Answer:
option D
Explanation:
given,
uniform length of cylinder = 1 m
diameter of the cylinder = 10 cm = 0.1 m
Eels have been recorded to spin = 14 rev/s
camera records at = 120 frames per second
time = [tex]\dfrac{1}{120}\ s/frame[/tex]
angle at which eel rotate = ?
ω = 14 rev/s
ω = 14 x 2 π rad/s
ω = 28 π rad/s
angle at which eel rotate
θ = ω t
θ = [tex]28\pi\times \dfrac{1}{120}[/tex]
θ = 0.733 rad
θ =[tex]0.733 \times \dfrac{180^0}{2\pi}[/tex]
θ =[tex]42^0[/tex]
Hence, the correct answer is option D
The angle of rotation is the angle at which the object is rotate about the fixed point.
The angle at which the eel rotates from one frame to next frame is 42 degree.
Given that, we can treat the eel as a uniform cylinder.
So, uniform length of cylinder = 1 [tex]\rm m[/tex] and diameter of the cylinder = 10 [tex]\rm cm[/tex] = 0.1 [tex]\rm m[/tex]. Eels have been recorded to spin at up to 14 rev/s when feeding in this way. The camera records at 120 frames per second.
Time taken by the camera to record one frame is [tex]t\;=\; \dfrac {1} {120} \;\rm s/\rm frame[/tex]
Revolution done by eel in radian per second is 14.
Angular Velocity is [tex]\omega\;=\; 14\;\times\;2\pi\;\times\;0.1\;\rm rad/s[/tex].
The angle of rotation can be calculated by the formula given below.
Angle of rotation [tex]\theta=\omega\;\times\;t[/tex].
Substituting the values in the above formula, the angle of rotation is,
[tex]\theta\;=\;14\;\times\;2\;\times\;3.14\;\times\;0.1\;\times\;\dfrac {1}{120}\;\rm rad[/tex]
[tex]\theta=0.0733\;\rm rad[/tex]
[tex]\theta=0.0733\;\times\;\dfrac {360^\circ}{2\pi}[/tex]
[tex]\theta=41.5^\circ[/tex] or [tex]42^\circ[/tex].
The angle at which the eel rotates from one frame to next frame is 42 degree.
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https://brainly.com/question/17671314.
A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a rate of 3 ft./sec. How fast is the height of the person's shadow changing when the person is 13 ft. from the projector
Answer:
y = 67.6 feet, y = 114.4/ (22 - 3t)
Explanation:
For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram
Large triangle Projector up to the screen
tan θ = y / L
For the small triangle. Projector up to the person
tan θ = y₀ / (L-d)
The angle is the same, so we equate the two equations
y₀ / (L -d) = y / L
y = y₀ L / (L-d)
The distance from the screen (d), we look for it with kinematics
v = d / t
d = v t
we replace
y = y₀ L / (L - v t)
y = 5.2 22 / (22 - 3 t)
y = 114.4 (22 - 3t)⁻¹
This is the equation of the shadow height change as a function of time
For the suggested distance the shadow has a height of
y = 114.4 / (22-13)
y = 67.6 feet
The edge of a cube was found to be 30 cm with a possible error in measurement of 0.5 cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in computing the volume of the cube and the surface area of the cube. (Round your answers to four decimal places.)
Answer:
In computing the volume of a cube,
Maximum possible error = +/-1350cm³
Relative error = 0.05
Percentage error = 5%
In computing the surface area of a cube,
Maximum possible error = +/-180cm²
Relative error = 0.0333
Percentage error = 3.33%
Explanation:
A cube is a three dimensional solid object with six (6) faces, twelve (12) edges and eight(8) vertices.
The volume of a cube = x³
Where x= length of the edge of a cube
X = 30cm +/- 0.5cm
Differentiate V with respect to x (V = Volume of a cube)
dV/dx = 3 x²
dV = 3 x² . dx
dV= 3 × 30² × (+/-0.5)
= 2700(+/-0.5)
= +/-1350cm³
Maximum possible error =
+/- 1350cm³
Relative error = Maximum error /surface area
= ΔV/V
Recall that V = x³
V= (30)³
A = 27000cm³
Substitute the values for and V into the formula for Relative error
Relative error = 1350 / 270000
Relative error = 0.05
% error = Relative error × 100
= 0.05× 100
= 5%
Surface Area of a cube = 6x²
A = 6x²
Differentiate A with respect to x
dA/dx= 12x
dA = 12x . dx
dA= 12 × 30 (0.5)
= +/- 180cm²
Maximum possible error =
+/- 180cm²
Relative error = Maximum error / total area
= dA/dx
Recall that A = 6x²
A = 6(30)²
A = 5400cm²
Substitute the values for and A into the formula for Relative error
Relative error = 180/ 5400
Relative error = 0.0333(4 decimal place)
% error = Relative error × 100
= 0.0333 × 100
= 3.33%
Final answer:
Use of differentials to estimate maximum and relative errors in volume and surface area calculations for a cube.
Explanation:
Differentials for Cube:
Maximum possible error in volume: 30*(0.5) = 15 cm³Relative error in volume: 15/30 = 0.5Percentage error in volume: (0.5)*100% = 50%Surface Area:
Maximum possible error in surface area: 6*(30)*(0.5) = 90 cm²Relative error in surface area: 90/(6*30) = 0.25Percentage error in surface area: (0.25)*100% = 25%a bike travels at a constant speed for 4.00 m/s for 5.00 seconds. How far does it go
Answer:
20 metres
Explanation:
Speed = distance ÷ time
[tex]s = \frac{d}{t} [/tex]
If we substitute the values:
[tex]4 = \frac{d}{5} [/tex]
[tex]20 = d[/tex]
Answer: 20m
Explanation:
Speed = Distance/time taken
Speed = 4.00m/s
Time taken = 5.00s
Distance = D = ?
