Answer:
10,200 Cal. per day
Explanation:
The mouse consumes 3.0 Cal each day, and has a mass of 20 grams. We can use this data to obtain a ratio of energy consumption per mass
[tex]\frac{3.0 \ Cal}{20 g} = 0.15 \frac{Cal}{g}[/tex].
For the human, we need to convert the 68 kilograms to grams. We can do this with a conversion factor. We know that:
[tex]1 \ kg = 1000 \ g[/tex],
Now, we can divide by 1 kg on each side
[tex]\frac{1 \ kg}{1 \ kg} = \frac{1000 \ g}{1 \ kg}[/tex],
[tex] 1 = \frac{1000 \ g}{1 \ kg}[/tex].
Using this conversion factor, we can obtain the mass of the human in grams, instead of kilograms. First, lets take:
[tex]mass_{human} = 68 \ kg[/tex]
We can multiply this mass for the conversion factor, we are allowed to do this, cause the conversion factor equals 1, and its adimensional
[tex]mass_{human} = 68 \ kg * \frac{1000 \ g}{1 \ kg} [/tex]
[tex]mass_{human} = 68,000 g [/tex]
Now that we know the mass of the human on grams, we can multiply for our ratio of energy consumption
[tex]68,000 \ g * 0.15 \frac{Cal}{g} = 10,200 \ Cal[/tex]
So, we would need 10,200 Cal per day.
If a 68 kg human used the same energy per kg of body mass as a mouse, they would need approximately 10,200 Calories each day. This high energy requirement is due to the greater relative metabolic rates of smaller mammals, as they experience higher heat loss and require more energy to maintain body temperature.
Explanation:The question asks for the approximate energy a 68 kg human would need each day if they used the same energy per kg of body mass as a mouse. To find this, we first calculate the energy per gram of the mouse, which is 3.0 Calories/20g = 0.15 Cal/g.
Converting this to per kg, we get 0.15 Cal/g * 1000g/kg = 150 Cal/kg. Then, multiplying this by the human's mass gives us the daily energy requirement for the human: 150 Cal/kg * 68 kg = 10,200 Calories a day, a significantly higher amount than the Basal Metabolic Rate (BMR) of a typical human.
This high energy requirement can be attributed to the higher metabolic rate of smaller mammals. Smaller animals have a greater surface area to mass ratio, which contributes to higher heat loss and thus higher energy demands to maintain body temperature. This results in a higher BMR per body weight in smaller mammals, as demonstrated by the mouse in the question.
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A sound mixer is impressed by the new equipment that was just installed in his recording studio. He says that now he will be able to help singers and musicians produce better CDs because he can eliminate unneeded and undesired wavelengths. What term describes the characteristic of sounds waves that the sound mixer is now able to alter?
a) amplitude
b) volleying
c) frequency
d) saturation
e) wavelength
Answer:
Wavelnght
Explanation:
He says undesired wavelengths can be eliminated with the new sound mixer. Musical sounds are spread over a medium through waves of different characteristics, one of them is wavelenght, that is the distance between begining and the end of an oscilation, normally measured on small distance units such as milimeters or even less. Sound may be higher or deeper depending on its wavelenght, so this sound mixer is able to filter all the sound the user does not want to include in his work, for example, noises from enviroment. Now the producer can make music focused on high notes, like the ones from violins, or focused on low ones like the ones from a bass.
An athlete crosses a 25-m-wide river by swimming perpendicular to the water current at a speed of 0.5 m/s relative to the water. He reaches the opposite side at a distance 40 m downstream from his starting point.
1) How fast is the water in the river flowing with respect to the ground?
2) What is the speed of the swimmer with respect to a friend at rest on the ground?
Answer:
1) Vx=0.8 m/s
2) V=0.94339 m/s
Explanation:
We know that the speed to cross the river is 0.5 m/s. This is our y axis. Vy
And we don't know the speed of the river. This is in our x axis. Vx
Since we know that the shore is 25m away, and we have a 0.5m/s of speed.
We can find with y=v.t
Since we know y=25m and v=0.5m/s
We can find that 50 seconds is the time we take to cross the river.
Now we need to calculate the velocity of the river, for that we use the same equation, x=v.t where x is 40m at downstream, and t now we know is 50 seconds.
We can find that v of the river, or Vx is 0.8 m/s.
With the two components of the velocity we use this equation to calculate the module or the velocity of the swimmer respect a fix point on the ground.
[tex]V=\sqrt{(Vx)^{2}+(Vy)^{2} }[/tex]
We replace the values of Vx and Vy and we find. V=0.94339 m/s.
The river current is flowing at a speed of 0.8 m/s concerning the ground, and the swimmer's speed concerning a friend on the ground is approximately 0.94 m/s.
Explanation:An athlete swims across a 25-m-wide river with a speed of 0.5 m/s perpendicular to the water current. The swimmer is carried 40 m downstream, indicating the presence of a river current. To find the speed of the river current, we use the Pythagorean theorem since the swimmer's motion relative to the river and the river's motion relative to the ground are perpendicular to each other.
The time taken to cross the river is the width of the river divided by the swimmer's speed concerning the water:
Time = 25 m / 0.5 m/s = 50 s. Now, the speed of the river current can be calculated using the downstream distance covered (40 m) and time (50 s): Speed of river = Distance downstream / Time = 40 m / 50 s = 0.8 m/s.
To calculate the swimmer's speed relative to the ground, we consider both the swimmer's speed across the river and the river's current speed. Using the Pythagorean theorem:
Swimmer's speed relative to the ground = √((0.5 m/s)^2 + (0.8 m/s)^2) = √(0.25 + 0.64) m/s = √(0.89) m/s ≈ 0.94 m/s.
An electron has an initial speed of 4.04 × 105 m/s. If it undergoes an acceleration of 3.3 × 1014 m/s 2 , how long will it take to reach a speed of 7.07 × 105 m/s?
Using the equation of motion, it is calculated that the electron will take approximately 9.18 x 10^-12 seconds to reach the specified speed.
Explanation:The question relates to the fundamental physics concept of motion, where the final speed of an object can be calculated from its initial speed, the acceleration it undergoes, and the time it accelerates. Specifically, we can use the formula v = u + at, where v is the final speed, u is the initial speed, a is the acceleration, and t is the time.
Here, the final speed v of the electron is 7.07 × 105 m/s, its initial speed u is 4.04 × 105 m/s, and its acceleration a is 3.3 × 1014 m/s2. We can rearrange the formula to solve for time t, which gives t = (v - u) / a.
