Answer:
1.78 m upward
Explanation:
We can find the displacement of the volleyball by using the SUVAT equation:
[tex]v^2 - u^2 = 2ad[/tex]
where, assuming upward as positive direction:
u = 6.0 m/s is the initial velocity
v = 1.1 m/s is the final velocity
a = g = -9.8 m/s^2 is the acceleration of gravity
d is the displacement
Solving the equation for d, we find:
[tex]d=\frac{v^2-u^2}{2a}=\frac{1.1^2-6.0^2}{2(-9.8)}=1.78 m[/tex]
And since it is positive, the displacement is upward.
The integral concepts applied include kinematics and gravitational acceleration. Given initial velocity, final velocity, and knowing gravitational due to acceleration, we can determine the time of flight up to its highest point and the maximal height or displacement through kinematic equations. In this particular case, displacement happens to be approximately 2.275 meters upwards.
Explanation:The subject of this question is Physics, specifically, it deals with the concept of kinematics. The first step to solve the problem is to use the final velocity equation in kinematics, v = u + at, where 'v' is the final velocity, 'u' is the initial velocity, 'a' is the acceleration (which would be gravity in this case, ‐9.8 m/s² since it's acting downward or against the direction of the initial velocity), and 't' time. From the given values, we know that v = 1.1 m/s (these are vector quantities, hence the direction is important, and in this case, both v and u are in the same direction, up the way), u = 6.0 m/s and a = -9.8 m/s². You should set up the equation as follows: 1.1 = 6 + (-9.8)t. By simplifying, we get t = (1.1 - 6) / -9.8 ≈ 0.5s. Next, to get the displacement, you can use another kinematic equation s = ut + (1/2)a*t² ('s' stands for displacement). Once you plug in the known values, you'll get s = 6*0.5 + 1/2*(-9.8)*(0.5)² ≈ 2.275 m.
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A new grill has a mass of 30.0 kg. You put 3.0 kg of charcoal in the grill. You burn all the charcoal and the grill has a mass of 30.0 kg. What is the mass of the gases given off? (Assume that the charcoal is pure carbon solid and that it burns completely in oxygen.)
Answer:
11 kg of gas are emmited to the atmosphere
Explanation:
The problem consists in calculating the total mass of the gases produced in the combustion reaction of the carcoal.
In this case we consider the simple combustion reaction:
C+O2 -> CO2
In relation 1:1 molar with the oxygen.
In this case, the limitant reactant is the carbon, so we make the calculations based in the 3kg of carbon combusted.
First we convert the mass of carbon into mol of carbon
m,C=3kg=3000 g of C
Then we have the mol
n,C=3000g/(12 g/mol)=249.77 mol C
This number also corresponds to the number of oxygen mol reacting with the charcoal. Therefore we calculate the mass of oxygen.
m,O2=249.77 mol *(32 g/mol) = 7990 g =7.99 kg, aproximately 8 kg
Then the total mass of gas is the mass of charcoal plus the mass of oxygen
m,total=3kg + 8kg=11kg
You are camping with two friends, Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close together. Joe's tent is 15.0 m from yours, in the direction 22.5^\circ north of east. Karl's tent is 42.0 m from yours, in the direction 34.0^\circ south of east
What is the distance between Karl's tent and Joe's tent?
Answer:
46 meters
Explanation:
In the figure you can see law of cosines.
In your case
α = 22.5° + 42 ° = 64.5 °
b = 15 m (distance between your's tent and Joe's tent)
c = 42 m (distance between your's tent and Karl's tent)
a is the distance between Joe's tent and Karl's tent
Replacing in the first equation:
[tex]a^2 = 15^2 m^2 + 42^2 m^2 - 2 \times 15 \times 42 \times cos(64.5) [/tex]
[tex] a = \sqrt{2111.46 m^2} [/tex]
[tex] a = 46 m [/tex]
A solid cylinder with a radius of 5.08 cm starts from rest at the top of a 12.0 meter long ramp inclined 20.3° above the horizontal. When it reaches the bottom of the ramp 3.25 seconds later the cylinder has a final linear velocity of 7.38 m/s .What was the average angular velocity of the cylinder?
Final answer:
The average angular velocity of the cylinder is calculated using the relationship between linear velocity and radius, with the formula ω = v / r. With a final linear velocity of 7.38 m/s and a radius of 5.08 cm, the average angular velocity is about 145.27 rad/s.
Explanation:
To calculate the average angular velocity of the cylinder, we must first understand the relationship between linear velocity and angular velocity. Angular velocity (ω) is related to linear velocity (v) through the radius of the object according to the equation ω = v / r, where r is the radius of the cylinder. Given that the final linear velocity is 7.38 m/s and the radius is 5.08 cm (or 0.0508 meters), we can calculate the average angular velocity.
