Answer:
(a) 2.154×10⁻¹¹ A.
(b) 98300000.
Explanation:
(a)
Using Ohm's law,
V = IR ........................ Equation 1
Where V = Potential difference between the walls, I = current, R = Resistance between the walls.
Make I the subject of the equation
I = V/R...................... Equation 2
Given: V = 84 mV, = 0.084 V, R = 3.9×10⁹ Ω.
Substitute into equation 2
I = 0.084/(3.9×10⁹)
I = 2.154×10⁻¹¹ A.
(b)
I = q/t
q = It .................... Equation 1
Where q = quantity of electric charge, t = time.
Given: I = 2.154×10⁻¹¹ A, t = 0.73 s.
q = 2.154×10⁻¹¹×0.73
q = 1.572×10⁻¹¹ C.
The charge on an electron e = 1.6×10⁻¹⁹ C
n = q/e
where n = number of ions.
n = 1.572×10⁻¹¹/1.6×10⁻¹⁹
n = 9.83×10⁷
n = 98300000.
Design at least two independent experiments to determine the wavelength of a microwave source. You have a microwave source, microwave detector, a barrier with different regions, and a goniometer to measure angles. (To avoid over-ranging the detector, please use the largest meter multiplier setting that allows you to see what you’re doing. Nevertheless, if you are seeing nothing at any angle, do go to a smaller multiplier.) One of your experiments should involve setting up a standing wave.
One of your experiments should involve Include the following in your report:
a. Design experiments to solve the problem and discuss how you will use the available equipment to make measurements.
b. Describe the mathematical procedures you will use.
c. List the assumptions are you making. Explain how each could affect the outcome.
d. What are the sources of experimental uncertainty? How could you minimize the uncertainties?
Answer:
The Double Slits Experiment and Michelson Interferometer
The questions will be answered for the double slits experiment.
(b) Mathematically, the double slits experiment equation can be given as: d sin 0 = m(Wavelength). d is the separation distance, 0 is the diffraction angle, m = 1,2,3,...
(c) assumptions
The width of the slits is lesser than the microwave's wavelength. This is to set up a standing wave between the microwave source and detector.
m is an positive integer. To obtain a constructive interference of the E-M wave(microwave)
(d) The uncertainties are:
(i)The zero error in the reading of the multiplier will disrupt the value of the wavelength by small percentage. This can be adjusted to obtain more accurate result.
(ii) The angle as obtained by the gionometer can not be measured to highest level of accuracy, as there are some approximations. High sensitive equipment should be used to obtain accurate result
Explanation:
Double Slits Experiment.
The double slits is a pair of 2cm wide slits with a 6cm separation, cut in a metal foil, with the double slits at the centre of turnable with the microwave source and microwave detector through the slits. The shunt on the microammeter is adjusted to
so that a large scale deflection is obtained. The interference pattern may then be explored by moving the receiver arm, whilst keeping the transmitter arm fixed. A graph of intensity (current) versus θ(Measured through the gionometer) should be plotted, and an estimate made of the microwave wavelength by measuring the angular separation of adjacent maxima in the interference pattern and the separation of the two slits.
Michelson Interferometer
Setting up the interferometer, the mirrors are metal sheets and the beamsplitter is a hardboard sheet. The metal sheets should be carefully positioned to be perpendicular to the microwave beam, and the beamsplitter positioned at the centre of the turntable and at 45° to the beam. Any slight change in the path length of either of the beams leaving the beamsplitter changes the interference pattern at the receiver. One of the metal sheets is mounted in a frame which slides along rails. If this sheet is slowly moved, maxima and minima of current will be recorded by the receiver and a plot of “intensity” against distance can be made. The distance between successive maxima (or minima) is one half of the microwave wavelength which can thus be measured.
A 1500 kg car traveling at 90 km/hr East collides with a 3000 kg truck traveling at 60 km/hr South. They stick together and move which direction?
Answer:
They both move towards South direction.
Explanation:
NB: The car traveling due east must be backing west and the car traveling due south must be backing north.
From their momentum in calculation below:
Momentum of the car traveling east is MU= 1500x90x1000/3600= 37,500kgm/s.
Momentum of car traveling due South =MU = 3000x60x1000/3600= 50,000. So, from here we see the the momentum of the car moving south has the high momemtum so they stick together and move towards the direction of south.