Substituting the given values into this formula, we get t = (7.07 × 105 - 4.04 × 105) / 3.3 × 1014 = 9.18 x 10-12 seconds. Therefore, it will take the electron approximately 9.18 x 10-12 seconds to reach a speed of 7.07 × 105 m/s.
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What number of moles of O2 are needed to produce 14.2 grams of P4O10 from P? (Molecular weight P4O10 = 284) (A) 0.0500 mole (B) 0.0625 mole (C) 0.125 mole (D) 0.250 mole (E) 0.500 mole
Answer:
(D) 0.250 mole
Explanation:
From molecular weight we know that 1 mole of P4O10 wights 284 grams. So
[tex]14.2 g P4O10 \times \frac{1 mole P4O10}{284 g P4O10} = 0.05 moles P4O10 [/tex]
In order to form 1 mole of P4O10 it is necessary 5 moles of O2, so that the number of oxygen atoms are the same.
Knowing that 0.05 moles of P4O10 is formed we need 5 x 0.05 = 0.25 moles of O2
A force causes a mass of 4 kg to have an acceleration of 8 m/s2. Suppose something causes the mass to be one-quarter of its original amount. What will happen to the acceleration if the amount of force used remains the same?
Explanation:
From Newton's second law:
F = ma
Given that m = 4 kg and a = 8 m/s²:
F = (4 kg) (8 m/s²)
F = 32 N
If m is reduced to 1 kg and F stays at 32 N:
32 N = (1 kg) a
a = 32 m/s²
So the acceleration increases by a factor of 4.
The anode heel affect 1. should be considered when the structure of interest is more than 43 inches long. 2. is a variation in photon intensity across the exposure field. 3. occurs because the heel of the anode absorbs more photons than the toe. 4. is best used with the thinner body toward the cathode.
Answer:
Option 2. and 3.
Explanation:
Anode heel effect or simply heel effect is observed in X-ray tubes.
It is the change in the intensity of the emitted X-rays from the anode and this intensity is direction dependent and depends on the direction of the emitted X-rays along the axis of anode-cathode.
This effect results from the absorption of X-rays prior to leaving the anode, their production source.
The absorption of photons is more in the anode heel than the toe of the anode which results in the anode heel effect.
A student is running at her top speed of 5.4 m/s to catch a bus, which is stopped at the bus stop. When the student is still a distance 38.5 m from the bus, it starts to pull away, moving with a constant acceleration of 0.171 m/s2.
a) For how much time and what distance does the student have to run at 5.4m/s before she overtakes the bus?
b) When she reaches the bus how fast is the bus traveling
c) Sketch an x-t graph for both the student and the bus. Take x=0 at the initial position of the student
d) the equations you used in part a to find the time have a second solution, corresponding to a later time for which the student and bus are again at the same place if they continue thier specified motion. Explain the significance of this second solution. How fast is the bus traveling at this point
e) If the students top speed is 3.5 m/s will she catch the bus?
f) is the minumun speed the student must have to just catch up with the bus? For what time and distance must she run in that case
Answer:
a) t=8.19s; x=44.2m
b) v=1.401 m/s
c) see attachment
d) The second solution is a later time at which the bus catches the student. v=9.40 m/s
e) No, she won't.
f) v=3.63m/s;t=21.2; x=77m
Explanation:
a) The motion of both the bus and the student can be explained by the equation [tex]x= v_{0} +\frac{1}{2}at^{2}[/tex]. Since the student is not accelerating, but rather maintaining a constant speed; the particular equation that describes the motion of the student is: [tex]x_{student} = 5.4 \frac{m}{s} * t[/tex]. Meanwhile, since the bus starts its motion at an initial velocity of zero, the equation that describes its motion is: [tex]x_{bus} =\frac{1}{2} * 0.171 \frac{m}{s^{2} } * t^{2}[/tex]. The motion of the student relative to that of the bus can be described by the equation: [tex]x_{s} =x_{bus} +38.5m[/tex]. By replacing terms in the last equation we end up with the following quadratic equation: [tex](0.0855 \frac{m}{s^{2} } * t^{2} )-(5.4\frac{m}{s} *t)+38.5m =0[/tex]. Solving the quadratic equation will yield two solutions; t1=8.19s and t2=55.0s. By plugging in t1 onto the equation that describes the motion of the student we will find the distance runned by her, x=44.2m.
b) The velocity of the bus can be modeled by the equation [tex]v^{2} = v_{0} ^{2} +2ax[/tex]. Since the initial velocity of the bus is zero, the first term of the equation cancels. Next, we solve for v and plug in the acceleration of 0.171 m/s2 and the distance 5.74m (traveled by the bus, note: this is equal to the 44.2m travelled by the student minus the 38.5m that separated the student from the bus at the beginning of the problem).
c) The equations that make up the x-t graph are: [tex] x = 5.4 \frac{m}{s} * t[/tex] and [tex]x =\frac{1}{2} * 0.171 \frac{m}{s^{2} } * t^{2} + 38.5[/tex]; as described in part a.
d) The first solution states that the student would have to run 44.2 m in 8.19 s in order to catch the bus. But, at that point, the student has a greater speed than that of the bus. So if both were to keep he same specified motion, the student would run past the bus until it reaches a velocity greater than that of the student. At which point, the bus will start to narrow the distance with the student until it finally catches up with the student 55 seconds after both started their respective motion.
e) No, because the quadratic equation [tex](0.0855 \frac{m}{s^{2} } * t^{2} )-(3.5\frac{m}{s} *t)+38.5m =0[/tex] has no solution. This means that the two curves that describe the distance vs time graph for both student and bus do not intersect.
f) This answer is reached by finding b of the quadratic equation. The minimum that b can be in order to find a real answer to the quadratic equation is found by solving [tex]b^{2} -4ac=0[/tex]. If we take a = 0.0855 and c = 38.5, then we find that the minimum speed that the student has to run at is 3.63 m/s. If we then solve the quadratic equation [tex](0.0855 \frac{m}{s^{2} } * t^{2} )-(3.63\frac{m}{s} *t)+38.5m =0[/tex],we will find that the time the student will run is 21.2 seconds. By pluging in that time in the equation that describes her motion: [tex]x_{student} = 3.63 \frac{m}{s} * t[/tex] we find that she has to run 77 meters in order to catch the bus.
To determine if and when the student catches the bus, one needs to use kinematic equations to calculate time and distance, as well as understand that the second solution of these equations is hypothetical and signifies the student overtaking the bus. The answer involves solving systems of equations and applying principles of kinematics.
