(a) Initial velocity (v₀) = 0 m/s, Final velocity (v) = 15.0 m/s.
(b) The dolphin rises approximately 11.5 meters above the water.
(c) The dolphin is in the air for approximately 1.5 seconds.
(a)
Initial velocity (v₀) = 0 m/s (since the dolphin jumps straight up from the water, its initial velocity is zero)
Final velocity (v) = 15.0 m/s (given)
(b)
To solve for the height, we can use the kinematic equation that relates the final velocity (v), initial velocity (v₀), acceleration (a), and displacement (Δy):
v² = v₀² + 2aΔy
v² = v₀² + 2gh
v² = 2gh
Solving for h:
h = (v²) / (2g)
h = (15.0)² / (2 × 9.8)
h ≈ 11.5 m
Therefore, the dolphin rises approximately 11.5 meters above the water.
(c) To determine the time the dolphin is in the air, we can use the equation for time (t) derived from the kinematic equation:
Δy = v₀t + (1/2)at²
Δy = (1/2)at²
t = √((2Δy) / a)
t = √((2 × 11.5) / -9.8)
t ≈ 1.5 s
Therefore, the dolphin is in the air for approximately 1.5 seconds.
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The dolphin rises 8.64 meters above the water and is in the air for 1.33 seconds.
Explanation:(a) The knowns in this problem are:
Initial velocity of the dolphin: 13.0 m/sAcceleration due to gravity: -9.8 m/s²(b) To find the height above the water, we need to find the time taken by the dolphin to reach that height and use the equation:
Final velocity = Initial velocity + Acceleration × Time
Solving for time, we have:
Time = (Final velocity - Initial velocity) / Acceleration
Substituting the known values into the equation, we get:
Time = (0 m/s - 13.0 m/s) / -9.8 m/s² = 1.33 s
Next, we can use the formula for height:
Height = Initial velocity × Time + 0.5 × Acceleration × Time²
Substituting the known values, we have:
Height = 13.0 m/s × 1.33 s + 0.5 × -9.8 m/s² × (1.33 s)² = 8.64 m
(c) The time the dolphin is in the air is equal to the time calculated above, which is 1.33 seconds.
_______ activities give the teacher the opportunity to deliver instruction on a more personal level than _______ activities
A. Independent, holistic
B. Small group, kinesthetic
C. whole group, independent
D. Small group, whole group
Answer:
D. Small group, whole group
Explanation:
Small group activities give the teacher the opportinity to deliver instruction on a more personal level than whole group activities because in a smaller group there is more time to work with individuals as opposed to have to dedicate all time to generalized lessons for the whole group.
"The correct answer is D. Small group, whole group. Small group activities give the teacher the opportunity to deliver instruction on a more personal level than whole group activities
When comparing the types of activities listed, it is clear that small group activities allow for a more personal level of instruction compared to whole group activities. Small group activities involve fewer students, which means that the teacher can focus more on the individual needs of each student, provide more targeted feedback, and facilitate more in-depth discussions. This setting is conducive to addressing specific questions, adapting the pace of instruction to the group's needs, and fostering a sense of community among the students.
On the other hand, whole group activities involve the entire class and are less personal by nature. In this setting, the teacher must address a larger audience, which can make it challenging to meet every student's individual needs. Instruction is typically more general and may not be as tailored to each student's learning style or pace.
To contrast, independent activities do not involve direct instruction from the teacher, so they do not provide the opportunity for personal-level instruction. Kinesthetic activities involve movement and can be done individually or in groups, but they are not inherently more personal than small group activities. Therefore, the comparison between independent and holistic activities (A), or whole group and independent activities (C), does not accurately reflect the level of personalization in instruction.
Thus, the correct pair that illustrates the difference in the level of personalization in instruction is small group activities, which are more personal, compared to whole group activities, which are less personal."
The nucleus of a 125 Xe atom (an isotope of the element xenon with mass 125 u) is 6.0 fm in diameter. It has 54 protons and charge q=+54e (1 fm = 1 femtometer = 1× 10 −15 m .) Hint: Treat the spherical nucleus as a point charge. Part A What is the electric force on a proton 3.0 fm from the surface of the nucleus? Express your answer in newtons. F nucleusonproton F n u c l e u s o n p r o t o n = nothing N SubmitRequest Answer Part B What is the proton's acceleration? Express your answer in meters per second squared. a proton a p r o t o n = nothing m/ s 2 SubmitRequest Answer Provide Feedback Next
Answer:
(a): 345.6 N.
