Answer:
(a) [tex]5.88\times 10^{12}mi[/tex]
Explanation:
We have given speed of light [tex]c=3\times 10^8m/sec[/tex]
Given time = 1 year =365 days
We know that in 1 minute = 60 sec
1 hour = 60×60 = 3600 sec
In one day = 24 hour = 24×60×60=86400 sec
So in 365 days = 365×86400[tex]=3.1536\times 10^7sec[/tex]
We know that distance = speed ×time[tex]=3.1536\times 10^7\times 3\times 10^8=9.46\times 10^{15}m[/tex]
We have given that 1 mi = 1609 m
So [tex]9.46\times 10^{15}m=\frac{9.46\times 10^{15}}{1609}=5.88\times 10^{12}mi[/tex]So option (a) is correct
The flywheel of a steam engine runs with a constant angular speed of 158 rev/min. When steam is shut off, the friction of the bearings and the air brings the wheel to rest in 2.6 h. What is the magnitude of the constant angular acceleration of the wheel in rev/min^2? Do not enter the units.
Answer:
α=1.013rev/min^2
Explanation:
Wo=initial angular speed=158rev/min
t=time=2.6h=156min
Wf=final angular speed=0rev /min , this happens because in the end the wheel stops
this is a circular motion with uniform acceleration therefore we can use the following equation
α=angular aceleration=(Wf-Wo)/t
α=(0-158)/156=1.013rev/min^2
A powerboat, starting from rest, maintains a constant acceleration. After a certain time 't', its displacement and velocity are 'r' and 'v'. At time '2t', what will be its displacement and velocity, assuming the acceleration remains constant?
a) 2r and 2v
b) 2r and 4v
c) 4r and 2v
d) 4r and 4v
Answer: Ok, so you know the acceleration, lets call it A.
now, the velocity will take the form of V= A*t + v0, where v0 is the inicial velocity, in this case the boat starts from the rest, so v0 = 0
integrating again you obtain R = (A*t*t)/2 + r0, and we will take r0 = 0.
so, at a time t₁ we have a velocity V = v = A*t₁
R = r = (A*t₁*t₁)/2
so a t₂=2*t₁
V= A*2*t₁= 2v
R= 0.5*A*t₁*t₁*4 = 4r
so the answer is c.
The mass of a radioactive substance follows a continuous exponential decay model, with a decay rate parameter of 1% per day. A sample of this radioactive substance has an initial mass of 2.5kg. Find the mass of the sample after five days.
Answer:
2,38kg
Explanation:
Mass in function of time can be found by the formula: [tex]m_{(t)} =m_{0} e^{-kt}[/tex], where [tex]m_{0}[/tex] is the initial mass, t is the time and k is a constant.
Given that a sample decay 1% per day, that means that after first day you have 99% of mass.
[tex]m_{(1)} =m_{0} e^{-k(1)}[/tex], but [tex]m_{(1)}=\frac{99m_{0} }{100}[/tex], so we have [tex]\frac{99m_{0} }{100}=m_{0}e^{-k}[/tex], then [tex]k=-ln(\frac{99}{100})=0.01[/tex]
Now using k found we must to find [tex]m_{(5)}[/tex].
[tex]m_{(5)}=m_{0}e^{-(0.01)5}=2.5kge^{-0.05} =2.5x0.951=2.38kg[/tex]
The formula for exponential decay can be used to calculate the mass of a radioactive sample after a certain time. The equation N = N0 * e^{(-decay rate*t)} can be used, with the decay rate expressed as a negative number. Substituting the given values will give the remaining mass after 5 days.
Explanation:The mass of a radioactive substance after a certain period can be calculated using the formula for exponential decay, which in this context states that the remaining mass of the substance is equal to its initial mass multiplied by e to the power of the decay rate multiplied by the time.
Given the initial mass (N0) is 2.5kg, the decay rate is 1% (expressed as -0.01 in the formula), and the time (t) is 5 days, the equation is:
N = N0 * e^{(-decay rate*t)}
This simplifies to:
N = 2.5 * e^{(-0.01*5)}
You can calculate this on a scientific calculator or use a programming language with a command for the base of natural logarithms (often designated by 'e').
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Mike's car, which weighs 1,000 kg, is out of gas. Mike is trying to push the car to a gas station, and he makes the car go 0.05 m/s2. Using Newton's Second Law, you can compute how much force Mike is applying to the car.
Answer:
The answer to your question is: F = 50 N
Explanation:
Data
mass = 1000 kg
acceleration = 0.05m/s2
F = ?
Formula
F = m x a
Substitution
F = 1000 kg x 0.05 m/s2 = 50 kgm/s2 = 50 N
Mike is applying a force of 50 N to the car.
The magnitude of applied force on the car by Mike is 50 N.
Given data:
The mass of car is, m = 1000 kg.
The magnitude of acceleration of car is, [tex]a = 0.05 \;\rm m/s^{2}[/tex].
According to Newton's second law of motion, the force applied on the object is expressed as the product of mass of object and magnitude of acceleration caused by the applied force on the object.
Therefore,
[tex]F = m \times a[/tex]
Here, F is magnitude of applied force on car.
Solving as,
[tex]F = 1000 \times 0.05\\F = 50 \;\rm N[/tex]
Thus, we can conclude that Mike is applying 50 N of force on his car.
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The electrical resistance of dry skin is 100 kΩ, but can be lowered to 20 Ω if electrode contact area is large and conducting gel is used on the skin. If Vmax of the defibrillatror 500 V and lasts 0.01 s, what is the maximum possible current delivered to the heart during defibrillation?
