Sheena can row a boat at 2.90 mi/h in still water. She needs to cross a river that is 1.20 mi wide with a current flowing at 2.00 mi/h. Not having her calculator ready, she guesses that to go straight across, she should head upstream at an angle of 25.0° from the direction straight across the river.What is her speed with respect to the starting point on the bank?How long does it take her to cross the river?How far upstream or downstream from her starting point will she reach the opposite bank? If upstream, enter a positive value and if downstream, enter a negative value.In order to go straight across, what angle upstream should she have headed?
Answers
Given
v = 2.90 mi/h
x = 1.20 mi
vs = 2.0 mi/h
θ = 25°
Procedure
Part a)
Velocity of the boat with respect to water stream is given as
[tex]\begin{gathered} v_b=v-v_s \\ v_b=(2.90\cdot\sin 25-2.00)\hat{j}+2.90\cdot\cos 25\hat{i} \\ v_b=-0.77\hat{j}+2.62\hat{i} \end{gathered}[/tex]magnitude of the speed is given as
[tex]\begin{gathered} v_b=\sqrt[]{0.77^2+2.62^2} \\ v_b=2.73mi/h \end{gathered}[/tex]Part b)
Time to cross the river is given as:
[tex]\begin{gathered} t=\frac{x}{v_x} \\ t=\frac{1.2mi}{2.90\cdot\cos25} \\ t=0.46h \end{gathered}[/tex]Part c)
Distance moved by boat in downstream is given as
[tex]\begin{gathered} x=v_yt \\ x=-0.77\cdot0.46 \\ x=-0.354mi \end{gathered}[/tex]Part d)
In order to go straight, we must net speed along the stream must be zero
so we will have
[tex]\begin{gathered} v\sin \theta=v_s \\ 2.90\sin \theta=2.00_{} \\ \sin \theta=\frac{2.00}{2.90} \\ \theta=43.60^{\circ} \end{gathered}[/tex]
Related Questions
which force is used when a rock fell
Answers
When something falls, it falls because of gravity. Because that object feels a force, it accelerates, which means its velocity gets bigger and bigger as it falls. The strength with which the Earth pulls on something in the form of gravity is a type of acceleration. Earth pulls on everything the exact same amount. A rockfall is a type of fast-moving landslide that happens when rock or earth falls, bounces, or rolls from a cliff or down a very steep slope. Rockfalls start from high outcrops of hard, erosion-resistant rock that become unstable for a variety of reasons. When a rock is thrown upward, if we exempt the air drag then the only force that is acting on the rock is the gravitational force, also known as the weight of the rock. The acceleration due to gravity is always acting downward.
Answer: The force is Gravity
Explanation:
Gravity is what makes us stay to the ground and if a rock fell it would go to the ground just like if i where to jump i would land on the floor/ground i wont be floating because of gavity.
Have a good day.
12. How could extreme heat (resulting from Climate Change) affect human andanimal life?
Answers
ANSWER:
The answer is given in the step by step of the question
STEP-BY-STEP EXPLANATION:
Extreme heat can affect human and animal life in the following ways:
• Animals are sensitive to changes in temperature, as much or more than humans. Think how bad you feel when you have a fever. They also suffer this type of discomfort as a consequence of climate change.
,• Climate change is pushing many species to the limit. They lack water to drink or suffer from temperatures in which they are not comfortable.
,• Provocation of forest fires damaging the habitat of animals and humans
A block of known mass hanging from an ideal spring of known spring constant is oscillating vertically. A motion detector records the position, velocity, and acceleration of the block as a function of time. Which of the following indicates the measured quantities that are sufficient to determine whether the net force exerted on the block equals the vector sum of the individual forces?A. Acceleration only B. Acceleration and position only C. Acceleration and velocity onlyD. Acceleration, position, and velocityPart 1. “Whether The net force exerted on the block equals the vector sum of the individual forces” really means “Newton’s Second Law”. The problem wants you to make measurements to show that the net force equals mass times acceleration. How would you find the force exerted by the spring? How do you find the force exerted by gravity? Part 2. Connect your answer to the previous question with the right answer. Clearly explain which quantities must be measured (between acceleration, velocity, and position) and explain what each quantity is used for to show Newton’s second law.
