acc. here means the applied acc. points *opposite* the direction of movement, thus making the box move backward which doesn't make sense. The coefficient of friction is too large for the given applied force to get the box moving. With µ = 0.75, the frictional force to overcome has mag. f ≈ 448 N. But the given push contributes a horizontal force of (485 N) cos(-35°) ≈ 397 N. This mag. needs to be increased in order to get the box moving.
(a) The time taken to move the box 4 meters is 2.14 s.
(b) The box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.
The given parameters;
weight of the book, W = 319 Napplied force, F = 485 Nangle of inclination, θ = 35 ⁰The mass of the books is calculated as;
[tex]m = \frac{W}{g} \\\\m = \frac{319}{9.8} \\\\m = 32.55 \ kg[/tex]
The normal force on the box is calculated as follows;
[tex]F_n = -W - Fsin\theta\\\\F_n = -319 - (485\times sin35)\\\\F_n = -597.18 \ N[/tex]
The frictional force when the coefficient of friction is 0.57;
[tex]F_f = \mu F_n\\\\F_f = 0.57 \times -597.18\\\\F_f = -340.39 \ N[/tex]
The acceleration of the box is calculated as follows;
[tex]F cos \theta - F_f = ma\\\\(485)cos(35) \ - 340.39 = 32.55 a\\\\56.899 = 32.55a\\\\a = \frac{56.899}{32.55} \\\\a = 1.75 \ m/s^2[/tex]
The time taken to move the box 4 meters is calculated as;
[tex]s = v_0t + \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} at^2\\\\t = \sqrt{\frac{2s}{a} } \\\\t = \sqrt{\frac{2\times 4}{1.75} } \\\\t = 2.14 \ s[/tex]
(b) The frictional force when the coefficient of friction is 0.75;
[tex]F_f = \mu F_n\\\\F_f = 0.75 \times -597.18\\\\F_f = -447.885 \ N[/tex]
The acceleration of the box is calculated as follows;
[tex]F cos \theta - F_f = ma\\\\(485)cos(35) \ -447.885 = 32.55 a\\\\-50.596 = 32.55a\\\\a = \frac{-50.596}{32.55} \\\\a = -1.55\ m/s^2[/tex]
Thus, the box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.
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In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give? Let's find out by doing a calculation in a very artificial situation. Suppose a man standing on frictionless ice fires a rifle horizontally. The mass of the man together with the rifle is 70 kg, and the mass of the bullet is 10 g. If the bullet leaves the muzzle at a speed of 500 m/s, what is the final speed of the man?
Answer:
Explanation:
m1v1=m2v2
m1=70 kg
m2=10 g=0.01 kg
v2=500 m/s
m1v1=m2v2
v1=m2v2/m1
v1=0.01*500/70
v1=0.07
A 715 kg car stopped at an intersection is rear-ended by a 1490 kg truck moving with a speed of 12.5 m/s. If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of both vehicles after the collision.
Answer:
The final velocity of the car is 16.893 m/s
The final velocity of the truck is 4.393 m/s
Explanation:
Given;
mass of the car, m₁ = 715 kg
mass of the truck, m₂ = 1490 kg
initial velocity of the car, u₁ = 0
initial velocity of the truck, u₂ = 12.5 m/s
let the final velocity of the car, = v₁
let the final velocity of the truck, = v₂
Apply the principle of conservation of linear momentum for elastic collision;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(715 x 0) + (1490 x 12.5) = 715v₁ + 1490v₂
18625 = 715v₁ + 1490v₂ -----equation (1)
Apply one-directional velocity formula;
u₁ + v₁ = u₂ + v₂
0 + v₁ = 12.5 + v₂
v₁ = 12.5 + v₂
Substitute v₁ into equation (1)
18625 = 715(12.5 + v₂) + 1490v₂
18625 =8937.5 + 715v₂ + 1490v₂
18625 - 8937.5 = 715v₂ + 1490v₂
9687.5 = 2205v₂
v₂ = 9687.5 / 2205
v₂ = 4.393 m/s
solve for v₁
v₁ = 12.5 + v₂
v₁ = 12.5 + 4.393
v₁ = 16.893 m/s
A 3520 kg truck moving north makes an INELASTIC collision with an 1480 kg car moving 13.0 m/s east. After colliding, they have a velocity of 9.80 m/s at 66.9 degrees. What was the initial velocity of the truck? (m/s)
Answer:
v = 12.8 m/s
Explanation:
Assuming no external forces acting during the collision, total momentum must be conserved.Since momentum is a vector, their components must be conserved too.Choosing a pair of axes coincident with the N-S and W-E directions, naming x to the W-E axis and y to the N-S one, we can write the following algebraic equations:[tex]p_{ox} = p_{fx} (1)[/tex]
[tex]p_{oy} = p_{fy} (2)[/tex]
Since we know all the information needed to solve (1), assuming a completely inelastic collision, we can focus in (2), writing both sides of the equation as follows:[tex]p_{oy} = m_{t} * v_{ot} = 3520 kg* v_{ot} (3)[/tex]
[tex]p_{fy} = m_{f} * v_{fy} = 5000 kg* 9.8 m/s * sin 66.9 = 45080 kg*m/s (4)[/tex]
Since (4) and (3) are equal each other, we can solve for vot, as follows:[tex]v_{ot} =\frac{45080kg*m/s}{3520kg} = 12.8 m/s (5)[/tex]
Need help ASAP..please help
Answer:
option 3
Explanation:
can i get brainliest
True or False when an object speeds up it gains momentum
Answer: True
Explanation:
Power is the rate at which work is done true or false
Answer:
false
Explanation:
How should the magnetic field lines be drawn for the magnets shown below?