Answer:
25 Ampers
Explanation:
As the current in a defibrillation is DC (Direct current), we can use simply the Ohm's law which shows:
V=I*R
Then if we want to know the current, we can follow the next step for I:
I=V/R
Using the information, and asumming electrode contact area is large and conducting gel is used on the skin (as is normal) , we can solve:
Imax=Vmax/R
Imax= 500 V / 20 ohms
Imax= 25 Ampers
*Fact: The time in this case doesn't make any difference in the result.
An aluminum clock pendulum having a period of 1.0000 s keeps perfect time at 20.0°C. When placed in a room at a temperature of -10.2°C, will the clock gain time (run fast) or lose time (run slow)? use 24 x 10^-6 for the expansion coefficient of aluminum.
The aluminum pendulum clock would gain time and run faster in a colder environment due to decreased thermal expansion. The period of the pendulum shortens due to the decrease in length of the pendulum. The exact time difference can be calculated using the expansion coefficient of aluminum and the given formulae.
Explanation:The behavior of the pendulum clock in different temperatures is subject to thermal expansion. The formula P=2π√L/g helps us understand that the period of the pendulum (P) is directly related to the pendulum's length (L). When the temperature decreases, the aluminum pendulum's length will also decrease due to the lower thermal expansion. Consequently, the pendulum's period will shorten causing the clock to gain time, or run faster.
For a more precise calculation, we can use the expansion coefficient of aluminum provided in the question which is 24 x 10^-6. The change in length due to temperature change can be calculated by using the formula ΔL = L0αΔT, where L0 is the original length, α is the linear coefficient of expansion and ΔT is the change in temperature. Hence, the time the clock gains or loses can be calculated with the help of these relations.
Note: It's crucial to recall that while this principle holds, in reality, pendulum clocks are often adjusted to maintain accuracy in varying temperatures.
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The drop in temperature will cause the aluminum pendulum to contract, making it shorter. This will result in the clock running faster and thus gain time.
Explanation:The question you're asking relates to the phenomenon of thermal expansion and its effect on a pendulum clock. In short, the temperature change will affect the accuracy of the clock, due to the expansion coefficient of aluminum.
When the temperature decreases from 20.0°C to -10.2°C, the aluminum pendulum will contract, due to the negative temperature change multiplied by the expansion coefficient of aluminum (24 x 10^-6). As a result, the length of the pendulum decreases which in turn, using the period equation P = 2π√L/g results in a shorter period.
As a result, the clock will run fast, or gain time. To understand why, consider that when the pendulum is shorter, it swings faster and hence takes less time to complete one swing. That means the clock ticks faster than normal, making it gain time.
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Which of the following have quantized values? Explain your selections. a. the pitch of a note played on a slide trombone b. the pitch of a note played on a flute c. the wavelengths of light produced by the heating elements in a toaster d. the wind speed at the top of Mt. Everest
Answer:
Option b and c
Explanation:
Quantized values are discrete in nature and are not continuous.
These values can be expressed as integral multiples or integers.
Therefore,
The pitch of notes played on trombone are of continuous form and are not quantized.The pitch of the notes that are played on a flute are dependent on the number of open and closed holes on the instrument, clearly discrete in nature thus quantized.The wavelength of light is produced as a result of transition of electron in between energy levels which occurs in discrete steps thus quantized.The wind speed at Mt. Everest's top is continuous and thus not quantized.Earth is slightly closer to the Sun in January than in July. How does the area swept out by Earth's orbit around the Sun during the 31 days of January compare to the area swept out during the 31 days of July?
a. Both areas are the same.
b. The area swept out in January is larger.
c. The area swept out in July is larger.
Answer: Option a.
Explanation:
Kepler's 2nd law of planetary motion states:
A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
It tells us that it doesn't matter how far Earth is from the Sun, at equal times, the area swept out by Earth's orbit it's always the same independently from the position in the orbit.
A ball is thrown horizontally from the top of a 60 m building and lands 100 m from the base of the building. How long is the ball in the air? What must have been the initial horizontal component of the velocity? What is the vertical component of the velocity just before the ball hits the ground? What is the velocity of the ball just before it hits the ground?