Explanation:We need to solve the kinematics equations for the student and the bus to find out when and if the student will catch the bus. From kinematics, the position of an object moving with constant acceleration is given by x = x0 + v0t + ½at2, where x0 is the initial position, v0 is the initial velocity, a is acceleration, and t is time.
Given the student's speed 5.4 m/s and the bus's acceleration of 0.171 m/s2, we set up two equations:
Student: xs = 5.4tBus: xb = 0 + ½(0.171)t2We find the time at which the student catches the bus by equating xs and xb and solving for t.
After finding the time, we can calculate the distance the student runs by plugging it back into the student's position equation. Then we find the speed of the bus at that point by using the equation of motion v = v0 + at.
The second solution represents another point in time where the student and bus would meet if they continued at the same speeds and acceleration, which physically represents the student overtaking the bus. Finally, if the student's top speed is 3.5 m/s, we again set up the equations similarly and solve to determine if she can catch the bus. The minimum speed required is found by setting the time derivative of the position equations equal and solving for the student's speed.
Hunter aims directly at a target (on the same level) 75.0 m away. (a) If the bullet leaves the gun at a speed of 180 m/sby how much will it miss the target? (b) At what angle should the gun be aimed so as to hit the target?
Answer:
Part a)
[tex]\delta y = 0.85 m[/tex]
Part b)
[tex]\theta = 0.65 degree[/tex]
Explanation:
Part a)
As we know that the target is at distance 75 m from the hunter position
so here we will have
[tex]x = v_x t[/tex]
here we know that
[tex]v_x = 180 m/s[/tex]
so we have
[tex]75.0 = 180 (t)[/tex]
[tex]t = 0.42 s[/tex]
now in the same time bullet will go vertically downwards by distance
[tex]\delta y = \frac{1}{2}gt^2[/tex]
[tex]\delta y = \frac{1}{2}(9.81)(0.42^2)[/tex]
[tex]\delta y = 0.85 m[/tex]
Part b)
In order to hit the target at same level we need to shot at such angle that the range will be 75 m
so here we have
[tex]R = \frac{v^2 sin(2\theta)}{g}[/tex]
[tex]75 = \frac{180^2 sin(2\theta)}{9.81}[/tex]
[tex]\theta = 0.65 degree[/tex]
A motorcycle traveling 95.0 km/hr approaches a car traveling in the same direction at 87.0 km/hr. When the motorcycle is 54.0 m behind the car, the rider accelerates and passes the car 17.0 s later. What is the acceleration of the motorcycle (in meters/second^2)?
Answer:
[tex]a = 0.1137 m/s^2[/tex]
Explanation:
Let Vc be the velocity of the car and Vm the velocity of the motorcycle. If we convert their given values, we get:
Vc = 87 km/h * 1000m / 1km * 1h / 3600s = 24.17m/s
Vm = 95 km/h * 1000m / 1km * 1h / 3600s = 26.38m/s
Since their positions are equal after 17s we can stablish that:
[tex]Xc = d + Vc*t = Xm = Vm*t + \frac{a*t^2}{2}[/tex]
Where d is the initial separation distance of 54m. Solving for a, we get:
[tex]a = \frac{d+Vc*t-Vm*t}{t^2}*2[/tex] Repacing the values:
[tex]a = 0.1137 m/s^2[/tex]
for the displacement vectors a=(3.0m)i+(4.0m)j and b=(5.0m)i+(-2.0m)j, gave a+b in(a) unit vector notation and as a magnitude and an angle (related to i) now give b-a in unit-vector notation and as magnitude and an angle
Answer:
Part (a) a + b = (8.0i - 2.0j) and [tex]\theta = 14.03^o[/tex] from the x-axis
Part (b) b - a = (2.0i - 6.0j) and [tex]\theta = -71.06^o[/tex] from the x- axis
Explanation:
Given,
[tex]\vec{a}\ =\ (3.0m)i\ +\ (4.0m)j[/tex][tex]\vec{b}\ =\ (5.0m)i\ +\ (-2.0m)j[/tex]From the addition of the two vectors,
[tex]\vec{a}\ +\ \vec{b}\ =\ (3.0i\ +\ 4.0j)\ +\ (5.0i\ -\ 2.0j)\\\Rightarrow \vec{a}\ +\ \vec{b}\ =\ (3.0\ +\ 5.0)i\ +\(4.0\ -\ 2.0)j\\\Rightarrow \vec{a}\ +\ \vec{b}\ =\ 8.0i\ +\ 2.0j[/tex]
Let [tex]\theta[/tex] be the angle of the resultant vector of the addition of the vectors with the x-axis (i).
[tex]\therefore Tan\theta\ =\ \dfrac{2.0}{8.0}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{2.0}{8.0}\ \right )\\\Rightarrow \theta\ =\ 14.03^o[/tex]
Part (b)
From the subtraction of the two vectors,
[tex]\vec{b}\ -\ \vec{a}\ =\ (5.0i\ -\ 2.0j)\ -\ (3.0i\ +\ 4.0j)\\\Rightarrow \vec{b}\ -\ \vec{a}\ =\ (5.0\ +\ 3.0)i\ +\(-2.0\ -\ 4.0)j\\\Rightarrow \vec{b}\ -\ \vec{a}\ =\ 2.0i\ -\ 6.0j[/tex]
Let [tex]\theta[/tex] be the angle of the resultant vector of the addition of the vectors with the x-axis (i).
[tex]\therefore Tan\theta\ =\ \dfrac{-6.0}{2.0}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{-6.0}{2.0}\ \right )\\\Rightarrow \theta\ =\ -71.06^o[/tex]
For an unknown sample of the experiment, students measure 1679 counts when they first receive their sample and 1336 counts four minutes later. Calculate the half-life t1/2of their sample.
Answer:
Half life of the sample, [tex]t_{\frac{1}{2}} = 12.15 min[/tex]
Given:
Initial amount, N = 1679
Final count of amount, N' = 1336
Time elapsed, t = 4 min = 240 s
Solution:
Now, To calculate the half life, using the relation:
[tex]N' = N(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}[/tex]
Now, substituting the given values in the above mentioned formula:
[tex]\frac{1336}{1679} = (\frac{1}{2})^{\frac{4}{t_{\frac{1}{2}}}}[/tex]
[tex]0.796 = 0.5^{\frac{4}{t_{\frac{1}{2}}}}[/tex]
Taking log on both the sides:
ln(0.796) = \frac{4}{t_{\frac{1}{2}}}ln(0.5)
[tex]0.329 = 4t_{\frac{1}{2}}[/tex]
[tex]t_{\frac{1}{2}} = 12.15 min[/tex]
The half-life of a substance, in nuclear physics, is the time it takes for half of the original substance to decay. The half-life for the unknown sample in the question can be calculated using the radioactive decay formula. The given values of the initial and final counts, along with the time taken, are plugged into this formula.