We insert the values in the formula
4.00m/s = D/5.00s
Multiply through by 5
D = 20m
Poiseuille's law remains valid as long as the fluid flow is laminar. For sufficiently high speed, however, the flow becomes turbulent, even if the fluid is moving through a smooth pipe with no restrictions. It is found experimentally that the flow is laminar as long as the Reynolds Number Re is less than about 2000: Re = 2v Normal 0 false false false IN X-NONE X-NONE MicrosoftInternetExplorer4 rhoR Normal 0 false false false IN X-NONE X-NONE MicrosoftInternetExplorer4 /η. Here Normal 0 false false false IN X-NONE X-NONE MicrosoftInternetExplorer4 v, rho, and η are, respectively, the average speed, density, and viscosity of fluid, and R is the radius of the pipe. Calculate the highest average speed that blood (rho = 1060 kg/m3, η = 4.0 x 10-3 Pa.s) could have and still remain in laminar flow when it flows through the aorta (R = 8.0 x 10-3 m)
Normal 0 false false false IN X-NONE X-NONE MicrosoftInternetExplorer4
Answer: 0.471 m/s
Explanation:
We are given the followin equation:
[tex]Re=\frac{D v \rho}{\eta}[/tex] (1)
Where:
[tex]Re[/tex] is the Reynolds Number, which is adimensional and indicates if the flow is laminar or turbulent
When [tex]Re<2100[/tex] we have a laminar flow
When [tex]Re>4000[/tex] we have a turbulent flow
When [tex]2100<Re<4000[/tex] the flow is in the transition region
[tex]D=2R[/tex] is the diameter of the pipe. If the pipe ha a radius [tex]R=8(10)^{-3} m[/tex] its diameter is [tex]D=2(8(10)^{-3} m)=0.016 m[/tex]
[tex]v[/tex] is the average speed of the fluid
[tex]\rho=1060 kg/m^{3}[/tex] is the density of the fluid
[tex]\eta=4(10)^{-3} Pa.s[/tex] is the viscosity of the fluid
Isolating [tex]v[/tex]:
[tex]v=\frac{Re \eta}{D \rho}[/tex] (2)
Solving for [tex]Re=2000[/tex]
[tex]v=\frac{(2000)(4(10)^{-3} Pa.s)}{(0.016 m)(1060 kg/m^{3})}[/tex] (3)
Finally:
[tex]v=0.471 m/s[/tex]
Calculate the electric potential V(h) inside the capacitor as a function of height h. Take the potential at the bottom plate to be zero.Express V(h) in terms of E and h.V(h) =
Answer:
V(h) = Eh
Explanation:
I will assume that the capacitor is a parallel-plate capacitor.
By Gauss' Law, electric field inside the capacitor is
[tex]E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}[/tex]
The relation between electric field and potential is
[tex]V_{ab} = -\int\limits^b_a {\vec{E}(h)} \, d\vec{h} = \int\limits^h_0 {\frac{Q}{\epsilon_0 A}} \, dh \\V(h) - V(0) = V(h) - 0 = Eh\\V(h) = Eh[/tex]
The important thing in this question is that the electric field inside the parallel plate is constant. So, the potential is also constant and proportional to the distance, h.
The electric potential V(h) inside a capacitor as a function of height h, with zero potential at the bottom plate, is V(h) = Eh, using the relationship between the electric field E and potential V where E is constant.
Explanation:To calculate the electric potential V(h) inside a capacitor as a function of height h, with the potential at the bottom plate taken to be zero, you can use a relation between the electric field E and the potential V. Given that E = V/d, where d is the separation between the plates, and that the electric field E is uniform, we have the relationship E = -dV/dh (the negative sign indicates the direction of the potential decrease). Integrating this equation from 0 to h, where V(0) = 0, gives us V(h) = -Eh. However, we can ignore the negative sign because we are interested in magnitude.
So, the final expression for V(h) inside the capacitor, in terms of the electric field E and the height h above the bottom plate, is:
V(h) = Eh
What type of wave is shown above?
A. transverse wave
B. longitudinal wave
C. surface wave
D. electromagnetic wave
Answer:
its a A. Transverse wave
The mass of a spacecraft is about 435 kg . An engine designed to increase the speed of the spacecraft while in outer space provides 0.09-N thrust at maximum power.By how much does the engine cause the craft's speed to change in 1 week of running at maximum power? Describe any assumptions you made.
Answer:
Δ v = 125 m/s
Explanation:
given,
mass of space craft = 435 Kg
thrust = 0.09 N
time = 1 week
= 7 x 24 x 60 x 60 s
change in speed of craft = ?
Assuming no external force is exerted on the space craft
now,
[tex]T= m_s a[/tex]
[tex]a=\dfrac{T}{m_s}[/tex]
[tex]a =\dfrac{0.09}{435}[/tex]
a = 2.068 x 10⁻⁴ m/s²
using equation of motion
Δ v = a t
Δ v = 2.068 x 10⁻⁴ x 7 x 24 x 60 x 60
Δ v = 125 m/s
Calculate the heat required when 2.50 mol of a reacts with excess b and a2b according to the reaction: 2a + b + a2b → 2ab + a2 given: 2a + b → a2b δh°
Answer:
Q = 12.5 kJ
Explanation:
The expression to use to calculate Heat is:
Q = H° * n
Where:
Q: heat (J or kJ)
H°: enthalpy of reaction (kJ/mol)
n: moles
Now, as it was stated in the comments, the question is incomplete, and here is the missing part:
Given:
2A + B A2B (1)
H° = – 25.0 kJ/mol
2A2B 2AB + A2 (2)
H° = 35.0 kJ/mol
With these two reactions, we can calculate the heat.
