Answer:
lower vapor pressure and freezing point and a higher boiling point
Explanation:
All the properties listed above are colligative properties, they depend on the amount of solute present in a solution.
It is known that a solution has a lower vapour pressure than the corresponding pure solvent due to the presence of a solute. Also the freezing point of a solution is decreased while its boiling point is increased when compared with those of the pure solvent.
A salt water solution, compared to pure water, has a lower vapor pressure and freezing point, but a higher boiling point due to the impact of the salt, a non-volatile solute.
Explanation:Compared to pure water, a salt water solution has lower vapor pressure, lower freezing point, and a higher boiling point. The presence of salt, a non-volatile solute, reduces the vapor pressure of the solution, leading to a decreased freezing point and an increased boiling point. This is due to a phenomenon known as colligative properties, which states that adding a solute to a solvent affects the physical properties of the solvent, regardless of the nature of the solute.
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You are traveling on an airplane. The velocity of the plane with respect to the air is 140 m/s due east. The velocity of the air with respect to the ground is 31 m/s at an angle of 30° west of due north. 1) What is the speed of the plane with respect to the ground?
Answer:
127.36 m/s
Explanation:
velocity of plane with respect to air = 140 m/s due east
velocity of air with respect to ground = 31 m/s 30° west of north
Write the velocities in the vector forms
[tex]\overrightarrow{V_{p/a}}=140\widehat{i}[/tex]
[tex]\overrightarrow{V_{a/g}}=31 \left ( -Sin30 \widehat{i}+Cos30\widehat{j} \right )[/tex]
[tex]\overrightarrow{V_{a/g}}= -15.5 \widehat{i}+26.85\widehat{j}[/tex]
Let velocity of plane with respect to ground is given by vp/g
According to the formula of relative velocities
[tex]\overrightarrow{V_{p/a}}=\overrightarrow{V_{p/g}}-\overrightarrow{V_{a/g}}[/tex]
[tex]\overrightarrow{V_{p/g}}=\overrightarrow{V_{p/a}}+\overrightarrow{V_{a/g}}[/tex]
[tex]\overrightarrow{V_{p/g}}= \left ( 140-15.5 \right )\widehat{i}+26.85\widehat{j}[/tex]
[tex]\overrightarrow{V_{p/g}}= \left ( 124.5 \right )\widehat{i}+26.85\widehat{j}[/tex]
The magnitude of the velocity of plane with respect to the ground is given by
[tex]V_{p/g} = \sqrt{124.5^{2}+26.85^{2}}=127.36 m/s[/tex]
Thus, the velocity of plane with respect to the ground is given by 127.36 m/s.
Problem: Hooke's law states that the force on a spring varies directly with the distance that it is stretched. If a spring has a k value of 100 newtons per meter and it is stretched 0.50 meters, what is the restoring force of the spring?
Answer:
Restoring force of the spring is 50 N.
Explanation:
Given that,
Spring constant of the spring, k = 100 N/m
Stretching in the spring, x = 0.5 m
We need to find the restoring force of the spring. It can be calculated using Hooke's law as "the force on a spring varies directly with the distance that it is stretched".
[tex]F=kx[/tex]
[tex]F=100\ N/m\times 0.5\ m[/tex]
F = 50 N
So, the restoring force of the spring is 50 N. Hence, this is the required solution.
. In a high school graduating class of 100 students, 47 studied mathematics, 61 studied physics, and 25 studied both mathematics and physics. If one of these students is selected at random, find the probability that (a) the student took mathematics or physics. (b) the student did not take either of these subjects. (c) the student took physics but not mathematics. Are studying mathematics and physics mutually exclusive events? Why or why not?
Answer:
given,
Probability of student studying math P(M)=[tex]\dfrac{45}{100} = 0.45[/tex]
Probability of student studying physicsP(P) = [tex]\dfrac{61}{100} = 0.61[/tex]
Probability of student studying both math and physics together P(M∩P) = [tex]\dfrac{25}{100} = 0.25[/tex]
a) student took mathematics or physics
P(M∪P) = P(M) + P(P) - P(M∩P)
= 0.45 + 0.61 - 0.25
= 0.81
b) student did not take either of the subject
P((M∪P)') = 1 - 0.81
= 0.19
c) Student take physics but not mathematics
P(P∩M') = P(P) - P(P∩M)
= 0.61 - 0.25
= 0.36
studying physics and mathematics is not mutually exclusive because we can study both the subjects.
The probability that a student took mathematics or physics is 83%, did not take either is 17%, and took physics but not mathematics is 36%. Mathematics and physics are not mutually exclusive since some students studied both.
The problem describes a situation where students in a high school graduating class have taken either mathematics, physics, or both. The task is to determine various probabilities based on this information. The solution requires an understanding of basic set theory and probability concepts.
Probability Calculations
(a) To find the probability that a student took mathematics or physics, we use the formula P(A \\u222a B) = P(A) + P(B) - P(A \\u2229 B). Substituting the given values, P(mathematics or physics) = 47/100 + 61/100 - 25/100 = 83/100.(b) The probability that a student did not take either of these subjects is the complement of the probability calculated in (a), which is 1 - 83/100 = 17/100.(c) The probability that a student took physics but not mathematics is the difference between the number of students who took physics and those who took both subjects, which is 61/100 - 25/100 = 36/100.Mathematics and physics are not mutually exclusive events because 25 students studied both. If they were mutually exclusive, no students would have studied both subjects.
A baseball is thrown straight downward with an initial speed of 40 ft=s from the top of the Washington Monu- ment (555 ft high). How long does it take to reach the ground, and with what speed does the baseball strike the ground?
To find the time it takes for the baseball to reach the ground, we can use the equations of motion for an object in free fall. We can solve a quadratic equation to find the time and then use it to find the final velocity of the baseball when it strikes the ground.
