Answers
Given displacement of a particle moving in a straight line,
[tex]S=t^2-7t+17[/tex](i) Calculate the average velocity in the interval [3,4]
[tex]\begin{gathered} \text{Average velocity = }\frac{S(4)-S(3)}{4-3} \\ \text{Average velocity = }\frac{\lbrack(4)^2-7\times4+17\rbrack-\lbrack(3)^2-7\times3+17\rbrack}{4-3} \\ \text{Average velocity =}\frac{5-5}{1} \\ \text{Average velocity = 0 m/s} \end{gathered}[/tex](iii) Calculate the average velocity in the time interval [4,5].
[tex]\begin{gathered} \text{Average velocity = }\frac{S(5)-S(4)}{5-4} \\ \text{Average velocity = }\frac{\lbrack(5)^2-7\times5+17\rbrack-\lbrack(4)^2-7\times4+17\rbrack}{5-4} \\ \text{Average velocity = }\frac{7-5}{1} \\ \text{Average velocity = }2 \end{gathered}[/tex](iv) Calculate the average velocity in the time interval [4,4.5].
[tex]\begin{gathered} \text{Average velocity = }\frac{S(4.5)-S(4)}{4.5-4} \\ \text{Average velocity = }\frac{\lbrack(4.5)^2-7\times4.5+17\rbrack-\lbrack(4)^2-7\times4+17\rbrack}{4.5-4} \\ \text{Average velocity = }\frac{5.75-5}{0.5} \\ \text{Average velocity = }1.5 \end{gathered}[/tex]Related Questions
I want to know if I’m doing this correctly, and if not, then what would be the best process to do solve it?
Answers
Given
[tex]3\times10^{-10}m^{}[/tex]
Factor Name Symbol
10-1 deci d
10-2 centi c
10-3 milli m
10-6 micro µ
10-9 nano n
10-12 pico p
10-15 femto f
10-18 atto a
10-21 zepto z
10-24 yocto y
a. nanometers
0.3 nm
[tex]3\times10^{-1}nm^{}[/tex]b. picometers
300 pm
[tex]3\times10^2pm[/tex]Assume a water strider has a roughly circular foot of radius 0.0203 mm. The surface tension of water is 0.0700 N/m.A. What is the maximum possible upward force on the foot due to surface tension of the water? NB. What is the maximum mass of this water strider so that it can keep from breaking through the water surface? The strider has six legs. mg
Answers
Part (A)
The maximum possible upward force acting on the foot is,
[tex]F=2\pi r\sigma[/tex]Substitute the known values,
[tex]\begin{gathered} F=2(3.14)(0.0203\text{ mm)(}\frac{10^{-3}\text{ m}}{1\text{ mm}})(0.0700\text{ N/m)} \\ =8.9\times10^{-6}\text{ N} \end{gathered}[/tex]Thus, the maximum possible upward force on the foot is
[tex]8.9\times10^{-6}\text{ N}[/tex]Part (B)
The maximum force due to six legs can be expressed as,
[tex]6F=mg[/tex]Substitute the known values,
[tex]\begin{gathered} 6(8.9\times10^{-6}N)=m(9.8m/s^2) \\ m=\frac{6(8.9\times10^{-6}\text{ N)}}{9.8m/s^2}(\frac{1kgm/s^2}{1\text{ N}}) \\ =(5.45\times10^{-6}\text{ kg)(}\frac{1\text{ mg}}{10^{-6}\text{ kg}}) \\ =5.45\text{ mg} \end{gathered}[/tex]Thus, the maximum mass of water strider is 5.45 mg.
Which shape fits a position vs. time graph of an object that is slowing down? Which shape fits a position vs. time graph of an object that is speeding up?
Answers
1.
In a position x time graph, if the velocity is constant, so the position increases at the same rate over the time, so we have a linear relation between the position and the time.
Therefore the shape that represents this relation is C.
2.
In a velocity x time graph, if the velocity is constant, its value is always the same over the time, it doesn't change. That is represented graphically by a horizontal line, therefore the shape that represents this relation is B.