Now, with the above two reactions, we need to get the general reaction (The one the question is giving), so, let's use (1) and (2) and do the sum of them:
2A + B -------> A2B H°1 = -25 kJ/mol
2A2B --------> 2AB + A2 H°2 = 35 kJ/mol
Now, we sum both equations, we can see that one A2B cancels out with one A2B from equation 2, so, the equation gives:
2A + B + 2A2B -------> 2AB + A2
And the enthalpy, it's just summed:
H°3 = -25 + 35 = 10 kJ/mol
Now with this value we can calculate heat:
Q = 10 * 2.5 = 25 kJ
However, in the reaction we have 2A, so it's not 1:1 mole ratio, but instead is 1:2, so this result we have to divide it between 2 so:
Q = 25 / 2 = 12.5 kJ
The heat required for the reaction is zero because the enthalpy changes for the two parts of the reaction cancel each other out.
To calculate the heat required for the reaction, we need to know the enthalpy change (ΔH°) for the reaction 2a + b → a2b. However, the enthalpy change for this reaction is not provided in the question. Assuming that the enthalpy change for this reaction is known, we can proceed with the calculation.
Let's denote the enthalpy change for the reaction 2a + b → a2b as ΔH°. The reaction of interest is:
[tex]\[ 2a + b + a_2b \rightarrow 2ab + a_2 \][/tex]
This reaction can be broken down into two parts:
1. The reaction of 2 moles of a with 1 mole of b to form 1 mole of a2b:
with an enthalpy change of H°.
[tex]\[ 2a + b \rightarrow a_2b \][/tex]
2. The reaction of 1 mole of a2b to form 2 moles of ab:
[tex]\[ a_2b \rightarrow 2ab \][/tex]
Since the reaction of a2b to form 2ab does not involve any additional reactants, it can be considered as the reverse of the first part, but with twice the amount of ab produced. Therefore, the enthalpy change for this part would be -H° (since the reverse reaction has the opposite sign of the forward reaction), and for 1 mole of a2b reacting, it would be -2ΔH° (because 2 moles of ab are formed).
Now, we are given 2.50 moles of a, and we assume there is excess b and a2b. The reaction will proceed until all of a is consumed. Since the reaction stoichiometry is 2 moles of a to 1 mole of a2b, 1.25 moles of a2b will be required to react with the 2.50 moles of a (since 2.50 moles of a require 1.25 moles of a2b).
The overall enthalpy change for the reaction will be the sum of the enthalpy changes for the two parts:
1. For the reaction of 2.50 moles of a with b to form a2b:
[tex]\[ \Delta H_1 = 2.50 \text{ moles} \times \Delta H^\circ \][/tex]
2. For the reaction of 1.25 moles of a2b to form ab:
[tex]\[ \Delta H_2 = 1.25 \text{ moles} \times (-2 \Delta H^\circ) \][/tex]
The total enthalpy change (ΔH_total) is the sum of ΔH_1 and ΔH_2:
[tex]\[ \Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 \][/tex][tex]\[ \Delta H_{\text{total}} = 2.50 \Delta H^\circ - 2 \times 1.25 \Delta H^\circ \][/tex]
[tex]\[ \Delta H_{\text{total}} = 2.50 \Delta H^\circ - 2.50 \Delta H^\circ \][/tex]
[tex]\[ \Delta H_{\text{total}} = 0 \][/tex]
Surprisingly, the total enthalpy change for the reaction is zero. This is because the amount of heat released in the formation of a2b is exactly equal to the amount of heat absorbed in the formation of ab from a2b. Therefore, no additional heat is required for the overall reaction to proceed.
The final answer is: [tex]\[ \boxed{0} \][/tex]
The heat required for the reaction is zero because the enthalpy changes for the two parts of the reaction cancel each other out. This conclusion is based on the assumption that the enthalpy change for the reaction 2a + b → a2b is known and that the reaction proceeds as written. If the enthalpy change for the formation of a2b is not known, then it would be necessary to look up the standard enthalpy of formation for a2b to perform the calculation.
A stationary source S generates circular outgoing waves on a lake. The wave speed is 5.0 m/s and the crest-to-crest distance is 2.0 m. A person in a motorboat heads directly toward S at 3.0 m/s. To this person, the frequency of these waves is:
Answer:4 Hz
Explanation:
Speed of wave [tex]v=5 m/s[/tex]
crest to crest distance [tex]\lambda =2 m[/tex]
velocity of observer [tex]v_0=3 m/s[/tex]
actual frequency [tex]f=\frac{velocity}{\lambda }[/tex]
[tex]f=\frac{5}{2}=2.5 Hz[/tex]
Apparent frequency [tex]f'=f(\frac{v+v_0}{v})[/tex]
[tex]f'=2.5\times \frac{5+3}{5}[/tex]
[tex]f'=4 Hz[/tex]
The frequency of these waves is 4hz.
Given information:The wave speed is 5.0 m/s and the crest-to-crest distance is 2.0 m. Alo, A person in a motorboat head directly toward S at 3.0 m/s.
Calculation of wave frequency:The actual frequency is [tex]= 5 \div 2[/tex] = 2.5Hz
Now
Apparent frequency is
[tex]= 2.5 \times (5 + 3) \div 5[/tex]
= 4hz
Hence, the frequency of these waves is 4hz.