Answers
Answer:
B. Acceleration and position only
Explanation:
We need to identify the measurements that show that the net force is equal to the sum of the force exerted by the spring and the force of gravity, so we want to know if the following equation is satisfied
[tex]\begin{gathered} F_{net}=F_s-mg \\ ma=k(\Delta x)-mg \end{gathered}[/tex]Where m is the mass, a is the acceleration, k is the spring constant, Δx is the stretched, and g is the gravity. The mass m, the spring constant k, and the gravity g are known. So, the measurement quantities that we need are the acceleration and the position.
So, the answer is
B. Acceleration and position only
Part 1.
How would you find the force exerted by the spring?
The force exerted by the spring is equal to k(Δx ), so to find Δx, we need to identify the position.
How do you find the force exerted by gravity?
The force exerted by gravity is calculated as mass times gravity, so it is known.
Part 2.
We need to measure Acceleration and position.
The acceleration to calculate the net force because by the second law of newton Fnet = ma
The position to calculate the force exerted by the spring.
Shown in the figure, are electric field lines around two point charges. The two charges might be which of these? A)two protons B)a proton and an electron C)two electrons D)two croutons E)two neutrons
Answers
Given:
A figure of the electric field lines.
To find:
The nature of the charges.
Explanation;
The electric field lines always start at a positive charge. And the electric field lines always end at a negative charge. If there is a single charge, the field lines may start at infinity or end at infinity.
Electric field lines also do not form closed loops. They do not cross each other.
In the figure, the field lines start at the blue charge and end at the charge represented by the color yellow.
Therefore the blue charge is positive and the yellow charge is negative.
Final answer:
Therefore the correct answer is option B.
Could someone please help me ASAP?
Answers
As shown in the picture a person is swinging a yo-yo in the circle then the direction of the velocity vector is given by the vector D.
What is Velocity?
The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object.
As given in the problem shown in the picture a person is swinging a yo-yo in a circle, we have to find the direction of the velocity vector,
As seen in the image, a person is swinging a yo-yo in a circle, and the vector D indicates the direction of the velocity vector.
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A helicopter takes off and travels forward at an angle of 59.4 above horizontal. After following this path for 294 meters, the pilot changes the angle of flight to 10.5 degrees above horizontal and follows this path for 849 meters. After these two legs, what is the helicopter’s horizontal distance from the point of take off?
985 m
447 m
408 m
964 m
Answers
After the two legs, the helicopter’s horizontal distance from the point of take off is 979 m
For the first leg,
d = 294 m
θ = 59.4°
[tex]d_{x}[/tex] = d cos θ
[tex]d_{x}[/tex] = 294 * cos 59.4°
[tex]d_{x}[/tex] = 147 m
For the second leg,
d = 849 m
θ = 10.5°
[tex]d_{x}[/tex] = d cos θ
[tex]d_{x}[/tex] = 849 * cos 10.5°
[tex]d_{x}[/tex] = 832 m
Total horizontal distance = [tex]d_{x}[/tex] ( 1st leg ) + [tex]d_{x}[/tex] ( 2nd leg )
Total horizontal distance = 147 + 832
Total horizontal distance = 979 m
Therefore, after the two legs, the helicopter’s horizontal distance from the point of take off is 979 m
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An Olympic long jumper is capable of jumping 8.0 m. Assuming his horizontal speed is 9.1 m/s as he leaves the ground, how long is he in the air and how high does he go?
Answers
Given data:
* The initial velocity of the jumper is u = 9.1 m/s.
* The horizontal range in the given case is 8 m.
Solution:
(a). By the kinematic equation, the time taken by the jumper in terms of the initial velocity and the horizontal range is,
[tex]R=ut+\frac{1}{2}at^2[/tex]where a is the acceleration of the jumper in the horizontal direction,
As there is no force acting on the jumper in the horizontal direction, thus, the value of acceleration is zero.