Explanation:The subject of this question is related to the concept of half-life in Physics, particularly Nuclear Physics. In nuclear physics, the half-life of a substance refers to the time taken for half the original elements to decay. To calculate the half-life of the unknown sample, we make use of the radioactive decay formula which says A = A0e^(-0.693/T1/2)T, where A is the final amount, A0 is the initial amount, T is the time elapsed, and T1/2 is the half-life we're trying to find.
In the given case, A0 = 1679 counts (the initial reading), A = 1336 counts (the reading after 4 minutes) and T = 4 minutes. Putting these into the equation, we would solve for T1/2, the half-life.
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Alice drops a rock down from a 40 meter tall building (vAlice,i = 0) at the same time that Bob (lying on the ground) throws a rock straight upwards from the ground with vBob,i = 20 m/s. At what height do the rocks pass each other?
Answer:
Rocks will pass each other at 20.4m.
Explanation:
Alice drops the rock from [tex]y_{a,i}=40m[/tex] (y is in the vertical axis) at [tex]v_{a,i}=0[/tex].
Bob throws the rock from [tex]y_{b,i}=0m[/tex] at [tex]v_{b,i}}=20\frac{m}{s}[/tex].
[tex]g[/tex] is gravity's acceleration.
Position equation for alice's rock:
[tex]y_{a}(t)=y_{a,i} +v_{a,i}t - \frac{1}{2}gt^{2} =40m-\frac{1}{2}gt^{2}[/tex]
Position equation for Bob's rock:
[tex]y_{b}(t)=y_{b,i} + v_{b,i}t - \frac{1}{2}gt^{2} =20\frac{m}{s}t -\frac{1}{2}gt^{2}[/tex]
We we'll first find the time [tex]t_{0}[/tex] at which the rocks meet. For this, we will equalize Bob's and Alice's equations:
[tex]y_{a}(t_{0})=y_{b}(t_{0})[/tex]
[tex]40m-\frac{1}{2}gt_{0}^{2}=20\frac{m}{s}t_{0} -\frac{1}{2}gt_{0}^{2}[/tex]
⇒[tex]20\frac{m}{s}t_{0}=40m[/tex] ⇒ [tex]t_{0}=2s[/tex]
So, we can take [tex]t_{0}=2[/tex] and just put it in any of the two laws of motion to see at what height the rocks meet. We will take Alice's equation (using g=9.8m/s):
[tex]y_{a}(2) =40m-\frac{1}{2}g2^{2}=20.4m[/tex]
Rocks will pass each other at 20.4m.
Two small heavy balls have the same diameter but one weighs twice as much as the other. The balls are dropped from a second-story balcony at the exact same time. The time to reach the ground below will be...(a) twice as long for the lighter ball as for the heavier one.
(b) longer for the lighter ball, but not twice as long.
(c) twice as long for the heavier ball as for the lighter one.
(d) longer for the heavier ball, but not twice as long.
(e) nearly the same for both balls.
Look through your class notes and your textbook and check out all the formulas and equations for the distance, speed and acceleration of falling objects. You won't find the weight of the object in any of those formulas because it doesn't matter. Without air resistance, all objects fall with the same speed and acceleration. If two objects are dropped from the same height at the same time, they both hit the ground at the same time. It doesn't matter what either one weighs. One can be a feather and the other one can be a school bus. Without air resistance it doesn't matter.
The answer is choice-(e).
Two small heavy balls have the same diameter but one weighs twice as much as the other. The time to reach the ground below will be nearly the same for both balls.
The mass of an object refers to the weight the object occupies in space.
When these two objects with different masses are dropped from a second-story building, the gravitational force acts on the objects in space.
In space, the objects fall with nearly the same acceleration regardless of the weight of their mass, the diameter, or the size of the object.
Therefore, we can conclude that for the two small heavy balls, the time to reach the ground below will be nearly the same for both balls.
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Which statement is FALSE?
a. ΔErxn is a measure of heat.
b. Enthalpy is the sum of a system's internal energy and the product of pressure and volume.
c. ΔHrxn is the heat of reaction.
d. An exothermic reaction gives heat off heat to the surroundings.
e. Endothermic has a positive ΔH.
Answer:
The false statement is a. ΔErxn is a measure of heat.
Explanation:
ΔErxn represents the energy change in a chemical reaction. However, in real processes the energy change is not only a measure of heat but also other forms like work and potential and kinetic energy.
Enthalpy is actually defined like the sum of a system's internal energy and the product of pressure and volume, so b is correct.
ΔHrxn is also the heat of reaction, which is also true
The last ones are also true
The false statement is a. ΔErxn is a measure of heat. ΔErxn refers to the change in energy in a reaction, not specifically heat.
Explanation:The false statement is a. ΔErxn is a measure of heat. This statement is not correct because ΔErxn refers to the change in energy in a reaction, which can be measured in different ways, such as heat, light, or electric potential. However, it is not solely a measure of heat.
Just to clarify the other options, b. Enthalpy is indeed the sum of a system's internal energy and the product of its pressure and volume, c. ΔHrxn is indeed the heat of reaction, d. An exothermic reaction does give off heat to the surroundings, and e. Endothermic reactions do have a positive ΔH, which represents the amount of energy needed for the reaction to occur.
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A Cessna aircraft has a liftoff speed of v = km/h = 33.3 m/s.
a) What minimum constant acceleration does the aircraft require if it is to be airborne after a takeoff run of S = 240 m?
b) How long does it take the aircraft to become airborne?
Answer:
a) Minimum acceleration is [tex]a=2.31\frac{m}{s^{2} }[/tex].
b) It will take [tex]t_{f}=14.41s[/tex].
Explanation:
Let's order the information.