Explanation:To solve this problem, we can use the equations of motion for an object in free fall. The object is thrown downward, so the initial velocity is negative. We can use the equation: h = ut + (1/2)gt² to find the time taken to reach the ground. In this equation, h is the height, u is the initial velocity, t is the time, and g is the acceleration due to gravity. Since the initial height is 555 ft and the initial velocity is 40 ft/s, the equation becomes 555 = -40t + (1/2)(32)(t²). We can solve this quadratic equation to find the time it takes for the baseball to reach the ground. Once we have the time, we can use the equation: v = u + gt to find the final velocity. Plugging in the values, we get v = 40 + (32)t. Substitute the value of t from the first equation to find the final velocity when the baseball strikes the ground.
A 2-g string that is 0.79 m long is fixed at both ends and is under tension. This string produces a 500-Hz tone when it vibrates in the third harmonic. The speed of sound in air is 344 m/s. The tension in the string, in is closest to __________
Answer:
[tex]T = 175.6 N[/tex]
Explanation:
As we know that string is vibrating in third harmonic
So we will have
[tex]L = 3\frac{\lambda}{2}[/tex]
so we have
[tex]0.79 = \frac{3}{2}\lambda[/tex]
so we have
[tex]\lambda = \frac{2}{3}(0.79)[/tex]
[tex]\lambda = 0.527[/tex]
we know that frequency of the wave is given as
f = 500 Hz
now we know that
speed of the wave is
[tex]v = frequency \times wavelength[/tex]
[tex]v = (500)(0.527)[/tex]
[tex]v = 263.3 m/s[/tex]
now we have
[tex]v = \sqrt{\frac{T}{m/L}}[/tex]
so we have
[tex]263.3 \sqrt{\frac{T}{(0.002/0.79)}}[/tex]
[tex]T = 175.6 N[/tex]
Please help!! 30 points!!
Describe the Net Force lab and explain what the lab demonstrated. You may use your notes from the lab to answer this question.
Answer:
The net force acting on an object is the sum of all the force acting on it, and the net force of an object is zero. I f the forces acting on it tend to cancel each other. For example you are sit in a chair, the earth's gravity is pulling you down, but the chair is pushing you up with an equal amount of force.
Explanation:
Answer:
The net force acting on an object is the sum of all the force acting on it, and the net force of an object is zero. I f the forces acting on it tend to cancel each other. For example you are sit in a chair, the earth's gravity is pulling you down, but the chair is pushing you up with an equal amount of force.
Explanation:
science teacher helped me
An anchor made of iron weighs 833 N on the deck of a ship. If the anchor is now suspended in seawater by a massless chain, what is the tension in the chain? (Note: The density of iron is 7800 kg/m3 and the density of seawater is 1025 km/m3)
Answer:
723.54 N
Explanation:
Weight of anchor = 833 N
density of iron = 7800 mg/m^3
density of sea water = 1025 kg/m^3
According to the Archimedes principle, if a body is immersed wholly or partly in a liquid it experiences a loss in weight of the body which is equal to the weight of liquid displaced by the body.
The loss in the weight of body is equal to the buoyant force acting on the body.
The formula for the buoyant force acting on the body
= volume of body x density of liquid x acceleration due to gravity
Weight of anchor = mass x acceleration due to gravity
833 = m x 9.8
m = 85 kg
mass of anchor = Volume of anchor x density of iron
85 = V x 7800
V = 0.01089 m^3
Buoyant force on the anchor = Volume of anchor x density of sea water x g
= 0.01089 x 1025 x 9.8 = 109.46 N
So, the tension in the chain = Apparent weight of the anchor
= Weight - Buoyant force
= 833 - 109.46 = 723.54 N
Thus, the tension in the chain is 723.54 N.
Final answer:
Around 13.14% of an iron anchor's weight will be supported by buoyant force when it is submerged in saltwater, calculated based on the densities of iron and seawater.
Explanation:
The question revolves around determining what fraction of an iron anchor's weight will be supported by buoyant force when it is submerged in saltwater. The buoyant force acting on any object submerged in a fluid is equal to the weight of the fluid displaced by the object. This principle is articulated by Archimedes' principle. For the iron anchor, the fraction supported by the buoyant force is directly linked to the densities of iron (7800 kg/m3) and the seawater (1025 kg/m3) in which it is submerged.
To find this fraction, we use the formula: Fraction Supported = Density of Seawater / Density of Iron.
Therefore, Fraction Supported = 1025 / 7800 ≈ 0.1314.
So, when an iron anchor is submerged in seawater, approximately 13.14% of its weight is supported by the buoyant force.
If the wind or current is pushing your boat away from the dock as you prepare to dock, which line should you secure first?
a. towing line
b. bow line
c. starboard (right) line
d. port (left) line
Answer:
Bow Line
Explanation:
If the wind or current is pushing your boat away from the dock, bow line should be secured first.
1- We should cast off the bow and stern lines.
2-With the help of an oar or boat hook, keep the boat clear of the dock.
3-Leave the boat on its own for sometime and let the wind or current carry the boat away from the dock.
4 - As you see there is sufficient clearance, shift into forward gear and slowly leave the area.
A negatively charged rod is brought near a neutral metal sphere. Which of the following is true? A negatively charged rod is brought near a neutral metal sphere. Which of the following is true? There is no electric force between the rod and sphere. There is a repulsive force between the rod and sphere. There is an attractive force between the rod and sphere.
Answer:
There is an attractive force between the rod and sphere.
Explanation:
When negatively charged rod is placed close to the metal sphere then due to the electric field of the rod the opposite free charge of metal sphere comes closer to the rod on one surface
While similar charge in the metal sphere move away from the rod due to repulsion of electric field of rod
This temporary charge distribution of the metal sphere is known as induction
And since opposite charge on the metal surface comes closer to the metal sphere so here we can say that the rod will attract the metal sphere
so here correct answer will be
There is an attractive force between the rod and sphere.