Calculating the average angular velocity:
Convert the radius to meters: 5.08 cm = 0.0508 m.Use the relationship ω = v / r to find the average angular velocity.ω = 7.38 m/s / 0.0508 m = 145.27 rad/s.Therefore, the average angular velocity of the cylinder is approximately 145.27 rad/s.
77. Two forces act on a parachutist falling in air: the force of gravity and air resistance. If the fall is steady, with no gain or loss of speed, then the parachutist is in dynamic equilibrium. How do the magnitudes of the gravitational force and air resistance compare?
Answer:
Explanation:
When a parachutist jump from an aeroplane, there are two forces acting on the parachutist.
One is the force called gravitational force which is acting downwards.
Other is the force of air resistance which is acting upwards.
When the parachute is not opened, the gravitational force is more than the air resistance force due to which the parachutist experiences a force with which he is coming down and thus , there is an acceleration in the body and the velocity of the body goes on increasing.
When the parachute is opened, the gravitational force is balanced by the air resistance force due to which there is no net force acting on the body of parachutist and thus, he comes down with uniform velocity as acceleration in the body is zero.
In this condition, the magnitude of gravitational force is equal to the air resistance force.
A meteor this past year was tracked while it re-entered the earth's atmosphere going 18,000 mph. The enormous air friction on it caused it to lose 90% of its velocity in about 10 seconds. How far did it travel in kilometers during this time?
Answer:
44257m
Explanation:
first we convert the initial velocity to m / s
Vo=initial velocity=18000Mph*(0.44m/s)/1mph=8046.7m/s
As the meteorite lost 90% of its speed, it means that it is moving with 10% of the initial speed
Vf=final velocity=(0.1)(8046.7)=804.67m/s
As the meteorite has a uniformly accelerated movement we can use the following equation
A=aceleration=(Vf-Vi)/t
A=(804.67-8046.7)/10=-724.2m/S^2
finally to calculate the displacement we use the following equation
(Vf^2-Vo^2)/2A=X=displacement
X=(804.67^2-8046.7^2)/(2*-724.2)=44257m
What is the best definition of "oxidation state"? The number of electrons an atom has The number of neutrons an atom has The number of electrons an atom has relative to a neutral atom of the same element The number of neutrons an atom has The number of electrons and protons an atom has
Answer:
The number of electrons an atom has relative to a neutral atom of the same element
Explanation:
the oxidation state is a number that indicates the ability to give or receive electrons. This number usually has a sign, which can be positive or negative.
If the oxidation state is a positive number, it means that this atom has the ability to receive such an amount of electrons. If the oxidation state is negative, the element can give that amount of electrons.
If an automobile had a 100%-efficient engine, transferring all of the fuel's energy to work, would the engine be warm to your touch? Would its exhaust heat the surrounding air? Would it make any noise? Would it vibrate? Would any of its fuel go unused? Discuss.
Answer:
The engine would be warm to touch, and the exhaust gases would be at ambient temperature. The engine would not vibrate nor make any noise. None of the fuel entering the engine would go unused.
Explanation:
In this ideal engine, none of these events would happen due to the nature of the efficiency.
We can define efficiency as the ratio between the used energy and the potential generable energy in the fuel.
n=W, total/(E, available).
However, in real engines the energy generated in the combustion of the fuel transforms into heat (which heates the exhost gases, and the engine therefore transfering some of this heat to the environment). Also, there are some mechanical energy loss due to vibrations and sound, which are also energy that comes from the fuel combustion.
Final answer:
This answer addresses the implications of a hypothetical 100%-efficient engine in an automobile, discussing its heat production, exhaust impact, noise and vibration levels, and fuel usage.
Explanation:
If an automobile had a 100%-efficient engine:
Would the engine be warm to your touch? - Yes, the engine would still produce heat as a byproduct of its operation, even with 100% efficiency, due to thermodynamics.Would its exhaust heat the surrounding air? - The exhaust gases would still carry thermal energy and heat the air around.Would it make any noise or vibration? - While noise and vibration are not related to engine efficiency, they stem from mechanical components and combustion processes.Would any of its fuel go unused? - With 100% efficiency, all the fuel's energy would convert to work, leaving no unused fuel.A solution contains a mixture of pentane and hexane at room temperature. The solution has a vapor pressure of 241 torr. Pure pentane and hexane have vapor pressures of 425 torr and 151 torr, respectively, at room temperature.
What is the mole fraction of hexane? (Assume ideal behavior.)
Answer:
The mole fraction of hexane is 0.67
Explanation:
This problem can be solved using Dalton´s law and Raoult´s law.
The vapor pressure of a mixture of gases is the sum of the partial pressures of each gas (Dalton´s law).