Initial velocity: [tex]v_{i}=0m/s[/tex]
Final velocity: [tex]v_{f}=33.3m/s[/tex]
Initial position: [tex]x_{i}=0m[/tex]
Final position: [tex]x_{f}=240m[/tex]
a) We can use velocity's equation:
[tex]v_{f}^{2} = v_{i}^{2} +2a(x_{f}-x_{i})[/tex]
⇒ [tex]a=\frac{v_{f}^{2}-v_{i}^{2}}{2(x_{f}-x_{i})}[/tex]
⇒ [tex]a=2.31\frac{m}{s^{2} }[/tex].
b) For this, equation for average acceleration will be helpful. Taking [tex]t_{i}=0[/tex] and having [tex]t_{f}[/tex] as the unknown time it becomes airborne:
[tex]a=\frac{v_{f}-v_{i}}{t_{f}-t_{i}} =\frac{v_{f} }{t_{f}}[/tex]
⇒ [tex]t_{f}=\frac{v_{f}}{a}=\frac{33.3\frac{m}{s}}{2.31\frac{m}{s^{2}}}[/tex]
⇒ [tex]t_{f}=14.41s[/tex].
To take off, the Cessna aircraft needs a minimum acceleration of 2.31 m/s² and it will take approximately 14.4 seconds to become airborne after a run of 240 meters.
To find the minimum constant acceleration required by a Cessna aircraft to become airborne after a takeoff run of 240 meters, and the time it takes to become airborne, follow these steps:
(a) Minimum Constant Acceleration
We use the kinematic equation: v² = u² + 2as, where:
v is the final velocity (33.3 m/s)u is the initial velocity (0 m/s, starting from rest)a is the accelerations is the distance (240 m)Substitute the known values:
[tex](33.3 m/s)^2 = 0 + 2a(240 m)\\1108.89 = 480a\\a = 2.31 m/s^2[/tex]
(b) Time to Become Airborne
We use the kinematic equation: v = u + at, where:
t is the timeOther variables are as previously definedSubstitute the known values:
[tex]33.3 m/s = 0 + (2.31 m/s^2)t\\t = 33.3 / 2.31\\t = 14.4 seconds[/tex]
Find x, the angle of depression from the top of a lighthouse that is 170 ft above water level to the waterline of a ship 1084 ft off shore. Round your answer to the nearest tenth of a degree.
Answer:
The answer to your question is: Ф = 81.1° = 278.9°
Explanation:
Data
angle, x, = ?
height = 170 ft
distance 1084 ft
Process
tan Ф = opposite side / adjacent side
tan Ф = 1084 / 170 = 6.37
Ф = tan ⁻¹ (6.37)
Ф = 81.1° or 360 - 81.1 = 278.9°
A car traveling 93 km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.90 m Part A What was the magnitude of the average acceleration of the driver during the collision? Express your answer using two significant figures a= m/s
Part B Express the answer in terms of "g's," where 1.00g 9.80 m/s2 Express your answer using two significant figures.
The magnitude of the average acceleration during the collision is 368.41 m/s^2. When this value is expressed in terms of 'g's, the acceleration becomes 37.6g.
Explanation:Part A: The magnitude of the average acceleration of the driver during the collision can be found using one of the equations of motion: a = v2 / 2d where v is the final velocity, and d is the distance traveled. Given that the car was traveling at 93 km/hr (or 25.83 m/s after conversion) and the driver came to rest after 0.90 m, the equation becomes a = (25.832)/ (2*0.90) which equals 368.41 m/s2.
Part B: To express the answer in terms of "g's", we divide the acceleration by the acceleration due to gravity (9.80 m/s2). Hence, the acceleration in g's = 368.41 / 9.80 = 37.6g
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Part A: Average acceleration is 370.6 m/s².
Part B: Acceleration is 38 g's.
We start by converting the car's speed from kilometers per hour to meters per second. The conversion factor is 1 km/h equals 0.27778 m/s. Thus, 93 km/h equals 93 multiplied by 0.27778, which equals approximately 25.83334 m/s.
Part A : Next, we apply the kinematic equation to find the acceleration:
Final velocity (v) equals initial velocity (u) plus acceleration (a) multiplied by time (t), where the driver comes to rest, so v equals 0. The displacement (s) is given as 0.90 m. The formula rearranges to:
a = (v - u) / t.
We can also use another kinematic equation:
v² = u² + 2as.
Substituting the known values, we get:
0 = (25.83334)² + 2a(0.90).
Solving for a gives us:
a = - (25.83334)² / (2 * 0.90).
Calculating this yields:
a = - (667.92856) / 1.80,
a = -370.6 m/s².
The magnitude is 370.6 m/s².
For Part B, to express the acceleration in terms of g's:
a in g's = 370.6 m/s² / 9.80 m/s²
= 37.8 g.
A drunken sailor stumbles 580 meters north, 530 meters northeast, then 480 meters northwest. What is the total displacement and the angle of the displacement? (Assume east is the +x-direction and north is the +y-direction.)
a) What's the magnitude?
b) What's the direction? (In degrees and counterclockwise to the x-axis.)
Answer:
(a) 1294.66 m
(b) 88.44°
Explanation:
d1 = 580 m North
d2 = 530 m North east
d3 = 480 m North west
(a) Write the displacements in vector forms
[tex]\overrightarrow{d_{1}}=580\widehat{j}[/tex]
[tex]\overrightarrow{d_{2}}=530\left ( Cos45\widehat{i}+Sin45\widehat{j} \right )[/tex]
[tex]\overrightarrow{d_{2}}=374.77\widehat{i}+374.77\widehat{j}[/tex]
[tex]\overrightarrow{d_{3}}=480\left ( - Cos45\widehat{i}+Sin45\widehat{j} \right )[/tex]
[tex]\overrightarrow{d_{3}}=-339.41\widehat{i}+339.41widehat{j}[/tex]
The resultant displacement is given by
[tex]\overrightarrow{d}\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}[/tex]
[tex]\overrightarrow{d}=\left ( 374.77-339.41 \right )\widehat{i}+\left ( 580+374.77+339.41 \right )\widehat{j}[/tex]
[tex]\overrightarrow{d}=35.36\widehat{i}+1294.18\widehat{j}[/tex]
magnitude of the displacement
[tex]d ={\sqrt{35.36^{2}+1294.18^{2}}}=1294.66 m[/tex]
d = 1294.66 m
(b) Let θ be the angle from + X axis direction in counter clockwise
[tex]tan\theta =\frac{1294.18}{35.36}=36.6[/tex]
θ = 88.44°
Which is the correct description of the Sun's orbit within the Milky Way Galaxy?
a. at the outer edge of the Galactic Bulge
b. at the outer edge of the Galactic Disk
c. roughly in the middle of the Galactic Disk
d. very close to the Galactic center
e. high above the Galactic Disk
f. inside the Galactic Halo
Answer:
b .at the outer edge of the Galactic Disk
Explanation:
The sun is located on the outeredge of the galaxy. So it has an orbit within our Milky Way galaxy that is also located arround the outer edge of the galactic disk, and goes around in about 30 million years.