Answer:
There is an attractive force between the rod and sphere.
Explanation:
Two cars are heading towards one another . Car a is moving with an acceleration of 11. And carb is moving at -4 m/s^2. The cars are at rest and seperated witha distance of s=1400m. What time do the cars meet?
Answer:
13.7 s
Explanation:
The position at time t of car A can be written as follows:
[tex]x_A (t) = \frac{1}{2}a_At^2[/tex]
where
[tex]a_A = 11 m/s^2[/tex] is the acceleration of car A
The position of car B instead can be written as
[tex]x_B(t) = d+\frac{1}{2}a_B t^2[/tex]
where
[tex]a_B = -4 m/s^2[/tex] is the acceleration of car B
d = 1400 m is the initial separation between the cars
The two cars meet when
[tex]x_A = x_B[/tex]
Using the two equations above,
[tex]\frac{1}{2}a_A t^2 = d + \frac{1}{2}a_B t^2\\\frac{1}{2}t^2 (a_A - a_B) = d\\t=\sqrt{\frac{2d}{a_A-a_B}}=\sqrt{\frac{2(1400)}{11-(-4)}}=13.7 s[/tex]
Final answer:
The two cars with given accelerations and an initial distance of 1400 meters between them when starting from rest will meet after 20 seconds.
Explanation:
To solve for the time at which the two cars meet, we must consider the accelerations of both cars and the initial distance between them. Assuming car A is moving with a positive acceleration of 11 m/s2 and car B is moving with a negative acceleration of -4 m/s², we need to find a common point in time where they both cover the total distance of 1400 m when starting from rest.
Let the time taken for the cars to meet be denoted by 't'. For car A, the displacement (sA) is given by the formula sA = 0.5 * aA * t², and for car B, the displacement (sB) is given by sB = 0.5 * aB * t². As they are moving towards each other, the sum of their displacements sA+sB should equal the initial separation distance which is 1400m.
This gives us the equation 0.5 * 11 * t2 + 0.5 * (-4) * t² = 1400. Simplifying, we get 3.5 * t² = 1400, and solving for 't' gives us t2 = 400, so t = 20 seconds. Hence, both cars will meet after 20 seconds.
Name two physical properties that characterize matter
Answer:
Matter have two essential types of properties, those are physical properties and chemical properties.
Explanation:
A quarterback back pedals 3.3 meters southward and then runs 5.7 meters northward. For this motion, what is the distance moved? What is the magnitude and direction of the displacement?
Answer:
The distance moved is 9 meters
The magnitude of the displacement is 2.4 meters
The direction of the displacement is northward
Explanation:
- Distance is the length of the actual path between the initial and the
final position. Distance is a scalar quantity
- Displacement is the change in position, measuring from its starting
position to the final position. Displacement is a vector quantity
The quarterback pedals 3.3 meters southward
That means it moves down 3.3 meters
Then runs 5.7 meters northward
That means it runs up 5.7 meters
The distance = 3.3 + 5.7 = 9 meters
The distance moved is 9 meters
It moves southward (down) for 3.3 meters and then moves northward
(up) for 5.7
It moves from zero to 3.3 down and then moves up to 5.7
The displacement = 5.7 - 3.3 = 2.4 meters northward
The magnitude of the displacement is 2.4 meters
The direction of the displacement is northward
Final answer:
The quarterback moved a total distance of 9.0 meters. The displacement of the quarterback was 2.4 meters northward, as displacement takes into account the direction of motion.
Explanation:
When considering the movement of a quarterback who backpedals 3.3 meters southward and then runs 5.7 meters northward, we need to determine both the total distance moved and the magnitude and direction of the displacement.
The distance is a scalar quantity that represents the total path length traveled, regardless of direction. In this case, the quarterback moved a total distance of 3.3 meters + 5.7 meters = 9.0 meters.
On the other hand, displacement is a vector quantity, which means it has both magnitude and direction. To find the quarterback's displacement, we subtract the southward movement from the northward movement, because these movements are in opposite directions. The displacement is thus 5.7 meters - 3.3 meters = 2.4 meters northward.
Suppose that the weight (in pounds) of an airplane is a linear function of the amount of fuel (in gallons) in its tank. When carrying 18 gallons of fuel, the airplane weighs 1999 pounds. When carrying 46 gallons of fuel, it weighs 2153 pounds. How much does the airplane weigh if it is carrying 58 gallons of fuel?
If the airplane is carrying 58 gallons of fuel, it will weigh 2452 pounds.
Given that the weight (in pounds) of an airplane is a linear function of the amount of fuel (in gallons) in its tank.
Carrying 18 gallons of fuel, the airplane weighs 1999 pounds.
Also carrying 46 gallons of fuel, it weighs 2153 pounds.
First we need to find the difference in gallons and the difference in the weights
46 - 12 = 34 gallons
2399 - 2176 = 223 pounds
Therefore, 34 gallons of gas weighs 223 pounds
Now the weight per pound:
223 pounds / 34 gallons = 6.5588 pounds per gallon
Similarly,
54 gallons - 46 gallons = 8 gallons
8 gallons x 6.6 pounds per gallon = 52.8 pounds, round to 53 pounds
Now Total weight of plane with 54 gallons
2399 + 53 = 2452
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The weight of the airplane is a linear function of the amount of fuel in its tank. The slope of this function is found to be 5.5 pounds per gallon of fuel. Substituting into the equation, we find that the airplane weighs 2217 pounds when it is carrying 58 gallons of fuel.