Example:
In a mixture of gases A and B
Pt = PA + PB
where:
Pt = total pressure
PA = partial pressure of A
PB = partial pressure of B
In an ideal solution, the vapor pressure of each component is equal to the vapor pressure of the pure component times the mole fraction of the component in the solution (Raoult´s law).
Example:
In a solution containing A and B, the vapor pressure of A in the solution will be:
PA = P(pure A) * Xa
Where
PA = vapor pressure of A in the solution
P(pure A) = vapor pressure of pure A
Xa = mole fraction of A
In our problem, we have that the vapor pressure of the solution is 241 torr.
Then, using Dalton´s law:
Pt = P(hexane) + P(Pentane)
Using Raoult´s law:
P(hexane) = P(pure hexane) * X(hexane)
P(pentane) = P(pure pentane) * X(pentane)
We also know that the sum of the molar fractions of each component in a solution equals 1:
X(hexane) + X(pentane) = 1
X(pentane) = 1 - X(hexane)
Replacing in the Dalton´s law in terms of X(hexane):
Pt = P(pure hexane) * X(hexane) + P(pure pentane) * (1 - X(hexane))
Solving for X(hexane):
Pt = P(pure hex) * X(hex) + P(pure pent) - P(pure pent) * X(hex)
Replacing with the data:
241 torr = 151 torr * X(hex) + 425 torr - 425 torr*X(hex)
-184 torr = -274 torr * X(hex)
-184torr/-274 torr = X(hex)
X(hex) = 0.67
The mole fraction of hexane in the solution is 1.59.
Explanation:To find the mole fraction of hexane in the solution, we can use Raoult's law, which states that the partial pressure of a component in a solution is equal to the product of the mole fraction of that component and its vapor pressure.
Let's calculate the mole fraction of hexane:
Partial pressure of hexane = mole fraction of hexane * vapor pressure of hexane
241 torr = x * 151 torr
241 torr / 151 torr = x = 1.59
The mole fraction of hexane in the solution is 1.59.
A 4.33 kg cat has 41.7 J of KE How fast is the cat moving?
Answer:
The answer to your question is:
Explanation:
Data
mass = 4.33 kg
E = 41.7 J
v = ?
Formula
Ke = (1/2)mv²
Clear v from the equation
v = √2ke/m
Substitution
v = √2(41.7)/4.33
v = 19.26 m/s Result
A factory conveyor belt rolls at 3 m/s. A mouse sees a piece of cheese directly across the belt and heads straight for the cheese at 4 m/s. What is the mouse's speed relative to the factory floor
A. 1m/s
B. 2m/s
C. 3m/s
D. 4m/s
E. 5m/s
Answer:
mouse speed is 5 m/s
Explanation:
given data
belt roll b = 3 m/s
head straight h = 4 m/s
to find out
mouse speed s
solution
we will apply here pythagorean theorem
that is
s = [tex]\sqrt{b^{2} +h^{2} }[/tex] ...........................1
put here value in equation 1 as b = 3 and h = 4
so
s = [tex]\sqrt{b^{2} +h^{2} }[/tex]
s = [tex]\sqrt{3^{2} +4^{2} }[/tex]
s = [tex]\sqrt{9 + 16 }[/tex]
s = [tex]\sqrt{25 }[/tex]
s = 5
so mouse speed is 5 m/s
The mouse's speed relative to the factory floor is calculated using the Pythagorean theorem, taking into account both the speed of the mouse and the conveyor belt. The correct answer is 5 m/s, option E.
Explanation:The speed of the mouse relative to the factory floor is calculated using the Pythagorean theorem given that the mouse's motion and the conveyor belt's motion form two sides of a right triangle. In this case, the speed of the mouse is 4 m/s and the speed of the conveyor belt is 3 m/s. The relative speed will therefore be the square root of (mousespeed)² + (conveyorbelt speed)². So, √(4² + 3²) = √(16 + 9) = √25 = 5 m/s. Therefore, the mouse's speed relative to the factory floor is 5 m/s, option E.
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-Our balance is maintained, at least in part, by the endolymph fluid in the inner ear. Spinning displaces this fluid, causing dizziness. Suppose a dancer (or skater) is spinning at a very fast 2.6 revolutions per second about a vertical axis through the center of his head. Although the distance varies from person to person, the inner ear is approximately 7.0 cm from the axis of spin. part A What is the radial acceleration (in m/s^2 ) of the endolymph fluid? part B What is the radial acceleration (in g's) of the endolymph fluid? 2-A model of a helicopter rotor has four blades, each of length 4.00m from the central shaft to the blade tip. The model is rotated in a wind tunnel at a rotational speed of 540rev/min . A-What is the linear speed of the blade tip? B-What is the radial acceleration of the blade tip expressed as a multiple of the acceleration of gravity, g?
