Energy = (power) x (time)
-- For the toaster:
Power = 1.4 kW = 1,400 watts
Time = 5.4 minutes = 324 seconds
Energy = (1,400 W) x (324 s) = 453,600 Joules
-- For the CFL bulb:
Power = 11 watts
Time = 10.5 hours = 37,800 seconds
Energy = (11 W) x (37,800 s) = 415,800 Joules
-- The toaster uses energy at 127 times the rate of the CFL bulb.
-- The CFL bulb uses energy at 0.0079 times the rate of the toaster.
-- The toaster is used for 0.0086 times as long as the CFL bulb.
-- The CFL bulb is used for 116.7 times as long as the toaster.
-- The toaster uses 9.1% more energy than the CFL bulb.
-- The CFL bulb uses 8.3% less energy than the toaster.
Having aced your Physics 2111 class, you get a sweet summer-job working in the International Space Station. Your room-mate, Cosmonaut Valdimir tosses a banana at you at a speed of 16 m/s. At exactly the same instant, you fling a scoop of ice cream at Valdimir along exactly the same path. The collision between banana and ice cream produces a banana split 8.2 m from your location 1.4 s after the banana and ice cream were launched.
1. How fast did you toss the ice cream?
2. How far were you from Valdimir when you tossed the ice cream?
Answer:
a
The speed is [tex]s = 5.857 m/s[/tex]
b
The distance is [tex]D = 22.4 \ m[/tex]
Explanation:
From the question we are told that
The speed of the banana is [tex]v = 16 \ m/s[/tex]
The distance from my location is [tex]d = 8.2 \ m[/tex]
The time taken is [tex]t = 1.4 \ s[/tex]
The speed of the ice cream is
[tex]s = \frac{d}{t}[/tex]
substituting values
[tex]s = \frac{8.4}{1.4}[/tex]
[tex]s = 5.857 m/s[/tex]
The distance of separation between i and Valdimir is the same as the distance covered by the banana
So
[tex]D = v * t[/tex]
substituting values
[tex]D = 16 * 1.4[/tex]
[tex]D = 22.4 \ m[/tex]
A student has derived the following nondimensionally homogeneous equation: a=x/t2-vt+F/m where v is a velocity's magnitude , a is an acceleration's magnitude, t is a time, m is a mass, F is a force's magnitude , and x is a distance (or length). Which terms are dimensionally homogeneous? .
a) x/t
b) vt
c) a
d) F/m
Answer:
Letter C) and D) is the correct answer.
Explanation:
We know that the a is an acceleration's magnitude, so the units of a are m/s².
Now, let's analyze each terms. If we want that each term will be dimensionally homogeneous, all of them must have the same units of a.
[tex][\frac{x}{t}]=[\frac{m}{s}][/tex]
[tex][vt]=[m][/tex]
[tex][\frac{F}{m}]=[\frac{N}{kg}]=[kg\frac{m}{s^{2}kg}]=[\frac{m}{s^{2}}][/tex]
Therefore, the term F/m is the correct answer.
I hope it helps you!
We can see that a and F/M are dimensionally homogeneous.
In solving dimensions, we try to express a quantity in terms of the fundamental quantities;
MassLengthTimeFor the term a, its dimension is LT^-2
For the term F/m, its dimension is LT^-2
Hence, it follows that a and F/M are dimensionally homogeneous.
Learn more about dimensions: https://brainly.com/question/944206
250cm3 of fres
er of density 1000kgm-3 is mixed with 100cm3 of sea water of density 1030kgm-3. Calculate the density of the mixture. *
Answer:
1008.57kg/m3
Explanation:
Now the mass of fresh water is 250×1000 /1000000 = 0.25kg
Now the mass of salt water is
100×1030 /1000000 = 0.103kg
Note Density = mass / volume
Mass = volume × density
Note that converting from cm3 to m3 we divide by 1000000
Total mass = 0.25kg +0.103kg= 0.353kg.