Explanation:This question can be solved by using the concept of linear function in mathematics. The weight of the airplane changes linearly with the amount of fuel in its tank. This gives us two points on the linear function (18, 1999) and (46, 2153). We can find the slope of this function by subtracting the y coordinates and dividing by the difference of the x coordinates, which gives us (2153-1999)/(46-18)=154/28 = 5.5 pounds/gallon. This is the amount the airplane's weight changes for each gallon of fuel. To find the weight of the airplane when it is carrying 58 gallons of fuel, we can use the point-slope form of a line, y-y1=m(x-x1), using one of our points and the slope we found. Plugging in gives us Weight - 1999 = 5.5×(Fuel - 18), which simplifies to Weight = 5.5*Fuel + 1999 - 5.5×18. Substituting 58 in place of Fuel, we get a weight of 2217 pounds.
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A combination of 0.250 kg of water at 20.0°C, 0.400 kg of aluminum at 26.0°C, and 0.100 kg of copper at 100°C is mixed in an insulated container and allowed to come to thermal equilibrium. Ignore any energy transfer to or from the container and determine;a. The final temperature of the mixture.b. The change in the entropy of the universe in this experiment.
Answer:
23.63 degree.
Explanation:
Specific heat of water = 4186 joule / kg degree
Specific heat of aluminium = 900 joule / kg degree
Specific heat of copper = 386 joule / kg degree
Let equilibrium temperature be T.
Heat will be gained by water and aluminium and lost by hot copper.
Heat gained or lost = mst , where m is mass , s is specific heat and t is rise or fall of temperature.
Heat gained by water = .250 X 4186 X ( T - 20 )
= 1046.5 ( T-20)
Heat gained by aluminium = .400 x 900 x ( T - 26 )
= 360 ( T - 26 )
Heat lost by copper = .100 x 386 x ( 100 - T )
38.6 ( 100 - T)
Heat gained = Heat lost
1046.5 ( T-20) + 360 ( T - 26 ) = 38.6 ( 100 - T)
T ( 1046.5 + 360 + 38.6 ) = 3860 + 9360 +20930
1445.1 T = 34150.
T = 23.63 degree.
Change in the entropy of the universe will be zero because no heat is exchanged with the universe. The container is insulated from outside .
Inside the container, entropy will be increased.
Final answer:
The student's question about thermal equilibrium and entropy change is a physics problem related to calorimetry in an adiabatic process. It requires calculating the final temperature using heat exchange equations and determining the total entropy change using the formula for entropy change.
Explanation:
The student's question involves the thermal equilibration of three different materials in an insulated container, a common scenario in calorimetry problems within physics. We assume no heat loss to the container or surroundings (adiabatic process), which simplifies the calculations.
Final Temperature Calculation
To find the final temperature, we need to set the heat lost by the warmer substances equal to the heat gained by the cooler substance:
∀Q = 0 = mcΔT (water) + mcΔT (aluminum) + mcΔT (copper)Here, m is mass, c is specific heat capacity, and ΔT is change in temperature.
For water, c = 4186 J/(kg°C), for aluminum, c = 900 J/(kg°C), and for copper, c = 387 J/(kg°C). Solving the equation ∀Q = 0 with the given masses and initial temperatures will yield the final equilibrium temperature.
Entropy Change
The change in entropy is calculated via:
ΔS = mc ln(Tx_{final} /Tx_{initial} )We apply this formula separately for each substance, summing the changes to find the total entropy change of the universe in the experiment.
The horizontal surface on which the block (mass 2.0 kg) slides is frictionless. The speed of the block before it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm? k = 2.0 kN/m
Answer:3.67 m/s
Explanation:
mass of block(m)=2 kg
Velocity of block=6 m/s
spring constant(k)=2 KN/m
Spring compression x=15 cm
Conserving Energy
energy lost by block =Gain in potential energy in spring
[tex]\frac{m(v_1^2-v_2^2)}{2}=\frac{kx^2}{2}[/tex]
[tex]2\left [ 6^2-v_2^2\right ]=2\times 10^3\times \left [ 0.15\right ]^2[/tex]
[tex]v_2=3.67 m/s[/tex]
Donna drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 8 hours. When Donna drove home, there was no traffic and the trip only took 4 hours. If her average rate was 45 miles per hour faster on the trip home, how far away does Donna live from the mountains?
Answer:
d=360 miles
Donna lives 360 miles from the mountains.
Explanation:
Conceptual analysis
We apply the formula to calculate uniform moving distance[
d=v*t Formula (1)
d: distance in miles
t: time in hours
v: speed in miles/hour
Development of problem
The distance Donna traveled to the mountains is equal to the distance back home, equal to d,then,we pose the kinematic equations for d, applying formula 1:
travel data to the mountains: t₁= 8 hours , v=v₁
d= v₁*t₁=8*v₁ Equation (1)
data back home : t₂=4hours , v=v₂=v₁+45
d=v₂*t₂=(v₁+45)*4=4v₁+180 Equation (2)
Equation (1)=Equation (2)
8*v₁=4v₁+180
8*v₁-4v₁=180
4v₁=180
v₁=180÷4=45 miles/hour
we replace v₁=45 miles/hour in equation (1)
d=8hour*45miles/hour
d=360 miles
Final answer:
To solve the problem, we used the relationship between speed, distance, and time to set up two equations and solve for the average speed to the mountains and the distance. Donna lives 360 miles away from the mountains.
Explanation:
Let's denote Donna's driving distance to the mountains as D miles. To find this distance, we can use the relationship between speed, distance, and time. When she drove to the mountains, the journey took her 8 hours. For the return trip with no traffic, it took 4 hours and the speed was 45 miles per hour faster.
Let v be the average speed on the way to the mountains. Therefore, the average speed on the way back is v + 45 mph.
Using the formula distance = speed × time, we have two equations:
D = v × 8 (to the mountains)D = (v + 45) × 4 (returning home)Since both equations are equal to D, we can set them equal to each other:
v × 8 = (v + 45) × 4
Expanding both sides:
8v = 4v + 180
Subtracting 4v from both sides we get:
4v = 180
Dividing by 4:
v = 45 mph (speed to the mountains)
So, the distance D is:
D = 45 mph × 8 h = 360 miles
Therefore, Donna lives 360 miles away from the mountains.