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The rumble feature on a video game controller is driven by a device that turns electrical energy into mechanical energy. This device is best referred to as
Answer:
MOTOR
Explanation:
The device which changes electrical energy into mechanical energy is known as motor
Loops of wire in a magnetic field make up motors. The magnetic field produces a torque on the loops when current flows through them, turning a shaft as a result.
What is electric motor?Electrical energy is transformed into mechanical energy by electric motors. When we turn on the fan, for instance, the electric motor begins to transform the electrical energy into mechanical energy.
The fan blades then begin whirling as a result of the mechanical energy, giving them the capacity to perform work.
On either side of the controller, there is a motor. This engine has an uneven weight linked to it.
This indicates that one side of it is heavier than the other. The imbalance of the weight causes the controller to vibrate when the motor turns.
Therefore, Mechanical work is created as a result of the conversion of electrical energy.
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A child pushes horizontally on a box of mass m which moves with constant speed v across a horizontal floor. The coefficient of friction between the box and the floor is μ. At what rate does the child do work on the box?
Answer:
Rate of child doing work on box = μmgv (Unit is Watt)
Explanation:
Rate of child doing work on box = Work done / time = Power
Power = Horizontal force x Velocity
We are aware that the Velocity in this case is v.
As the object is moving with constant velocity, the acceleration would be zero and the applied horizontal force will be equal to friction force. So in our case,
Horizontal force = friction force
We know that the coefficient of friction is the ratio of friction force to Normal force,
μ = friction force / Normal force
Normal Force = mg, where m is the mass and g is the gravitational acceleration
Friction force = μ x Normal Force
Friction force = μmg
Power = μmgv (Unit of power is Watt)
Answer:
P = μ*mg*v
Explanation:
A child pushes horizontally on a box of mass m which moves with constant speed v across a horizontal floor. The coefficient of friction between the box and the floor is μ. The rate at which the child works is calculated as shown below:
mass of the box = m; coefficient of friction is μ; speed = v.
In order to push the box, the child must exert a force equal to or more than the frictional force.
force = coefficient of friction*weight of the box
f = μ*mg
In addition, to calculate the rate of work (i.e. power). We have:
Power = force*velocity (or speed)
Therefore:
P = μ*mg*v
A motorcycle rides on the vertical walls around the perimeter of a large circular room. The friction coefficient between the motorcycle tires and the walls is µ. How does the minimum µ needed to prevent the motorcycle from slipping downwards change with the motorcycle’s speed, s?
a) µ ∝ s0b) µ ∝ s−1/2c) µ ∝ s−1d) µ ∝ s−2e) none of these
Answer:
option D
Explanation:
given,
coefficient of friction between wall and tire = µ
speed of motorcycle = s
friction force = f = μ N
where normal force will be equal to centripetal force
[tex]N = \dfrac{mv^2}{r}[/tex]
for motorcycle to not to slip weight should equal to the centripetal force
now,
[tex]m g =\mu \dfrac{mv^2}{r}[/tex]
[tex]\mu =\dfrac{rg}{s^2}[/tex]
where "rg" is constant
[tex]\mu\ \alpha \ \dfrac{1}{s^2}[/tex]
[tex]\mu\ \alpha \ s^{-2}[/tex]
Hence, the correct answer is option D
The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the moon and that the mass of the earth is 5.98 x 10^24 kg. Find the period for the moon's motion around the earth. Express the answer in days and compare it to the length of a month.
Answer:
27.5 days
0.92 month
Explanation:
[tex]r[/tex] = radius of the orbit of moon around the earth = [tex]3.85\times10^{8} m[/tex]
[tex]M[/tex] = Mass of earth = [tex]5.98\times10^{24} m[/tex]
[tex]T[/tex] = Time period of moon's motion
According to Kepler's third law, Time period is related to radius of orbit as
[tex]T^{2} = \frac{4\pi ^{2} r^{3} }{GM}[/tex]
inserting the values, we get
[tex]T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3} }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec[/tex]
we know that
1 day = 24 hours = 24 x 3600 sec = 86400 s
[tex]T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days[/tex]
1 month = 30 days
[tex]T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month[/tex]
Final answer:
The period for the moon's motion around the earth is approximately 0.59 days, which is much shorter than the length of a month.
Explanation:
To find the period for the moon's motion around the earth, we can use Kepler's third law. According to Kepler's third law, the square of the period of a planet's orbit is directly proportional to the cube of its average distance from the center of the orbit.
We are given that the moon orbits the earth at a distance of 3.85 x 10^8 m. We can use this information to calculate the period as follows:
Convert the given distance to meters: 3.85 x 10^8 m.Calculate the period using Kepler's third law equation:Hence, the period for the moon's motion around the earth is approximately 0.59 days. This is much shorter than the length of a month, which is about 30 days. Therefore, the moon completes multiple orbits around the earth in one month.
Two technicians are discussing oil leaks. Technician A says that an oil leak can be found using a fluorescent dye in the oil with a black light to check for leaks. Technician B says a white spray powder can be used to locate oil leaks. Which technician is correct?a. Technician A only
b. Technician B only
c. Both Technicians A and B
d. Neither Technician A nor B
Answer:
B
Explanation:
cause it makes the most sence
which cruising altitude is appropriate for vfr flight on a magnetic course of 135°
Answer:odd thousands plus 500 feet.
Explanation:
On a magnetic course of zero through 179 degrees, select an odd thousand foot cruising altitude plus 500 feet, such as 3,500, 5,500, up to and including 17,500. Even and odd thousands are reserved for those aircraft on an active instrument flight plan. Even thousands plus 500 feet are for aircraft flying between 180 and 359 degrees.