A tank with a volume of 0.150 m3 contains 27.0oC helium gas at a pressure of 100 atm. How many balloons can be blown up if each filled balloon is a sphere 30.0 cm in diameter at 27.0oC and absolute pressure of 1.20 atm? Assume all the helium is transferred to the balloons.
Answer:
884 balloons
Explanation:
Assume ideal gas, since temperature is constant, then the product of pressure and volume is constant.
So if pressures reduces from 100 to 1.2, the new volume would be
[tex]V_2 = \frac{P_1V_1}{P_2} = \frac{100*0.15}{1.2} = 12.5 m^2[/tex]
The spherical volume of each of the balloon of 30cm diameter (15 cm or 0.15 m in radius) is
[tex]V_b = \frac{4}{3}\pir^3 = \frac{4}{3}\pi 0.15^3 = 0.014 m^3[/tex]
The number of balloons that 12.5 m3 can fill in is
[tex]V_2/V_b = 12.5 / 0.014 = 884[/tex]
Final answer:
Based on the volume of the tank, temperature, and pressure, we can calculate the number of moles of helium. By dividing this number by the number of moles of helium in one balloon, we can determine how many balloons can be blown up.
Explanation:
To find out how many balloons can be blown up with the given amount of helium, we need to calculate the total volume of helium in the tank and divide it by the volume of each balloon.
Given:
The volume of the tank is 0.150 m³
The temperature of the helium in the tank is 27.0°C
The pressure of the helium in the tank is 100 atm
The volume of each balloon is 30.0 cm in diameter, which is equivalent to a radius of 15.0 cm or 0.15 m
The temperature of the filled balloon is 27.0°C
The absolute pressure of the filled balloon is 1.20 atm
First, we need to convert the volume of the tank to liters:
0.150 m³ * 1000 L/m³ = 150 L
Next, we need to calculate the number of moles of helium in the tank using the ideal gas law:
P * V = n * R * T
n = (P * V) / (R * T)
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Using the given values:
Pressure (P) = 100 atm
Volume (V) = 150 L
Ideal gas constant (R) = 0.0821 L·atm/K·mol
Temperature (T) = 27.0°C + 273.15 = 300.15 K
Calculating the number of moles:
n = (100 atm * 150 L) / (0.0821 L·atm/K·mol * 300.15 K) = 6.19 moles
Now we can calculate the number of balloons that can be blown up:
Number of balloons = (Number of moles of helium) / (Number of moles of helium in one balloon)
The number of moles of helium in one balloon can be found using the ideal gas law:
P * V = n * R * T
n = (P * V) / (R * T)
Using the given values:
Pressure (P) = 1.20 atm
Volume (V) = (4/3) * π * (0.15 m)³ = 0.141 m³
Ideal gas constant (R) = 0.0821 L·atm/K·mol
Temperature (T) = 27.0°C + 273.15 = 300.15 K
Calculating the number of moles:
n = (1.20 atm * 0.141 m3) / (0.0821 L·atm/K·mol * 300.15 K) = 0.006 moles
Finally, calculating the number of balloons:
Number of balloons = 6.19 moles / 0.006 moles = 1031
Therefore, 1031 balloons can be blown up with the given amount of helium.
On a ring road, 12 trams are spaced at regular intervals and travel at a constant speed. How many trams need to be added to the circuit so that, maintaining the same speed, the intervals between them will decrease by 1 5
3 trams must be added
Explanation:
In this problem, there are 12 trams along the ring road, spaced at regular intervals.
Calling L the length of the ring road, this means that the space between two consecutive trams is
[tex]d=\frac{L}{12}[/tex] (1)
In this problem, we want to add n trams such that the interval between the trams will decrease by 1/5; therefore the distance will become
[tex]d'=(1-\frac{1}{5})d=\frac{4}{5}d[/tex]
And the number of trams will become
[tex]12+n[/tex]
So eq.(1) will become
[tex]\frac{4}{5}d=\frac{L}{n+12}[/tex] (2)
And substituting eq.(1) into eq.(2), we find:
[tex]\frac{4}{5}(\frac{L}{12})=\frac{L}{n+12}\\\rightarrow n+12=15\\\rightarrow n = 3[/tex]
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Answer: 3
Explanation:
We need to do (1/(5-1))x12
A ball is thrown at an angle of to the ground. If the ball lands 90 m away, what was the initial speed of the ball?