1)
Answer:
[tex]a = 18.68 m/s^2[/tex]
Part b)
[tex]a = 1.9 g[/tex]
Explanation:
Rate of the spinning of the dancer is given as
[tex]f = 2.6 rev/s[/tex]
angular speed is given as
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 2\pi(2.6) = 16.33 rad/s[/tex]
distance of the ear is given as
[tex]r = 7 cm = 0.07 m[/tex]
Part a)
Radial acceleration is given as
[tex]a = \omega^2 r[/tex]
[tex]a = (16.33)^2(0.07)[/tex]
[tex]a = 18.68 m/s^2[/tex]
Part b)
also we know that
[tex]g = 9.81 m/s[/tex]
so now we have
[tex]\frac{a}{g} = \frac{18.68}{9.81}[/tex]
[tex]a = 1.9 g[/tex]
2)
Answer:
Part a)
[tex]v = 226.2 m/s[/tex]
Part b)
[tex]a = 1.304 \times 10^3 g[/tex]
Explanation:
Length of the blades = 4.00 m
frequency of the blades = 540 rev/min
[tex]f = 540 \times \frac{1}{60} = 9 rev/s[/tex]
so angular speed is given as
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 2\pi(9) = 56.5 rad/s[/tex]
Part a)
Linear speed of the tip of the blade is given as
[tex]v = r\omega[/tex]
[tex]v = (4.00)(56.5)[/tex]
[tex]v = 226.2 m/s[/tex]
Part b)
Radial acceleration of the tip of the blade
[tex]a = \frac{v^2}{r}[/tex]
[tex]a = \frac{226.2^2}{4}[/tex]
[tex]a = 1.28 \times 10^4 m/s^2[/tex]
also we know
[tex]\frac{a}{g} = \frac{1.28 \times 10^4}{9.81}[/tex]
[tex]a = 1.304 \times 10^3 g[/tex]
Technician A says a power stroke in a modern common rail engine only uses a single injection of fuel. Tecnician B says a late-model common rail diesel has multiple injection events during a combustion cycle. Who is correct?
Answer:
B
Explanation:
In a common rail system, the fuel is distributed to the injectors from the rail where there is high pressure accumulation.High pressure fuel pump feds the rail and with the help of the start and end signals the injector is activated.For example in diesel common rail direct injection systems, the rail is connected to injectors using individual pipes in that the injectors work hand in hand with the fuel pump which ensures fuel injection timing and amount. Comparing this with early models, the fuel pump was responsible for timing, quantity and pressure.So power stroke in a modern common rail system does not only rely on single injection of pump but also has multiple injection events during combution process.
A motorboat maintained a constant speed of 21 miles per hour relative to the water in going 14 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.
a) 3 miles per hour
b) 5 miles per hour
c) 7 miles per hour
d) 4 miles per hour
e) 2 miles per hour
Answer:
c)7 mph
Explanation:
Going downstream the speed of the boat is added to the speed of the water, on the way back it is opposite to it, then the speed of the water is substracted to that of the boat, and we have the total time of the trip, then we can write down a system of equations as follows:
x:=Speed of water that is being asked
t1:= time going downstream
t2:= time going upstream
(21+x)*t1=14 (1)
(21-x)*t2=14 (2)
t1+t2=1.5 (3)
then t1=1.5-t2, replacing this in (1) we have (21+x)*(1.5-t2)=14, then t2=1.5-(14/(21+x)) (4). Doing the same in (2) we get t2=14/(21-x) (5), this way we can use (4) and (5) and get that:
1.5-(14/(21+x))=14/(21-x), doing the operations we get that 588/(441-x^2)=1.5 or equivalently 1.5x^2-73.5=0 or x^2-49=0 or (x-7)(x+7)=0, then the possible answers are x=7 or x= -7, here we consider the positive answer, that in magnitude is the same as the other, because we already considered the direction of the water into the equations. Then the answer is c) 7mph.
Final answer:
The speed of the current is 3 miles per hour, obtained by setting up an equation based on the total time taken for an upstream and downstream trip and solving for the current's speed. So the correct option is a.
Explanation:
To find the speed of the current, we can set up an equation considering the speed of the boat in still water (which we know is 21 miles per hour) and the speed of the current, which we will call 'c'.
When the boat goes upstream, the effective speed is reduced by the speed of the current, so it's 21 - c miles per hour. Similarly, when the boat goes downstream, the effective speed is increased by the speed of the current, making it 21 + c miles per hour.