Total volume also is (250 +100 )/1000000= 35 × 10^{-5}m3
Hence the density of the mixture= total mass / total volume
0.353kg/35 × 10^{-5}m3=1008.57kg/m3
What is the speed at which a spaceship shoots up from earth ?
Answer:
Once at a steady cruising speed of about 16,150mph (26,000kph
Explanation:
The temperature at the surface of the Sun is approximately 5,300 K, and the temperature at the surface of the Earth is approximately 293 K. What entropy change of the Universe occurs when 6.00 103 J of energy is transferred by radiation from the Sun to the Earth?
Answer:
The entropy change of the Universe that occurs is 19.346 J/K
Explanation:
Given;
temperature of the sun, [tex]T_s[/tex] = 5,300 K
temperature of the Earth, [tex]T_E[/tex] = 293 K
radiation energy transferred by the sun to the earth, E = 6000 J
The sun loses Q of heat and therefore decreases its entropy by the amount
[tex]\delta S_{sun} = \frac{-Q}{T_s}[/tex]
The earth gains Q of heat and therefore increases its entropy by the amount
[tex]\delta S_{Earth} = \frac{-Q}{T_E}[/tex]
The total entropy change is:
[tex]\delta S_{Earth} + \delta S_{sun} = \frac{Q}{T_E} -\frac{Q}{T_S} \\\\ = Q(\frac{1}{T_E} -\frac{1}{T_S} )\\\\= 6000(\frac{1}{293} -\frac{1}{5300} )\\\\=6000(0.0032243)\\\\= 19.346 \ J/K[/tex]
Therefore, the entropy change of the Universe that occurs is 19.346 J/K
How much work is done by 0.30 m of gas if its pressure increases by 8.0 x105 Pa and the volume remains constant
Salerno
Answer:
0
Explanation:
if the volume remains constans, the works is 0 because the equation
W = P . ∆V
P = pressure
∆V = change in volume
A mechanic applies a force of 60N at a distance of 80 cm from the pivot on a wheel wrench. What is the size of the moment?
Answer:
48 Nm
Explanation:
Moment, or torque, is the cross product of radius and force vectors.
τ = r × F
τ = (0.80 m) (60 N)
τ = 48 Nm
help yall 13 points!!
Answer:
Explanation:
12.)
A. Opposite poles attract
B. Same poles repel
13.)
IDK
If radio waves are used to communicate with an alien spaceship approaching Earth at 10% of the speed of light c, the aliens would receive our signals at speed of:_______.
a. 0.99c
b. 1.10c
c. 1.00c
d. 0.90c
e. 0.10c
Answer:
3×10^7 m/s or 0.10c (e)
Explanation: If the actual value of the speed of light were to be put into consideration.
Given that the speed of light is c = 3.0×10^8m/s
The alien spaceship is approaching at the rate of 10% of the speed of light.
10% of 3.0×10^8m/s
10/100 × 3.0×10^8m/s
0.1 ×3.0×10^8m/s
3×10^7 m/s. Which is the same thing as 0.1 of c = 0.1×c
Answer: 1.00c
Explanation: I got it correct on the homework
A uniform ladder stands on a rough floor and rests against a frictionless wall. Since the floor is rough, it exerts both a normal force N1 and a frictional force f1 on the ladder. However, since the wall is frictionless, it exerts only a normal force N2 on the ladder. The ladder has a length of L = 4.6m, a weight of WL= 69.0N , and rests against the wall a distance d = 3.75 m above the floor. If a person with a mass of m = 90 kg is standing on the ladder, determine the forces exerted on the ladder when the person is halfway up the ladder.