Substituting the known values,
[tex]\begin{gathered} 8=9.1\times t \\ t=\frac{8}{9.1} \\ t=0.88\text{ s} \end{gathered}[/tex]Thus, the time for which the jumper remains in the air is 0.88 seconds.
(b). By the kinematics equation, the initial velocity of the jumper in the upward direction is,
[tex]v_y-u_y=gt^{\prime}_{}\ldots\ldots\ldots(1)[/tex]where u_y is the initial velocity, v_y is the final velocity of the jumper at the top of vertical displacement, g is the acceleration due to gravity, and t' is the time taken by the jumper to reach the top of vertical displacement,
The jumper will come to rest at the higher position, thus, the final velocity of the jumper at the highest position is zero.
The time taken by the jumper to reach the maximum height is equal to the time taken by the jumper to reach the ground from the maximum height.
As the total time for the complete motion of the jumper is t, thus, the time taken by the jumper to reach the maximum height from the ground is,
[tex]\begin{gathered} t^{\prime}=\frac{t}{2} \\ t^{\prime}=\frac{0.88}{2} \\ t^{\prime}=0.44\text{ s} \end{gathered}[/tex]Substituting the known values in the equation (1),
[tex]\begin{gathered} 0-u_y=-9.8\times0.44_{} \\ u_y=4.312\text{ m/s} \end{gathered}[/tex]By the kinematics equation, the maximum height reached by the jumper is,
[tex]h=u_yt^{\prime}+\frac{1}{2}gt^{\prime}^2[/tex]Substituting the known values,
[tex]\begin{gathered} h=4.312\times0.44+\frac{1}{2}\times(-9.8)\times(0.44)^2 \\ h=1.9-0.95 \\ h=0.95\text{ m} \end{gathered}[/tex]Thus, the maximum height reached by the jumper is 0.95 meters.
What is the normal force of a 0.0037 kg tennis ball rolling at a constant speed of 3 m/s
across a desk?
Answers
The normal force of the rolling ball is 0.037 N
Mass of tennis ball= 0.0037 kg
Constant speed= 3m/s
we need to apply the concept of laws of motion
Since the ball is rolling at a constant speed, it is an example of uniform motion.
So,
Normal force=weight of the body
= 0.0037 x 10 ( acceleration of gravity= 10 m/s²)
Normal force= 0.037 N
Therefore the normal force on the ball is 0.037 N
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The inductor in the RLC tuning circuit of an AM radio has a value of 9 mH. What should be the value of the variable capacitor, in picofarads, in the circuit to tune the radio to 94 kHz
Answers
We are asked to determine the capacitance of an RLC circuit given the frequency. To do that we will use the following formula:
[tex]C=\frac{1}{4\pi^2f^2L}[/tex]Where:
[tex]\begin{gathered} C=\text{ capacitance} \\ f=\text{ frequency} \\ L=\text{ inductance} \end{gathered}[/tex]Now, We plug in the values:
[tex]C=\frac{1}{4\pi^2(94\times10^3Hz)^2(9\times10^{-3}H)}[/tex]Now, we solve the operations:
[tex]C=3.19\times10^{-10}F[/tex]The capacitance is 3.19x10^-10 farads. In Picofarads this is equivalent to:
[tex]C=0.0319pF[/tex]Point charges create equipotential lines that are circular around the charge (in the plane of the paper). What is the potential energy, in nJ, of a 1 nC charge located 1.99 m from a 2 nC charge ?
Answers
The potential energy between two charges can be written as:
[tex]U_e=\frac{kq_1q_2}{r}[/tex]In our case, it'll be equal to:
[tex]U_e=\frac{9*10^9*1*10^{-9}*2*10^{-9}}{1.99}=9.045nJ[/tex]Then, our answer is PE=9.045nJ
Which of these uses digital signals to store, send, or receive information?
Answers
Answer:
here you go
Explanation:
For storing information we use digital signals since that’s what our digital computer storage uses.