A speedboat starts from rest and accelerates at - 2.01 m/s2 for 7.00 s. At the end of this time, the boat continues for an additional 6.00 s with an acceleration of +0.518 Following this, the boat accelerates at 1.49 m/s2 for 8.00 s (a) What is the velocity of the boat at t = 21.0 s? (b) Find the total displacement of the boat.
Answer:
(a) 52.724 m/s
(b) Total displacement, d = 551.25 m
Solution:
As per the question:
Initial acceleration of the speed boat, a = - 2.01 [tex]m/s^{2}[/tex]
Time duration, t = 7.00 s
Additional time, t' =6.00 s
Acceleration for additional time, a' = 0.518 [tex]m/s^{2}[/tex]
The followed up acceleration, a'' = 1.49 [tex]m/s^{2}[/tex]
Time duration, t'' = 8.00 s
(a) Now, to calculate the velocity of the boat at timer, t = 21.0 s, we have:
After the initial 7.00 s, the velocity of the boat, from eqn-1 of motion:
v = u + at
v = 0 - 2.1(7.00) = - 14.7 m/s
After t + t' = 13 s:
v' = v + at
v' = 14.7 + 0.518(13) = 21.434 m/s
Now, velocity of the boat after t = 21 s:
v'' = v' + a''t
v'' = 21.434 + 1.49(21) = 52.724 m/s
(b) Now, the total displacement, d:
For the first case:
d = ut + [tex]\frac{1}{2}at^{2} = 0 - 0.5\times 2.1\times 7^{2} = - 51.45 m[/tex]
For the second case:
d = v't' = 21.434(6) = 128.6 m
For the third case:
d = ut + [tex]\frac{1}{2}a't'^{2} = 0 + 0.5\times 0.518\times 6^{2} = 4.65 m[/tex]
For the fourth case:
d = v''t'' = 52.724(8) = 421.79 m
For the last case:
d = ut + [tex]\frac{1}{2}a't'^{2} = 0 + 0.5\times 1.49\times 8^{2} = 47.68 m[/tex]
Total displacement, d = -51.45 + 128.6 + 4.65 + 421.79 + 47.68 = 551.25 m
how high can you throw a rock if you can throw the ball 58.8 m/s straight up?
-- Gravity slows the ball by 9.8 m/s every second. So it slows to zero and reached its highest point in
(58.8 / 9.8) = 6 seconds .
-- Its average speed all the way up is
(1/2) (58.8 + 0) = 29.4 m/s
-- Traveling for 6 seconds at an average speed of 29.4 m/s, the ball covers
(6) (29.4) = 176.4 meters .
Consider a slab of face area A and thickness L. Suppose that L = 33 cm, A = 55 cm2, and the material is copper. If the faces of the slab are maintained at temperatures TH = 129°C and TC = 21°C, and a steady state is reached, find the conduction rate through the slab. The thermal conductivity of copper is 401 W/m·K.
Answer:
721.8 Joule per second
Explanation:
L = 33 cm = 0.33 m
A = 55 cm^2 = 0.0055 m^2
Th = 129°C
Tc = 21°C
k = 401 W/mK
The rate of flow of heat is given by the formula
[tex]H =\frac{K A \left ( T_{h}-T_{c} \right )}{L}[/tex]
[tex]H =\frac {401 \times 0.0055 \times \left (129-21\right )}{0.33}[/tex]
H = 721.8 Joule per second
Thus, the rate of flow of heat is given by 721.8 Joule per second.
Final answer:
To find the conduction rate through the slab, use the formula Q/t = kA(TH - TC)/L. With the given values for copper's thermal conductivity, area, thickness, and temperature difference, the conduction rate is calculated to be 74,454 Watts or 74.454 kW.
Explanation:
To find the conduction rate through a copper slab, we can use the formula for heat transfer through conduction: Q/t = kA(TH - TC)/L. In this formula, Q/t is the rate of heat transfer (conduction rate), k represents the thermal conductivity of copper, A is the surface area of the slab, TH and TC are the temperatures of the hot and cold faces respectively, and L is the thickness of the slab.
Using the given values, L = 0.33 m (converted from 33 cm), A = 0.0055 m2 (converted from 55 cm2), k = 401 W/m·K, TH = 129°C, and TC = 21°C, the conduction rate can be calculated as follows:
Q/t = 401 W/m·K × 0.0055 m2 × (129°C - 21°C) / 0.33 m
Q/t = 401 W/m·K × 0.0055 m2 × 108 K / 0.33 m
Q/t = 74,454 W/m·K × m2 / m
Q/t = 74,454 W
Therefore, the conduction rate through the copper slab is 74,454 Watts or 74.454 kW.
A child, who is 45 m from the bank of a river, is being carried helplessly downstream by the river's swift current of 1.0 m/s. As the child passes a lifeguard on the river's bank, the lifeguard starts swimming in a straight line (Fig. 3–46) until she reaches the child at a point downstream. If the lifeguard can swim at a speed of 2.0 m/s relative to the water, how long does it take her to reach the child? How far downstream does the lifeguard intercept the child?
Answer:
The lifeguard takes 25.9 seconds to reach the child, at 25.9 meters from the start point downstream.
Explanation:
As the image shows, the child trajectory, the lifeguard trajectory and the distance from the bank form a triangle. This triangle is formed by the distances, an we already know the distance from the bank and the speed of child, and the speed of the lifeguard. So we have unknom time in common. Lets see the equations:
Using phitagoras theorem
[tex]45^{2}+(1*t_{1} )^{2} =(2*t_{2} )^{2}\\\\but\\t_{1} =t_{2} , then\\\\45^{2} =3t^{2} \\\\t=\sqrt{\frac{45^{2}}{3} } = 25.9seconds\\and replacing in X1= 25.9 meters[/tex]
Final answer:
The lifeguard takes 22.5 seconds to reach the child, intercepting 22.5 meters downstream due to the river's current. This solution involves calculating relative speeds and distances, showcasing principles of physics.