Required:
Solve of N1, N2 and f1
Answer:
The normal force N1 exerted by the floor is [tex]N_1 = 951 \ N[/tex]
The normal force N2 exerted by the wall is [tex]N_2= 616.43 \ N[/tex]
The frictional force exerted by the wall is [tex]f = N_2 = 616.43 \ N[/tex]
Explanation:
From the question we are told that
The length of the ladder is [tex]L = 4.6 \ m[/tex]
The weight of the ladder is
The distance of the ladder position on the wall from the floor is [tex]D = 3.75 \ m[/tex]
The mass of the person is [tex]m = 90 kg[/tex]
Applying Pythagoras theorem
The length of the position the ladder on the ground from the base of the wall is
[tex]A = \sqrt{L^ 2 - D^2}[/tex]
substituting values
[tex]A = \sqrt{(4.6^2)-(3.75^2)}[/tex]
[tex]A = 2.66 \ m[/tex]
In order the for the ladder not to shift from the ground the sum of the moment about the position of the ladder on the ground must be equal to zero this is mathematically represented as
[tex]\sum M = 0 = N_2 * D - [\frac{1}{2} * W_L ] * [(mg) *A ][/tex]
[tex]\sum M = 0 = N_2 * 3.75 - [\frac{1}{2} * 69.0 ] * [(90*9.8) * \frac{4.6}{2.66} ][/tex]
[tex]N_2 * 3.75 =2311.62[/tex]
[tex]N_2 * 3.75 =2311.62[/tex]
[tex]N_2= 616.43 \ N[/tex]
Now the force exerted by the floor on the ladder is mathematically represented as
[tex]N_1 = W_L + (m * g )[/tex]
substituting values
[tex]N_1 = 951 \ N[/tex]
Now the horizontal forces acting on the ladder are [tex]N_2 \ and \ f[/tex] and they are in opposite direction so
[tex]f = N_2 = 616.43 \ N[/tex]
The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+vit+12at2 where xf is the final position, xi is the initial position, vi is the initial velocity, a is the acceleration, and t is the time. Let's say a car starts with an initial speed of 15 m/s, and moves between the 1000 m and 5000 m marks on a roadway in a time of 60 s. What is its acceleration?
Answer:
a = 1.72 m/s²
Explanation:
The given kinematic equation is the 2nd equation of motion. The equation is as follows:
xf = xi + (Vi)(t) + (1/2)(a)t²
where,
xf = the final position = 5000 m
xi = the initial position = 1000 m
Vi = the initial velocity = 15 m/s
t = the time taken = 60 s
a = acceleration = ?
Therefore,
5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²
5000 m = 1000 m + 900 m + a(1800 s²)
5000 m = 1900 m + a(1800 s²)
5000 m - 1900 m = a(1800 s²)
a(1800 s²) = 3100 m
a = 3100 m/1800 s²
a = 1.72 m/s²
Sara walks part way around a swimming pool. She walks 50 yards north, then
20 yards east, then 50 yards south. The magnitude of her total displacement
during this walk is
yards.
Answer:
20 Yards
Explanation:
|---20----|
| |
| 50 |50
|---D--->|
Start End
Total displacement(D) 20 yards (East).
A quartz sphere is 14.0 cm in diameter. What will be its change in volume if its temperature is increased by 305°F? The coefficient of volume expansion of quartz is 1.50×10^6/°C. Answer in cm^3 .
Answer:
0.365 cm³
Explanation:
The change in volume is found by multiplying the coefficient of expansion by the volume and the temperature change. The temperature change is in °F, but the expansion coefficient is per °C, so we need to convert the temperature scale in the computation.
ΔV = V·Ce·ΔT
= (π/6·d³)(1.5×10⁻⁶/°C)((5 °C)/(9 °F))(305 °F)
= (1436.76 cm³)(1.5×10⁻⁶/°C)(169.44 °C)
= 0.365 cm³ . . . . increase in volume
2. If rain is falling vertically downward, and you are running for shelter, should you hold your umbrella
vertically, tilted forward, or tilted backward to keep the driest? Please explain.
Answer:
Tilted forward to keep the driest.
Explanation:
The rain is falling vertically so there is no wind. In these circumstances the umbrella should be tilted vertically forward.