For transmitting signals in a computer we use digital signals by simply changing the voltage on the data route or bus.
For wirelessly transmitting signals we use analog signals since electromagnetic radiation is analog itself. Although, the information can be analog modulated OR digitally modulated, depending on the application.
Ellipses have only one focus. Is this true or false?
Answers
In any ellipse there are two foci. This two points are fixed and are fundamental for the construction opf the ellipse. Therefore the statement is false.
Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h.(a) Assuming they start at the same point, how much sooner does the faster car arrive at a destination 14 mi away?_____ min(b) How far must the faster car travel before it has a 15-min lead on the slower car?_____ mi
Answers
We will have the following:
a) We first determine the time it takes to travel the distance to both vehicles:
*
[tex]t_1=\frac{14mi\ast1h}{55}\Rightarrow t_1=\frac{14}{55}h[/tex]*
[tex]t_2=\frac{14mi\ast1h}{60mi}\Rightarrow t_2=\frac{11}{12}h[/tex]So, we determine now the difference in time:
[tex]\frac{11}{12}h-\frac{14}{55}h=\frac{437}{660}h\approx0.66h[/tex]So, the fastest car will arrive approximately 0.66 hours sooner.
b) We determmine the distance it must travel the fastest car to have a 15 minute lead on the other one as follows:
First, we determine the time difference required:
[tex]t=\frac{15min\ast1h}{60min}\Rightarrow t=0.25h[/tex]Then, since both vehicles will move relative to each other, we will have that:
[tex]d_{c1}=(60mi/h)(0.25h)\Rightarrow d_{c1}=15mi[/tex]So, the fastest car must be 15 miles ahead of the other car in order to have a 15 minute lead with respect to the second car.
Two different masses have equal, non-zero kinetic energies. The momentum of the smaller mass isSelect one:A) smaller than the momentum of the larger mass.B) There is not enough information to answer the question.C) equal to the momentum of the larger mass.D) zero.E) Larger than momentum of the larger mass.
Answers
Let the mass of smaller object be m and mass of larger object be M.
As the kinetic energies are equal,
[tex]\frac{p_m^2}{2m}=\frac{p^2_{_M}}{2M}[/tex]On solving further,
[tex]\frac{p_m}{p_M}=\text{ }\sqrt[]{\frac{m}{M}}[/tex]Hence, option A is correct.
The density of copper is 8.94 g/cm^3. What is the mass of a rectangular sheet of copper: 10 cm wide, 45 cm long, and 0.2 cm thick? I have no idea how to solve for mass or where to start. :(
Answers
Mass = 804.6 g
Explanation:The dimension of the rectangular copper = 10 cm wide, 45 cm long, and 0.2 cm thick
The volume of the rectangular copper = 0.1m x 0.45m x 0.002m
The volume of the rectangular copper = 0.00009 m³
The density of copper = 8.94 g/cm³ = 8.94 x 1000 kg/m³
The density of copper = 8940 kg/m³
Mass = Density x Volume
Mass = 8940 x 0.00009
Mass = 0.8046 kg
Mass = 804.6 g
1. Consider the example problem, but with the lower pressure reduced to 100 Pa. How much work would be done in a single cycle?1. 29,970 J2. 0 J3. -30,030 J4. 27,000 J
Answers
29970 J
Explanation:Given:
[tex]\begin{gathered} P_A=P_B=10^5Pa \\ P_c=P_D=100\text{ Pa\lparen The new lower pressure\rparen} \\ V_A=V_D=0.1m^3 \\ V_B=V_C=0.4m^3 \end{gathered}[/tex]Work done across AB
[tex]\begin{gathered} W_{AB}=P_{AB}(V_B-V_A) \\ \\ W_{AB}=10^5(0.4-0.1) \\ \\ W_{AB}=0.3\times10^5 \\ \\ W_{AB}=30000J \end{gathered}[/tex]Work done across BC = 0 J (Since there is no change in volume)
Work done across DA = 0 J (Since there is no change in volume)
Work done across CD
[tex]\begin{gathered} W_{CD}=P_{CD}(V_D-V_C) \\ \\ W_{CD}=100(0.1-0.4) \\ \\ W_{CD}=100(-0.3) \\ \\ W_{CD}=-30J \end{gathered}[/tex]Work done in one cycle:
[tex]\begin{gathered} W_{cycle}=W_{AB}+W_{BC}+W_{DA}+W_{CD} \\ \\ W_{cycle}=3000J+0+0-30 \\ \\ W_{cycle}=29970J \end{gathered}[/tex]29970J of work will be done in a single cycle
Multiple Choice: A car with a mass of 825 kg moves along the roadway at aspeed of 15 m/s to the east. What impulse is required to decrease the speed of theboat to 10 m/s east?