Answer:
The ball is in the air for 3.5 seconds
The initial horizontal component of velocity is 28.6 m/s
The vertical component of the final velocity is 34.3 m/s downward
The final velocity is 44.7 m/s in the direction 50.2° below the horizontal
Explanation:
A ball is thrown horizontally
That means the vertical component of the initial velocity [tex]u_{y}=0[/tex]
The initial velocity is the horizontal component [tex]u_{x}[/tex]
The ball is thrown from the top of a 60 m
That means the vertical displacement component y = 60 m
→ y = [tex]u_{y}[/tex] t + [tex]\frac{1}{2}[/tex] gt²
where g is the acceleration of gravity and t is the time
y = -60 m , g = -9.8 m/s² , [tex]u_{y}=0[/tex]
Substitute these values in the rule
→ -60 = 0 + [tex]\frac{1}{2}[/tex] (-9.8)t²
→ -60 = -4.9t²
Divide both sides by -4.9
→ 12.2449 = t²
Take √ for both sides
∴ t = 3.5 seconds
* The ball is in the air for 3.5 seconds
The initial velocity is the horizontal component [tex]u_{x}[/tex]
The ball lands 100 meter from the base of the building
That means the horizontal displacement x = 100 m
→ x = [tex]u_{x}[/tex] t
→ t = 3.5 s , x = 100 m
Substitute these values in the rule
→ 100 = [tex]u_{x}[/tex] (3.5)
Divide both sides by 3.5
→ [tex]u_{x}[/tex] = 28.57 m/s
The initial horizontal component of velocity is 28.6 m/s
The vertical component of the final velocity is [tex]v_{y}[/tex]
→ [tex]v_{y}[/tex] = [tex]u_{y}[/tex] + gt
→ [tex]u_{y}[/tex] = 0 , g = -9.8 m/s² , t = 3.5 s
Substitute these values in the rule
→ [tex]v_{y}[/tex] = 0 + (-9.8)(3.5)
→ [tex]v_{y}[/tex] = -34.3 m/s
The vertical component of the final velocity is 34.3 m/s downward
The final velocity v is the resultant vector of [tex]v_{x}[/tex] and [tex]v_{y}[/tex]
→ Its magnetude is [tex]v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}[/tex]
→ Its direction [tex]tan^{-1}\frac{v_{y}}{v_{x}}[/tex]
→ [tex]v_{y}[/tex] = 28.6 , [tex]v_{y}[/tex] = -34.3
Substitute this values in the rules above
→ [tex]v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66[/tex]
→ Its direction [tex]tan^{-1}\frac{-34.3}{28.6}=-50.18[/tex]
The negative sign means the direction is below the horizontal
The final velocity is 44.7 m/s in the direction 50.2° below the horizontal
The ball remained in the air for 3.49 seconds. The initial horizontal velocity was approximately 28.65 m/s. Just before hitting the ground, the ball had a vertical velocity of approximately 34.21 m/s downwards and an overall velocity of approximately 45.1 m/s at an angle of roughly 49.8 degrees below the horizontal.
Explanation:The time a projectile, such as the ball in this case, is in the air is completely determined by vertical motion. To calculate this, we can use the equation, y = y0 + v0yt - 0.5gt^2, where y is the final position, y0 is the initial position (the height of the building, 60 m in this case), v0y is the initial vertical velocity (which is 0 since the ball is thrown horizontally), g is acceleration due to gravity (approximated as 9.8 m/s^2), and t is the time we are looking for. By setting y to 0 (the base of the building where the ball lands), we can solve for t which gives us 3.49 seconds.
Next, the initial horizontal component of the velocity can be calculated using distance/time, which in this case is 100 m (distance traveled by the ball) divided by 3.49 seconds (the time the ball was in the air), giving us approximately 28.65 m/s.
The vertical component of the velocity just before the ball hit the ground can be calculated using v = v0y + gt, where v0y is still 0 and t is the time the ball is in the air. This calculation gives us approximately 34.21 m/s, in the downward direction, taking into account that acceleration due to gravity acts downward.
By combining the horizontal and vertical components, we can determine the velocity of the ball just before it hits the ground. The magnitude of this velocity can be found by using the Pythagorean theorem (v = sqrt((v0x)^2 + (v0y)^2)) which will give us approximately 45.1 m/s and the angle it makes with the horizontal can be calculated through the formula theta = tan^-1(v0y/v0x) giving us approximately 49.8 degrees below the horizontal.
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Calculate the amount of heat transferred when 710 grams of water warms from an initial temperature of 4.0 ºC to a final temperature of 25.0 ºC. The specific heat capacity of liquid water is 4.184 J/g ºC.
Answer:
Q = 62383.44 Joules
Explanation:
Given that,
Mass of water, m = 710 gm
Initial temperature of water, [tex]T_i=4^{\circ} C[/tex]
Final temperature of water, [tex]T_f=25^{\circ} C[/tex]
The specific heat capacity of liquid water is, [tex]c=4.184\ J/g\ ^oC[/tex]
Heat transferred is given by :
[tex]Q=mc(T_f-T_i)[/tex]
[tex]Q=710\times 4.184\times (25-4)[/tex]
Q = 62383.44 Joules
So, the amount of heat transferred is 62383.44 Joules. Hence, this is the required solution.
If a child pulls a sled through the snow on a level path with a force of 60 N exerted at an angle of 39° above the horizontal, find the horizontal and vertical components of the force.
Explanation:
Given that,
Force with which a child pulls a sled, F = 60 N
It is at an angle of 39° above the horizontal. We need to find the horizontal and vertical components of the force.
The horizontal component is given by :
[tex]F_x=30\ cos(39)=23.31\ N[/tex]
The vertical component is given by :
[tex]F_y=30\ sin(39)=18.87\ N[/tex]
So, the horizontal and vertical components of the force are 23.31 N and 18.87 N. Hence, this is the required solution.
To find the horizontal and vertical components of the force, use the formulas Fx = F * cos(angle) and Fy = F * sin(angle), respectively.
Explanation:To find the horizontal and vertical components of the force, we can use trigonometry. The horizontal component of the force can be found using the formula: Fx = F * cos(angle), where F is the magnitude of the force and angle is the angle above the horizontal. Plugging in the values, we get Fx = 60 N * cos(39°) = 45.85 N.
The vertical component of the force can be found using the formula: Fy = F * sin(angle), where F is the magnitude of the force and angle is the angle above the horizontal. Plugging in the values, we get Fy = 60 N * sin(39°) = 36.34 N.
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Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m . If the track is completely flat and the race car is traveling at a constant 27.5 m/s (about 62 mph ) around the turn, what is the race car's centripetal (radial) acceleration?
What is the force responsible for the centripetal acceleration in this case?
friction, normal, gravity, or weight?
Answer:
[tex]a_c = 13.26 m/s^2[/tex]
Friction Force
Explanation:
As we know that centripetal force is the product of mass and centripetal acceleration
so we know that
[tex]a_c = \frac{v^2}{R}[/tex]
so here we have
[tex]v = 27.5 m/s[/tex]
[tex]R = 57 m[/tex]
so we have
[tex]a_c = \frac{27.5^2}{57}[/tex]
[tex]a_c = 13.26 m/s^2[/tex]
This acceleration is given by the force which may be towards the center of the circular path
Here in the above case it is possible due to friction force.