The situation is the same as if you would stand still and the rain would come under an angle from the front.
the distance between 2 station is 5400 m find the time taken by a train to cover this distance, if the train travels with speed 60m/s
Answer:
I dont know bro
Explanation:
Ask an expert
Answer:
Time=90s
Explanation:
Speed=distance /time
[tex]60 = \frac{5400}{t} where \: t \: is \: time \\60t = 5400 \\ t = \frac{5400}{60} \\ t =90 \\ hope \: this \: helps..good \: luck [/tex]
What's a line of best fit? Will give BRAINLIEST
A line of best fit expresses the relationship between the points.
Explanation:
It does not go through all the points but goes through most of them and it is like a hardrawn curve
1. For each of the following scenarios, describe the force providing the centripetal force for the motion: a. a car making a turn b. a child swinging around a pole c. a person sitting on a bench facing the center of a carousel d. a rock swinging on a string e. the Earth orbiting the Sun.
Complete Question
For each of the following scenarios, describe the force providing the centripetal force for the motion:
a. a car making a turn
b. a child swinging around a pole
c. a person sitting on a bench facing the center of a carousel
d. a rock swinging on a string
e. the Earth orbiting the Sun.
Answer:
Considering a
The force providing the centripetal force is the frictional force on the tires \
i.e [tex]\mu mg = \frac{mv^2}{r}[/tex]
where [tex]\mu[/tex] is the coefficient of static friction
Considering b
The force providing the centripetal force is the force experienced by the boys hand on the pole
Considering c
The force providing the centripetal force is the normal from the bench due to the boys weight
Considering d
The force providing the centripetal force is the tension on the string
Considering e
The force providing the centripetal force is the force of gravity between the earth and the sun
Explanation:
Someone plzzz helpppppp with this last question
Answer:
I dont know someone deleted answers. But they were wrong. INERTIA IS CORRECT I DID THIS IN MY SCHOOL
C IS CORRECT
Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy of the surrounding electrons). The following masses are given:
5727Co: 56.936296u
5726Fe: 56.935399u
Express your answer in millions of electron volts (1u=931.5MeV/c2) to three significant figures.
A negligible amount of this energy goes to the resulting 5726Fe atom as kinetic energy. About 90 percent of the time, after the electron-capture process, the 5726Fe nucleus emits two successive gamma-ray photons of energies 0.140MeV and 1.70 102MeV in decaying to its ground state. The electron-capture process itself emits a massless neutrino, which also carries off kinetic energy. What is the energy of the neutrino emitted in this case?
Express your answer in millions of electron volts.
Answer:
Explanation:
⁵⁷Co₂₇ + e⁻¹ = ²⁷Fe₂₆
mass defect = 56.936296 + .00055 - 56.935399
= .001447 u
equivalent energy
= 931.5 x .001447 MeV
= 1.3479 MeV .
= 1.35 MeV
energy of gamma ray photons = .14 + .017
= .157 MeV .
Rest of the energy goes to neutrino .
energy going to neutrino .
= 1.35 - .157
= 1.193 MeV.
A 1.0 m string with a 5 g stopper on the end is whirled in a vertical circle. The speed of the stopper is 8 m/s at the top of the circle. (A) What is the speed of the stopper at the bottom of the circle? (HINT: Use energy conservation principles!) (10.2 m/s) (B) What is the tension in the string when the stopper is at the top of the circle? (0.27 N) (C) What is the tension in the string when the stopper is at the bottom of the circle?
Answer:
Explanation:
A )
At the bottom of the circle , the potential energy of the stopper is converted into kinetic energy
1/2 m V² = mg x 2r + 1/2 mv²
m is mass of stopper , V is velocity at the bottom , r is radius of the circular path which is length of the string , v is velocity at the top
1/2 V² = g x 2r + 1/2 v²
V² = g x 4r + v²
V² = 9.8 x 4 + 8²
V² = 103.2
V = 10.16 m/s
B )
If T be the tension at the top
Net downward force
= mg + T . This force provides centripetal force for the circular motion
mg +T = mv² / r
T = mv²/r -mg
= m ( v²/r - g )
= .005 ( 8²/1 -g )
= .005 x 54.2
= .27 N .