Let's calculate the time taken to travel upstream and downstream using the distance and the effective speed. For upstream, it is 14 / (21 - c) hours, and for downstream, it is 14 / (21 + c) hours. The total time for the trip is given as 1.5 hours, so we sum the upstream and downstream times to set up the equation:
14 / (21 - c) + 14 / (21 + c) = 1.5
By solving this equation for 'c', we find that the speed of the current is 3 miles per hour, which corresponds to option (a).
The stars in the sky are organized into groups of stars called constellations which appear near each other in the sky but are not necessarily close together in space. How many constellations are currently accepted by the IAU?
Answer:
The International Astronomical Union (IAU) has accepted 88 constellations in the sky.
Explanation:
Constellations has been used since the beginnings of civilizations and each one of them named them as they considered appropiate. It means Greeks' constellations were different than the ones described by Chinese, so it was necessary to gather all these constellations and make a great record with all of them, but there was a problem: Some constellations from different civilizations overlaped because they shared the same stars. There was necessary to put some order on this and that is when in 1922 the International Astronomical Union (IAU) defned a set of 88 moderm constellations that would become the international standard to look at the night sky. Each one of them is unique and does not share stars with the other constellations.
If the initial velocity is 100 m/s at 42.8 ◦ above the horizontal and the initial height of the target is 95.5 m , how high above the ground does the collision take place? The acceleration due to gravity is 9.8 m/s 2 .
Answer:
Explanation:
Since it only mentions the height we only have to concern ourselves with the y direction, up and down, so that's nice. Also, acceleration due to gravity is -9.8 m/s^2
First we want the formula for the position of the object, for that we need the initial speed. We have an initial speed but we need the speed in the y direction. To do that we break the speed we have into its x and y components. Imagine a right triangle where the hypotenuse is 100 and the angle between that and the horizontal leg is 42.8 degrees. Notice that the legs of the right triangle are vertical and horizontal, they represent the y and x components of the speed respectively. Anyway, we want the y component, so the vertical. Relative to the given angle that is the "opposite" side, when we use SOH CAH TOA. Let me know if this doesn't make sense and I can draw it out.
Now, using trig, we can find the y component with SOH. so sin(42.8)=y/100. Make sure your calculator is in degree mode of course. we get
y = 100*sin(42.8) = 67.94 so the vertical speed is 67.94 m/s.
With that now we can use the displacement formula s = vi*t + .5a*t^2 if you don't remember this formula you definitely should have some kind of paper or something with all your physics formula. Anyway, we can plug in all values except for t, which is our variable.
s = 95.5
vi=67.94
a= -9.8
95.5 = 67.94t + .5(-9.8)t^2
Then if you subtract 95.5 from both sides you have a traditional quadratic equation you can solve with the quadratic formula, or completing the square or however you solve them. If you need help with this step let me know, or if something else didn't make sense.
Your bedroom air conditioner blows very cold air at night but only cool air during the day. Your bedroom gets lots of direct sunlight all day long. What's the hypothesis and experiment.
The hypothesis could be stated as follows: "Since there is no direct sunlight at night, the temperature of the air blown by the bedroom air conditioner is colder at night than it is during the day."
You may devise an experiment to gauge the temperature of the air being blasted by the air conditioner both at night and during the day in order to verify this theory. This is a general description of the experiment.
Place a thermometer or temperature sensor close to the air conditioner's outlet, which is where the room's cool air is blown. Make sure the sensor is in the same spot for the measurements during the day and at night.
The exact relationship between the air temperature blasted by the air conditioner and the presence or absence of direct sunshine is what the experiment attempts to investigate. According to the theory, cooler air may be blasted by the air conditioner during the daytime when there is sunlight but warmer air during the night when there is no direct sunlight.
Hence, the hypothesis could be stated as follows: "Since there is no direct sunlight at night, the temperature of the air blown by the bedroom air conditioner is colder at night than it is during the day."
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diver uses a sonar device to determine the distance to her diving partner. The device sends a sound wave and records the time it takes for the wave to reach the diving partner and return to the device. Suppose the wave travels at a rate of about 4000 feet per second. The wave returns 0.8 second(s) after it was sent. How far away is the diving partner?
Answer:
Assuming the divers are not moving. The diving partner is at 1600 feet.
Explanation:
The sound wave have to go to the diving partner and return, so its traveling the double of the total distance.
Been X= distance, V= speed, and t= time
[tex]x=v*t\\[/tex]
[tex]WaveTravelDistance=4000*0.8\\\\Distance to the partner=WaveTravelDistance/2\\\\WaveTravelDistance=3200feet\\\\Distance to the partner=1600feet[/tex]
A 1400 kg aircraft going 40 m/s collides with a 1500 kg aircraft that is parked and they stick together after the collision and are going 19.3 m/s after the collision. If they skid for 8.2 seconds before stopping, how far did they skid?