(b): [tex]\rm 2.069\times 10^{29}\ m/s^2.[/tex]
Explanation:
Given:
Charge on the 125 Xe nucleus, [tex]\rm q = +54e.[/tex]Mass of the 125 Xe nucleus, [tex]\rm m = 125\ u.[/tex]Diameter of the 125 Xe nucleus, [tex]\rm d=6.0\ fm = 6.0\times 10^{-15}\ m.[/tex]Distance of the proton from the surface of the nucleus, [tex]\rm a=3.0\ fm = 3.0\times 10^{-15}\ m.[/tex]Part A:According to Coulomb's law, the electric field due to a charged sphere of charge Q at a point r distance away from its center is given by
[tex]\rm E=\dfrac{kQ}{r^2}.[/tex]
where, k is the Coulomb's constant, having value = [tex]9\times 10^9\ \rm Nm^2/C^2.[/tex]
Therefore, the electric field due to the 125 Xe nucleus at the proton is given by
[tex]\rm E=\dfrac{kq}{r^2}=\dfrac{(9\times 10^9)\times (+54 e)}{r^2}[/tex]
Here,
e is the elementary charge, having value = [tex]\rm 1.6\times 10^{-19}\ C.[/tex]
r is the distance of the proton from the center of the nucleus = [tex]\rm a + \dfrac d2 = 3.0+\dfrac{6.0}2=6.0\ fm = 6.0\times 10^{-15}\ m.[/tex]
Using these values,
[tex]\rm E=\dfrac{(9\times 10^9)\times (+54 \times 1.6\times 10^{-19})}{(6.0\times 10^{-15})^2}=2.16\times 10^{21}\ N/C.[/tex]
Now, the electric force on a charge q due to an electric field is given as
[tex]\rm F=qE[/tex]
For the proton, [tex]\rm q = e =1.6\times 10^{-19}\ C.[/tex]
Thus, the electric force on the proton is given by
[tex]\rm F = 1.6\times 10^{-19}\times 2.16\times 10^{21}=345.6\ N.[/tex]
Part B:According to Newton's second law,
[tex]\rm F=ma[/tex]
where, a is the acceleration.
The mass of the proton is [tex]\rm m_p=1.67\times 10^{-27}\ kg.[/tex]
Therefore, the proton's acceleration is given by
[tex]\rm a = \dfrac{F}{m_p}=\dfrac{345.6}{1.67\times 10^{-27}}=2.069\times 10^{29}\ m/s^2.[/tex]
(a) The electric force of the proton is 2.56 x 10⁻²⁹ N.
(b) The acceleration of the proton is 0.0153 m/s².
Electric force of the protonThe electric force of the proton is calculated using Coulomb's law of electrostatic force.
F = kq²/r²
F = (9 x 10⁹ x 1.6 x 10⁻¹⁹ x 1.6 x 10⁻¹⁹)/(3)²
F = 2.56 x 10⁻²⁹ N
Acceleration of the protonThe acceleration of the proton is determined by applying Newton's second law of motion;
F = ma
a = F/m
a = (2.56 x 10⁻²⁹ )/(1.67 x 10⁻²⁷)
a = 0.0153 m/s²
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How would you use the parallelogram method of vector addition when more than two forces are added?
Answer:
In order to use the parallelogram method, we have to select first two vectors. We move the vectors until their initial points coincide. Then we draw lines to form a complete parallelogram, as is shown in the figure annexed. The diagonal from the initial point to the opposite vertex of the parallelogram is the resultant.
We use the last vector resultant with a third vector, and we use again the parallelogram method to add them. We use the new resultant and we add it with a 4th vector. We repeat this task until all the vectors are used.
A student applies a horizontal force of 100 N to a 25 kg box causing the block to accelerate at a rate of 1.5 m/s^2. Which of the following is the best approximation of the magnitude of the friction force on the box?
Answer:
By Newton's law Friction= -62.5N
Explanation:
Applying second law of Newton (which is the the sum of the forces is equal to the mass of the object who feels the forces times the acceleration)
Push+Friction= [tex]m_{box}a_{box}[/tex]
[tex]100+Friction=25*1.5=37.5\\Friction=37.5-100=-62.5 N[/tex]
Which indicates that the friction is opposed to the push and it is really appreciable, that is to say, there is a lot of friction between the box and the floor.
Final answer:
The best approximation of the magnitude of the friction force on the box, when a 25 kg box is accelerated by a 100 N force at 1.5 m/s^2, is 62.5 N.
Explanation:
The student's question involves understanding Newton's second law of motion and the concept of friction. Since a horizontal force of 100 N is applied to a 25 kg box, causing it to accelerate at a rate of 1.5 m/s2, we can calculate the net force using Newton's second law (F = ma). The net force is the difference between the applied force (100 N) and the friction force which we are trying to find.
First, calculate the net force:
Fnet = m × a = 25 kg × 1.5 m/s2 = 37.5 N
Next, calculate the friction force:
Ffriction = Fapplied - Fnet = 100 N - 37.5 N = 62.5 N
Therefore, the best approximation of the magnitude of the friction force on the box is 62.5 N.
Many people believe that a vacuum created inside a
vacuumcleaner causes particles of dirt to be drawn in. Actually,
however,the dirt is pushed in. Explain.
Answer:Due to the pressure difference created by rotating fans.
Explanation:
In most of the vacuum cleaners, there is an area which is of disc shape and it is in right next to the motor. There are several fans within the disc that spin at a very high velocity.
The blades will push the air outside of the disk.There is no air in inside of the disc and air pressure creates which pushes air inside the disk to replace the missing air.
So motor is pushing the air outside and to maintain this pressure the air is pushing toward inside with dirt.
A body moving at .500c with respect to an observer
disintigratesinto two fragments
that move in opposite directions relative to their center
ofmass along the same line of motion as the original body.
Onefragment has a velocity of .600c in the backward direction
relativeto the center of mass and the other has a velocity of .500c
in theforward direction. What velocities will the observer
find?
Answer:
0.8c and -0.14c
Explanation:
The first fragment will have a speed of +0.5c respect of a frame of reference moving at +0.5c
Lest name v the velocity of the frame of reference, and u' the velocity of the object respect of this moving frame of reference.