Answer:
The initial speed of the ball is 30 m/s.
Explanation:
It can be assumed that the ball is thrown at an angle of 45 degrees to the ground. The ball lands 90 m away. We need to find the initial speed of the ball. We know that the horizontal distance covered by the projectile is called its range. It is given by :
[tex]R=\dfrac{u^2\ sin2\theta}{g}[/tex]
u is the initial speed of the ball.
[tex]v=\sqrt{\dfrac{Rg}{sin2\theta}}[/tex]
[tex]v=\sqrt{\dfrac{90\times 9.8}{sin2(45)}}[/tex]
v = 29.69 m/s
or
v = 30 m/s
So, the initial speed of the ball is 30 m/s. Hence, this is the required solution.
The exact initial speed requires the angle [tex]\( \theta \)[/tex]; assuming [tex]\( \theta = 45^\circ \)[/tex], the initial speed is 29.71 m/s.
To solve this problem, we need to use the kinematic equations for projectile motion. The horizontal distance (range) \( R \) that the ball travels can be found using the following equation:
[tex]\[ R = \frac{{v_0^2 \sin(2\theta)}}{g} \][/tex]
Given that the ball lands 90 m away and assuming that the angle of projection [tex]\( \theta \)[/tex] is known (but not provided in the question), we can rearrange the equation to solve for [tex]\( v_0 \)[/tex]:
[tex]\[ v_0^2 = \frac{R \cdot g}{\sin(2\theta)} \][/tex]
[tex]\[ v_0 = \sqrt{\frac{R \cdot g}{\sin(2\theta)}} \][/tex]
However, since the angle [tex]\( \theta \)[/tex] is not given, we cannot calculate the exact initial speed [tex]\( v_0 \)[/tex] without additional information. The problem as stated is incomplete because it requires the value of [tex]\( \theta \)[/tex] to find [tex]\( v_0 \)[/tex].
If we assume that the angle [tex]\( \theta \)[/tex] is such that [tex]\( \sin(2\theta) \)[/tex] is maximized (which occurs at [tex]\( \theta = 45^\circ \)[/tex], then [tex]\( \sin(2 \cdot 45^\circ) = \sin(90^\circ) = 1 \)[/tex], and the equation simplifies to:
[tex]\[ v_0 = \sqrt{R \cdot g} \][/tex]
[tex]\[ v_0 = \sqrt{90 \, \text{m} \cdot 9.81 \, \text{m/s}^2} \][/tex]
[tex]\[ v_0 = \sqrt{882.9 \, \text{m}^2/\text{s}^2} \][/tex]
[tex]\[ v_0 =29.71 \, \text{m/s} \][/tex]
Therefore, under the assumption that [tex]\( \theta = 45^\circ \)[/tex], the initial speed of the ball would be [tex]\( 29.71 \, \text{m/s} \)[/tex].
In conclusion, without the value of [tex]\( \theta \)[/tex], we cannot determine the exact initial speed of the ball. However, if we assume the most favorable condition for maximum range, where [tex]\( \theta = 45^\circ \)[/tex], the initial speed [tex]\( v_0 \)[/tex] would be [tex]\( 29.71 \, \text{m/s} \)[/tex].
At the instant that you fire a bullet horizontally from a rifle, you drop a bullet from the height of the gun barrel. If there is no air resistance, which bullet hits the level ground first? Explain.
Answer:
Both hit at the same time
Explanation:
If air resistant is ignored, then gravitational acceleration g is the only thing that affect the bullets vertical motion, no matter what their horizontal motion is. Since both bullets are starting from rest, vertically speaking their speeds are 0 initially, they are both subjected to the same acceleration g, then they travel at the same rate and would reach the ground at the same time.
Find the magnitude of the net electric force exerted on this charge. Express your answer in terms of some or all of the variables q, R, and appropriate constants.
Answer:
Th steps is as shown in the attachment
Explanation:
from the diagram, its indicates that twelve identical charges are distributed evenly on the circumference of the circle. assuming one of gthe charge is shifted to the centre of he circle alomng the x axis, as such the charge is unbalanced and there is need ot balanced all the identical charges for the net force to be equal to zero.