Explanation:
The question describes a scenario where a lifeguard needs to save a child being carried downstream by a river's current. The lifeguard can swim at a speed of 2.0 m/s relative to the water, and the child is 45 meters away from the riverbank, being carried by a current of 1.0 m/s. To solve for the time it takes for the lifeguard to reach the child and the distance downstream where the interception occurs, we must analyze the relative velocities and distances involved.
Step 1: Determine the Time to Reach the Child
The lifeguard's effective speed towards the child in the direction perpendicular to the current is 2.0 m/s. Since the child is 45 m from the bank and the lifeguard swims directly towards the child, the time it takes will be Time = Distance / Speed = 45 m / 2.0 m/s = 22.5 seconds.
Step 2: Determine the Distance Downstream
During this time, both the child and the lifeguard are being carried downstream by the current. The distance covered downstream can be calculated by the current's speed multiplied by the time: Distance downstream = Current speed × Time = 1.0 m/s × 22.5 s = 22.5 meters.
Therefore, it takes the lifeguard 22.5 seconds to reach the child, and they intercept 22.5 meters downstream from the point directly opposite the lifeguard's starting position.
A member of the marching band tunes their trombone inside their school. When they walk out to the field the instrument goes out of tune. What is the best explanation for this?
The trombone becomes out of tune when moving from indoors to the field because of the temperature change affecting the air density and speed of sound within the instrument's tubing, thus altering the pitch.
When the marching band member tunes their trombone indoors and then finds it out of tune upon moving to the field, the primary cause is likely due to a change in temperature. Brass instruments, like trombones, are sensitive to temperature changes because they directly affect the air's density inside the instrument's tubing. Warmer air makes the instrument sound sharper, while colder air will cause it to sound flatter. Since the trombone changes its pitch by varying the length of the tube through its slide mechanism, a change in outdoor temperature can significantly alter its tuning. This is because the warmer or colder air affects the speed of sound in the air column, changing the resonant frequencies of the tube and thus the pitch produced.
A certain dog whistle operates at 23.5 kHz, while another (brand X) operates at an unknown frequency. If humans can hear neither whistle when played separately, but a shrill whine of frequency 5000 Hz occurs when they are played simultaneously, estimate the operating frequency of brand X.
The unknown frequency of the other whistle (brand x) is either 28.5 kHz or 18.5 kHz. This is because when both are played together, they create a beat frequency of 5000 Hz due to wave interference.
Explanation:The perception of frequency is called pitch. Humans, typically, can distinguish between two sounds if their frequencies differ by roughly 0.3%, but they cannot hear frequencies beyond the 20,000 Hz range. If a 5000 Hz frequency is created when the two dog whistles are played together, it suggests that some form of wave interference is occurring. This particularly seems like a case of beat frequency, which is the resultant frequency produced due to the superposition of two sound waves of different frequencies.
Given that one whistle operates at 23.5 kHz (or 23500 Hz), and the beat frequency when both whistles are played together is 5000Hz, the other whistle (brand x) likely operates at either 23500 + 5000 = 28.5 kHz (or 28500 Hz) or 23500 - 5000 = 18.5 kHz (or 18500 Hz). Either of these frequencies would yield a 5000 Hz difference when superimposed with the 23500 Hz frequency, thereby creating a beat frequency of 5000 Hz when the two are played together.
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The situation described is a phenomenon called beat frequency. Since humans cannot hear the individual sounds of the whistles which must be above 20kHz but only a beat frequency of 5000 Hz, the frequency of the brand X whistle is calculated to be 28.5 kHz.
Explanation:The scenario described in this situation is an example of beat frequency, which is a phenomenon in physics. When two waves of slightly different frequencies interfere with each other, they produce a beat frequency which is the difference of the initial two frequencies. Considering the provided information, we know a shrill whine of frequency 5000 Hz is heard when the two dog whistles are played together. One whistle operates at a frequency of 23.5 kHz (or 23500 Hz).
Since the beat frequency is the difference between the two initial frequencies, the frequency of the whistle brand X (denoted as f) can be calculated by subtracting or adding the beat frequency from the known whistle frequency. This means that the unknown frequency could be either 23500 Hz - 5000 Hz = 18500 Hz, or 23500 Hz + 5000 Hz = 28500 Hz. However, since the upper limit of human hearing is 20 kHz and humans can't hear the shrill whine produced by either whistle individually, the frequency of brand X must be higher than 20 kHz and therefore is 28500 Hz.
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A motorboat maintained a constant speed of 15 miles per hour relative to the water in going 10 miles upstream and then returning. The total time for the trip was 1.5 hours. Use this information to find the speed of the current.
Answer:
speed of current is 5 mile/hr
Explanation:
GIVEN DATA:
speed of motorboat = 15 miles/hr relative with water
let c is speed of current
15-c is speed of boat at upstream
15+c is speed of boat at downstream
we know that
travel time=distance/speed
[tex]\frac{10}{15-c} +\frac{10}{15+c} = 1.5[/tex]
150+10c+150-10c=1.5(15-c)(15+c)
300=1.5(225-c^2)
300=337.5-1.5c^2
200=225-c^2
c^2=25
c = 5
so speed of current is 5 mile/hr
Final answer:
To find the speed of the current for a boat trip upstream and downstream taking a total of 1.5 hours at a constant speed of 15 mph relative to the water, we derive and solve an equation based on time taken for each part of the journey. The solution reveals that the current's speed is 3 mph.
Explanation:
The student asked how to find the speed of the current when a motorboat, traveling at a constant speed of 15 miles per hour relative to the water, went 10 miles upstream and then returned downstream, with the total trip taking 1.5 hours.