.
You decide to roll a 0.10 kg ball across the floor so slowly that it will have a small momentum and a large de Broglie wavelength. If you roll it at 1.4x1^-3 m/s, what is its wavelength? Express your answer to two significant figures and include the appropriate units. How will the answer from above compare with the de Broglie wavelength of the high-speed electron that strikes the back face of one of the early models of a TV screen at 1/10 the speed of light (2x10^-11 m) ?
Answers
ANSWER:
(a) 4.73*10^-30 m
(b) 2.37*10^-19 times smaller
STEP-BY-STEP EXPLANATION:
Given:
Mass of ball = m = 0.10 kg
Speed of ball = v = 1.4x10^-3 m/s
(a)
Since, de Broglie wavelength is given by:
[tex]\lambda=\frac{h}{mv}[/tex]Where, h is the Plank's Constant ( h = 6.626x10^-34 kg m^2/s). Therefore, de Broglie wavelength of the ball will be:
[tex]\begin{gathered} \lambda=\frac{6.626\cdot10^{-34}}{0.10\cdot1.4\cdot10^{-3}} \\ \lambda_{\text{ball}}=4.73\cdot10^{-30} \end{gathered}[/tex](b)
[tex]\begin{gathered} \lambda_{electron}=2\cdot10^{-11} \\ \frac{\lambda_{\text{ball}}}{\lambda_{electron}}=\frac{4.73\cdot10^{-30}}{2\cdot10^{-11}} \\ \frac{\lambda_{\text{ball}}}{\lambda_{electron}}=2.37\cdot10^{-19} \end{gathered}[/tex]It means that the wavelength of the ball is 2.37*10^-19 times smaller
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Which of the following best represents R= A - B ?
Please help, it’s due soon!
Answers
The option (C) best represents the R = A- B
What is vector in mathematics?A vector in mathematics is a quantity that not only expresses magnitude but also motion or position of an object in relation to another point or object. Euclidean vector, geometric vector, and spatial vector are other names for it.
In mathematics, a vector's magnitude is defined as the length of a segment of a directed line, and the vector's direction is indicated by the angle at which the vector is inclined.
What are the components in vector quantity?A vector primarily consists of two elements, the horizontal component and the vertical component. The horizontal component's value is cosθ, and the vertical component's value is sinθ.
There are two types of vector multiplication, they are dot products and cross products.
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Calculate the kinetic energy of a roller coaster that has a mass of 1,778.6 kg and is traveling with a velocity to 24.5 m/s at the bottom of the first hill.
Answers
Given:
The mass of the roller coaster is m = 1778.6 kg
The velocity of the roller coaster is v = 24.5 m/s
To find the kinetic energy of the roller coaster.
Explanation:
The formula to calculate the kinetic energy is
[tex]K\mathrm{}E\text{. =}\frac{1}{2}mv^2[/tex]Substituting the values, the kinetic energy will be
[tex]\begin{gathered} K\mathrm{}E\text{. =}\frac{1}{2}\times1778.6\times(24.5)^2 \\ =533802.325\text{ J} \end{gathered}[/tex]Final Answer: The kinetic energy of the roller coaster is 533802.325 J.
What is the wavelength of the sound produced by a bat if the frequency of the sound is 90 kHz on a night when the air temperature is 22°C?
Answers
We know that the wavelength is related to the speed of the wave by:
[tex]v=\lambda f[/tex]where f is the frequency.
The speed of sound on air at a given temperature is given by:
[tex]v=331\sqrt[]{1+\frac{T}{273}}[/tex]so in this case the speed is:
[tex]v=331\sqrt[]{1+\frac{22}{273}}=344.08[/tex]Plugging this and the frequency in the first expression above we have:
[tex]\begin{gathered} 344.08=90\times10^3\lambda \\ \lambda=\frac{344.08}{90\times10^3} \\ \lambda=3.82\times10^{-3} \end{gathered}[/tex]Therefore the wavelength is:
[tex]3.82\times10^{-3}\text{ m}[/tex]PLEASE HELPPPPPPPPPPPPPPPPPPPPPPP
Answers
Answer:
right answer is death valley
Explanation:
because it is close to surface gravitational field
mexico
cus its the farthest
Megan kicks a soccer ball with a mass of 2 kg. The ball leaves the ground moving 50 meters per second. What is the kinetic energy of the ball?