Answers
Given:
The mass of the car is m = 825 kg
The intial speed of the car is
[tex]v_i\text{ = 15 m/s}[/tex]towards east.
The final speed of the car is
[tex]v_f\text{ = 10 m/s}[/tex]To find the impulse of the car.
Explanation:
The impulse can be calculated by the formula
[tex]Impulse\text{ = mv}_f-mv_i[/tex]On substituting the values, the impulse will be
[tex]\begin{gathered} Impulse\text{ = \lparen825}\times10\text{\rparen-\lparen825}\times15\text{\rparen} \\ =\text{ -4125 kg m/s} \end{gathered}[/tex]The impulse will be 4125 kg m/s due
When a 6.0-F capacitor is connected to a generator whose rms output is 25 V, the current in the circuit is observed to be 0.40 A. What is the frequency of the source?
Answers
The frequency of the source is 1.66 Hz.
What is the impedance?Let us recall that the impedance of the circuit is the opposition that is offered to the flow of current by a circuit component that is not a resistor. Now let us find the impedance.
I = V/Z
I = current
V = voltage
Z = impedance
Z = V/I
Z = 25/0.4
Z = 62.5 ohm
Z^2 = R^2 + Xc^2
Z^2 = Xc^2
Xc= Z
Xc = 2πfC
f = frequency
C = capacitance
f= Xc/2πC
f = 62.5/2 * 3.142 * 6
f = 1.66 Hz
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Students conducted an experiment in which a ball was thrown into the air and then caught at the same height from which it was released. They determined that the ball had 100 J of kinetic energy when it was released. They calculated the ball's energy just before it was caught and determined that it's kinetic energy was less than 100 J.Which of the following statements MOST LIKELY accounts for the difference in the ball's initial and final kinetic energy?
Answers
A ball was thrown into the air and then caught at the same height from which it was released.
The initial kinetic energy of the ball was 100 J and the ball's kinetic energy just before it was caught was less than 100 J.
Marc is sitting at his desk with good posture. What MOST likely is Marc doing?
Answers
It is most likely that Marc is studying while assuming a good sitting posture.
What is posture?Posture is defined as the the various ways an individual carry their body in order to assume a particular position which may be while sitting of standing up.
There different types of postures which are classified under good or bad posture. They include the following:
Healthy Posture.Kyphosis Posture.Flat Back Posture.Swayback Posture.Forward Head Posture.To sit correctly at a dest with a good posture is an example of a healthy posture which was assumed by Marc.
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A traffic light hangs from a pole as shown in (Figure 1). The uniform aluminum pole AB is 7.20 m long and has a mass of 11.0 kg . The mass of the traffic light is 22.0 kg .
-Determine the tension in the horizontal massless cable CD.
-Determine the vertical component of the force exerted by the pivot A on the aluminum pole.
-Determine the horizontal component of the force exerted by the pivot A on the aluminum pole.
Answers
There is a 409.30 N force in the horizontal massless cable CD.
The vertical component of the pivot A's force on the aluminum pole, which is 323.73N.
The force that the pivot A applies to the aluminum pole has a horizontal component of 409.30 N.
Force is any influence to the motion of the body .