Explanation:
It is given that,
Mass of the aircraft 1, m₁ = 1400 kg
Mass of aircraft 2, m₂ = 1500 kg
Initially, the aircraft 2 is at rest, u₂ = 0
Initial speed of the aircraft 1, u₁ = 40 m/s
After the collision, the total speed of the system, V = 19.3 m/s
Time, t = 8.2 s
Since, two objects stick together it is a case of inelastic collision. The acceleration of the system is given by :
[tex]a=\dfrac{V}{t}[/tex]
[tex]a=\dfrac{19.3\ m/s}{8.2\ s}[/tex]
[tex]a=2.35\ m/s^2[/tex]
Distance covered by the system before stopping is given by :
[tex]x=\dfrac{1}{2}\times a\times t^2[/tex]
[tex]x=\dfrac{1}{2}\times 2.35\times (8.2)^2[/tex]
x = 79.007 meters
Hence, this is the required solution.
Two small frogs simultaneously leap straight up from a lily pad. Frog A leaps with an initial velocity of 0.551 m/s, while frog B leaps with an initial velocity of 1.23 m/s. When the first frog to return to the lily pad does so, what is the position and velocity of the other frog? Take upwards to be positive, and let the position of the lily pad be zero.
Answer:
d = .076 m
Explanation:
The time for frog A can be calculated from equation of motion
[tex]v_f = v_o + at[/tex]
where v_f is final velocity, a is acceleration due to gravity
so from given data we have
[tex]-0.551= 0.551 + (-9.8)(t)[/tex]
t = 0.112 sec
Now we will use that time for frog B
[tex]v_f = v_o + at[/tex]
[tex]v_f = 1.23 + (-9.8)(0.112)[/tex]
[tex]v_f = 0.128 m/s [/tex](Note its positive)
For the displacement
[tex]s = v_o t + 0.5at^2[/tex]
[tex]s = (1.23)(0.112) + (.5)(-9.8)(0.112)^2[/tex]
d = .076 m
The position of an object is given by x = at3 - bt2 + ct,where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI units. What is the instantaneous acceleration of the object when t = 4.1 s?
Answer:
The answer to your question is: 15 m/s2
Explanation:
Equation x = at3 - bt2 + ct
a = 4.1 m/s3
b = 2.2 m/s2
c = 1.7 m/s
First we find x at t = 4.1 s
x = 4.1(4.1)3 - 2.2(4.1)2 + 1.7(4.1)
x = 4.1(68.921) - 2.2(16.81) + 6.97
x = 282.58 - 36.98 + 6.98
x = 252.58 m
Now we find speed
v = x/t = 252.58/ 4.1 = 61.6 m/s
Finally
acceleration = v/t = 61.6/4.1 = 15 m/s2
The acceleration of the object is the change in the velocity of the object within given time interval.
The acceleration of the object at time t = 4.1 seconds is 15.26 m/s2.
How do you calculate the acceleration of the object?Given that the position of an object is x = at3 - bt2 + ct, where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s.
The position of the object at time t = 4.1 s is calculated as given below.
[tex]x = at^3-bt^2+ct\\[/tex]
[tex]x = 4.1\times (4.1)^3 - 2.2 \times (4.1)^2 + 1.7\times 4.1[/tex]
[tex]x = 252.56\;\rm m[/tex]
The velocity v of the object at time t = 4.1 s is given below.
[tex]v = \dfrac{x}{t}[/tex]
[tex]v = \dfrac {252.56}{4.1}[/tex]
[tex]v = 62.67\;\rm m/s[/tex]
The acceleration of the object at time t =4.1 s is given below.
[tex]a = \dfrac {v}{t}[/tex]
[tex]a = \dfrac {62.67}{4.1}[/tex]
[tex]a = 15.26\;\rm m/s^2[/tex]
Hence we can conclude that the acceleration of the object at time t = 4.1 seconds is 15.26 m/s2.
To know more about the acceleration, follow the link given below.
https://brainly.com/question/12134554.
A rigid tank whose volume is unknown is divided into two parts by a partition. One side of the tank contains an ideal gas at 935°C. The other side is evacuated and has a volume twice the size of the part containing the gas. The partition is now removed and the gas expands to fill the entire tank. Heat is now transferred to the gas until the pressure equals the initial pressure. Determine the final temperature of the gas.
Final answer:
The final temperature of the gas, after expanding and being heated until the pressure returns to the initial pressure, is 3351°C.
Explanation:
The question involves thermodynamics and the behavior of an ideal gas when it expands and is then heated to reach the initial pressure. Initially, the gas occupies one-third of the tank's volume at a temperature of 935°C. When the partition is removed, the gas expands to fill the entire tank, which is three times its initial volume. Assuming an ideal gas and the process to be isobaric (constant pressure) during heating, we apply the ideal gas law and the concept of absolute temperature to determine the final temperature.