Answers
ANSWER
[tex]2500J[/tex]EXPLANATION
The kinetic energy of a body is the energy it possesses due to its motion and it can be found by applying the formula:
[tex]E=\frac{1}{2}mv^2[/tex]where m = mass, v = velocity
From the question:
[tex]\begin{gathered} m=2\text{ kg} \\ v=50\text{ m/s} \end{gathered}[/tex]Therefore, the kinetic energy of the ball is:
[tex]\begin{gathered} E=\frac{1}{2}\cdot2\cdot50^2 \\ E=2500J \end{gathered}[/tex]A sailboat starts from rest and accelerates at a rate of 0.13 m/s? over a distance of 344 m.(a) Find the magnitude of the boat's final velocitym/s (b)find the time it takes the boat to travel this distance
Answers
We know that
• It starts from rest. (The initial velocity is zero).
,• The acceleration rate is 0.13 m/s^2.
,• The distance covered is 344 m.
To find the magnitude of the boat's final velocity, we have to use the following formula.
[tex]v^2_f=v^2_0+2ad[/tex]Let's use the given magnitudes.
[tex]\begin{gathered} v^2_f=0^2+2(0.13(\frac{m}{s^2}))(344m) \\ v^2_f=89.44(\frac{m^2}{s^2}) \\ v_f=\sqrt[]{89.44(\frac{m^2}{s^2})} \\ v_f\approx9.46(\frac{m}{s}) \end{gathered}[/tex](a) Therefore, the final velocity is 9.46 meters per second.Now, to find the time it takes the boat to travel this distance, we use the following formula.
[tex]d=v_0\cdot t+\frac{1}{2}at^2[/tex]Using the given magnitudes, we have the following.
[tex]344m=\frac{1}{2}(0.13(\frac{m}{s^2}))t^2[/tex]Let's solve for t.
[tex]\begin{gathered} t=\sqrt[]{\frac{2\cdot344m}{0.13(\frac{m}{s^2})}} \\ t=\sqrt[]{\frac{688m}{0.13}}\sec \\ t\approx72.75\sec \end{gathered}[/tex](b) Therefore, it takes 72.75 seconds.40) Climbing the Empire State Building A new record for running the stairs of the Empire State Building was set on February 4, 2003. The 86 flights, with a total of 1576 steps, was run in 9 minutes and 33 seconds. If the height gain of each step was 0.20 m, and the mass of the runner was 70.0 kg, what was his average power output during the climb? Give your answer in both watts and horsepower.
Answers
The power is given by:
[tex]P=\frac{W}{t}[/tex]where W is the work and t is the time.
We know that the work done is related to the gravitational potential energy by:
[tex]W=mg(y_f-y_0)[/tex]where yf is the final height of the object and y0 is the initial height.
Now, in this case we have a total of 1576 steps, each of them with a height of 0.2 meters that means that the total height gained is:
[tex](1576m)\cdot0.2=315.2\text{ m}[/tex]This in turns means that the work done is:
[tex]W=(70\operatorname{kg})(9.8\frac{m}{s^2})(315.2m)=216227.2\text{ J}[/tex]Now, the time it takes to achieve this is 9 minutes 33 seconds, this is the same as:
[tex]9(60)+33=573\text{ s}[/tex]Finally we use the power formula:
[tex]P=\frac{216227.2\text{ J}}{573\text{ s}}=377.36\text{ W}[/tex]Now we need to remember that 1 Hp is equal to 745.7 W, then we have that this is the same as:
[tex]377.36\text{ W}\cdot\frac{1\text{ Hp}}{745.7\text{ W}}=0.506[/tex]Therefore the power to make that climb is 377.36 W or 0.506 Hp
name the quatity whose SI unit is j/kg/°c
Answers
Answer:
Explanation:
Specific heat:
[ c ] = 1 J / (kg·°C)
Jacob Grena raises a spoon 0.210 m above a table . If the spoon and its contents have a mass of 30.0 g, what is the gravitational potential energy associated with the spoon at that height relative to the table's surface?