The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface (1) has an area of 1.70 m², while surface (2) has an area of 4.00 m². The electric field in the drawing is uniform and has a magnitude of 210 N/C. Find the magnitude of the electric flux through surface (1) if the angle θ made between the electric field with surface (2) is 34.0°.
Answer:
Ф1=295.96Nm^2/C
Ф2=469.73Nm^2/C
Explanation:
The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface (1) has an area of 1.70 m², while surface (2) has an area of 4.00 m². The electric field in the drawing is uniform and has a magnitude of 210 N/C. Find the magnitude of the electric flux through surface (1) if the angle θ made between the electric field with surface (2) is 34.0
we have two surfaces where we know angle of surface 1 lets call it
s1= 34.
Therefore to find s2
s2 (the angle from surface 2) we have that
s2=180-(90+s1),
so s2=180-(90+34),
thus s2=56 degrees.
Flux equation reads as Φ=ΕΑ,
where Φ is the flux,
E is the electric field and
A is the surface area.
So with respect to the angles and the figure provided,
we have Φ=EAcos(s).
So we can solve further by writing
. For Surface 1 we have
Φ1=EAcos(s1)=210 x 1.7 x cos 34,
so Φ1=295.96,
approximately to an whole number
Φ1=296 Nm^2/c.
Similarly for Φ2, we have
Φ2=EAcos(s2)
=210 x 4 x cos34=469.7,
thus Φ2=469.7Nm^2/c.
A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C at an average speed of 60 kilometers per hour. If the train did not stop at Station B, what was the average speed at which the train traveled from Station A to C?
(1) The distance that the train traveled from Station A to Station B was 4 times the distance that train traveled from Station B to Station C.
(2) The amount of time it took to the train to travel from Station A to Station B is 3 times the amount of time that it took the train to travel from Station B to Station C.
Answer:
1)
75 kmh⁻¹
2)
75 kmh⁻¹
Explanation:
1)
[tex]v_{ab}[/tex] = Speed of train from station A to station B = 80 kmh⁻¹
[tex]d_{ab}[/tex] = distance traveled from station A to station B
[tex]t_{ab}[/tex] = time of travel between station A to station B
we know that
[tex]Time = \frac{distance}{speed}[/tex]
[tex]t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}[/tex]
[tex]d_{bc}[/tex] = distance traveled from station B to station C
[tex]v_{bc}[/tex] = Speed of train from station B to station C = 60 kmh⁻¹
[tex]t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}[/tex]
Total distance traveled is given as
[tex]d = d_{ab} + d_{bc}[/tex]
Total time of travel is given as
[tex]t = t_{ab} + t_{bc}[/tex]
Average speed is given as
[tex]v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }[/tex]
Given that :
[tex]d_{ab} = 4 d_{bc}[/tex]
So
[tex]v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}[/tex]
2)
[tex]v_{ab}[/tex] = Speed of train from station A to station B = 80 kmh⁻¹
[tex]t_{ab}[/tex] = time of travel between station A to station B
[tex]d_{ab}[/tex] = distance traveled from station A to station B
we know that
[tex]distance = (speed) (time)[/tex]
[tex]d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}[/tex]
[tex]d_{bc}[/tex] = distance traveled from station B to station C
[tex]v_{bc}[/tex] = Speed of train from station B to station C = 60 kmh⁻¹
[tex]t_{bc}[/tex] = time of travel for train from station B to station C
we know that
[tex]distance = (speed) (time)[/tex]
[tex]d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}[/tex]
Total distance traveled is given as
[tex]d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}[/tex]
Total time of travel is given as
[tex]t = t_{ab} + t_{bc}[/tex]
Average speed is given as
[tex]v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}[/tex]
Given that :
[tex]t_{ab} = 3 t_{bc}[/tex]
So
[tex]v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}[/tex]
A 5.0-kg cannonball is fired over level ground with a velocity of 2.00 ⨯ 102 m/s at an angle of 25° above the horizontal. Just before it hits the ground its speed is 150 m/s. Over the entire trip, find the change in the thermal energy of the cannonball and air.
Answer:
E=147898.01J
Explanation:
A 5.0-kg cannonball is fired over level ground with a velocity of 2.00 ⨯ 102 m/s at an angle of 25° above the horizontal. Just before it hits the ground its speed is 150 m/s. Over the entire trip, find the change in the thermal energy of the cannonball and air
firstly , we look for the time of flight it takes to make the projectile path
T=2Usin∅/g
take g= 9.81m/s
T=2*200sin25/(9.81)
T=17.23Secs
energy is force *distance
E=f*d
f=m*g
f=5*9.81
f=49.05N
s=distance
s=(v+u)T/2
s=(150+200)17.23/2
s=3015.25m
49.05N*3015.25m
E=147898.01J
Approximate the work required to lift a 2.5-kg object to a height of 6.0 meters. A student applies a force to a cart to pull it up an inclined plane at a constant speed during a physics lab. A force of 20.8 N is applied parallel to the incline to lift a 3.00-kg loaded cart to a height of 0.450 m along an incline which is 0.636-m long. Determine the work done upon the cart and the subsequent potential energy change of the cart. Eddy, whose mass is 65-kg, climbs up the 1.6-meter high stairs in 1.2 s. Approximate Eddy's power rating.