The Lorentz transform for velocity is:
u = (u' + v) / (1 + (u' * v) / c^2)
u = (0.5c + 0.5c) / (1 + (0.5c * 0.5c) / c^2) = 0.8c
The other fragment has a velocity of u' = -0.6c respect of the moving frame of reference.
u = (-0.6v + 0.5c) / (1 + (0.5c * 0.5c) / c^2) = -0.14c
You irradiate a crystalline sample that has 2.9344 Å between atoms with electrons to do an electron diffraction experiment in reflection mode. You observe a first order (m = 1) diffraction peak at θ = 12.062°. What is the wavelength of the electrons?
Answer:
[tex]\lambda=1.23[/tex]Å
Explanation:
Bragg's Law refers to the simple equation:
[tex]n\lambda = 2d sin(\theta)[/tex]
In this case:
n=1
θ = 12.062°
d=2.9344 Å
[tex]\lambda = 2*2.9344sin(12.062)=1.23[/tex]Å
A ball is thrown upward at time t=0 from the ground with an initial velocity of 4 m/s (~ 9 mph). Assume that g = 10 m/s^2. What is the greatest height (in meters) reached by the ball?
Answer:
The greatest height reached by the ball is 0.8 m.
Explanation:
Given that,
Initial velocity = 4 m/s
We need to calculate the greatest height reached by the ball
Using equation of motion
[tex]v^2=u^2+2gh[/tex]
Where, v = final velocity
u = initial velocity
g = acceleration due to gravity
Put the value in the equation
[tex]0=4^2+2\times(-10)\times h[/tex]
[tex]h =\dfrac{16}{20}[/tex]
[tex]h =0.8\ m[/tex]
Hence, The greatest height reached by the ball is 0.8 m.
An insulated Thermos contains 134 g of water at 70.7°C. You put in a 13.8 g ice cube at 0.00°C to form a system of ice + original water. The specific heat of liquid water is 4190 J/kg*K; and the heat of fusion of water is 333 kJ/kg. What is the net entropy change of the system from then until the system reaches the final (equilibrium) temperature?
Answer:
[tex]\Delta s\ =\ 21.33\ J/K[/tex]
Explanation:
Given,
Mass of the ice = [tex]m_i\ =\ 13.8\ kg\ =\ 0.0138\ kg[/tex]Temperature of the ice = [tex]T_i\ =\ 0^o\ C[/tex]Mass of the original water = [tex]m_w\ =\ 134\ g\ =\ 0.134\ kg[/tex]Temperature of the original water = [tex]T_w\ =\ 70.0^o\ C[/tex]Specific heat of water = [tex]S_w\ =\ 4190\ J/kg K[/tex]Latent heat of fusion of ice = [tex]L_f\ =\ 333\ kJ/kg[/tex]Let T be the final temperature of the mixture,
Therefore From the law of mixing, heat loss by the water is equal to the heat gained by the ice,
[tex]m_iL_f\ +\ m_is_w(T_f\ -\ 0)\ =\ m_ws_w(T_w\ -\ T_f)\\\Rightarrow 333000\times 0.0138\ +\ 0.0138\times 4190T_f\ =\ 0.134\times 4190\times(70.7\ -\ T_f)\\\Rightarrow 4595.4\ +\ 57.96T_f\ =\ 39789.96\ -\ 562.8T_f\\\Rightarrow 620.76T_f\ =\ 35194.56\\\Rightarrow T_f\ =\ \dfrac{35194.56}{620.76}\\\Rightarrow T_f\ =\ 56.69^o\ C[/tex]
Now, We know that,
Change in the entropy,
[tex]\Delta s\ =\ s_f\ -\ s_i\ =\ \dfrac{Q}{T}\\\Rightarrow \displaystyle\int_{s_i}^{s_f} ds\ =\ \displaystyle\int_{T_i}^{T_f}\dfrac{msdT}{T}\\\Rightarrow \Delta s =\ ms \ln \left (\dfrac{T_i}{T_f}\ \right )\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)[/tex]
Now, change in entropy for the ice at 0^o\ C to convert into 0^o\ C water.
[tex]\Delta s_1\ =\ \dfrac{Q}{T}\\\Rightarrow \Delta s_1\ =\ \dfrac{m_iL_f}{T}\ =\ \dfrac{0.0138\times 333000}{273.15}\ =\ 16.82\ J/K.[/tex]
Change in entropy of the water converted from ice from [tex]273.15\ K[/tex] to water 330.11 K water.
From the equation (1),
[tex]\therefore \Delta s_2\ =\ ms \ln \left (\dfrac{T_i}{T_f}\ \right )\\\Rightarrow \Delta s_2\ =\ 0.0138\times 4190\times \ln \left (\dfrac{273.15}{330.11}\ \right )\\\Rightarrow \Delta s_2\ =\ 6.88\ J/K[/tex]
Change in entropy of the original water from the temperature 342.85 K to 330.11 K
From the equation (1),
[tex]\therefore \Delta s_3\ =\ ms \ln \left (\dfrac{T_i}{T_f}\ \right )\\\Rightarrow \Delta s_3\ =\ 0.134\times 4190\times \ln \left (\dfrac{330.11}{343.85} \right )\\\Rightarrow \Delta s_3\ =\ -2.36\ J/K[/tex]
Total entropy change = [tex]\Delta s\ =\ \Delta s_1\ +\ \Delta s_2\ +\ \Delta s_3\\\Rightarrow \Delta s\ =\ (16.82\ +\ 6.88\ -\ 2.36)\ J/K\\\Rightarrow \Delta s\ =\ 21.33\ J/K.[/tex]
Hence, the change in entropy of the system form then untill the system reaches the final temperature is 21.33 J/K
A mass of 222g of Helium gas at an initial temperature of 54.43°C and at an initial absolute pressure of 4.45 atm undergoes an isothermal expansion until its volume increases by a factor of 2.25. (a) What is the final pressure? (Pa)
(b) How much work is done on the gas?