Answers
Explanation:
The spoon is raised so it gains Gravitational potential energy. Formula to find Gravitational potential energy;
Gravitational potential energy = mass × Gravitational field strength × height of the body from the surface (table in this scenario)
In symbols; E = m×g×h
Substitute values:
m = 30g = 0.03kg (don't forget to convert grams to kg)
g = 10N/kg
h = 0.210 m
So it's;
0.03kg × 10N/kg × 0.210m = 0.063 Joules
SI Unit of energy is joules
carts, bricks, and bands
8. What acceleration results when 2 rubber bands stretched to 20 cm are used to pull a cart with three bricks?
a. About 0.25 m/s2
b. About 0.33 m/s2
c. About 0.50 m/s2
d. About 1.00 m/s2
Answers
A. The acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with three bricks is 0.25 m/s².
What is the applied force on an object?The force applied on object is obtained by multiplying the mass and acceleration of the object.
According to Newton's second law of motion, the force exerted on an object is directly proportional to the product of mass and acceleration of the object.
Also, the applied force is directly proportional to the change in the momentum of the object.
Mathematically, the force acting on object is given as;
F = ma
a = F/m
where;
a is the acceleration of the objectm is the mass of the objectF is the applied forceFrom the trials, the acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with three bricks is 0.25 m/s².
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What are the units for buoyant force?
Answers
Answer Unit of Buoyant Force The unit of the buoyant force is the Newton (N). = Here, F= buoyant force
Explanation:
i hope this helpes
For each letter, write a word to describe its role:A + B = C D x E = F
Answers
A is an addend
B is an addend
C is a sum
D is a factor
E is a factor
F is a product
Explanations:Note:Numbers(or characters) that are added together with the addition operator are called addends
The result of an addition operation is called sum
Numbers that are multiplied together are called factors
The result of a multiplication operator is called product
Considering the definitions above:In A + B = C
A is an addend
B is an addend
C is a sum
In D x E = F
D is a factor
E is a factor
F is a product
In a chosen coordinate system, the position of an object in motion can have negative values.Question 2 options:TrueFalse
Answers
True
Explanations:The position of an object can take any sign, it can be negative, zero or positive. This depends on the coordinate system chosen.
The position of an object (whether moving or static) is specified with respect to a frame of reference, and can be positive, zero or negative depending on the coordinate system chosen
Therefore, we can conclude that, in a chosen coordinate system, the position of an object in motion can have negative values.
After reading your textbook, you are able to maintain the bold, key words in coded representations in a network of neurons in your brain. In memory, this process is called
Answers
In memory, the process in which after reading your textbook, you are able to maintain the bold, key words in coded representations in a network of neurons in your brain is called storage.
In memory there are three phases. They are:
Encoding StorageRetrievalIn the given scenario, the process in which converting the bold, key words to coded representations is encoding process. The process which maintains the coded representations in a network of neurons in your brain is storage. The process of remembering the information stored when needed is retrieval.
Therefore, in memory, this process is called Storage
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A student fires a cannonball vertically upwards. The cannonball returns to the ground after a 3.80s flight. Determine all unknowns and answer the following questions. Neglect drag and the initial height and horizontal motion of the cannonball. Use regular metric units (ie. meters).How long did the cannonball rise? What was the cannonball's initial speed? What was the cannonball's maximum height?