Initially, the gas temperature is 935°C, which is 1208K (since absolute temperature in Kelvin = temperature in Celsius + 273). Upon expansion, the volume triples without specifying pressure change, but we'll assume adiabatic free expansion for this initial step, meaning temperature remains unchanged. The gas is then heated to return to its initial pressure. Since the volume tripled and pressure returns to initial, using the relation P1V1/T1 = P2V2/T2, where P1=P2 (initial and final pressures are equal), V2=3V1 (final volume is three times the initial volume), and T1=1208K, we find that the final temperature (T2) can be obtained by rearranging the equation to T2 = T1(V2/V1). Thus, the final temperature is 3 times the initial temperature in Kelvin, or 3624K, which is 3351°C.
When a board with a box on it is slowly tilted to a larger and larger angle, common experience shows that the box will at some point "break loose" and start to accelerate down the board. The box begins to slide once the component of its weight parallel to the board, , equals the maximum force of static friction. Which of the following is the most general explanation for why the box accelerates down the board after it begins to slide (rather than sliding with constant speed)?Once the box is moving, w|| is greater than the force of static friction but less than the force of kinetic friction.Once the box is moving, w|| is less than the force of static friction but greater than the force of kinetic friction.The coefficient of kinetic friction is less than the coefficient of static friction.When the box is stationary, w|| equals the force of static friction, but once the box starts moving, the sliding reduces the normal force, which in turn reduces the friction.
Answer:
the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic frictionExplanation:
This exercise uses Newton's second law with the condition that the acceleration is zero, by the time the body begins to slide. At this point the balance of forces is
fr- w || = 0
The expression for friction force is that it is proportional to the coefficient of friction by normal.
fr = μ N
When the system is immobile, the coefficient of friction is called static coefficient and has a value, this is due to the union between the surface, when the movement begins some joints are broken giving rise to coefficient of kinetic friction less than static.
In consequence a lower friction force, which is why the system comes out of balance and begins to accelerate.
μ kinetic <μ static
In all this movement the normal with changed that the angle of the table remains fixed.
Consequently, the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic friction
The cart has a velocity vC = 2.5 ft/sec to the right. Determine the angular speed N (positive if counterclockwise, negative if clockwise) of the wheel so that point A on the top of the rim has a velocity
(a) equal to 2.5 ft/sec to the left,
(b) equal to zero, and
(c) equal to 5.0 ft/sec to the right.
Answer:
Part a)
[tex]\omega = 5 ft/s[/tex] counterclockwise
Part b)
[tex]\omega = 2.5 ft/s[/tex] counterclockwise
Part c)
[tex]\omega = 2.5 ft/s[/tex] clockwise
Explanation:
Let the cart has radius R = 1 ft
so here we have speed of the center of the cart is
[tex]v_c = 2.5 ft/s[/tex]
let the angular speed is given as
[tex]\omega[/tex] counter clockwise
Part a)
if the top most point of the rim has same speed as that of speed of the center but it is towards left
so we have
[tex]v = v_c + r\omega[/tex]
[tex]-2.5 = 2.5 + 1(\omega)[/tex]
so we have
[tex]\omega = -5 ft/s[/tex]
Part b)
if the speed of the top point on the rim is zero
[tex]v = v_c + 1(\omega)[/tex]
[tex]0 = 2.5 + \omega[/tex]
[tex]\omega = -2.5 ft/s[/tex]
Part c)
if the speed at the top position on the rim is 5 ft/s
[tex]5 ft/s = 2.5 ft/s + 1(\omega)[/tex]
[tex]\omega = 2.5 ft/s[/tex]
A 44 kg child jumps off a 2.2 kg skateboard that was moving at 8.0 m/s. The skateboard comes to a stop as a result. Find the speed at which the child jumped from the board. Show all your work. Assume a frictionless, closed system.
Answer:
[tex]v_f=8.4m/s[/tex]
Explanation:
We need to use the conservation of momentum Law, the total momentum is the same before and after the kid leaves the skate.
In the axis X:
[tex](m_{kid}+m_{skate})*v_{o}=m_{kid}*v_f[/tex] (1)
[tex]v_f=(m_{kid}+m_{skate})*v_{o}/m_{kid}=(44+2.2)*8/44=8.4m/s[/tex]
A bullet is fired and a bullet is dropped simultaneously from the same height. The bullet that is fired hits the ground first.
True False
Answer: false, they hit the ground at the same time.
Explanation:
You are standing on a straight stretch of road and watching the motion of a bicycle; you choose your position as the origin. At one instant, the position of the bicycle is negative and its velocity is positive. Is the bicycle getting closer to you or farther away? Explain.