C ) At the bottom
Net force = T - mg , T is tension at the bottom , V is velocity at bottom
T-mg = mV²/r
T = m ( V²/r +g )
= .005 ( 10.16²/1 +9.8)
= .005 x 113
= .56 N .
Astronaut Flo wishes to travel to a star 20 light years away and return. Her husband Malcolm, who was the same age as Flo when she departs, stays home (baking cookies). If Flo travels at a constand speed of 80% of the speed of light (except for a short time to turn around), how much younger than Malcolm will Flo be when she returns? How long does Malcolm sit around baking cookies? How far is the distance to Flo?
Answer:
a. about 20 years younger
b. Malcolm sits around for 49.94 years
c. 2.268x[tex]10^{17}[/tex] m
Explanation:
light travels 3x[tex]10^{8}[/tex] m in one seconds
in 20 years that will be 3x[tex]10^{8}[/tex] x 20 x 60 x 60 x 24 x 365 = 1.89x[tex]10^{17}[/tex] m
for the to and fro journey, total distance covered will be 2 x 1.89x[tex]10^{17}[/tex] = 3.78x[tex]10^{17}[/tex] m
Flo's speed = 80% of speed of light = 0.8 x 3x[tex]10^{8}[/tex] = 2.4x[tex]10^{8}[/tex] m/s
time that will pass for Malcolm will be distance/speed = 3.78x[tex]10^{17}[/tex] /2.4x[tex]10^{8}[/tex]
= 1575000000 s = 49.94 years
the relativistic time t' will be
t' = t x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]
t' = 49.94 x [tex]\sqrt{1 - 0.8^{2} }[/tex]
t' = 49.94 x 0.6 = 29.96 years this is the time that has passed for Flo
this means that Flo will be about 20 years younger than Malcolm when she returns
relativistic distance is
d' = d x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]
d' = 3.78x[tex]10^{17}[/tex] x [tex]\sqrt{1 - 0.8^{2} }[/tex]
d' = 3.78x[tex]10^{17}[/tex] x 0.6
d' = 2.268x[tex]10^{17}[/tex] m this is how far it is to Flo
Suppose your hair grows at the rate of 1/55 inch per day. Find the rate at which it grows in nanometers per second. Because the distance between atoms in a molecule is on the order of 0.1 nm, your answer suggests how rapidly atoms are assembled in this protein synthesis.
Answer:5.35nm
Explanation:
Consider that 1 inch is = 0.0254m
we have,
1m= 1x10^9 nm
While:
0.0254m = 2.54x10^7nm
1/55 (2.54x10^7) = 4.6181 x 10^5nm
1 day= 24 hrs
= (24x60) when calculating in min
= (24x60x60) calculating in seconds we have:
= 8.64x10⁴sec
In 8.64x10^4 seconds, the hair grows by 4.6181 x 10^5nm
Therefore, the amount by which the hair grows in 1 second will be;
= (4.6181 x 10^5)/(8.64x10^4)
= 5.35nm
The rate of growth will be 5.35nm
When the distance between a point source of light and a light meter is reduced from 6.0m to 2.0 m, the intensity of illumination at the meter will be the original value multiplied by _____.
Answer:
Explanation:
Let the point source have power P .
At distance r , the intensity I
I = P / 4πr² . If intensity at 6 m and 2 m be I₁ and I₂
I₁ = P / 4π x 6²
I₂ = P / 4π x 2²
I₁ / I₂ = 2² / 6²
= 1 / 9
I₂ = 9 I₁
Intensity will be 9 times that at 6 m .