Answers
The time taken by ball to rise is given as,
[tex]t=3.80\text{ s}[/tex]Therefore, the total time taken to move the ball in the upward direction is calculated as,
[tex]\begin{gathered} t_0=\frac{3.80\text{ s}}{2} \\ =1.90\text{ s} \end{gathered}[/tex]Therefore, the ball rise for 1.90 s.
The final speed of the ball can be given as,
[tex]v=u+gt[/tex]At the maximum height the final speed of ball is zero.
Substitute the known values,
[tex]\begin{gathered} 0m/s=u+(9.8m/s^2)(1.90\text{ s)} \\ u=-(9.8m/s^2)(1.90\text{ s)} \\ =-18.62\text{ m/s} \end{gathered}[/tex]Therefore, the initial speed of the ball is -18.62 m/s where negative sign indicates the direction of ball.
The final speed of the ball can be given as,
[tex]v^2=u^2-2gh_m[/tex]At the maximum height the final speed is zero. Substitute the known values,
[tex]\begin{gathered} (0m/s)^2=(-18.62m/s)^2-2(9.8m/s^2)h_m \\ h_m=\frac{(-18.62m/s)^2}{2(9.8m/s)^2} \\ =17.7\text{ m} \end{gathered}[/tex]Therefore, the maximum height of the canon is 17.7 m.
Carol Gillian theorized that when it comes to a perspective of Justice,males per socialized for a blank environment while females are socialized for a blank environment
Answers
Answer: men = work environment , women = home environment
Explanation: Gillian proposed that women come to prioritize as “ethics of care” and men as “ethics of justice”.
a projectile starting from ground hits a target on the ground located at the distance of 1000m after 40 sec
Answers
A projectile starting from ground hits a target on the ground located at a distance of 1000 meters after 40 seconds.
a) The size of the angle θ is = 83°
b)The initial velocity was the projectile launched is (u)=205.13 m/s.
What is velocity?The velocity is a physical term that is refer to how much the object has covered the distance in a given time. It can be measured in m/s and cm/s.
How can we calculate the velocity?a) To calculate the angle we are using two formulas, they are
r= u cos(θ)T
T= 2u sin (θ)/g
Here we are given,
r= The distance covered in the motion = 1000m.
T = The time covered in the motion = 40 Seconds.
g= The acceleration due to gravity = 9.8 m/s²
We have calculate the values of angle = θ
Now we put the values in above equation we get,
1000= u cosθ*40...(1)
40*g= 2u sinθ.....(2)
Equation(2) divided by equation(1) we get,
2tanθ=40*40*g/1000
Or, tanθ=7.84
Or, θ= 82.7°≈83°
From the above calculation we can say that, The size of the angle θ is = 83°
b) The initial velocity was the projectile launched is = u
As we know, r= u cos(θ)T
Now we put the values in the equation we get,
1000= u cos(83)*40
Or, u=205.13 m/s
From the above calculation we can say that, The initial velocity was the projectile launched is (u)=205.13 m/s.
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A projectile starting from ground hits a target on the ground located at a distance of 1000 meters after 40 seconds.
a) What is the size of the angle θ?
b) At what initial velocity was the projectile launched?
2. A 14000 kg air jet accelerates from rest to 70 m/s before it takes off. What is the changein momentum of the jet?
Answers
Answer:
980,000 kg m/s
Explanation:
The change in momentum can be calculated as
[tex]\begin{gathered} \Delta p=m\Delta v \\ \Delta p=m(v_f-v_i) \end{gathered}[/tex]Where m is the mass, vf is the final velocity and vi is the initial velocity. Replacing m = 14000 kg, vf = 70 m/s and vi = 0 m/s, we get
[tex]\begin{gathered} \Delta p=14000\text{ kg \lparen}70\text{ m/s - 0 m/s\rparen} \\ \Delta p=14000\text{ kg \lparen70 m/s\rparen} \\ \Delta p=980000\text{ kg m/s} \end{gathered}[/tex]Therefore, the change in momentum of the jet is 980,000 kg m/s
Block 1 has a mass of 12 kg is moving to the right on a levelsurface at a speed of 2 m/s. Block 2 has a mass of 2.5 kg andis at rest on the surface. Block 1 collides with block 2, causingblock 2 to move to the right with a speed of 4 m/s. How fast,and in what direction, is block 1 moving after the collision?