Final answer:
The centripetal acceleration of the race car is 3.45 m/s². Centripetal force is calculated to be 1725 N, and the normal force is responsible for the centripetal acceleration.
Explanation:
The centripetal acceleration of the race car is 3.45 m/s². Centripetal force on the race car can be calculated using the formula F = m * a, where m is the mass of the car and a is the centripetal acceleration. This makes the centripetal force 1725 N. In this case, the force responsible for the centripetal acceleration is the normal force.
Starting from home, you bicycle 25 km north in 2.4 h and then turn around and pedal straight home in 1.4 h . What are your (a) displacement at the end of the first 2.4 h , (b) average velocity over the first 2.4 h , (c) average velocity for the homeward leg of the trip, (d) displacement for the entire trip, and (e) average velocity for the entire trip
Answer:
Part a)
[tex]d = 25 km[/tex]
Part b)
[tex]v_{avg} = 10.42 km/h[/tex]
Part c)
[tex]v_{avg} = 17.86 km/h[/tex]
Part d)
Displacement for entire trip = 0
Part e)
Average velocity for entire trip will be zero
Explanation:
Part a)
Displacement after t = 2.4 hours is the straight line distance between initial and final positions
so we have
[tex]d = 25 km[/tex]
Part b)
Average velocity is defined as
[tex]v_{avg} = \frac{displacement}{time}[/tex]
[tex]v_{avg} = \frac{25 km}{2.4 h}[/tex]
[tex]v_{avg} = 10.42 km/h[/tex]
Part c)
During his return journey the displacement will be same
[tex]displacement = 25 km[/tex]
[tex]time = 1.4 h[/tex]
so average velocity is defined as
[tex]v_{avg} = \frac{25 km}{1.4 h}[/tex]
[tex]v_{avg} = 17.86 km/h[/tex]
Part d)
Displacement for entire trip = 0
as initial and final position will be same
Part e)
Average velocity for entire trip will be zero
Final answer:
The displacement after 2.4 hours of cycling is 25 km north. The average velocity is 10.42 km/h north for the first 2.4 hours, 17.86 km/h south for the return trip, and the average velocity for the entire trip is zero because the displacement is zero.
Explanation:
To address the questions posed by the student, let's examine each part of the cycling trip:
(a) Displacement at the end of the first 2.4 h
The displacement after the first 2.4 hours is the straight-line distance from the starting point to the turnaround point, which would be 25 km directly north. Displacement is a vector quantity that refers to the change in position of an object.
(b) Average velocity over the first 2.4 h
The average velocity is the displacement divided by the time taken. Since the cyclist traveled 25 km north in 2.4 hours, we calculate the average velocity as 25 km / 2.4 h = 10.42 km/h north.
(c) Average velocity for the homeward leg of the trip
On the way back, the cyclist travels 25 km to the south (back home) in 1.4 h. The average velocity for this would be 25 km / 1.4 h = 17.86 km/h south.
(d) Displacement for the entire trip
Since the cyclist returns straight home, the displacement for the entire trip is zero, because there is no net change in position from the starting point.
(e) Average velocity for the entire trip
The total distance travelled is 25 km out and 25 km back, for a total of 50 km. The total time taken is 2.4 h for the first leg and 1.4 h for the return, which adds up to 3.8 h. However, the displacement is zero, therefore the average velocity for the entire trip is zero.
Which of the following is the best example of a proper hypothesis? Orange popsicles taste sweeter than grape popsicles. If two objects are dropped at the same height, the heavier object will hit the ground first. Children who wear backpacks do better academically. I think there is no effect on hearing if you listen to loud music only three hours a day.
Answer:
If two objects are dropped at the same height, the heavier object will hit the ground first.
Explanation:
A proper hypothesis has the form:
"If ... , then ... ".
Now, why the other sentence aren't hypothesis?
Orange popsicles taste sweeter than grape popsicles.This sentence its an OPINION. It may be true that, for the speaker/writer, orange popsicles taste sweeter, but this is an subjective opinion, and difficult to test.
Children who wear backpacks do better academically.This sentence its an STATEMENT. If we wish to make an proper hypothesis, it should take the form:
If children wear backpacks, then they do better academically.
I think there is no effect on hearing if you listen to loud music only three hours a day.This sentence its another OPINION. This opinion COULD be tested, but the proper way of write it as an hypothesis would be:
If someone listed to loud music only three hours a day, then there shoul be no effect on his/her hearing.
Answer:
Answer is:If two objects are dropped at the same height, the heavier object will hit the ground first.
Explanation:
After robbing a bank, a criminal tries to escape from the police by driving at a constant speed of 55 m/s (about 125 mph). A police officer, initially at rest, starts chasing the criminal when he drives by, accelerating at a constant rate of 6.1m/s2. How long will it take the officer to catch the criminal?
Answer:
18.03 s
Explanation:
We have two different types of motions, the criminal moves with uniform motion while the police do it with uniformly accelerated motion. Therefore we will use the equations of these cases. We know that by the time the police reach the criminal they will have traveled the same distance.