(c) How much heat does the gas absorb?
(d) What is the change in the total internal energy of the gas?
Explanation:
Given that,
Mass of gas = 222 g
Temperature = 54.43°
Pressure = 4.45 atm
Final volume = 2.25 initial volume
For isothermal expansion
(a). We need to calculate the pressure
Using relation of pressure and volume
[tex]P_{i}V_{i}=P_{f}V_{f}[/tex]
[tex]P_{f}=\dfrac{P_{1}V_{i}}{V_{f}}[/tex]
Put the value into the formula
[tex]P_{f}=\dfrac{4.45\timesV_{i}}{2.25V_{i}}[/tex]
[tex]P_{f}=\dfrac{4.45}{2.25}[/tex]
[tex]P_{f}=1.97\ atm[/tex]
[tex]P_{f}=199610.3\ pa[/tex]
[tex]P_{f}=1.99\times10^{5}\ Pa[/tex]
The pressure is [tex]1.99\times10^{5}\ Pa[/tex]
(b). We need to calculate the work done
1 mole of Hg is 200.59 gram
222 g of Hg is
[tex]n =\dfrac{200}{200.59}[/tex]
[tex]n =0.997[/tex]
Using formula of work done
[tex]W=nRTln(\dfrac{V_{f}}{V_{i}})[/tex]
Put the value into the formula
[tex]W=0.997\times8.314\times(54.43+273)ln2.25[/tex]
[tex]W=2200.9\ J[/tex]
[tex]W=2.2009\ kJ[/tex]
The work done is 2.20 kJ.
(c). We need to calculate the gas absorb
Heat absorbed by the gas is the work done
[tex]Q=W[/tex]
[tex]Q=2.20 kJ[/tex]
The absorb heat is 2.20 kJ.
(d). We need to calculate the change in the total internal energy of the gas
Change in internal energy in an isothermal process is zero.
So, [tex]U=0 [/tex]
Hence, This is the required solution.
Our 12 V car battery does not appear to be functioning correctly, so we measure the voltage with a volt meter and find that the voltage on the battery is only 9V. To fix the problem, we connect the battery to a charger which delivers a constant current of i = 15 A. After t = 53 min on the charger, we find that voltage on the car battery is now 12.6V. Assuming that the voltage changed linearly during the charging process, how much energy was delivered to the car battery.
Answer:
507599.78 J
Explanation:
Charge input = current x time
=15 x 53 x 60
= 47000 coulomb
increase in voltage
= 12.6 - 9 = 3.6
capacity of the battery C = Charge input / increase in voltage
= 47000 / 3.6 = 13055.55
energy of the capacitor = 1/2 CV²
Initial energy of car battery = .5 x 13055.55 x 9 x 9
= 528749.77 J
Final energy of car battery
= .5 x 13055.55 x 12.6 x 12.6
= 1036349.55
Increase in energy = 507599.78 J
Final answer:
The total energy delivered to the car battery during its charging process from 9 V to 12.6 V over 53 minutes with a charging current of 15 A is 515,160 Joules.
Explanation:
To calculate the energy delivered to the car battery while it is being charged, we need to consider the change in voltage, the charge current, and the time it was charged. Since the voltage changed linearly from 9 V to 12.6 V over 53 minutes with a constant charging current of 15 A, we first convert the charging time to seconds (53 min imes 60 s/min = 3180 s) and then use the formula Energy (E) = Power (P) imes Time (t), where Power (P) is the product of voltage (V) and current (I).
Assuming a linear voltage increase, we take the average voltage (9 V + 12.6 V)/2 during the charge time. The average voltage is then 10.8 V. The energy delivered is given by:
E = P imes t = V imes I imes t = 10.8 V imes 15 A imes 3180 s
E = 515160 Joules
Therefore, the energy delivered to the car battery during the charging process is 515,160 Joules.
What is necessary to designate a position? A. a reference point B. a direction C. fundamental units D. motion E. all of these
Answer:
E. all of these
Explanation:
The designation of a point in space all the points that necessary
- reference point
- a direction
- fundamental units
- a direction
- motion
all are necessary to designate a point in space. Hence option E is correct.
For example in simple harmonic motion we need to specify all the above factors of the object in order to designate the position of the object.
To designate a position, the necessary elements are A. A reference point, B. A direction, C. Fundamental units, D. Motion. Therefore option E is correct.
A. A reference point: A reference point is needed to establish a starting point or a fixed location from which the position is measured. It serves as a point of comparison and allows for consistent measurements.
B. A direction: The direction specifies the orientation or path followed from the reference point to the object or location being described. It provides information on the relative position or displacement of the object.
C. Fundamental units: Fundamental units, such as meters or feet, are used to quantify and measure the distance or displacement between the reference point and the object. These units provide a standardized way to express the position.
D. Motion: While motion itself is not necessary to designate a position, it can be relevant in certain cases when describing the position of a moving object. In such situations, the position is defined with respect to both the reference point and the object's movement.