Answer:
Explanation:
(a)mass [tex]m= 2.5 kg [/tex]
height [tex]h=6 m[/tex]
work required to raise
[tex]W=mgh[/tex]
[tex]W=2.5\times 9.8\times 6=147 J[/tex]
(b)Force [tex]F=20.8 N[/tex]
mass of cart [tex]m=3 kg[/tex]
length of track [tex]s=0.636 m[/tex]
[tex]Work\ done=F\cdot s[/tex]
Work done[tex]=20.8\cdot 0.636=13.22 J[/tex]
(c)mass of eddy [tex]m_e=65 kg[/tex]
height climbed [tex]h=1.6 m[/tex]
time [tex]t=1.2 s[/tex]
Energy required [tex]E=mgh=65\times 9.8\times 1.6=1019.2 J[/tex]
[tex]power=\frac{E}{t}=\frac{1019.2}{1.2}=849.33 W[/tex]
The work done in lifting the 2.5 kg object is 147 J, in moving the 3-kg cart is 13.2 J, and the power used by Eddy in climbing stairs is 851.7 W.
Explanation:To solve these problems, we need to apply principles of physics, specifically related to work, energy, and power. For the first question, we use the concept of gravitational potential energy, which is calculated by multiplying together the object's mass, the acceleration due to gravity (~9.81 m/s² on Earth), and the height to which the object is lifted. Thus for the 2.5-kg object, the work done or energy required to lift it to a height of 6.0 meters is: W = m * g * h = 2.5 kg * 9.8 m/s² * 6.0 m = 147 J.
Next, for the 3-kg cart, since the cart moves at constant speed, we can say the work done is equal to the change in potential energy. Therefore, the work done is W = m * g * h = 3.0 kg * 9.8 m/s² * 0.450 m = 13.2 J.
Fianlly, for Eddy's case, power is defined as the work done per unit time. If Eddy lifts his own mass to the height of 1.6 m, the work done (again considering as change in gravitational potential energy) is W = m * g * h = 65 kg * 9.8 m/s² * 1.6 m = 1022 J. Given that he does this work in 1.2 seconds, the power expended would be P = W / t = 1022 J / 1.2 s = 851.7 W.
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A pelican flying along a horizontal path drops a fish from a height of 4.7 m. The fish travels 9.3 m horizontally before it hits the water below. What was the pelican’s initial speed? The acceleration of gravity is 9.81 m/s 2 . Answer in units of m/s. If the pelican was traveling at the same speed but was only 1.5 m above the water, how far would the fish travel horizontally before hitting the water below?
Answer:
(A) 9.5 m/s
(B) 5.225 m
Explanation:
vertical height (h) = 4.7 m
horizontal distance (d) = 9.3 m
acceleration due to gravity (g) = 9.8 m/s^{2}
initial speed of the fish (u) = 0 m/s
(A) what is the pelicans initial speed ?
lets first calculate the time it took the fish to falls = ut + [tex](\frac{1}{2}) at^{2}[/tex]
since u = 0
s = [tex](\frac{1}{2}) at^{2}[/tex]
t = [tex]\sqrt{\frac{2s}{a} }[/tex]
where a = acceleration due to gravity and s = vertical height
t = [tex]\sqrt{\frac{2 x 4.7 }{9.8} }[/tex] = 0.98 s
pelicans initial speed = speed of the fishspeed of the fish = distance / time = 9.3 / 0.98 = 9.5 m/s
initial speed of the pelican = 9.5 m/s
(B) If the pelican was traveling at the same speed but was only 1.5 m above the water, how far would the fish travel horizontally before hitting the water below?
vertical height = 1.5 m
pelican's speed = 9.5 m/s
lets also calculate the time it will take the fish to falls = ut + [tex](\frac{1}{2}) at^{2}[/tex]
since u = 0
s = [tex](\frac{1}{2}) at^{2}[/tex]
t = [tex]\sqrt{\frac{2s}{a} }[/tex]
where a = acceleration due to gravity and s = vertical height
t = [tex]\sqrt{\frac{2 x 1.5 }{9.8} }[/tex] = 0.55 s
distance traveled by the fish = speed x time = 9.5 x 0.55 = 5.225 m
The initial speed of the pelican is 9.6 m/s. If the pelican is flying at the same speed but only 1.5 m above the water, the fish will travel approximately 5.3m horizontally before it hits the water.
Explanation:We can begin by solving for the time the fish takes to hit the water. Using the equation of motion, h = 0.5gt², where h represents height and g represents the acceleration due to gravity, we find it takes approximately 0.97 seconds for the fish to hit the water. During this time, the fish has traveled 9.3m horizontally. Therefore, the initial speed of the pelican can be determined by v = d/t, which gives us a value of 9.6 m/s.
Next, if the initial speed remained the same, but the height was reduced to 1.5 meters, we can use the same process. The time it would take for the fish to hit the water would be approximately 0.55 seconds (obtained from the equation h = 0.5gt²), and using this time in the equation v = d/t, we find the fish would travel approximately 5.3m horizontally before it hits the water.
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When testing an PNP transistor with an ohmmeter, what are the high or low resistance values expected for a good transistor?
Answer:
0.45 V and 0.9 V.
Explanation:
To test a PNP transistor with an ohmmeter,
Plug the positive lead from the multimeter to the transistor EMITTER (E). Plug the negative meter into the transistor BASE (B). If you are using a PNP resistor you must watch OL that is over limit, the voltage decrease will indicate between 0.45V and 0.9V if you are measuring it.
Testing a PNP transistor with an ohmmeter should yield high resistance values when the base is negative relative to the emitter or collector, and low resistance values when the base is positive relative to the emitter or collector.
Explanation:When testing a PNP transistor with an ohmmeter, the high or low resistance values that are expected for a good transistor are as follows: When you measure between the base and the emitter or collector, you should see a high resistance value (in the range of megaohms) if the base is negative with respect to the emitter or collector. Conversely, you should see a low resistance value (in the range of a few hundred Ohms) when the base is positive with respect to the emitter or collector.