[tex]x=vt\\x=x_{0}+v_{0}t+\frac{a}{2}t^2[/tex]
The distance between the police and the criminal when the first one starts the persecution is 0, its initial speed is also zero. So:
[tex]x=(55m/s)t\\x=\frac{6.1m/s^2}{2}t^2=(3.05m/s^2)t^2[/tex]
Equalizing these two equations and solving for t:
[tex](55m/s)t=(3.05m/s^2)t^2\\(3.05m/s^2)t^2-(55m/s)t=0\\t((3.05m/s^2)t-55m/s)=0\\t=0 \\(3.05m/s^2)t-55m/s=0\\t=\frac{55m/s}{3.05m/s^2}=18.03 s[/tex]
The cost of producing q newspaper articles is given by the function C = f(q) = 100 + 2q. a.) Find a formula for the inverse function. b.) Explain in practical terms what the inverse function tells you.
Answer:
a.) [tex]f^{-1} (q) =\frac{C - 100}{2}[/tex]
b.) The inverse function allows me to understand the number of newspapers articles I can produce using different amounts of money to spend on production costs.
Explanation:
Finding the inverse function
The cost of production function is the mathematical formula that tells how much cost is required to produce certain amount of newspaper articles. The invere function then is the one that tell us the reversed concept: how many newspapers article you can produce if you have certain amount of money for production. How do we get the inverse function? As the function is solved for C (cost of production), we need now to solve it for q (quantity of newspapers articles). This process is as follows:
[tex]C = 100 + 2q[/tex] (That is the cost function)Substract 100 from the function: [tex]C - 100 = 2q[/tex]Divide by 2: [tex]\frac{c - 100}{2} = q[/tex]Reorganizing: [tex]q = \frac{C - 100}{2}[/tex]That is okay, but what does it mean to us?
The inverse function would at first glance seem to mean the same than the original function but said in other way, but this function it is very useful by itself. Let's think about this case: If you are the head of the printing house and you have a limited budget to produce the articles. How do you know you have enough money to produce the number of articles for all your readers in the city? Just replace C for your budget and see if the amount of articles will be enough. Just as simple as replacing values. Mathematics at service of printing houses.
High-speed stroboscopic photographs show that the head of a 160-g golf club is traveling at 43 m/s just before it strikes a 46-g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 31 m/s. Find the speed of the golf ball just after impact.
Answer:
Vb = 41.74m/s
Explanation:
Let Voc be the velocity of the club before the collision, Vfc the velocity of the club after the collision, Vfb the velocity of the ball after the collision.
The masses of the club and the ball expressed in kg are:
mc = 0.16kg mb = 0.046kg
By conservation of the moment:
Po = Pf
mc*Voc + mb *0 = mc*Vfc + mb*Vfb Solving for Vfb:
[tex]Vfb = \frac{mc}{mb}*(Voc - Vfc) = 41.74m/s[/tex]
High‑speed ultracentrifuges are useful devices to sediment materials quickly or to separate materials. An ultracentrifuge spins a small tube in a circle of radius 13.1 cm at 50000.0 rev/min. What is the centripetal force experienced by a sample that has a mass of 0.00310 kg?
The force needed to move a body in a curved way is understood as centripetal force. The centripetal force experienced by a sample will be 7.14 N.
What is centripetal force?The force needed to move a body in a curved way is understood as centripetal force. This is a force that can be sensed from both the fixed frame and the spinning body's frame of concern.
The direction of centripetal force is always in the path of the center of the course.
The given data in the problem;
r is the radius =13.1 cm=0.13 m
n is the rpm = 50000.0 rev/min
To find velocity first we have to go to angular velocity
[tex]\omega = \frac{2\pi n}{60} \\\\ \omega = \frac{2\times3.14\times50000}{60} \\\\\omega = 132.63\; rad/sec.[/tex]
The centripetal force is given as
[tex]\rm F_C= m\omega^{2} r\\\\F_C= 0.00310(132.63)^2 \times 0.131\\\\\rm{F_C=7.14 N[/tex]
Hence The centripetal force experienced by a sample will be 7.14 N.
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Theophylline, an asthma medicine, is to be prepared from an elixir with a drug concentration of 5 mg/mL and a cherry-flavored syrup that is to be added to hide the taste of the drug. How much of each must be used to prepare 100 milliliters of solution with a drug concentration of 2 mg/mL?
Answer:
[tex]V_{syrup}= 60ml[/tex] and [tex]V_{elixir}= 40ml[/tex]
Explanation:
Well, we need 100ml of solution, with a concentration of 2mg/ml. So the total amount of drug we need is :
[tex]M_{drug} =2\frac{mg}{ml}\times100ml = 200mg[/tex]
But the drug concentration is 5mg/ml, so the amount of elixir needed to get 200mg of drug is:
[tex]V_{elixir}= \frac{200mg}{5mg/ml}=40ml[/tex]
And so the amount of cherry flavoured syrup needed is:
[tex]V_{syrup}= 100-40 = 60ml[/tex]
Hope my answer helps.
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A 2 kg ball of putty moving to the right at 3 m/s has a perfectly inelastic, head-on collision with a 1 kg ball of putty moving at 2 m/s. What is the final magnitude and direction (left or right) of the velocity of the balls after the collision?
Answer:
[tex]V=1.33m/s[/tex] to the right
Explanation:
The balls collide in a completely inelastic collision, in other words they have the same velocity after the collision, this velocity has a magnitude V.
We need to use the conservation of momentum Law, the total momentum is the same before and after the collision.
In the axis X:
[tex]m_{1}*v_{o1}-m_{2}*v_{o2}=(m_{1}+m_{2})V[/tex] (1)
[tex]V=(m_{1}*v_{o1}-m_{2}*v_{o2})/(m_{1}+m_{2})=(2*3-1*2)/(2+1)=1.33m/s[/tex]
A car accelerates at a constant rate from zero to 33.7 m/s in 10 seconds and then slows to 17.6 m/s in 5 seconds. What is its average acceleration to the nearest tenth of a m/s2 during the 15 seconds?