Answers
Given:
The mass of block 1, m₁=12 kg
The mass of block 2, m₂=2.5 kg
The velocity of block 1 before the collision, u=2 m/s
The velocity of block 2 after the collision, v₂=4 m/s
To find:
The velocity of block 1 after the collision.
Explanation:
From the law of conservation of momentum, the total momentum of the blocks before the collision will be equal to the total momentum of the blocks after the collision.
Thus,
[tex]m_1u=m_1v_1+m_2v_2[/tex]Where v₁ is the velocity of block 1 after the collision.
On rearranging the above equation,
[tex]v_1=\frac{m_1u-m_2v_2}{m_1}[/tex]On substituting the known values,
[tex]\begin{gathered} v_1=\frac{12\times2-2.5\times4}{12} \\ =1.17\text{ m/s} \end{gathered}[/tex]The positive sign of the velocity indicates that block 1 will continue to move to the right.
Final answer:
The velocity of block 1 after the collision will be 1.17 m/s and its direction is to the right.
Two children are riding on a merry-go-round. Child A is at a greater distance from the axis of rotation than child B. Which child hasthe larger linear displacement?Select one:a. There is not enough information given to answer the question.b. Child Ac. They have the same non-zero linear displacementd. Child Be. They have the same zero linear displacement.
Answers
The linear displacement in a rotation is given by:
[tex]s=r\theta[/tex]where r is the distance from the axis of rotation and theta is the angular displacement.
Since the linear displacement is proportional to the radius we conclude that child A has a larger linear displacement.
A block of mass 0.500 kg slides on a flat smooth surface with a speed of 2.80 m/s. It then slides over a rough surface with μk and slows to a halt. While the block is slowing, (a) What is the frictional force on the block? (b) What is the magnitude of the block’s acceleration? (c) How far does the block slide on the rough part before it comes to a halt?
Answers
a ) The frictional force on the block = 1.47 N
b ) The magnitude of the block’s acceleration = - 2.94 m / s²
c ) Distance travelled on rough part before it comes to a halt = 1.33 m
m = 0.5 kg
v = 2.8 m / s
Since there is no vertical motion,
∑ [tex]F_{y}[/tex] = 0
N - mg = 0
N = 0.5 * 9.8
N = 4.9 N
μ = 0.3
[tex]f_{k}[/tex] = μ N
[tex]f_{k}[/tex] = 0.3 * 4.9
[tex]f_{k}[/tex] = 1.47 N
The net force acting on the block is due to friction,
F = [tex]f_{k}[/tex] = 1.47 N
F = m a
1.47 = 0.5 * a
a = 2.94 m / s²
Since, acceleration is towards the opposite of motion,
a = - 2.94 m / s²
v² = u² + 2 a s
0 = 2.8² + ( 2 * - 2.94 * s )
s = 1.33 m
Therefore,
a ) The frictional force on the block = 1.47 N
b ) The magnitude of the block’s acceleration = - 2.94 m / s²
c ) Distance travelled on rough part before it comes to a halt = 1.33 m
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A penguin runs 29,000 m/s how far will it travel in 10 seconds
Answers
Speed is measured as the ratio of distance to the time in which the distance was covered. Speed is a scalar quantity as it has only direction and no magnitude.
[tex]s=\frac{d}{t}[/tex]s = speed
d = distance
t = time
[tex]\begin{gathered} d=st \\ d=29000\text{ m/s}\cdot10s \\ d=290000\text{ m} \end{gathered}[/tex]The distance would be 290,000 m
When x = –4, what is the value of g(x)? g(x)=-5x+3 help please
Answers
Replace the value x=-4 into the expression for g(x) and simplify, as follow:
[tex]\begin{gathered} g(-4)=-5(-4)+3 \\ g(-4)=20+3 \\ g(-4)=23 \end{gathered}[/tex]Hence, g(-4) = 23
Please help me with this! (Sadly my previous tutor couldn't help me with this)
Answers
a.