In reallity, allong its trayectory, the sun goes above and below the gallactic disk, but it stays in the outer edge of the galaxy, far away from the galactic bulge and the galactic center, but closer to it than the galactic halo.
A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how far would the cannonball travel before it lands on the ground? (Show all work)
Answer: 473.640 m
Explanation:
This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:
x-component:
[tex]x=V_{o}cos\theta t[/tex] (1)
Where:
[tex]V_{o}=48.1 m/s[/tex] is the cannonball's initial velocity
[tex]\theta=0[/tex] because we are told the cannonball is shot horizontally
[tex]t[/tex] is the time since the cannonball is shot until it hits the ground
y-component:
[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] (2)
Where:
[tex]y_{o}=1.5m[/tex] is the initial height of the cannonball
[tex]y=0[/tex] is the final height of the cannonball (when it finally hits the ground)
[tex]g=9.8m/s^{2}[/tex] is the acceleration due gravity
We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it [tex]X_{max}[/tex] and this occurs when [tex]y=0[/tex].
So, firstly we will find the time with (2):
[tex]0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2[/tex] (3)
Rearranging the equation:
[tex]0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m[/tex] (4)
[tex]-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0[/tex] (5)
This is a quadratic equation (also called equation of the second degree) of the form [tex]at^{2}+bt+c=0[/tex], which can be solved with the following formula:
[tex]t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}[/tex] (6)
Where:
[tex]a=-4.9 m/s^{2}[/tex]
[tex]b=48.1 m/s[/tex]
[tex]c=1.5 m[/tex]
Substituting the known values:
[tex]t=\frac{-48.1 \pm \sqrt{48.1^{2}-4(-4.9)(1.5)}}{2(-4.9)}[/tex] (7)
Solving (7) we find the positive result is:
[tex]t=9.847 s[/tex] (8)
Substituting this value in (1):
[tex]x=(48.1 m/s)cos(0\°) (9.847 s)[/tex] (9)
[tex]x=473.640 m[/tex] This is the horizontal distance the cannonball traveled before it landed on the ground.
Two identical grasshoppers jump into the air with the same initial speed and experience no air resistance. Grasshopper A goes straight up, but grasshopper B goes up at a 66° angle above the horizontal. Which of the following statements about these grasshoppers are correct? (There could be more than one correct choice.
Choices below:
a.) At their highest point, both of them have the same amount of mechanical energy.
b.) At their highest point, grasshopper B is moving faster than grasshopper A.
c.) At their highest point, both of them have the same amount of kinetic energy.
d.) At their highest point, both of them have the same amount of gravitational potential energy.
e.)At their highest point, grasshopper A has more gravitational potential energy than grasshopper B.
At their highest point, grasshoppers A and B have the same mechanical and gravitational potential energy. Grasshopper B has a non-zero total velocity due to its horizontal motion at its highest point, while grasshopper A does not. They both reach the same height, so they have the same gravitational potential energy.
Explanation:To answer your question about the two identical grasshoppers that jump into the air with the same initial speed, experiencing no air resistance: At their highest point, grasshopper A going straight up and grasshopper B going up at a 66° angle, we need to understand some basic concepts of physics first.
At the highest point of their jump, the vertical component of their velocity becomes zero. There is no kinetic energy associated with vertical motion for both of them.
Therefore, the answer to your multiple-choice question is:
a.) At their highest point, both of them have the same amount of mechanical energy. This is true because mechanical energy is the sum of kinetic and potential energy which remains constant due to the conservation of energy. d.) At their highest point, both of them have the same amount of gravitational potential energy. It is true because potential energy depends solely on the height from the ground and both reached the same height. b.) At their highest point, grasshopper B is moving faster than grasshopper A. It's not correct as the vertical velocity component of both grasshoppers is zero at their highest points, but grasshopper B still has horizontal velocity component, making its total velocity non-zero. c.) At their highest point, both of them have the same amount of kinetic energy. It's not correct as at the highest point, grasshopper A has zero kinetic energy but grasshopper B still has kinetic energy due to its horizontal motion. e.) At their highest point, grasshopper A has more gravitational potential energy than grasshopper B. It's not correct because they reached the same height, so they have same potential energy. Learn more about Physics of Motion here:
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Big cockroaches can run as fast as 1.50 m/s over short distances. Suppose you turn on the light in an inexpensive motel and see one run directly away from you at a constant speed of 1.50 m/s. If you run after it to squish it beginning 0.80 m behind it with an initial velocity of 0.70 m/s, what minimum acceleration must you have to catch it before it reaches safety 1.30 m away from where it starts running?
Answer:
a = 4.054 m/ s2
Explanation:
[tex]safety\ Time\ for\ cockroach = \frac{Distance\ travelled\ by\ cockroach}{ speed of cockroach}[/tex]
[tex]t = \frac{1.3}{1.5}[/tex]
t = 0.86 seconds
total distance travelled by you = 0.8 + 1.3 = 2.1 m
Initial velocity u = 0.7 m/s
let a is acceleration,
from equation of motion we have
s = ut + 0.5 at^2
[tex]2.1 = 0.7 \frac{1.3}{1.5} + 0.5 a [\frac{1.3}{1.5}]^2[/tex]
2.1 = 0.60 + 0.37 a
1. 5 = 0.37a
a = 4.054 m/ s2
The distance from the Sun to the nearest star is about 4 ✕ 1016 m. The Milky Way galaxy is roughly a disk of diameter ~ 1021 m and thickness ~ 1019 m. Find the order of magnitude of the number of stars in the Milky Way. Assume the distance between the Sun and our nearest neighbor is typical.