Answer:
[tex]a_{avg} = 1.17 m/s^{2}[/tex]
Given:
initial velocity, u = 0
final velocity, v = 33.7 m/s
t = 10 s
final velocity, v' = 17.6 m/s
t' = 5 s
Total time, T = 10 + 5 = 15 s
Solution:
The rate of change of velocity of an object is referred to as the acceleration of that object.
Average accelaeration, [tex]a_{avg} = \frac{\Deta v}{\Delta t}[/tex]
Now,
Initial acceleration of the body, [tex]a = \frac{v - u}{t}[/tex]
[tex]a = \frac{33.7 - 0}{10} = 3.37 m/s^{2}[/tex]
Now, average acceleration during the 15 seconds:
[tex]a = \frac{v' - u}{T}[/tex]
[tex]a = \frac{17.6 - 0}{15} = 1.17 m/s^{2}[/tex]
Imagine that you are driving in your car with a heavy bag of groceries on the front seat and you slam on your brakes to stop. The car stops but the heavy bag continues to move forward. The concept that best explains why your bag of groceries moves forward is _
Answer:
The correct answer is inertia.
Explanation:
The heavy bag of groceries is initially within the inertia frame of the car. This indicates that the heavy bag acquires the same speed as the car.
When the car stops, the heavy bag continues to move forward with the speed it had due to the principle of inertia, which states the property that the bodies cannot modify by themselves the state of rest or movement in which they are.
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The Bohr model was determined ultimately to be flawed. Knowing that "opposites attract (and 'likes' repel)," can you think of any flaws in the Bohr model that disobey these statements?
Answer:
Yes, there are 2 flaws
Explanation:
Electromagnetics say that, given two particles with charge, they will be attracted if they have opposite charge or repelled if they have identical charge.
Then, we can find two flaws in the Bohr model. The first one is that the electrons move in energy layers (or just layers to get the idea) far from the proton compared with the distance between the electrons themselves. So, if the model was right, how can it be that they don't repel each other? With the same logic, the protons don't repel each other either even though they are all together in the nucleus.
The second flaw, related to what we've just said, is that the electrons can move from one layer to another, but they will always stay at a minimum distance to the proton. How can it be so if it is known that they attract because of their charge signs?
You measure water in two containers: a 10-mL graduated cylinder with marks at every mL, and a 1-mL pipet marked at every 0.1 mL. If you have some water in each of the containers and add them together, to what decimal place could you report the total volume of water?
Answer:
To a tenth of a mililiter (0.1mL)
Explanation:
When considering the precision of a measurement system, a more precise intrument needs to record smaller intervals of data; a better resolution. In this case we have a cylinder with a resolution of 1 mL. When we pour the amount of water we can't precisely measure anything less than 1 mL UNLESS it is added to the pipet described in this problem which records measurements of up to 0.1 mL (a tenth of a mililiter). Thus the total measurement system can only report a resolution of up to 0.1 mL
A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .
(a) How much later does it reach the bottom of the cliff?
(b) What is its speed just before hitting?
(c) What total distance did it travel?AAI8yoZA helicopter is ascending vertically with a speed of 5.40 m/s. At a height of 105 m above the Earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground?
Answer:
(a) 5.8s
(b) 41.36m/s
(c) 99.52m
5.21s
Explanation:
(a) This is the total time it takes the stone to reach its maximum height above the cliff and strike the ground at the base of the cliff after projection.
let the height attained by the stone above the cliff be [tex]h_1[/tex] and the time taken to attain this height be [tex]t_1[/tex]. We can safely assume acceleration due to gravity to be taken as [tex]g=9.8m/s^2[/tex].
We use the first equation of motion under free fall to obtain [tex]t_1[/tex] as follows;
[tex]v=u-gt_1............(1)[/tex]
given: u = 15.5m/s
Where [tex]v[/tex] is the final velocity and u is the initial velocity. The negative sign in the equation indicates the fact that the stone is moving upwards against gravitational pull. The final velocity [tex]v=0[/tex] at height [tex]h_1[/tex] because the stone will momentarily at the maximum height come to rest before it begins to fall back downwards.
Hence from equation (1) we obtain the following,
[tex]0=15.5-9.8t_1\\9.8t_1=15.5\\hence\\t_1=15.5/9.8\\t_1=1.58s[/tex]
To get [tex]h_1[/tex] we use the third equation as follows;
[tex]v^2=u^2-2gh_1[/tex] ( the body is moving upward so g is negative)
[tex]0^2=15.5^2-2*9.8*h_1\\0=240.25-19.6h_1\\19.6h_1=240.25\\therefore\\h_1=240.25/19.6\\h_1=12.26m[/tex]
Next we obtain the time it takes to fall back from the maximum height downwards to the base of the cliff. Let this time be [tex]t_2[/tex]. We use the second equation of motion.
[tex]H=ut+gt_2^2/2............(2)[/tex]
( g is positive because the stone is falling downwards)
However in this case, u = 0 because the stone is falling freely from rest downwards.
[tex]H=h_1+75m=12.26+75\\H=87.26m\\[/tex]
Substituting into equation (2), we obtain;
[tex]87.26=(0*t_2)+9.8t_2^2/2[/tex]
Simplifying further we obtain;
[tex]4.9t_2^2=87.26\\t_2^2=87.26/4.9\\ =17.81\\t_2=\sqrt{17.81}=4.22s[/tex]
Hence the total time spent in air = 1.58+4.22 = 5.8s
(b) We use the third equation of motion to find the velocity with which the stone strikes the ground.