Therefore, the correct answer is E. All of these elements - a reference point, a direction, fundamental units, and in some cases, motion - are necessary to designate a position accurately.
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You know that a point charge is located somewhere along the x-axis. When you measure the electric field at 2.00 m the result is 3.000 N/C and points left while the electric field at 5.00 m is 0.750 N/C also pointing left. (a) Where is the charge located?
(b) What is the size of the charge, including sign?
The electric field measurements indicate a point charge located to the right of both measurement positions on the x-axis, being more than 5.00 m away. Using the electric field values (3.000 N/C at 2.00 m and 0.750 N/C at 5.00 m), and applying the equation E = kQ/r^2, we can solve for both the point charge's magnitude and location.
The student's question revolves around the electric field produced by a point charge and involves finding both the location of the point charge and its magnitude. To solve this, we need to apply Coulomb's Law and the principle that the intensity of the electric field (E) from a point charge can be described by the equation E = kQ/r^2, where k is Coulomb's constant (8.988 imes 10^9 N m^2/C^2), Q is the charge, and r is the distance from the charge to the point where the electric field is measured.
Given the electric field measurements at two different points both pointing to the left, we can deduce that the point charge is located to the right of both points on the x-axis. The point charge must thus be more than 5.00 m to the right of the origin.
By using the measured electric field values and the distances to set up two equations, we can determine the charge's value:
E1 = kQ/r1^2 (at 2.00 m)
E2 = kQ/r2^2 (at 5.00 m)
Where E1 = 3.000 N/C, E2 = 0.750 N/C, r1 = 2.00 m, and r2 = 5.00 m. By solving these equations simultaneously, we can find the value of Q and the exact location along the x-axis.
Air is compressed in a cylinder such that the volume changes from 100.0 to 10.0 in^3. The initial pressure is 50.0 psia and the temperature is held at a constant 100.0 °F. How much work is required to change the volume? (Hint you will need to identify what psia means)
Answer:
5953.42 J
Explanation:
Given:
Initial volume, [tex]V_i[/tex]= 100 in³
Final Volume, [tex]V_f[/tex] = 10 in³
Initial pressure = 50 psia
Temperature = 100° F = 310.93 K
For isothermal reversible process, work done is given as:
Work done = [tex]-2.303RTlog_{10}\frac{V_f}{V_i}[/tex]
Where,
R is the ideal gas constant = 8.314 J/mol.K
or
Work done = [tex]-2.303\times8.314\times310.93log_{10}\frac{10}{100}[/tex]
or
Work done = 5953.42 J
While following a treasure map, you start at an old oak tree. You first walk 825 m directly south, then turn and walk 1.25 km at 30.0° west of north, and finally walk 1.00 km at 32.0° north of east, where you find the treasure: a biography of Isaac Newton! (a) To return to the old oak tree, in what direction should you head and how far will you walk? Use components to solve this problem. (b) To see whether your calculation in part (a) is reason- able, compare it with a graphical solution drawn roughly to scale.
To return to the tree after finding the treasure, you need to calculate the total displacement vector from the tree to the treasure by adding up the vectors of each leg of the journey. The direction and magnitude of the opposite of this vector will provide the direction and distance to get back to the tree. A graphical representation of the vectors will help confirm this.
Explanation:The problem is about determining the direction and distance to return to the original location after walking certain distances at specific angles. This involves the understanding of vectors and how they are applied in real life.
Given the three legs of the journey:
1st leg: 825 m directly south, which can be represented as a vector with a magnitude of 825 m and a direction of 180 degrees.2nd leg: 1.25 km at a 30-degree angle west of north, which can be represented as a vector with north and west components. Using trigonometry, we get a north-component of 1.25 km * cos(30), and a west-component of 1.25 km * sin(30).3rd leg: 1.00 km at a 32-degree angle north of east, which can be represented with east and north components. Calculating similarly, we get the east-component as 1.00 km * cos(32), and the north-component as 1.00 km * sin(32).
Summing up these components for all the legs, we get the total displacement vector from the tree to the treasure. To get back to the tree, you would need to walk in the opposite direction of this vector. The magnitude of this vector gives the total distance to be covered.
Comparing this calculated result with a graphical solution will help validate the projected direction and distance.
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The question ask:
A block is hung by a string from the inside roof of a van.When
the van goes straight ahead at a speed of 28m/s, theblock hangs
vertically down. But when the van maintains this samespeed around
an unbanked curve (radius=150m), the block swingstoward the outside
of the curve. Then the string makes anangle theta with the
vertical. Find theta.
Answer:
[tex]\theta=28.07^{\circ}[/tex]
Explanation:
Speed of van, v = 28 m/s
Radius of unbanked curve, r = 150 m
Let [tex]\theta[/tex] is the angle with the vertical. In case of banking of road,
[tex]T\ cos\theta=mg[/tex].............(1)
And
[tex]T\ sin\theta=\dfrac{mv^2}{r}[/tex]..........(2)
From equation (1) and (2) :
[tex]tan\theta=\dfrac{v^2}{rg}[/tex]
[tex]tan\theta=\dfrac{(28)^2}{150\times 9.8}[/tex]
[tex]\theta=28.07^{\circ}[/tex]
So, the string makes an angle of 28.07 degrees with the vertical. Hence, this is the required solution.