Answer:
Magnetic field lines can be drawn by moving a small compass from point to point around a magnet. At each point, draw a short line in the direction of the compass needle.When opposite poles of two magnets are brought together, the magnetic field lines join together and become denser between the poles.
Explanation:
What would happen if there is more male hyenas than female hyenas in a population?
Choices:
Male hyenas will compete to mate with the females.
Some male hyenas will die.
Male hyenas for wait for more females to join the population.
Answer:
Option 1
Explanation:
I always see animals do that
How much work is done by the gravitational force on the block?
Answer:
Work = Mass * Gravity * Height and is measured in Joules. Imagine you find a 2 -Kg book on the floor and lift it 0.75 meters and put it on a table. Remember, that “force” is simply a push or a pull. If you lift 100 kg of mass 1-meter, you will have done 980 Joules of work.
Explanation:
4. A substance has a density of 0.79 g/cm'. It is soluble in water. List all the possibilities of what it might be How could you determine the actual identity?
Answer:
See explanation
Explanation:
Given that the density of the unknown substance is 0.79 g/cm3 and is soluble in water, the possible substances it could be are;
i) t-butanol
ii) ethanol
iii) 2-propanol
iv) acetone
However, the actual identity of the unknown substance can be obtained by carrying out a boiling point test. The four substances listed above have different boiling points. Hence the boiling point of the unknown substance ultimately discloses its identity.
A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. How far away is the cliff? The speed of sound in air a ordinary temperature is 343 m/s.
Answer:
480.2 m
Explanation:
The following data were obtained from the question:
Speed of sound (v) = 343 m/s.
Time (t) = 2.8 s
Distance (x) of the cliff =?
The distance of the cliff from the woman can be obtained as follow:
v = 2x /t
343 = 2x /2.8
Cross multiply
2x = 343 × 2.8
2x = 960.4
Divide both side by the coefficient of x i.e 2
x = 960.4/2
x = 480.2 m
Therefore, the cliff is 480.2 m away from the woman.
The distance should be 480.2 m
The calculation is as follows:Since A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. And, there is the velocity of 343 m/s
[tex]v = 2x \div t\\\\343 = 2x \div 2.8\\\\2x = 343 \times 2.8[/tex]
2x = 960.4
x = 480.2 m
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A bug crawls 2.25 m along the base of a wall. Upon reaching a corner, the bugs direction of travel changes from south to west. THe bug that crawls 3.15 m before stopping. What is the magnitude of the bugs displacment?A) 5.40 m.B) 2.72m.C) 3.45 m.D) 3.87 in.E) 4.29 m.
Answer:
The magnitude of the bugs displacement is 3.87 m
Explanation:
An illustrative diagram for the scenario is given in the attachment below.
In the diagram, the bug's displacement is given by x. The diagram shows a right angle triangle with x as the hypotenuse. We can determine x from the Pythagorean theorem which states that " the square of the hypotenuse equals sum of squares of the other two sides". That is
x² = 2.25² + 3.15²
x² = 5.0625 + 9.9225
x² = 14.985
x = √14.985
x = 3.87 m
Hence, the magnitude of the bugs displacement is 3.87 m.
Dolphins rely on echolocation to be able to survive in the ocean. In a 20 °C ocean, a dolphin produces an ultrasonic sound with a frequency of 125 kHz. What is the wavelength of this sound, in meters?While remaining stationary, the dolphin emits a sound pulse and receives an echo after 0.220 s. How far away, in meters, is the reflecting object from the dolphin?
Answer:
wavelength = 0.01 m
distance = 162.8 m
Explanation:
Given that;
Speed of sound in water = 1,480 meters per second
Frequency of ultrasound = 125KHZ
From=
v=λf
v= speed of sound
λ= wavelength of sound
f= frequency of sound
λ= 1,480 ms-1/125 * 10^3 Hz
λ= 0.01 m
From
v = 2x/t
where;
v= velocity of sound in water
x= distance traveled
t = time taken
x = vt/2
x = 1,480 ms-1 * 0.220 s/2
x= 162.8 m
How many significant figures are in 0.0067?
Answer:
2
Explanation:
there are 2 significant figures in there
Measurements of the radioactivity of a certain isotope tell you that the decay rate decreases from 8255 decays per minute to 3110 decays per minute over a period of 4.50 days.
What is the half-life (T1/2) of this isotope?
Answer:half-life (T1/2) of this isotope =
Explanation:
The number of nuclei of any radioactive substance at a given time is expressed by
Nt = N0e⁻kt
Nt=decay of material at a time t, =3110 decays per minute
N=decays at t=0, 8255 decays per minute
k=constant
Nt=N0e−kt
3110= 8255 e⁻k(4.50)
3110/ 8255=e−k(4.50)
0.3767 = e−k(4.50)
In 0.3767 = -k (4.50)
0.976=-4.5k
k=0.976/4.5
=0.2159
Also we know that t 1/2= time that it takes half the original material to decay.it is related to the rate constant by
T₁/₂=ln 2 / k
Therefore half-life (T1/2) of this isotope
T₁/₂=ln 2/0.2159
T₁/₂=3.12 days