The mathematical interpretation is as shown in the attachment.
A rod of mass M = 146 g and length L = 47 cm can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 19 g, moving with speed V = 5 m/s, strikes the rod at angle θ = 29° a distance D = L/2 from the end and sticks to the rod after the collision.What is the angular speed ωf of the system immediately after the collision, in terms of system parameters and I?
Answer:
[tex]\omega = \frac{mv\frac{L}{2}\sin(29^\circ)}{\frac{1}{3}ML^2 + m(\frac{L}{2})^2} = 0.91~{\rm rad/s}[/tex]
Explanation:
The angular speed of the system can be found by conservation of angular momentum. Since the ball and the rod are stick together, the collision is completely inelastic, ergo kinetic energy is not conserved.
[tex]L_{initial} = L_{final}\\m\vec{v}\times \vec{r} = I\omega[/tex]
The moment of inertia of the combined objects is equal to the sum of moment of inertia of the separate objects.
[tex]I = \frac{1}{3}ML^2 + m(\frac{L}{2})^2[/tex]
The cross product in the left-hand side can be written as a sine of the angle.
Therefore;
[tex]mv\frac{L}{2}\sin(29^\circ) = (\frac{1}{3}ML^2 + m(\frac{L}{2})^2)\omega\\(19\times 10^{-3})(5)(\frac{47\times 10^{-2}}{2})\sin(29^\circ) = (\frac{1}{3}(146\times 10^{-3})(47\times 10^{-2})^2 + (19\times 10^{-3})(\frac{47\times 10^{-2}}{2})^2)\omega\\\omega = 0.91~{\rm rad/s}[/tex]In terms of system parameters:
[tex]\omega = \frac{mv\frac{L}{2}\sin(29^\circ)}{\frac{1}{3}ML^2 + m(\frac{L}{2})^2}[/tex]
Facilitated diffusion and simple diffusion are processes that decrease the potential energy stored across a membrane true or false
Answer:
It's true. Potential energy is actually the concentrations of different elements. Diffusion is the process of moving the elements or materials from the area of high concentration (high potential energy) to the low concentration. So in bot facilitated and simple types of diffusion the level of potential energy decreases across the membrane.
Explanation:
A water droplet of mass ‘m’ and net charge ‘-q’ remains stationary in the air due to Earth’s Electric field.
(a) What must be the direction of the Earth’s electric field?
(b) Find an expression for the Earth’s electric field in terms of the mass and charge of the droplet.
Answer:
(a) the electric field of the Earth will be directed towards the negatively charged water droplet.
(b) E = (9.8*m)/q
Explanation:
Part (a) the direction of the Earth’s electric field
Electric field is always directed towards negatively charged objects.
The water droplet has a negative charge ‘-q’, therefore the electric field of the Earth will be directed towards the negatively charged water droplet.
Part (b) an expression for the Earth’s electric field in terms of the mass and charge of the droplet
The magnitude of Electric field is given as;
E = F/q
where;
f is force and q is charge
Also from Newton's law, F = mg
where;
m is mass and g is acceleration due to gravity = 9.8 m/s²
E = F/q = mg/q
E = (9.8*m)/q, Electric field in terms of mass and charge of the droplet.
Two cars, one in front of the other, are travelling down the highway at 25 m/s. The car behind sounds its horn, which has a frequency of 640 Hz. What is the frequency heard by the driver of the lead car? (Vsound=340 m/s).
The answer choices are:
A) 463 Hz
B) 640 Hz
C)579 Hz
D) 425 Hz
E) 500 Hz
Answer:
[tex] f_s = 640 Hz[/tex]
Explanation:
For this case we know that the speed of the sound is given by:
[tex] V_s = 340 m/s[/tex]
And we have the following info provided:
[tex] v_c = 25 m/s [/tex] represent the car leading
[tex] v_s= 25 m/s[/tex] represent the car behind with the source
[tex] f_o = 640 Hz[/tex] is the frequency for the observer
And we can find the frequency of the source [tex] f_s[/tex] with the following formula:
[tex] f_s = \frac{v-v_o}{v-v_s} f_o [/tex]
And replacing we got:
[tex] f_s = \frac{340-25}{340-25} *640 Hz = 640 Hz[/tex]
So then the frequency for the source would be the same since the both objects are travelling at the same speed.