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How long would it take for 3.5 C of charge to pass through a cross-sectional area of a wire in which a current of 5 mA passes?
Answer:
The time taken for the charge to pass through the wire = 700 s or 11.67 minutes or 0.194 hours
Explanation:
Electric Charge: This is defined as the product of electric current to time in a an electric circuit. The S.I unit of charge is Coulombs (C).
Mathematically, it is represented as
Q = it.......................... Equation 1
Where Q =quantity of charge, I = current, t = time
making t the subject of formula in equation 1,
t = Q/I ....................... Equation 2
Given: Q = 3.5 C, I = 5 mA.
Conversion: We convert from mA to A
I.e 5 mA = (5 × 10⁻³) A = 0.005 A.
Substituting these values into equation 2
t = 3.5/0.005
t = 700 seconds
or
(700/60) minutes = 11.67 minutes
or
(700/3600) hours = 0.194 hours.
Therefore the time taken for the charge to pass through the wire = 700 s or 11.67 minutes or 0.194 hours
700 second will take for 3.5 C of charge to pass through a cross-sectional area of a wire in which a current of 5 mA passes.
Partical charge and current based problem:What information do we have?
Charges of partical = 3.5 C
Current = 5 mA = (5 × 10⁻³) A
Charge = (Curernt)(time)
Q = iT
T = Q / i
Time taken = 3.5 / (5 × 10⁻³)
Time taken = 700 seconds or 11.67 minutes
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A space station in the form of a large wheel, 283 m in diameter, rotates to provide an "artif icial gravity" of 9.5 m/s 2 for people located at the outer rim. What is the frequency of the rotational motion for the wheel to produce this effect? Answer in units of rev/min.
Answer:
The frequency of the rotational motion for the wheel to produce this effect is 2.473 rev/min.
Explanation:
Given that,
Acceleration = 9.5 m/s²
Diameter = 283 m
We need to calculate the frequency of the rotational motion for the wheel to produce this effect
Using formula of rotational frequency
[tex]a= r\omega^2[/tex]
[tex]\omega=\sqrt{\dfrac{a}{r}}[/tex]
Where, r = radius
a = acceleration
[tex]\omega[/tex] = rotational frequency
Put the value into the formula
[tex]\omega=\sqrt{\dfrac{9.5\times2}{283}}[/tex]
[tex]\omega=0.259\ rad/s[/tex]
The frequency in rev/min
[tex]\omega=0.259\times\dfrac{60}{2\pi}[/tex]
[tex]\omega=2.473\ rev/min[/tex]
Hence, The frequency of the rotational motion for the wheel to produce this effect is 2.473 rev/min.
A mass resting on a shelf 10.0 meters above the floor has a gravitational potential energy of 980. joules with respect to the floor. The mass is moved to a shelf 8.00 meters above the floor. What is the new gravitational potential energy of the mass?
a. 960. J
b. 784 J
c. 490. J
d. 196 J
Answer:
New potential energy will be 784 J
So option (b) will be correct answer
Explanation:
We have given height of the shelf h = 10 m
Potential energy E = 980 J
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
We know that potential energy [tex]E=mgh[/tex]
So [tex]980=m\times 9.8\times 10[/tex]
m = 10 kg
Now new height h = 8 m
So new potential energy [tex]E=mgh=10\times 9.8\times 8=784J[/tex]
So option (B) will be the correct answer
Answer:
option (b) is correct.
Explanation:
h1 = 10 m
U 1 = 980 J
Let mg be weight of mass.
U1 = m x g x h1
980 = mg x 10
mg = 98
Now h = 8 m
U2 = mg x h = 98 x 8
U2 = 784 J
thus, the potential energy is 784 J.
Which statement is true regarding radioactive particles subjected to an electric field?
Answer:There are three types of radiation
Alpha, Beta and Gamma radiation
Explanation: In an electric field produced by two parallel charged plates alpha particle would be deflected toward a - plate following a parabolic path, beta rays toward a +plate following a parabolic path and gamma radiation either - or + source.
The force that attracts earth to an object is equal to and opposite the force that earth exerts on the object. Explain why earth's acceleration is not equal to and opposite the object's acceleration.
Answer:
Because of heavy mass
Explanation:
When force acts on a body it tends to accelerate the body. The acceleration produced in the body depends on two things:
1). Magnitude of force
2). Mass of the body
F= ma
⇒ a = F/m
As the force exerted on earth and another object are the equal in magnitude but opposite in direction. This forces will accelerate the object toward the earth but can't accelerate the earth as earth has very high mass.
a = F/m
This force tends to accelerate the earth but but due to earth's inertia the earth does not accelerate.
the decimal reduction time (DRT) is the time it takes to kill 90% of cells present. Assume that a DRT value for autoclaving a culture is 1.5 minutes. How long would it take to kill all the cells if 10^6 cells were present? What would happen if you stopped the heating process at 9 minutes?
Answer:
It takes 10.5 minutes to kill all the bacteria.
Only 1 cell would remain after 9 minutes.
Explanation:
It will take 1.5 minutes to kill 90% of the cells. So, after 1.5 minutes, only 10% would remain. After 3 minutes, only 1% remain. So, to figure out how long it would take to kill a million cells, we have to multiply 1 million by 0.1 repeatedly until the final value is less than 1 that is because when the value is less than 1, it means there are no more bacteria.
So:
[tex]10^6 \times (0.1)^7[/tex] = 0.1
So, you need 10.5 minutes of killing to kill one million cells.
Time taken= 7 x 1.5 minutes = 10.5 minutes.