[tex]\boxed{n=2.92\times 10^{10} \ stars}[/tex]
Explanation:Order-of-magnitude estimates are useful when we want to get a very crude estimate of a quantity. Sometimes, to get the exact calculation of a problem is very difficult, or impossible, so we use order of magnitud in order to get a rough approximation. In this exercise, we need to find the order of magnitude of the number of stars in the Milky Way. We know:
The distance from the Sun to the nearest star is:
[tex]4 \times 10^{16}m[/tex]
The Milky Way galaxy is roughly a disk of diameter:
[tex]10^{21}[/tex]
The Milky Way galaxy is roughly a disk of thickness:
[tex]10^{19}[/tex]
So we can approximate volume of the Milky Way:
[tex]V=\pi r^2 h \\ \\ r=\frac{10^{21}}{2}m=5 \times 10^{20}m \\ \\ h=10^{19}m \\ \\ \\ V=\pi(5 \times 10^{20})^2(10^{19}) \\ \\ V=7.85\times 10^{60}m^3[/tex]
Now, let's estimate a rough density for the Milky Way. It is well known that in a sphere with a radius of [tex]4\times 10^{16}m[/tex] there is a star, which is the sun. So the density of the Milky way is:
[tex]\rho =\frac{n}{V} \\ \\ \rho:Density \ of \ Milky \ way \\ \\ n: Number \ of \ stars \\ \\ V:Volume[/tex]
For one star [tex]n=1[/tex] so we know the data at the neighborhood around the Sun, so the volume is a sphere:
[tex]V=V_{sphere}=\frac{4}{3}\pi r^3 \\ \\ V_{sphere}=\frac{4}{3}\pi (4\times 10^{16})^3 \\ \\ V_{sphere}=2.68\times 10^{50}m^3 \\ \\ So: \\ \\ \rho=\frac{1}{2.68\times 10^{50}}=3.73\times 10^{-51}stars/m^3[/tex]
Finally, the numbers of stars can be found as:
[tex]n=\rho V \\ \\ V:Volume \ of \ the \ Milky \ Way \\ \\ p:Density \ of \ the \ Milky \ Way \\ \\ n:Number \ of \ stars \\ \\ n=(3.73\times 10^{-51})(7.85\times 10^{60}) \\ \\ \boxed{n=2.92\times 10^{10} \ stars}[/tex]
A lead ball is dropped into a lake from a diving board 5.0 meters above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 3.0 seconds after it is released. How deep is the lake?
Answer:
28.8 meters
Explanation:
We must first determine at which velocity the ball hits the water. To do so we will:
1) Assume no air resistance.
2) Use the Law of conservation of mechanical energy: E=K+P
Where
E is the mechanical energy (which is constant)
K is the kinetic energy.
P is the potential energy.
With this we have [tex]\frac{m}{2} *v^{2} = m*g*h[/tex]
Where:
m is the balls's mass <- we will see that it cancels out and as such we don't need to know it.
v is the speed when it hits the water.
g is the gravitational constant (we will assume g=9.8[tex]\frac{m}{s^{2} }[/tex].
h is the height from which the ball fell.
Because when we initially drop the ball, all its energy is potential (and [tex]P = - m*g*h[/tex]) and when it hits the water, all its energy is kinetic ([tex]K=\frac{m}{2} *v^{2}[/tex]. And all that potential was converted to kinetic energy.
Now, from [tex]\frac{m}{2} *v^{2} = m*g*h[/tex] we can deduce that [tex]v=\sqrt{2*g*h}[/tex]
Therefore v=9.6[tex]\frac{m}{s}[/tex]
Now, to answer how deep is the lake we just need to multiply that speed by the time it took the ball to reach the bottom.
So D=9.6[tex]\frac{m}{s}[/tex]*3[tex]s[/tex]=28.8[tex]m[/tex]
Which is our answer.
Final answer:
The depth of the lake can be calculated by determining the time it takes for the ball to reach the bottom. Using the equation d = 1/2gt^2, where d is the distance fallen, g is the acceleration due to gravity, and t is the time taken, we can solve for t. The time taken to fall from the diving board is subtracted from the total time taken to sink to the bottom, and then the distance fallen from the diving board to the bottom of the lake is calculated using the equation d = 1/2gt^2. The depth of the lake is found to be 19.42 meters.
Explanation:
To determine the depth of the lake, we first need to calculate how long it takes for the ball to reach the bottom. Since the ball sinks with a constant velocity, the time it takes to reach the bottom is the same as the time it takes to fall from the diving board. Using the equation d = 1/2gt^2, where d is the distance fallen, g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken, we can solve for t: 5 = 1/2 * 9.8 * t^2. Rearranging the equation, we have t^2 = 5 / (1/2 * 9.8), which simplifies to t^2 = 1.02. Taking the square root of both sides, we find t = 1.01 s. Since the ball reaches the bottom after 3.0 s, we subtract the time taken to fall from the diving board to get the time taken to sink to the bottom: 3.0 s - 1.01 s = 1.99 s.
Next, we calculate the distance fallen from the diving board to the bottom of the lake using the equation d = 1/2gt^2. The time taken is 1.99 s, and the acceleration due to gravity is 9.8 m/s^2. Plugging in the values, we get d = 1/2 * 9.8 * (1.99)^2. Simplifying the equation, we find d = 19.42 m. Therefore, the depth of the lake is 19.42 meters.
The horn on a commuter train emits sound with a frequency of 342.5 Hz when the train is at rest. The train sounds its horn as it passes a passenger platform at a constant speed of 41 m/s. What overall change in frequency is detected by a person on the platform as the train moves from approaching to receding? Remember that change of frequency means the final frequency minus the initial frequency. Be sure to include the correct sign on the answer. (Take the speed of sound to be 345 m/s.)
Answer:
Δf = -82.57 Hz
Explanation:
For this problem we will use Doppler effect formula:
[tex]f = \frac{C+Vr}{C+Vs}*fo[/tex] where:
C is the speed of sound
Vr is the velocity of the person = 0m/s
Vs is the velocity of the source (train): When approaching Vs = -41m/s and when receding Vs = 41m/s
fo = 342.5 Hz
When the train is approaching:
[tex]fa = \frac{C}{C+Vs}*fo = \frac{345}{345-41}*342.5=388.69Hz[/tex]
And when the train is receding:
[tex]fr = \frac{C}{C+Vs}*fo = \frac{345}{345+41}*342.5=306.12Hz[/tex]
So, the change of frequency will be:
Δf = fr - fa = 306.12 - 388.69 = -82.57 Hz
Matthias schleiden contribution to the cell theory
Answer:
In 1838, Matthias Schleiden, a German botanist, concluded that all plant tissues are composed of cells and that an embryonic plant arose from a single cell. He declared that the cell is the basic building block of all plant matter.
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Explanation:
Final answer:
Matthias Schleiden, along with Theodor Schwann, contributed greatly to cell theory by proposing that all plants are composed of cells and are the basic unit of life, which was fundamental in stating that all living things are made up of cells.
Explanation:
Matthias Schleiden, a notable German botanist, had a pivotal role in the development of the cell theory. In 1838, Schleiden made significant microscopic observations of plant tissues and proposed that plants are composed of cells, which he believed to be the fundamental unit of life. Along with Theodor Schwann, a zoologist who studied animals, Schleiden advanced the concept that all living things are made of cells. Although Schleiden initially thought cells formed through crystallization, later discoveries, including those by Rudolf Virchow, confirmed that new cells arise from the division of existing cells. This collaborative work laid the foundations for our understanding that cells are the basic building blocks of life, constituting all organisms, and perpetuating life through cell division.