[tex]v^2=u^2+2gH....... (3)[/tex]
the stone is falling downwards in this case from height H from rest, u = 0, v is the final velocity with which is strikes the ground. Equation (3) can therefore be reduced to the following form by putting u = 0;
[tex]v=\sqrt{2gH}\\v=\sqrt{2*9.8*87.26} \\v=41.36m/s[/tex]
(c) The total distance travelled is given as follows;
[tex]H_{total}=h_1+H\\H_{total}=12.26+87.26=99.52m[/tex]
When the package was dropped from the ascending helicopter, it will be projected upwards with an initial velocity equal to that of the helicopter, attain a maximum height and then fall back downwards. The total time spent in air by the package is the sum of the time it takes to attain maximum height and the time it takes to fall to the ground from the maximum height. This solution is similar to that of part (a) of this question.
To find the time it takes to attain maximum height, we use equation (1): v = 0, u = 5.4m/s and g is negative since the package is moving upward against gravity. Hence;
[tex]0^2=5.4^2+9.8t_1\\9.8t_1=5.4\\t_1=5.4/9.8\\t_1=0.55s[/tex]
Similarly to the previous solution, we obtain the maximum height as follows;
[tex]v^2=u^2-2gh_1\\0^2=5.4^2-2*9.8*h_1\\19.6h_1=29.16\\h_1=29.16/19.6\\h_1=1.49m[/tex]
therefore maximum height is
H = 105+1.49 = 106.49m
The time taken by the package to fall from H to the ground is given by equation (2), where u = 0 since the package is falling from rest; g is positive in this case.
[tex]106.49=(0*t)+ 9.8t^2/2\\106.49=4.9t^2\\t^2=106.49/4.9=21.73\\t=\sqrt{21.73}=4.66s[/tex]
therefore the total time spent by the package before striking the ground is given by;
[tex]t_{total}=0.55s+4.66s=5.21s[/tex]
The time it takes for a stone thrown upward from a cliff to reach the bottom, and its speed just before impact, are found using kinematic equations that consider initial velocity, height of the cliff, and acceleration due to gravity.
Explanation:When a stone is thrown vertically upward with an initial speed from the edge of a cliff, we can calculate how long it takes to reach the bottom by using the equations of motion under constant acceleration due to gravity. In this case, we can use the kinematic equation:
s = ut + ½at²
Where s is the displacement, u is the initial velocity, t is time, and a is the acceleration due to gravity (9.81 m/s²). Let's break down the initial question into parts (a) and (b):
Part (a):
Determine how much later the stone reaches the bottom of the cliff by solving the equation for t, considering that the stone must travel the height of the cliff plus the additional distance it ascends before starting to fall back down.
Part (b):
The speed just before hitting can be calculated using the equation v = u + at, where v is the final velocity. For a comprehensive calculation, more elaborate explanations and use of kinematic formulas are required to solve for the time of flight, final velocity, and total distance traveled.
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Suppose 16.0 g of oxygen (O2) is heated at constant atmospheric pressure from 26.4°C to 111°C. (a) How many moles of oxygen are present? (Take the molar mass of oxygen to be 32.0 g/mol) (b) How much energy is transferred to the oxygen as heat? (The molecules rotate but do not oscillate.) (c) What fraction of the heat is used to raise the internal energy of the oxygen?
Answer:
a)n=0.5 mol
b)n=0.5 mol
c)U/Q=0.714
Explanation:
Given that
Mass of oxygen m= 16 g
Heated from 26.4°C to 111°C
Molar weight M=32 g/mol
a)
Number of mole,n
n=m/M
n=16/32 mol
n=0.5 mol
b)
Heat,Q
Q= n Cp ΔT
As we know that it is diatomic gas so
Cp=(7/2)R
Q=7 R n ΔT/2
Now by putting the values
[tex]\Delta U=\dfrac{7nR\Delta T}{2}[/tex]----1
[tex]Q=\dfrac{7\times 8.314\times 0.5\times (111-26.4)}{2}[/tex]
Q=1230.88 J
We know that internal energy
[tex]\Delta U=\dfrac{5nR\Delta T}{2}[/tex]--------2
From equation 1 and 2
U/Q= 5/7
U/Q=0.714
In 16.0 g of oxygen, there are 0.500 moles of oxygen. To calculate the energy transferred to the oxygen as heat and the fraction of heat used to raise the internal energy, specific heat capacity and the degree of freedom for rotational motion are required, which are not provided.
Explanation:A student asked how many moles of oxygen are present in 16.0 g of O₂ and how much energy is transferred when the oxygen is heated from 26.4°C to 111°C at constant pressure, given that the molecules rotate but do not oscillate. The following steps provide the answers:
Determine the number of moles of oxygen: Use the molar mass of oxygen (32.0 g/mol) to convert grams to moles.Calculate the energy transferred as heat: Use the specific heat capacity of oxygen and the temperature change to find the amount of energy transferred to the oxygen (this could require accessing the specific heat capacity from a table or reference).Find the fraction of heat used to raise the internal energy: Since the oxygen molecules rotate but do not oscillate, we use the degrees of freedom for a diatomic molecule in the rotational energy mode to determine the internal energy change.To answer (a), we use the formula:
16.0 g O₂ x (1 mol O₂ / 32.00 g O₂) = 0.500 mol O₂
The student now knows that there are 0.500 moles of oxygen in 16.0 g of O₂.