Answer: Ok, if the position of the bicycle is negative, then you can think is in the -x range, lets call it -r.
And the velocity is positive, you know that the movement equation of something is x= v*t + x0
where v is te velocity, x0 the position and t the time.
so in our case x = v*t - r.
so when v*t = r, for some time, the bicycle will be on your position.
So the bicycle was getting closer to you.
A soccer ball is kicked from the ground with an initial speed of 20.1 m/s at an upward angle of 47.8˚. A player 53.0 m away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground? Neglect air resistance.
Answer:3.95 m/s
Explanation:
Given
Initial velocity(u)[tex]=20.1 m/s[/tex]
launch angle[tex]=47.8^{\circ}[/tex]
Player is 53 m away
Range of projectile
[tex]R=\frac{u^2sin2\theta }{g}[/tex]
[tex]R=\frac{20.1^2\times sin95.6}{9.8}[/tex]
R=41.02 m
so he need to run 53-41.02=11.98 m
Time of flight of projectile [tex]=\frac{2usin\theta }{g}[/tex]
T=3.03 s
thus average speed of boy [tex]s_{avg}=\frac{11.98}{3.03}=3.95 m/s[/tex]
Two identical balls are at rest and side by side at the top of a hill. You let one ball, A, start rolling down the hill. A little later you start the second ball, B, down the hill by giving it a shove. The second ball rolls down the hill along a line parallel to the path of the first ball and passes it. At the instant ball B passes ball A:
a. it has the same position and the same velocity as A.b. it has the same position and the same acceleration as A.c. it has the same velocity and the same acceleration as A.d. it has the same displacement and the same velocity as A.e. it has the same position, displacement and velocity as A.
Answer:
Option b. it has the same position and the same acceleration as A
Explanation:
Let's analyze every statement:
a. it has the same position and the same velocity as A
In the instant where B passes A, they Do have the same position. Velocity however, cannot be the same because if they were, ball B would never pass ball A. So, this is false.
b. it has the same position and the same acceleration as A
As we said in the previous option, the position is the same. The acceleration is gravity for both balls, so this is true.
c. it has the same velocity and the same acceleration as A
Acceleration is the same but velocities are not, so this is false.
d. it has the same displacement and the same velocity as A
The distance they have traveled is the same, so the displacement is the same, but the velocity is not, so this is false.
e. it has the same position, displacement and velocity as A
The position and displacement is the same but not velocity, so this is false.
Only option b is true.
You have a pure (24-karat) gold ring with mass Gold has an atomic mass of and an atomic number of 79. (a) How many protons are in the ring, and what is their total positive charge? (b) If the ring carries no net charge, how many electrons are in it?
Answer:
Explanation:
(a)
Gold has an atomic mass of 197 g/mol means 197 g of gold contains 6.023×10^{-23} atoms.
therefore number of atoms in 17.4 g is
[tex]n= \frac{6.23\times10^{23}\times17.4}{197}[/tex]
[tex]5.32\times10^{22}[/tex]
Number of protons in each atom is 79.
therefore Total number of protons is 5.32×10^{22}×79 =4.2×10^24. protons
Their total positive charge is 4.2×10^{24}×1.6×10^{-19) =6.72×10^5 C.
(b)
equal number of protons should be there to neutralize the material therefore number of electrons is 4.2×10^(24.)
An uncharged molecule of DNA (deoxyribonucleic acid) is 2.14 µm long. The ends of the molecule become singly ionized so that there is a charge of −1.6 × 10−19 C on one end and +1.6 × 10−19 C on the other. The helical molecule acts like a spring and compresses 1.4% upon becoming charged. Find the effective spring constant of the molecule. The value of Coulomb’s constant is 8.98755 × 109 N · m2 /C 2 and the acceleration due to gravity is 9.8 m/s 2 . Answer in units of N/m.
Answer:
K = 1.72 *10^{-11} N/m
Explanation:
given data:
L= 2.14*10^{-6} m
[tex]\Delta = 1.4%* of L[/tex]
[tex]= \frac{1.4}{100} *2.14*10^{-6}[/tex]
= 2.99*10^{-8} m
[tex]L' = L - \Delta L[/tex]
[tex]L ' = 2.14*10^{-6} -2.99*10^{-8}[/tex]
[tex]L = 2.11*10^{-6} m[/tex]
electrostatic force = spring force
[tex]\frac{ kq^2}{L'^2} = K \Delta L[/tex]
putting all equation to get spring constant K value
[tex]\frac{8.98*10^9 *1.6*10^{-19}}{2.11*10^{-6}} = K 2.99*10^{-8}[/tex]
K = 1.72 *10^{-11} N/m