Step-by-Step Solution
Let c represent the speed of the current. Hence, the speed of the motorboat upstream is (15 - c) mph, and downstream is (15 + c) mph.The time to travel upstream is the distance/speed = 10 / (15 - c) hours, and the time to travel downstream is 10 / (15 + c) hours.Since the total time for both parts of the trip is 1.5 hours, we can write the equation: 10 / (15 - c) + 10 / (15 + c) = 1.5.To solve for c, multiply each term by the common denominator, which is (15 - c)(15 + c), to clear the fractions and simplify the equation to find the value of c.By solving the equation, we find that the speed of the current is 3 miles per hour.
In the sum →A+→B=→C, vector →A has a magnitude of 12.0 m and is angled 40.0° counterclockwise from the +x direction, and vector →C has a magnitude of 15.0 m and is angled 20.0° counterclockwise from the −x direction. What are (a) the magnitude and (b) the angle (relative to +x) of →B?
Answer:
Explanation: Ok, first caracterize the two vectors that we know.
A = ax + ay = (12*cos(40°)*i + 12*sin(40°)*j) m
now, see that C is angled 20° from -x, -x is angled 180° counterclockwise from +x, so C is angled 200° counterclockwise from +x
C = cx + cy = (15*cos(200°)*i + 15*sin(200°)*j) m
where i and j refers to the versors associated to te x axis and the y axis respectively.
in a sum of vectors, we must decompose in components, so: ax + bx = cx and ay + by = cy. From this two equations we can obtain B.
bx= (15*cos(200°) - 12*cos(40°)) m = -23.288 m
by = (15*sin(200°) - 12*sin(40°)) m = -12.843 m
Now with te value of both components of B, we proceed to see his magnitude an angle relative to +x.
Lets call a to the angle between -x and B, from trigonometry we know that tg(a) = by/bx, that means a = arctg(12.843/23.288) = 28.8°
So the total angle will be 180° + 28.8° = 208.8°.
For the magnitude of B, lets call it B', we can use the angle that we just obtained.
bx = B'*cos(208.8°) so B' = (-23.288 m)/cos(208.8°) = 26.58 m.
So the magnitude of B is 26.58 m.
Final answer:
The magnitude of vector B is 3.0 m and its angle relative to +x direction is 60.0°.
Explanation:
To find the magnitude and angle of vector →B, we can use vector addition. The magnitude of →B can be found using the equation:
|→B| = |→C| - |→A|
Substituting the given magnitudes, we have:
|→B| = 15.0 m - 12.0 m = 3.0 m
Next, we can find the angle of →B using trigonometry. Since →A is angled 40.0° counterclockwise from the +x direction, and →C is angled 20.0° counterclockwise from the −x direction, the angle of →B can be found by subtracting these angles:
θ = (20.0° - (-40.0°)) = 60.0°
Therefore, the magnitude of →B is 3.0 m and its angle relative to the +x direction is 60.0°.
Which of the following is a statement of the first law of thermodynamics?
[A] Ek = ½ mV2
[B] A negative ΔH corresponds to an exothermic process. [C] ΔE = Efinal -Einitial
[D] Energy lost by the system must be gained by the surroundings.
[E] 1 cal = 4.184 J (exactly)
Answer:
Option D.
Explanation:
The first law of thermodynamics is a law of conservation of energy. This automatically tells us that energy lost by a system won't dissapear, but it will be gained by the surroundings.
Mathematically is stated this way:
ΔU=Q-W
Where ΔU is the change in the internal energy of a closed system, Q is the amount of heat supplied and W the amount of work done by the system on its surroundings. It means that if the internal energy U decreases, then that energy lost must have been converted to work W in the surroundings.
The first law of thermodynamics is best represented by the statement: Energy lost by the system must be gained by the surroundings.
Explanation:The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed but it can be transformed from one form to another. The most accurate representation of the first law of thermodynamics from the provided options is [D] Energy lost by the system must be gained by the surroundings. This statement effectively communicates the balance of energy transfer which is fundamental to understanding the concept of first law of thermodynamics.
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a vector points 1280 units along the x-axis, and -847 units along the y-axis. Find the magnitude and direction of the vector
Answer:
The answer to your question is: magnitude: 1534.9 units
direction: 326.5°
Explanation:
Data
x = 1280 u
y = -847 y
Then as x is positive and y is negative this vector is quadrangle
To find the magnitude, we use the pythagorean theorem
c2 = a2 + b2
c2 = 1280² + (-847)² = 1638400 + 717409 = 2355809
c = 1534.9 units
to find the direction we use the tangent function
tanФ = os/as = -847/1280 = -0.661
Ф = 33.49
But it is in the 4th quadrangle, then
Ф = 360 - 33-49 = 326.5°
Answer:
1530 units, -33.5°
Explanation:
Given the x-component and the y-component, the magnitude can be found with Pythagorean theorem:
v² = vx² + vy²
And the direction can be found with trigonometry:
θ = atan(vy / vx)
Given that vx = 1280 and vy = -847:
v² = (1280)² + (-847)²
v ≈ 1530
θ = atan(-847 / 1280)
θ ≈ -33.5° or 146.5°
θ is in the fourth quadrant, so θ = -33.5°.
An object of mass m is sliding with speed vi at some instant across a level tabletop, with which its coefficient of kinetic friction is µ. It then moves through a distance d and comes to rest. Which of the following equations for the speed vi is reasonable?vi = v2µdvi = v2µmgdvi = v-2µmgdvi = v-2µgdvi = v2µgd
Answer:
√2µgd = vi
Explanation:
Given that the mass of object is m
Sliding speed of object is vi
coefficient of kinetic friction is µ
Distance covered is d
Final speed, vf is 0
Equation to apply will be;
v²f=v²i-2µgd where g is acceleration due to gravity
v²f+2µgd=v²i----------------------but v²f = 0
2µgd=v²i
√2µgd = vi
In the physical scenario described, the most reasonable formula for calculating the object's initial speed across a level tabletop, given the coefficient of kinetic friction, the object's mass, and the distance moved, would be vi is equal to the square root of twice the coefficient of friction, the mass of the object, gravity, and the distance traveled.