A charge Q= 2 C is distributed uniformly through out a bar of length L=2.5 m. The bar is placed horizontally in free space. A second charge q = 10-°C is placed along the line of the bar a distance d= 2m measured from the right end of the bar. What force is exerted on charge q by the charged bar?
Answer:
The force exerted by the charge q on the rod is [tex]5\times 10^{10}\ \rm N[/tex]
Explanation:
Given:
Charge on the rod=Q=2 CLength of the rod=L=2.5 mmagnitude of the point charge q=-10 CThe distance of the point charge from the right end of the rod d=2 mWe have to find the force exerted by the charge on the rod. this will be equal to the force exerted by the rod on the charge according to coulombs law.
The Electric field due the the rod at the location of the charge is given by
[tex]E=\dfrac{kQL}{d(d+L}\\E=\dfrac{9\times10^9\times2\times2.5}{2\times4.5}\\\\E=5\times 10^9\ \rm N/C\\[/tex]
Force between them is given by F
[tex]F=qE\\\\=10\times5\times10^9\\=5\times10^{10}\ \rm N[/tex]
A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helicopter is traveling horizontally with a speed of 70 m/s (250 km/h), (a) how far in advance of the recipients (horizontal distance) must the package be dropped? (b) Suppose, instead, that the helicopter releases the package a horizontal distance of 400 m in advance of the mountain climbers. What vertical velocity should the package be given (up or down) so that it arrives precisely at the climbers’ position? (c) With what speed does the package land in the latter case?
Answer:
a) 447.21m
b) -62.99 m/s
c)94.17 m/s
Explanation:
This situation we can divide in 2 parts:
⇒ Vertical : y =-200 m
y =1/2 at²
-200 = 1/2 *(-9.81)*t²
t= 6.388766 s
⇒Horizontal: Vx = Δx/Δt
Δx = 70 * 6.388766 = 447.21 m
b) ⇒ Horizontal
Vx = Δx/Δt ⇒ 70 = 400 /Δt
Δt= 5.7142857 s
⇒ Vertical:
y = v0t + 1/2 at²
-200 = v(5.7142857) + 1/2 *(-9.81) * 5.7142857²
v0= -7 m/s ⇒ it's negative because it goes down.
v= v0 +at
v= -7 + (-9.81) * 5.7142857
v= -62.99 m/s
c) √(70² + 62.99²) = 94.17 m/s
The horizontal distance at which the the package must be dropped and the vertical velocity are; 316.23 m and -22.08 m/s
What is the vertical velocity?
A)To fall 200 m, the time required is given by the formula;
t = √(2H/g)
t= √(200/9.8)
t = 4.518 seconds.
The supplies will travel forward by a distance of;
Δx = 4.518 * 70
Δx = 316.23 m
B) Time in the horizontal direction is;
t = Δx/V
t = 316.23/70
t = 4.52 s
In the vertical direction;
y = ut + ¹/₂at²
-200 = 4.52u + ¹/₂(-9.81)(4.52)²
100.21 - 200 = 4.52u
u = -99.79/4.52
u = -22.08 m/s
C) Speed at which the package lands in the latter case is;
v = √(70² + (-22.08²))
v = 73.4 m/s
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A sparrow is flying around in a circle at a constant speed and height. there is air resistance. In what direction is the net force of the sparrow on the air as the sparrow flies. 1. outward and downward 2.Inward and Downward 3.Downward and Backward 4. outward, downward and backward
Final answer:
The sparrow exerts a net force on the air in the downward and backward directions as per Newton's Third Law, as it pushes against the air to get lift and move forward.
Explanation:
The sparrow flying around in a circle at a constant speed and height with air resistance is subject to several forces. However, when we consider the force exerted by the sparrow on the air, we need to apply Newton's Third Law of motion which states that for every action, there is an equal and opposite reaction.
Therefore, as the sparrow's wings push the air downward and backward to get lift and forward movement, the air will exert an equal and opposite force on the sparrow. The sparrow's net force on the air would be downward and backward because the sparrow pushes the air in these directions to maintain flight.
A boy in a sleigh glides on a 40 angle snow-covered hill. The coefficient of kinetic friction between the surface and the sled is 0.12. The acceleration of the sled is: a. 3.6 m / s^2
b. 0.76 m / s^2
c. 2.3 m / s^2
d. 5.4 m / s^2
Answer:
option D
Explanation:
given,
angle of the snow-covering hill (θ) = 40°
coefficient of kinetic friction = 0.12
acceleration of the shed = ?
we know,
F = m a...................(1)
now,
[tex]F = m g sin\theta -\mu_k N sin\theta[/tex].......(2)
comparing equation (1) with (2)
[tex]m a =m g sin\theta -\mu_k m g sin\theta[/tex]
[tex]a = g sin \theta - \mu_k sin\theta[/tex]
[tex]a = 9.8\times sin 40^0 - 0.12\times 9.8\times sin40^0[/tex]
a = 5.54 m/s²
Hence, the correct answer is option D
An emission spectrum shows : a. Brightly colored lines
b. A rainbow of colors merging into each other
c. Only red and orange bands
d. Only green and blue bands
Answer: a)
Explanation: The phenomenon of emission is related to electronic transitions with the atom so brightly emission lines can represent the most important electronics transitions. They cover the whole spectrum from UV to IR.