After 9 minutes you would have:
[tex]10^{6} \times (0.1)^{6}[/tex] = 1 cell left
The decimal reduction time (DRT) is the time it takes to kill 90% of cells present. If a DRT value for autoclaving a culture is 1.5 minutes, it means that it takes 1.5 minutes to kill 90% of the cells. To calculate how long it would take to kill all the cells if 10^6 cells were present, you can use the DRT value as a basis. If you stopped the heating process at 9 minutes, it means that you haven't reached the time required to kill 100% of the cells.
Explanation:The decimal reduction time (DRT) is the time it takes to kill 90% of cells present. If a DRT value for autoclaving a culture is 1.5 minutes, it means that it takes 1.5 minutes to kill 90% of the cells. To calculate how long it would take to kill all the cells if 10^6 cells were present, you can use the DRT value as a basis. Since the DRT value represents the time it takes to kill 90% of the cells, you can calculate the time to kill 100% of the cells by dividing the DRT value by 90 and then multiplying it by 100. In this case, it would be as follows:
T = (1.5 minutes / 90) * 100 = 1.67 minutes
If you stopped the heating process at 9 minutes, it means that you haven't reached the time required to kill 100% of the cells. As a result, some cells would still be alive.
The temperature inside a vehicle can rise ____ degrees higher than the outside temperature.
Answer:
Generally, 40 to 50 degrees
Explanation:
About the heat-up over time, whether the windows of a vehicle are locked or partially open makes very little difference. In both situations, in an internal temperature of a vehicle, even at an outside temperature of only 72 ° F, it may exceed approximately 40 ° F within one hour. This happens mainly due to the greenhouse effect that is the heat inside the car is trapped and not allowed to escape. Thus, raising the temperature of the vehicle.
Final answer:
The greenhouse effect causes the temperature inside a parked car to be much higher than outside, sometimes 20 to 30 degrees higher, due to sunlight being trapped as heat.
Explanation:
The temperature inside a vehicle can rise significantly higher than the outside temperature due to a phenomenon known as the greenhouse effect. When a car is parked in the sun with windows closed, the sunlight passes through the glass and is absorbed by interior surfaces such as dashboard and seats. These surfaces then emit infrared radiation, which cannot escape back through the glass, and consequently, the trapped heat raises the temperature of the air inside the car. This effect can cause interior temperatures to be much higher—often 20 to 30 degrees higher—than the external air temperature.
A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is __________ atm. A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is __________ atm. 1.5 7.5 0.67 3.3 15
Answer:
The final pressure will be 0.67 atm.
Explanation:
Using Boyle's law
[tex] {P_1}\times {V_1}={P_2}\times {V_2}[/tex]
Given ,
V₁ = 10 L
V₂ = 15 L
P₁ = 1.0 atm
P₂ = ?
Using above equation as:
[tex]{P_1}\times {V_1}={P_2}\times {V_2}[/tex]
[tex]{1.0}\times {10}={P_2}\times {15} atm[/tex]
[tex]{P_2}=\frac{{1.0}\times {10}}{15} atm[/tex]
[tex]{P_2}=0.67\ atm[/tex]
The final pressure will be 0.67 atm.
Answer:
A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is 0.67 atm.
Explanation:
Boyle's law states that in constant temperature the variation volume of gas is inversely proportional to the applied pressure.
The formula is,
[tex]\rm \bold{ P_1\times V_1= P_2\times V_2}[/tex]
Where,
[tex]\rm \bold P_1[/tex] is initial pressure = 1 atm
[tex]\rm \bold P_2[/tex] is final pressure = ? (Not Known)
[tex]\rm \bold V_1[/tex] is initial volume = 10 L
[tex]\rm \bold V_2[/tex] is final volume = 15 L
Now put the values in the formula,
[tex]\rm 1\times 10 = P_2\times 15\\\\\rm P_2 = \frac{10}{15\\} \\\\\rm P_2 = 0.67[/tex]
Therefore, the answer is 0.67 atm.
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A 0.47 kg mass is attached to a spring with a spring constant of 130 N/m so that the mass is allowed to move on a horizontal friction-less surface. The mass is released from rest when the spring is compressed 0.12 m. A) Find the force on the mass at the instant the spring is released. B) Find the acceleration of the mass at the instant the spring is released.
Answer:
F= 15.6 N, a= 33.2 m/s^2
Explanation:
mass= m = 0.47 kg
spring constant= k = 130N/m
spring compression = x = 0.12 m
a).
force on the mass= F = k*x
F = 130 * 0.12 N
F= 15.6 N
b).
Acceleration of mass= a=?
F= ma
a=F / m
a= 15.6/ 0.47 m/s^2
a= 33.2 m/s^2
Final answer:
The force on the mass when the spring is released is 15.6 N, and the acceleration of the mass at that instant is approximately 33.19 m/s^2, calculated using Hooke's Law and Newton's second law, respectively.
Explanation:
Understanding Spring Force and Acceleration
To find the force on the mass at the instant the spring is released (in part A), we use Hooke's Law, which states that the force exerted by a spring (F) is equal to the negative spring constant (k) times the displacement from equilibrium (x), so F = -kx. Here, k = 130 N/m and x = 0.12 m, so the force is F = 130 N/m * 0.12 m = 15.6 N. The negative sign indicates that the force is in the opposite direction of the displacement.
To find the acceleration of the mass at the instant the spring is released (in part B), we apply Newton's second law, which relates force (F), mass (m), and acceleration (a) as F = ma. Rearranging for acceleration, we get a = F/m. Substituting the values, we have a = 15.6 N / 0.47 kg = approximately 33.19 m/s2.