Explanation:The object in question moves with initial speed vi across a level tabletop under the influence of kinetic friction, and eventually comes to a stop after moving through a distance of d. According to the principles of physics, specifically those pertaining to motion and kinetic friction, the initial speed of an object can be inferred from an equation that takes into account the weight of the object (mass m times gravity g), the distance traveled d, and the coefficient of kinetic friction µ. Considering this, the equation which makes the most sense in this situation would be vi = √2µmgd, where the initial speed vi is equal to the square root of twice the coefficient of friction µ times the mass of the object m, gravity g, and the distance d.
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n object is undergoing uniform circular motion. Which of the following is/are true? I. The object is moving at constant velocity. II. The object is moving at constant speed. III. The object is accelerating.
Explanation:
When an object is moving in a circular path, the motion of the object is called uniform circular motion. The object moves under the action of centripetal acceleration. It is given by :
[tex]a=\dfrac{v^2}{r}[/tex]
r is the radius of circular path
v is the speed of the object
In uniform circular motion, the object moves with constant speed. Also, the velocity of the object keeps on changing because it changes direction at every instant of time. Also, the object is accelerating due to change in velocity.
So, the correct options are (b) and (c).
In uniform circular motion, an object travels in a circular path at a constant speed, but its velocity is not constant because the direction of motion changes. Thus, the object is accelerating due to the continuous change in direction, despite the speed being constant.
Explanation:Understanding Uniform Circular Motion
Uniform circular motion refers to the motion of an object traveling in a circular path at a constant speed. This situation presents a peculiar form of acceleration. Even though the object maintains a constant speed, its velocity is not constant because its direction is continually changing.
The three statements provided, when evaluated, give us the following insights:
Statement I: The object is moving at constant velocity - This is not true. In uniform circular motion, though the speed is constant, the direction changes continually, resulting in a changing velocity.Statement II: The object is moving at constant speed - This is true. The object travels at a constant speed along the circular path.Statement III: The object is accelerating - This is also true. The constant change in direction means there is a centripetal acceleration acting towards the center of the circle, changing the direction of the velocity vector.Therefore, the correct statements about an object in uniform circular motion are that it moves at a constant speed and it is indeed accelerating.
In 1999, Robbie Knievel was the first to jump the Grand Canyon on a motorcycle. At a narrow part of the canyon (65 m wide) and traveling 36 m/s off the takeoff ramp, he reached the other side. What was his launch angle?
Final answer:
Robbie Knievel's launch angle for his jump across the Grand Canyon was approximately -5.35 degrees.
Explanation:
To calculate the launch angle of Robbie Knievel's jump across the Grand Canyon, we can use the horizontal distance and initial velocity of the motorcycle. The horizontal distance is given as the width of the canyon, which is 65 meters. The initial velocity is given as 36 m/s. We can use the equation for horizontal motion to find the time of flight:
d = v0xt
Where d is the horizontal distance, v0x is the initial velocity in the x-direction, and t is the time of flight. Solving for t, we get:
t = d / v0x
Substituting the given values, we have:
t = 65 m / 36 m/s = 1.81 s
Next, we can use the equation for vertical motion to find the launch angle:
y = v0yt + (1/2)gt2
Where y is the vertical distance, v0y is the initial velocity in the y-direction, g is the acceleration due to gravity, and t is the time of flight. The vertical distance is zero since Robbie reached the other side of the canyon. Solving for v0y and using the given value of t, we get:
v0y = - (1/2)gt
v0y = - (1/2)(3.7 m/s2)(1.81 s) = -3.34 m/s
Finally, we can use the launch angle formula:
tan(θ) = v0y / v0x
Substituting the calculated values, we have:
tan(θ) = -3.34 m/s / 36 m/s
θ = atan(-3.34 m/s / 36 m/s) = -5.35 degrees
Therefore, Robbie Knievel's launch angle was approximately -5.35 degrees.
A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be independently adjusted. When the engines are fired simultaneously and each applies its force in the same direction, the probe, starting from rest, takes 21.2 s to travel a certain distance. How long does it take to travel the same distance, again starting from rest, if the engines are fired simultaneously and the forces that they apply to the probe are perpendicular?
Answer:
25.2 s
Explanation:
In an uniformly accelerated motion starting from rest, the time taken for the motion is given by the equation
[tex]d=\frac{1}{2}at^2 \rightarrow t=\sqrt{\frac{2d}{a}}[/tex]
where t is the time, d is the distance covered, a is the acceleration.
We also know that the acceleration can be found by using Newton's second law:
[tex]\sum F = ma \rightarrow a=\frac{\sum F}{m}[/tex]
where [tex]\sum F[/tex] is the net force on the object and m its mass. Substituting into the previous equation,
[tex]t=\sqrt{\frac{2md}{\sum F}}[/tex] (1)
So we see that the time taken for the motion is inversely proportional to the square root of the net force.
Let's consider now the space probe in the problem. Let's call F the magnitude of the force generated by each engine.
When the two forces are applied in the same direction, the net force on the space probe is
[tex]\sum F = F+F = 2F[/tex]
But when the two forces are applied perpendicularly, the net force is
[tex]\sum F' = \sqrt{F^2+F^2}=\sqrt{2} F[/tex]
Using eq.(1) we can write:
[tex]\frac{t}{t'}=\sqrt{\frac{\sum F'}{\sum F}}[/tex]
where t' is the new duration of the motion. Solving for t',
[tex]t'=\sqrt{\frac{\sum F}{\sum F'}}t=\sqrt{\frac{2F}{\sqrt{2}F}}(21.2 s)=25.2 s[/tex]