Basically the product of mass and acceleration.Length of the pole = 7.20 m
Mass of the pole = m = 11.0 kg
Mass of the traffic light = M = 22.0 kg
Let the length of the rod AD be L
Also [tex]Lsin\alpha = 3.8[/tex]
[tex]\alpha = 37[/tex]°
[tex]Sin\alpha = 0.602[/tex]
L = [tex]\frac{3.8}{0.602}[/tex]
L = 6.312 m
In order to determine the tension (T) in the cable, the free body diagram will give the detailed information.
On balancing all the forces in both the 'x' and 'y' direction, i.e. summation of all the forces in 'x' and 'y' direction must equal to zero.∑[tex]F_{x}[/tex] = 0
[tex]R_{x}[/tex] - T = 0
[tex]R_{x}[/tex] = T
∑[tex]F_{y}[/tex] = 0
[tex]R_y- W_p - W_l = 0[/tex]
[tex]W_p[/tex] = 9.81 * 11 = 107.91 kg
[tex]W_l\\[/tex] = 9.81 * 22 = 215.82 kg
[tex]R_y\\[/tex] = [tex]W_p + W_l[/tex]
[tex]R_y\\[/tex] = 107.91 + 215.82
[tex]R_y\\[/tex] = 323.73 N
The vertical component of the force exerted by the pivot A on the aluminum pole is 323.73 N
Taking moment along x-axis
M = 0
[tex]Th-mgCos\alpha* \frac{1}{2} - mglCos\alpha[/tex] = 0
T = [tex]\frac{glCos\alpha }{h} *\frac{m}{2}*M[/tex]
T = 409.30 N
The tension in the horizontal massless cable CD is 409.30 N
[tex]R_{x}[/tex] = T = 409.30 N
The horizontal component of the force exerted by the pivot A on the aluminum pole is 409.30 N.
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How is kinetic energy and pontential energy alike ?
Answers
The kinetic energy of the particle is associated with the velocity of the particle where as the potential energy of the particle depneds upon the position ofthe particle.
Both the energies can transform into each other.
For example,
When a ball is dropped from the building's top floor, the potential energy of the ball is maximum at the top of the builiding and transform into the kinetic energy while moving in the downward direction.
The kinetic energy of the ball is maximum just before hitting the ground.
This shows the tranformation of the potential energy to the kinetic energy.
Thus, the kinetic energy and potential can transform into each other is one the similarity in behaviour.
A bungee jumper jumps off a bridge and bounces up and down several times.She finally comes to rest 30 m below the bridge from which she just jumped.If her mass is 50 kg and the spring constant of the bungee cord is 10 N/m,how much energy was lost due to air resistance while she was bouncing?(Recall that g = 9.8 m/s2)A. 9200 NB. 7330 Nc. 10,200 ND. 8605 N
Answers
Given data
*The given height is h = 30 m
*The given mass is m = 50 kg
*The spring constant of bungee cord is k = 10 N/m
*The value of the acceleration due to the gravity is g = 9.8 m/s^2
The net change in potential energy is calculated as
[tex]\begin{gathered} \Delta U_p=mgh \\ =(50)(9.8)(30) \\ =14700\text{ J} \end{gathered}[/tex]The spring stretch is calculated by using the relation as
[tex]\begin{gathered} F=mg \\ kx=mg \\ x=\frac{mg}{k} \\ =\frac{(50)(9.8)}{10} \\ =49\text{ m} \end{gathered}[/tex]The energy stored in spring is calculated as
[tex]\begin{gathered} U_s=\frac{1}{2}kx^2 \\ =\frac{1}{2}(10)(49)^2 \\ =12005 \end{gathered}[/tex]The energy was lost due to air resistance while she was bouncing is calculated as
[tex]\begin{gathered} \Delta E=\Delta U_p-U_s_{}_{} \\ =14700-12005 \\ =2695\text{ J} \end{gathered}[/tex]Hence, the energy was lost due to air resistance while she was bouncing is 2695 J
how many joules does a lamp marked 12 volts, 24 w consumed in an hour . And also what is the current?
Answers
Given data:
* The voltage given is 12 volts.