Parts (b) and (c) would require additional information such as the specific heat capacity of oxygen, which was not provided in the question. Generally, the energy transferred depends on the specific heat capacity, the change in temperature, and the number of moles. The internal energy for gases increases with temperature, but for a more detailed answer, specific values are needed.
A car travels in the +x-direction on a straight and level road. For the first 3.00 s of its motion, the average velocity of the car is υav−x = 6.31 m/s .
How far does the car travel in 4.00s?
Answer:
25.24 m
Explanation:
We are given that a car travels in the positive x direction on a straight and level road.
We have to find the distance travel by car in 4 s.
Average velocity of car=6.31 m/s
Time =3 s
Distance traveled by the car in 3 sec=[tex]velocity\times time[/tex]
Distance traveled by the car in 3 sec=[tex]6.31\times 3=18.93 m[/tex]
Distance traveled by the car in 1 sec=6.93 m
Distance traveled by the car in 4 sec=[tex]6.93\times 4=25.24 m[/tex]
Hence,distance traveled by the car in 4 sec=25.24 m
Using the constant velocity of 6.31 m/s, the car travels a total of 25.24 meters in 4.00 seconds.
Explanation:To determine how far the car travels in 4.00 seconds, we can use the equation for displacement given that the car is moving with a constant velocity. Since the average velocity over the first 3.00 seconds is given as νav-x = 6.31 m/s, and there's no indication that the velocity changes, we can assume that the car continues to move at this velocity for the remaining 1.00 second.
The total displacement x, can thus be calculated using the equation x = xo + νavt. Assuming the car starts from position xo = 0, the displacement after 4.00 seconds is x = (6.31 m/s) × (4.00 s) = 25.24 m. Therefore, the car travels 25.24 meters in 4.00 seconds.
A model of a helicopter rotor has four blades, each of length 3.80 m from the central shaft to the blade tip. The model is rotated in a wind tunnel at a rotational speed of 450 rev/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?
Answer:
a)V= 179.056 m/s
b)[tex]a_{r}=843.71 g \ m/s^2[/tex]
Explanation:
Given that
Length of blade = 3.8 m
Rotational speed N= 450 rev/min
We know that
[tex]\omega =\dfrac{2\pi N}{60}\ \frac{rad}{s}[/tex]
[tex]\omega =\dfrac{2\pi \times 450}{60}\ \frac{rad}{s}[/tex]
ω=47.12 rad/s
Linear velocity
We know that linear velocity V = ω x r
Here r = 3.8 m
So by putting the values
V = ω x r
V = 47.12 x 3.8 m/s
V= 179.056 m/s
Radial acceleration
[tex]a_{r}=\omega ^2r\ m/s^2[/tex]
[tex]a_{r}=47.12^2\times 3.8 \ m/s^2[/tex]
[tex]a_{r}=8437.12 \ m/s^2[/tex]
[tex]a_{r}=843.71 g \ m/s^2[/tex]
The linear speed of the blade tip is approximately 178.6 m/s. The radial acceleration of the blade tip is approximately 864 times the acceleration of gravity.
Explanation:The two parts of this question relate to basic concepts in physics, specifically relating to circular motion and the relationship between linear speed, radial acceleration, and the acceleration of gravity. The model of a helicopter rotor in question has four blades, each of length 3.80 m, and rotates at a speed of 450 rev/min.
a. What is the linear speed of the blade tip? The linear velocity (v) of the tip of the blade can be calculated using the formula v = rω, where r is the radial distance (half the length of the blade) and ω is the angular velocity. The angular velocity can be converted to rad/s by multiplying the rotational speed (450 rev/min) by 2π rad/rev and then by 1/60 min/s, giving approximately 47 rad/s. Then, v = (3.8 m) * (47 rad/s) gives a linear velocity of approximately 178.6 m/s.
b. What is the radial acceleration of the blade tip expressed as a multiple of the acceleration of gravity? The radial (or centripetal) acceleration (a_r) can be calculated using the formula a_r = ω²r. Substituting the known values gives a_r = (47 rad/s)² * 3.8 m = approximately 8460 m/s². As the acceleration of gravity is approximately 9.8 m/s², the radial acceleration is therefore about 864 times the acceleration of gravity.
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What happens to dry ice when its temperature rises
When heated, dry ice undergoes sublimation, transitioning directly from a solid to a gas. It bypasses the liquid phase, turning directly into carbon dioxide gas.
Explanation:When the temperature of dry ice, which is the solid form of carbon dioxide, rises, it undergoes a process called sublimation. This is a phase transition where a substance moves directly from the solid phase to the gas phase, without passing through the liquid phase. So instead of melting and becoming liquid when heated, dry ice turns directly into carbon dioxide gas.
This is quite a unique characteristic, as most forms of matter have to pass through a liquid state before becoming a gas. The phenomenon of sublimation can also be observed in other substances like iodine and naphthalene (the main ingredient in mothballs).
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In the previous question, the person had an initial velocity of 8m/s and a constant acceleration of −4m/s2. How would the maximum distance he travels to the right of the origin change if instead his initial velocity were doubled (vx,0=16m/s)?
Final answer:
The maximum distance traveled to the right of the origin will change if the initial velocity is doubled.
Explanation:
To find the maximum distance traveled by the person to the right of the origin, we can use the equation:
d = v0t + 0.5at2
Given the initial velocity (vx,0) of 16 m/s, we can substitute it into the equation along with the constant acceleration (-4 m/s2) and the time (t) to calculate the maximum distance.
d = (16 m/s)(t) + 0.5(-4 m/s2)(t2)
The maximum distance will change as we vary the time, but it will be greater than the distance traveled with an initial velocity of 8 m/s.