A starship travels to a planet that is 20 light years away. The astronauts stay on the planet for 2.0 years before returning at the same speed and they are greeted back on earth 52 years after they left. Assume that the time needed to accelerate and decelerate is negligible. How much have the astronauts aged? (a) 15 years, (b) 20 years, (c) 22 years, (d) 30 years, (e) 32 years.
Answer:
astronauts age is 32 years
correct option is e 32 years
Explanation:
given data
travels = 20 light year
stay = 2 year
return = 52 years
to find out
astronauts aged
solution
we know here they stay 2 year so time taken in traveling is
time in traveling = ( 52 -2 ) = 50 year
so it mean 25 year in going and 25 years in return
and distance is given 20 light year
so speed will be
speed = distance / time
speed = 20 / 25 = 0.8 light year
so time is
time = [tex]\frac{t}{\sqrt{1-v^2} }[/tex]
time = [tex]\frac{25}{\sqrt{1-0.8^2} }[/tex]
time = 15 year
so age is 15 + 2 + 15
so astronauts age is 32 years
so correct option is e 32 years
During the execution of a play, a football player carries the ball for a distance of 33 m in the direction 58° north of east. To determine the number of meters gained on the play, find the northward component of the ball's displacement.
Answer:28 m
Explanation:
Given
Direction is [tex]58^{\circ}[/tex] North of east i.e. [tex]58 ^{\circ}[/tex] with x axis
Also ball moved by 33 m
therefore its east component is 33cos58=17.48 m
Northward component [tex]=33sin58=27.98 m\approx 28 m[/tex]
If two particles have equal kinetic energies, are their momenta necessarily equal? explain.
No, the momenta of two particles with equal kinetic energies are not necessarily equal.
Explanation:No, the momenta of two particles with equal kinetic energies are not necessarily equal. Momentum is given by the equation:
p = mv
Where p is momentum, m is mass, and v is velocity. Kinetic energy is given by the equation:
K = mv²/2
In order for two particles to have equal kinetic energies, their masses and velocities must satisfy the equation:
mv₁²/2 = mv₂²/2
But this does not necessarily mean that their momenta are equal, as the masses and velocities can still differ.
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Two satellites, A and B are in different circular orbits
aboutthe earth. The orbital speed of satellite A is twice that
ofsatellite B. Find theratio (Ta/Tb) of the
periods ofthe satellites.
The ratio of the orbital periods (Ta/Tb) of two satellites, where the orbital speed of satellite A is twice that of satellite B, can be found using Kepler's third law, which relates the orbital period squared to the radius of the orbit cubed. By understanding the relationship between orbital speed and radius, the ratio of the orbital periods can be calculated.
Explanation:When comparing two satellites, A and B, with orbital speeds such that the orbital speed of satellite A is twice that of satellite B, we are tasked with finding the ratio of their orbital periods (Ta/Tb). This problem is grounded in the principles of classical mechanics and specifically relates to Kepler's laws of planetary motion.
According to Kepler's third law, the square of the orbital period (T) is proportional to the cube of the radius of the orbit (r). Mathematically, this is expressed as T² ≈ r³ for a satellite orbiting a much larger body, such as the Earth. Since the gravitational attraction provides the necessary centripetal force for the satellite's circular motion, the gravitational force is also centrally involved in this relationship.
To compare the periods of two satellites, we use this proportionality. If the orbital speed of satellite A is twice that of satellite B, the radius of the orbit is also related to the speed. Specifically, speed is directly related to the square root of the radius of the orbit based on centripetal force considerations. Since the speed of satellite A is twice that of satellite B (VA = 2*VB), it follows that the radius of A's orbit would be four times that of B's orbity (rA = 4*rB). Applying Kepler's law then allows us to find the period ratio as (Ta/Tb)² = (rA/rB)³, and after substituting the relation for the radii, we can solve for (Ta/Tb).
An antelope moving with constant acceleration covers the distance 70.0 m between two points in time 6.60 s. Its speed as it passes the second point is 14.3 m/s. Part A
What is its speed at the first point?
Express your answer with the appropriate units. Part B
What is the acceleration?
Express your answer with the appropriate units.
Answer:
A) The speed at the first point is 6.91 m/s.
B) The acceleration is 1.12 m/s²
Explanation:
The equations of velocity and position of the antelope will be as follows:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position of the antelope at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
v = velocity at time t
A) If we place the origin of the frame of reference at the first point, we can say that at t = 6.60 the position of the antelope is 70.0 m and its velocity is 14.3 m/s. In this way, we will have 2 equations with 2 unknowns, the initial velocity (the velocity at the first point) and the acceleration.
Let´s start finding the speed at the first point:
v = v0 + a · t (solving for "a")
(v - v0)/t = a
Replacing a = (v - v0)/t in the equation of the position:
x = x0 + v0 · t + 1/2 · (v - v0)/t · t² (x0 = 0)
x = v0 · t + 1/2 · v · t - 1/2 · v0 · t
x - 1/2 · v · t = 1/2 · v0 · t
2/t · (x - 1/2 · v · t) = v0
2/6.60 s · (70.0 m - 1/2 · 14.3 m/s · 6.60s) = v0
v0 = 6.91 m/s
The speed at the first point is 6.91 m/s.
B) Using the equation of velocity
a = (v - v0)/t
a = (14,3 m/s - 6.91 m/s) / 6.60 s
a = 1.12 m/s²
The acceleration is 1.12 m/s²
An aircraft cruises at an altitude of 10,700 meters above sea level. Estimate the atmospheric pressure in bar at cruise altitude. Assume the acceleration of gravity is constant at g = 9.8 m/s". The average specific volume of the air is 1.334 m®/kg.