The free body diagram (not at scale) for each crate is shown below:
b.
In this case we know that the force is just sufficient to keep the crates from sliding, this means that the acceleration of the system is zero.
From the free body diagram and Newton's second law we have that for the 45 kg crate that:
[tex]\begin{gathered} T-W=0 \\ T=W \\ T=(45)(9.8) \\ T=441 \end{gathered}[/tex]For the 35 kg crate the equations of motion would be:
[tex]\begin{gathered} F+f_f-T=0 \\ N-W^{\prime}=0 \end{gathered}[/tex]but we know that the force of friction is given by:
[tex]F_f=\mu N[/tex]and from the second equation of motion we have that:
[tex]\begin{gathered} N=W^{\prime} \\ N=(35)(9.8) \\ N=343 \end{gathered}[/tex]Then we have that:
[tex]\begin{gathered} F+F_f-T=0 \\ F+343\mu-441=0 \end{gathered}[/tex]Since the crates are not moving we need to use the static coefficient of friction, then:
[tex]\begin{gathered} F+343\mu-441=0 \\ F+343(0.5)-441=0 \\ F+171.5-441=0 \\ F-269.5=0 \\ F=269.5 \end{gathered}[/tex]Therefore the force applied is 269.5 N
c.
The diagram in this case is:
d.
In this case we know that the 35 kg is sliding to the right at constant velocity, this means that the acceleration for the system is zero. (Notice that the difference with the previous case is that the friction points to the left)
From the discussion in part b we know that for the 45 kg block:
[tex]T-W=0[/tex]and then:
[tex]T=441[/tex]For the 35 kg we have that:
[tex]\begin{gathered} F-F_f-T=0 \\ N-W^{\prime}=0 \end{gathered}[/tex]from the previos discussion we know that:
[tex]N=343[/tex]and since in this case the crates are moving we need to use the kinetic coefficient of friction, then we have:
[tex]\begin{gathered} F-F_f-T=0 \\ F-(0.3)(343)-441=0 \\ F-102.9-441=0 \\ F-543.9=0 \\ F=543.9 \end{gathered}[/tex]Therefore in this case the force applied is 543.9 N
e.
In this case the free body diagram is:
f.
Since the crates are moving with an accelearion of 0.5 m/s^2 we have for the 45 kg crate that:
[tex]T-W=ma[/tex]from where:
[tex]\begin{gathered} T-(9.8)(45)=(45)(0.5) \\ T-441=22.5 \\ T=441+22.5 \\ T=463.5 \end{gathered}[/tex]For the 35 kg crate we have that:
[tex]\begin{gathered} F-F_f-T=m^{\prime}a \\ N-W^{\prime}=0 \end{gathered}[/tex]from the previous discussion we know that N=343, plugging this in the first equation we have:
[tex]\begin{gathered} F-(0.3)(343)-463.5=(35)(0.5) \\ F-102.9-463.5=17.5 \\ F-566.4=17.5 \\ F=566.4+17.5 \\ F=583.9 \end{gathered}[/tex]Therefore the force applied in this case is 583.9 N
You turn a corner and are driving up asteep hill. Suddenly, a small boy runs out on the street chasing a ball. You slam on the brakes and skid to astop leaving a 50-foot-long skid mark on the street. The boy calmly walks away but a policemen watching fromthe sidewalk walks over and gives you a speeding ticket. He points out that the speed limit on this street is 25mph. After you recover your wits, you begin to examine the situation. You determine that the street makes anangle of 25◦with the horizontal and that the coefficient of static friction between your tires and the street is0.80. You also find that the coefficient of kinetic friction between your tires and the street is 0.60. Your car’sinformation book tells you that the mass of your car is 1600 kg. You weigh 140 lbs.