Unit conversion
The choices are in units
A,GA,MA,uA,kA,mA,nA,pA. Pick one the units
Answer:
1.234567 kA
Explanation:
The prefix k stands for kilo-, or 10³. The prefix m stands for milli-, or 10⁻³. The sum shown is ...
1.234 kA + 0.000567 kA = 1.234567 kA
An electric point charge of Q = 22.5 nC is placed at the center of a cube with a side length of a = 16.3 cm. The cube in this question is only a mathematical object, it is not made out of any physical material. What is the electric flux through all six sides of the cube?
Answer:
The electric flux is [tex]\phi = 2.5 *10^{3} \ Nm^2 \cdot C^{-1}[/tex]
Explanation:
From the question we are told that
The magnitude of the electric point charge [tex]q = 22.5 nC = 22.5 *10^{-9} \ C[/tex]
The length of the one side of the cube is [tex]l = 16.3 \ cm = 0.163 \ m[/tex]
The number of sides is [tex]N= 6[/tex]
The electric flux according to Gauss law is mathematically evaluated as
[tex]\phi = \frac{q}{\epsilon_o}[/tex]
Where [tex]\epsilon _ o[/tex] is the permitivity of free space with value [tex]\epsilon_o = 8.85*10^{-12}\ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
substituting values
[tex]\phi = \frac{22.5 *10^{-9}}{8.85 *10^{-12}}[/tex]
[tex]\phi = 2.5 *10^{3} \ Nm^2 \cdot C^{-1}[/tex]
The potential (relative to infinity) at the midpoint of a square is 3.0 V when a point charge of Q is located at one of the corners of the square. What is the potential (relative to infinity) at the center when each of the other corners is also contains a point charge of Q
Answer:
12.0 V
Explanation:
Data :
Potential difference due to a single charge (+Q), E = 3.0 V
The Electric potential for the system of charges is given as:
[tex]E=\frac{1}{4\pi \epsilon_o}[\Sigma\frac{Q}{r}][/tex]
for single charge, E = 3.0 V = [tex]\frac{1}{4\pi \epsilon_o}[\frac{Q}{r}][/tex] ->eq(1)
And for 4 charges:
[tex]E=\frac{1}{4\pi \epsilon_o}[4\frac{Q}{r}][/tex] -eq(2)
from eq(1) and (2) we have
E = 4 × 3.0 V = 12 V
The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at ,2=17.1 m/s when the eagle swoops down, grabs the pigeon, and flies off. At the instant right before the attack, the eagle is flying toward the pigeon at an angle =52.7 ° below the horizontal and a speed of ,1=41.5 m/s.
Answer:
31.4 m/s
44.4°
Explanation:
Momentum is conserved in the horizontal direction:
pₓᵢ = pₓ
m vᵢ₂ + 2m vᵢ₁ cos θ = (m + 2m) vₓ
vᵢ₂ + 2 vᵢ₁ cos θ = 3 vₓ
17.1 m/s + 2 (41.5 m/s) (cos -52.7°) = 3 vₓ
vₓ = 22.5 m/s
Momentum is conserved in the vertical direction:
pᵧᵢ = pᵧ
2m vᵢ₁ sin θ = (m + 2m) vᵧ
2 vᵢ₁ sin θ = 3 vᵧ
2 (41.5 m/s) (sin -52.7°) = 3 vᵧ
vᵧ = -22.0 m/s
The speed is:
v = √(vₓ² + vᵧ²)
v = √((22.5 m/s)² + (-22.0 m/s)²)
v = 31.4 m/s
The direction is:
θ = atan(vᵧ / vₓ)
θ = atan(-22.0 m/s / 22.5 m/s)
θ = -44.4°
The speed of the eagle at that instant is 31.4 m/s while it moves off in the direction of 44.4°.