The cornerstone principles that Schleiden helped to formulate include: all living things are composed of one or more cells, the cell is the basic unit of life, and all new cells are produced from preexisting cells. These principles remain central to modern biology and medicine and provide a critical framework for understanding the structure and function of all living organisms.
A cartridge electrical heater is shaped as a cylinder of length L = 200 mm and outer diameter D = 20 mm. Under normal operating conditions the heater dissipates 2 kW while submerged in a water flow that is at 20°C and provides a convection heat transfer coefficient of h = 5000 W/m2 · K. Neglecting heat transfer from the ends of the heater, determine its surface temperature Ts. If the water flow is inadvertently terminated while the heater continues to operate, the ____
Answer:
Ts=51.83C
Explanation:
First we calculate the surface area of the cylinder, neglecting the top and bottom covers as indicated by the question
Cilinder Area= A=πDL
L=200mm=0.2m
D=20mm=0.02m
A=π(0.02m)(0.2m)=0.012566m^2
we use the equation for heat transfer by convection
q=ha(Ts-T)
q= heat=2Kw=2000W
A=Area=0.012566m^2
Ts=surface temperature
T=water temperature=20C
Solving for ts
Ts=q/(ha)+T
Ts=2000/(5000*0.012566m^2)+20=51.83C
Final answer:
The surface temperature Ts of the cylindrical cartridge electrical heater is approximately 51.85°C under normal operating conditions. If the water flow is stopped, the temperature will rise sharply and could lead to overheating and damage.
Explanation:
To determine the surface temperature Ts of the cylindrical cartridge electrical heater, we apply the concept of steady-state heat transfer. The heater dissipates power P of 2 kW and is submerged in water with a convection heat transfer coefficient h of 5000 W/m2·K. The formula for heat dissipation per unit area due to convection is:
q = h(Ts - Tinf)
where:
q is the heat flux, the rate of heat transfer per unit area (W/m2),
h is the heat transfer coefficient,
Ts is the surface temperature of the heater,
Tinf is the temperature of the water, which is 20°C.
The total heat dissipated by the heater is equal to the heat transfer rate per unit area multiplied by the surface area of the cylinder A, which can be calculated using the formula:
A = πDL
Given that L = 200 mm = 0.2 m and D = 20 mm = 0.02 m, we can calculate the surface area. Substituting the values:
A = π (0.02 m) (0.2 m) = 0.01256 m2
Now, knowing that P = qA, we can solve for q:
q = № / A = 2000 W / 0.01256 m2 = 159235.67 W/m2
Using q, we then calculate Ts:
q = h(Ts - Tinf) → 159235.67 W/m2 = 5000 W/m2·K (Ts - 20°C)
Ts = (159235.67 W/m2 / 5000 W/m2·K) + 20°C
Ts = 31.85°C + 20°C
Ts = 51.85°C
If the water flow is inadvertently terminated while the heater continues to operate, the temperature of the heater surface will rise sharply. This will likely lead to overheating, which can damage the heater and pose safety risks.
The burner on an electric stove has a power output of 2.0 kW. An 810 g stainless steel tea kettle is filled with water at 20∘ C and placed on the already hot burner. If it takes 2.4 min for the water to reach a boil, what volume of water was in the kettle? Stainless steel is mostly iron, so you can assume its specific heat is that of iron.
Answer:
volume of water in the kettle, V = [tex]774 cm^{3}[/tex]
Given:
Power output of burner, P = 2.0 kW = 2000 W
Mass of kettle, m = 810 g = 0.81 kg
Temperature of water in the kettle, T = [tex]20^{\circ}C[/tex]
Time taken by water to boil, t = 2.4 min = 144 s
Temperaturre at boiling, T' = [tex]100^{\circ}C[/tex]
Solution:
Now, we know that:
Iron's specific heat capacity, [tex]c = 0.45 J/g ^{\circ}C[/tex]
Water's specific heat capacity, [tex]c' = 4.18 J/g ^{\circ}C[/tex]
Now,
Total heat, q = Pt
q = [tex]2000\times 144 = 288 kJ[/tex]
Now,
q = (mc +m'c')(T' - T)
[tex]288\times 10^{3} = (0.81\times 0.45 + m'\times 4.18)(100^{\circ} - 20^{\circ})[/tex]
Solving the above eqn m', we get:
m' = 774 g
Now, the volume of water in the kettle, V:
[tex]V = \frac{m'}{\rho}[/tex]
where
[tex]\rho = density of water = 1 g/cm^{3}[/tex]
Now,
[tex]V = \frac{774}{1}[/tex]
Volume, V = [tex]774 cm^{3}[/tex]
"Sequence the kinetic energy, temperature, and density of most solids, liquids, and gases. Use 1 to represent the lowest amount and 3 to represent the highest."
Answer:
Kinetic energy
Solids 1, liquids 2, gases 3
Temperature
Solids 1, liquids 2, gases 3
Density
Solids 3, liquids 2, gases 1
Explanation:
Most of elements can be in the three different states at different temperatures, solid in lower temperatures then liquid and gas in higher temperatures.
When it's gaseous it is more expanded, it's volume is bigger then it's density is lower. As the molecules are more separated they can move more easily.
On the opposite hand in solid state, it's more compressed, the volume is smaller and the density is higher. The molecules are more compressed an they can't move easily.
What is the gravitational potential energy of a two-particle system with masses 4.5 kg and 6.3 kg, if they are separated by 1.7 m? If you triple the separation between the particles, how much work is done (b) by the gravitational force between the particles and (c) by you?
The gravitational potential energy of a two-particle system can be calculated with a specific formula that uses the gravitational constant and both masses. The work done by the gravitational force and an external force when separation is tripled can be found by calculating the change in gravitational potential energy.
Explanation:The gravitational potential energy between two particles can be calculated with the formula: U = -G (M₁M₂/R) , where U is the gravitational potential energy, G is the gravitational constant (6.67 × 10^-11 Nm²/kg²), M₁ and M₂ are the masses of the two bodies, and R is the distance of separation between them.
To find the work done by gravitational force and by you when the separation is tripled, we will need to calculate the change in gravitational potential energy which is given by ΔU = U_final - U_initial. The negative of this value gives the work done by the gravitational force, while the magnitude of ΔU gives the work done by an external force (you in this case).
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