Answers
Given data:
Total displacement of the car;
[tex]s=50\text{ ft}[/tex]Speed limit;
[tex]v_m=25\text{ mph}[/tex]The angle of street from horizontal;
[tex]\theta=25\degree[/tex]Coefficient of static friction;
[tex]\mu_s=0.80[/tex]Coefficient of kinetic friction;
[tex]\mu_k=0.60[/tex]Mass of the car;
[tex]M=1600\text{ kg}[/tex]Weight of the man;
[tex]W=140\text{ lbs}[/tex]The kinetic friction force is given as,
[tex]F_k=\mu_k(M+m)g\cos \theta[/tex]Here, m is the mass of the man and g is the acceleration due to gravity.
The acceleration of the car driving up a steep hill is given as,
[tex]\begin{gathered} (M+m)g\sin \theta+F_k=(M+m)a \\ (M+m)g\sin \theta+\mu_k(M+m)g\cos \theta=(M+m)a \\ g\sin \theta+\mu_kg\cos \theta=a \end{gathered}[/tex]Substituting all known values,
[tex]\begin{gathered} (32\text{ ft/s}^2)\times\sin (25\degree)+0.6\times(32\text{ ft/s}^2)\times\cos (25\degree)=a \\ \approx30.92\text{ ft/s}^2 \end{gathered}[/tex]The velocity of the car is given as,
[tex]v^2=u^2-2as[/tex]Here, v is the final velocity (v=0, as the car stops), and u is the initial velocity.
The initial velocity of the car is given as,
[tex]u=\sqrt[]{v^2+2as}[/tex]Substituting all known values,
[tex]\begin{gathered} u=\sqrt[]{0^2+2\times(30.92\text{ ft/s}^2)\times(50\text{ ft})} \\ \approx55.61\text{ ft/s} \\ \approx37.91\text{ mph} \end{gathered}[/tex]Therefore, your speed is greater than the speed limit. Thus, you can not fight the ticket in the court.
b. Andrea has a particularly old bike, and she has not lubricated the chain orgears. This causes friction inside the gears as she rides. She begins coasting onflat ground at a speed of 10 m/s, and after a few seconds is only going 8 m/s.How much energy has been converted to heat? (1 point) (The combined mass is 50kg)c. Andrea is going 10 m/s toward a hill. She coasts up the hill without pedaling. Iffriction causes 10% of her energy to be converted to heat inside the gears, howhigh up the hill will she be able to coast? (2 points)
Answers
b.
The energy converted in heat is the change in kinetic energy, that is, the kinetic energy lost is the same as the heat.
The kinetic energy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]The change in kinetic energy is given by:
[tex]K_f-K_0=\frac{1}{2}mv^2_f-\frac{1}{2}mv^2_0[/tex]Plugging the values given we have that:
[tex]\begin{gathered} K_f-K_0=\frac{1}{2}(50)(8)^2-\frac{1}{2}(50)(10)^2 \\ =-900 \end{gathered}[/tex]Hence Andrea has lost 900 J of energy and therefore 900 J is converted to heat.
c.
We know that the energy is conserved, in this case this means that the initial kinetic energy has to be equal to the energy lost by heat in the gears and the potential energy gained by climbing up the hill, that is:
[tex]K=Q+U[/tex]where K is the kinetic energy, Q is the heat and U is the potential energy.
We know that 10% of her energy is converted in heat, this means that:
[tex]Q=0.1K[/tex]and hece we have:
[tex]\begin{gathered} K=0.1K+U \\ U=0.9K \end{gathered}[/tex]the potential energy is given by:
[tex]U=mgh[/tex]then we have that:
[tex]mgh=0.9(\frac{1}{2}mv^2)[/tex]Plugging the values given and solving for h we have:
[tex]\begin{gathered} (50)(9.8)h=0.9(\frac{1}{2}(50)(10^2)) \\ 490h=2250 \\ h=\frac{2250}{490} \\ h=4.59 \end{gathered}[/tex]Therefore Andrea will be able to climp up to 4.59 meters