* The value of the power given is 24 Watt.
* The time till which the lamp is used is,
[tex]\begin{gathered} t=1\text{ hr} \\ t=60\times60\text{ s} \\ t=3600\text{ s} \end{gathered}[/tex]Solution:
(a). The energy consumed by the lamp in time t is,
[tex]E=P\times t[/tex]where P is the power of lamp, t is the time, and E is the energy consumed,
Substituting the known values,
[tex]\begin{gathered} E=24\times3600 \\ E=86400\text{ J} \\ E=86.4\times10^3\text{ J} \\ E=86.4\text{ kJ} \end{gathered}[/tex]Thus, the energy consumed by the lamp in 1 hour is 86.4 kJ.
(b). The power of the lamp in terms of voltage and current is,
[tex]\begin{gathered} P=VI \\ I=\frac{P}{V} \end{gathered}[/tex]where P is the power, V is the voltage and I is the current,
Substituting the known values,
[tex]\begin{gathered} I=\frac{24}{12} \\ I=2\text{ A} \end{gathered}[/tex]Thus, the current flowing through the lamp is 2 A.
The potential energy of a system can be changed by varying the position of objects in the system. At which point do the coaster cars have the most potential energy?
Answers
To find:
At which point the potential energy of the coaster car is the highest?
Explanation:
The potential energy of an object is the energy possessed by the object due to the position of the object. The gravitational potential energy of the object is directly proportional to the height of the object.
Thus an object will have the highest potential energy when it is at the highest point.
Final answer:
Thus the coaster cars will have the highest potential energy when it is at the highest point on the roller coaster.
Therefore, the coaster cars will have the most potential energy at point A.
a piece of steel expands 10 cm when it is heated from 20 to 40 deg C. How much would it expand if it was heated from 20 to 60 deg.C?
Answers
ANSWER
20 cm
EXPLANATION
Given:
• The change in length of a piece of steel, ΔL = 10 cm, when its temperature change is ΔT = 20°C
Find:
• The change in length of the same piece of steel, ΔL, when its temperature is changed ΔT = 40°C
The change in length of a solid material is,
[tex]\Delta L=\alpha\cdot L_o\cdot\Delta T[/tex]Where L₀ is the original length, α is the linear expansion coefficient and ΔT is the change in temperature.
In this case, we know how much is the piece of steel expanded when the temperature change is 20°C, so we can find the product αL₀,
[tex]\alpha L_o=\frac{\Delta L}{\Delta T}[/tex]Replace the known values and solve,
[tex]\alpha L_o=\frac{10\text{ }cm}{20\degree C}=0.5cm/\degree C[/tex]If the temperature change now is 40°C,
[tex]\Delta L=\alpha L_o\Delta T=0.5cm/\degree C\cdot40\degree C=20cm[/tex]Hence, when the piece of steel is heated from 20°C to 60°C it will expand 20 centimeters.
The displacement (in meters) of a particle moving in a straight line is given by S= t^2 - 7t + 17 ii)iii)iv)
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]What conditions must be met in order for work to be done?
A. The applied force must make the object move.
B. The output force must be greater than the input force.
C. At least part of the applied force must be in the same direction as the movement of the object.
D. The work must be greater than the momentum.
Answers
At least part of the applied force must be in the same direction as the movement of the object must be met in order for work to be done.
What are the conditions to work?The following are the two prerequisites for working: To do the work, the body must be subjected to a force, or F 0. The body must move in the direction of the applied force, or S 0, as a result of the applied force.There must be a force used. The displacement is the distance over which the force must act. The displacement must be a component of the force.A legal term known as a condition precedent refers to an event that must occur before a certain contract is regarded as being in effect or before either party is obliged to fulfill any obligations.
Therefore, the correct answer is option C. At least part of the applied force must be in the same direction as the movement of the object.
To learn more about conditions to work refer to:
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this is a 3 part question20) A 9.50-g bullet has a speed of 1.30 km>s. (a) What is its kinetic energy in joules? (b) What is the bullet’s kinetic energy if its speed is halved? (c) If its speed is doubled?