Answer:
We can calculate the atmospheric pressure using the given formula:
[tex]P = P_{o} - \rho\times g\times h[/tex]
where;
[tex]P_{o} = 1.013 \times 10^{5} pa[/tex]
[tex]\rho = \frac{1}{V}[/tex]
[tex]\rho = \frac{1}{1.334} = 0.75 kg/m^{3}[/tex]
h = 10,700 m
equating the following variables in above equation, we get;
[tex]P = 1.013\times 10^{5} - 0.75 \times 9.8 \times 10700[/tex]
P = 0.227 bar
Walk 42 miles due north, deviate 78 degrees to east, and walk 65miles. What is the displacement? ( magnitude and direction with respect to North). A) Show work through calculations for predictions
Answer:84.405m,[tex]\theta =48.876^{\circ}[/tex]
Explanation:
Given
Person walk 42 miles due to north so its position vector is
[tex]r_1=42\hat{j}[/tex]
Now he deviates [tex]78^{\circ}[/tex] to east and walk 65 miles
so its new position vector
[tex]r_2=42\hat{j}+65cos78\hat{j}+65sin78\hat{i}[/tex]
[tex]r_2=65sin78\hat{i}+\left ( 42+65cos78\right )\hat{j}[/tex]
So magnitude of acceleration is
[tex]|r_2|=\sqrt{\left ( 65sin78\right )^2+\left ( 42+65cos78\right )^2}[/tex]
[tex]|r_2|=\sqrt{63.58^2+55.514^2}[/tex]
[tex]|r_2|=84.405 m[/tex]
for direction
[tex]tan\theta =\frac{42+65cos78}{65sin78}[/tex]
[tex]tan\theta =0.8731 [/tex]
[tex]\theta =41.124^{\circ}with\ respect\ to\ east[/tex]
[tex]\theta =48.876^{\circ}with\ respect\ to\ North[/tex]
A delivery truck starts it’s run by driving 5.20 km due west before turning due north and driving an additional 2.10 km. Finally, the truck turns 30.0 degrees north of east and drives for 3.70 km before reaching its first dropoff point. What is the magnitude of the total displacement of the truck from where it started to its first dropoff point?
Answer:
4.427 m
Explanation:
We shall consider east as + x- axes and north as + ve y- axes. .
We shall represent every displacement in vector form as follows
D₁ = 5.2 km due west = - 5.2 i
D₂ = 2.1 km due north = 2.1 j
D₃ = 3.7 km towards north east at 30 degree from east
= 3.7 cos30 i + 3.7 sin 30 j = 3.2 i + 1.85 j
Total displacement D = D₁ + D₂ +D₃ +D₄.
- 5.2 i + 2.1 j + 3.2 i + 1.85 j
= - 2 i + 3.95 j
Magnitude of D
D² = (2)² + (3.95)²
= 4 + 15.6025
D = 4.427 m
You and your family take a trip to see your aunt who lives 100 miles away along a straight highway. The first 60 miles of the trip are driven at 55 mi/h but then you get stuck in a standstill traffic jam for 20 minutes. In order to make up time, you then proceed at 75 mi/h for the rest of the trip. What is the magnitude of your average velocity for the whole trip?
The magnitude of your average velocity for the whole trip is 102.19 miles per hour.
First, let's calculate the time taken for each segment of the trip:
1. For the first 60 miles at 55 mi/h:
Time = Distance / Speed
= 60 miles / 55 mi/h
= 1.0909 hours
2. During the traffic jam, you're not moving, so the time is 20 minutes, which needs to be converted to hours:
Time = 20 minutes / 60 minutes/hour
= 1/3 hours
3. For the last 40 miles at 75 mi/h:
Time = 40 miles / 75 mi/h
= 0.5333 hours
Now, calculate the total time for the trip:
Total Time = 1.0909 hours + 1/3 hours + 0.5333 hours
= 1.9572 hours
Since you traveled 100 miles to your aunt's house and then returned,
the total displacement = 2 * 100 miles = 200 miles.
Now, calculate the average velocity:
Average Velocity = Total Displacement / Total Time
= 200 miles / 1.9572 hours
= 102.19 mi/h
So, the average velocity is 102.19 miles per hour.
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If you and your family take a trip to see your aunt who lives 100 miles away along a straight highway. The magnitude of your average velocity for the whole trip is 51.04 miles per hour.
What is the magnitude?To determine the magnitude of your average velocity for the whole trip, we can use the formula for average velocity:
Average Velocity = Total Displacement / Total Time
Time for the first 60 miles at 55 mi/h:
Time = Distance / Speed
= 60 miles / 55 mi/h
= 1.0909 hours
Time spent in the traffic jam:
20 minutes = 20 / 60
= 1/3 hours
Time for the remaining 40 miles at 75 mi/h:
Time = Distance / Speed
= 40 miles / 75 mi/h
= 0.5333 hours
Total Time = 1.0909 + 1/3 + 0.5333
= 1.9572 hours
Now we can calculate the average velocity:
Average Velocity = Total Displacement / Total Time
Average Velocity = 100 miles / 1.9572 hours
≈ 51.04 mi/h
Therefore the magnitude of your average velocity for the whole trip is 51.04 miles per hour.
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