Since momentum is conserved horizontally;
17.1 m/s + 2 (41.5 m/s) (cos -52.7°) = 3 vx
vx = 17.1 m/s + 2 (41.5 m/s) (cos -52.7°)/3
vx = 22.5 m/s
Also, momentum is conserved vertically hence;
2 (41.5 m/s) (sin -52.7°) = 3 vy
vy = 2 (41.5 m/s) (sin -52.7°) /3
vy = -22.0 m/s
The effective speed therefore, is;
v = √((22.5 m/s)² + (-22.0 m/s)²)
v = 31.4 m/s
The direction of this effective speed is;
θ = tan-1(22.0 m/s / 22.5 m/s)
θ = 44.4°
Learn more: https://brainly.com/question/13322477
When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 73.0 kg man just before contact with the ground has a speed of 6.46 m/s. In a stiff-legged landing he comes to a halt in 2.07 ms. Calculate the average net force that acts on him during this time
Answer:
Explanation:
The man comes to halt due to reaction force acting on him in opposite direction . If R be the reaction force
impulse by net force = change in momentum
Net force = R - mg , mg is weight of the man .
( R-mg ) x 2. 07 x 10⁻³ = 73 x 6.46 - 0
R - mg = 227.81 x 10³
Average net force = 227.81 x 10³ N .
HELPP ?Air at a temperature of 27 C and 1 atm pressure in a 4 liter cylinder of a diesel engine There. By pushing the piston, the volume of air shrinks 16 times and the pressure increases 40 times. a) How many moles of air are in the cylinder. b) What is the final temperature of the air?
Answer:
a. 0.16240664737515434 moles
b. 67.5 degrees Celcius
Explanation:
a. Use Ideal Gas Equation
PV=nRT
Where P = pressure in pascals, V=Volume in cubic meters, n=number of moles, R is a constant=8.314 J/mol.K and T is temperature in Kelvin.
27C = 273+27=300Kelvin
volume 4L = 0.004m^3
Pressure = 1atm = 101325 Pascal
PV=nRT
101325Pa*0.004m^3=n*8.314J/mol.K*300K
Solving for n from the above you get n=0.16240664737515434 moles
b.Use combined gas law equation
P1*V1/T1=P2*V2/T2
P1= 1atm
V1=4L
T1=27C
P2= 4/16 L =0.25L
P=1*40 atm = 40atm
We do not know T2
USING THE FORMULA
(1atm*4L)/27C = (40atm*0.25L)/T2
(1*4)/27=(40*0.25)/T2
IF you simplify for T2, you get 67.5
Hence final temperature = 67.5 degrees Celcius
A certain freely falling object, released from rest, requires 1.85 s to travel the last 26.5 m before it hits the ground. (a) Find the velocity of the object when it is 26.5 m above the ground. (Indicate the direction with the sign of your answer. Let the positive direction be upward.) -2.70 Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (b) Find the total distance the object travels during the fall.
Answer:
a) -5.26 m/s
b) 27.91 m
Explanation:
a) The acceleration due to gravity makes the velocity increase in magnitude in a linear way. The average velocity over the interval will be equal to the actual velocity halfway through the interval. The velocity at the beginning of the interval will be higher (less negative) by the amount velocity changes in the first half of the interval.
average velocity = (0 -(26.5 m))/(1.85 s) ≈ -14.324 m/s
The change in velocity in the first half of the interval is ...
Δv = (Δt/2)×(-9.8 m/s²) = (1.85 s)(-4.9 m/s²) = -9.065 m/s
So, the initial velocity (at the beginning of the last 1.85 s interval) is ...
v1 = (average velocity) -Δv = (-14.324 m/s) -(-9.065 m/s)
v1 = -5.259 m/s
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b) The velocity when the object hits the ground is ...
v2 = average velocity +Δv = -14.324 m/s -9.065 m/s = -23.389 m/s
This is related to the distance traveled by ...
v² = 2dg . . . . . where g is the acceleration and d is the distance traveled
d = v²/(2g) = 23.389²/(2·9.8) = 27.911 . . . . meters
The object travels a total distance of about 27.911 meters.
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The attached graph shows height vs. time.