Answers
Given that mass of bullet, m = 9.50 g = 0.0095 kg
speed of bullet, v is 1.30 km/s
(a) Kinetic energy is given by the formula
[tex]K\mathrm{}E\text{. = }\frac{1}{2}mv^2[/tex]Substituting the values in the above formula, we get
[tex]K\mathrm{}E\mathrm{}=\frac{1}{2}(0.0095)(1.30)^2=8.0275\times10^{-3}\text{ J}[/tex](b) Speed of bullet is v/2
Sustituting this value in the formula of kinetic energy, we get
[tex]\begin{gathered} K.E._1=\text{ }\frac{1}{2}m(\frac{v}{2})^2 \\ =\frac{1}{2}\times(0.0095)\times(\frac{1.30}{2})^2 \\ =2.0068\times10^{-3\text{ }}J \end{gathered}[/tex](c) Speed of bulllet becomes 2v
Sustituting this value in the formula of kinetic energy, we get
[tex]\begin{gathered} K.E._2=\text{ }\frac{1}{2}m(2v)^2 \\ =\frac{1}{2}\times(0.0095)\times(2\times1.30)^2 \\ =0.03211J \end{gathered}[/tex]A small mailbag is released from a helicopter that is descending steadily at 2.82 m/s.(a) After 3.00 s, what is the speed of the mailbag?v = m/s(b) How far is it below the helicopter?d = m(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 2.82 m/s?v = m/sd = m
Answers
a)
When the package is released from the moving helicopter, the package and the helicopter has a common speed. The package is in freefall. We would calculate the speed of the helicopter after a given time t by applying the formula,
v = vo + gt
where
vo is the initial velocity of of the package and it is equal to the speed of the helicopter
v is the final velocity of the package after time t
g is th acceleration due to gravity
From the information given
vo = 2.82
t = 3
g = 9.8
Thus,
v = 2.82 + 9.8 * 3 = 2.82 + 29.4
v = 32.22 m/s
After 3.00 s, the speed of the mailbag is 32.22 m/s
b) We want to calculate the distance covered by the mailbag in 3 s. We would apply the formula which is expressed as
s = vot + 1/2 x g x t^2
where
s is the distance
vo = 2.82
g = 9.8
t = 3
Thus,
s = 2.82 x 3 + 1/2 x 9.8 x 3^2 = 8.46 + 44.1
s = 52.56 m
Since we want to calculate the distance from the helicopter, we would calculate the diatance that the helicopter also travelled downwards in 3 s by applying the formula for calculating distance which is expressed as
distance = speed x time
Thus
distance = 2.82 x 3 = 8.46 m
Difference in distance = 52.56 - 8.46 = 44.1
The package is 44.1 m from the helicopter
c) If the helicopter is moving upwards, it would be thrown out and it would attain a certain height before it starts descending. The height is calculated by the formula,
h = vo^2/2g
By substituting the values,
h = 2.82^2/2 x 9.8 = 0.406 m
When the mail bag attains this height, it will start moving downwards. At this height, the final velocity is zero. We would calculate the time taken to attain this height by applying the formula,
v = vo - gt
v = 0
Thus,
0 = 2.82 - 9.8 x t
9.8t = 2.82
t = 2.82/9.8 = 0.288
The time left for freefall within the first 3 seconds is
3 - 0.288 = 2.712 s
The height attained by the mailbag in 2.712s is calculated by the formula,
h = gt^2/2
h = 9.8 x 2.712^2/2 = 36.04 m
Distance travelled by helicopter by ascending upward in 3 s is
distance = 2.82 x 3 = 8.46
Height of mailbag from final position after 3 seconds is
36.04 - 0.406 = 35.634
Difference in distance = 35.634 + 8.46 = 44.094
The package is 44.094 m from the helicopter
For the velocity of the mailbag after 3 seconds,
v = - vo + gt
v = - 2.82 + 9.8 x 3 = - 2.82 + 29.4
v = 26.54 m/s