[tex] f_s = 640 Hz[/tex]
Explain why an egg cooks more slowly in boiling water in Denver than in New York City. (Hint: Consider the effect of temperature on reaction rate and the effect of pressure on boiling point.)
Answer:
In New York is at ocean level while Denver is at an altitude of 1 mile from ocean level. In this way, the breaking point of water is lower in Denver than in New York, that is, the water will boil at lower temperatures in Denver than in New York. If the breaking point of the water decreases, at this point it will put aside more effort to cook an egg. Now the time required to cook the egg is higher in Denver than in New York.
An egg cooks slower in Denver because water boils at a lower temperature due to the city's higher altitude and lower atmospheric pressure. This lower boiling point decreases the rate of the cooking chemical reactions.
Explanation:An egg cooks more slowly in boiling water in Denver than in New York City due to differences in atmospheric pressure and the effect this has on boiling point temperatures. Denver, known as the Mile-High City, is approximately one mile above sea level. This altitude results in a lower atmospheric pressure than New York City which is virtually at sea level. Because of this, water boils at a slightly lower temperature in Denver (approximately 202 degrees Fahrenheit) than it does in New York City (approximately 212 degrees Fahrenheit).
Since cooking is essentially a series of chemical reactions, and these reactions occur faster at higher temperatures, an egg will cook more slowly in Denver’s boiling water than in New York's. This is because the boiling water in Denver is at a lower temperature due to the effect of pressure on boiling point.
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A certain substance has a heat of vaporization of 37.51 kJ / mol. At what Kelvin temperature will the vapor pressure be 3.50 times higher than it was at 307 K?
Answer:
T2=336K
Explanation:
Clausius-Clapeyron equation is used to determine the vapour pressure at different temperatures:
where:
In(P2/P1) = ΔvapH/R(1/T1 - 1/T2)
p1 and p2 are the vapour pressures at temperatures
T1 and T2
ΔvapH = the enthalpy of vaporization of the liquid
R = the Universal Gas Constant
p1=p1, T1=307K
p2=3.50p1; T2=?
ΔvapH=37.51kJ/mol=37510J/mol
R=8.314J.K^-1moL^-1
In(3.50P1/P1)= (37510J/mol)/(8.314J.K^-1)*(1/307 - 1/T2)
P1 and P1 cancelled out:
In(3.50)=4511.667(T2 - 307/307T2)
1.253=14.696(T2 - 307/T2)
1.253=(14.696T2) - (14.696*307)/T2
1.253T2=14.696T2 - 4511.672
Therefore,
4511.672=14.696T2 - 1.253T2
4511.672=13.443T2
So therefore, T2=4511.672/13.443=335.61
Approximately, T2=336K
An experiment is performed in the lab, where the mass and the volume of an object are measured to determine its density. Two completely different valid methods are used. Each experimental method to measure the density is performed, two considerable sets of data are taken on each and the results are compared. The results of the density measurement by each method should be:______.
a. dependent on the mass and the volume of the object
b. completely different, because it was measured different ways.
c. correct on the most accurate method and wrong in the other
d. the same within the uncertainty of each measurement method
Answer:
d. the same within the uncertainty of each measurement method
Explanation:
The density of an object and in general any physical property, has the same value regardless of the method used to measure it, either directly or indirectly. Since two completely different valid methods are used, the results must be the same, taking into account the level of precision of each of the methods.
If you find two stars with the same Right Ascension, are they necessarily close together in the sky? Why or why not?
In space, spatial coordinates can be roughly divided into measures of Right ascension and declination. The declination is measured in degrees while the ascent is measured in hours, minutes, seconds. When you have objects in space such as those of the characteristics presented we will have to they are not necessarily close together in the sky because we can find two stars on the same right ascension but on different declination lines (Which means they can be very far apart from each other)
A uniformly charged ring of radius 10.0 cm has a total charge of 78.0 μC. Find the electric field on the axis of the ring at the following distances from the center of the ring. (Choose the x-axis to point along the axis of the ring.)(a) 1.00 cm(b) 5.00 cm(c) 30.0 cm(d) 100 cm
Answer:
6908316.619 N/C
25087609.3949 N/C
6652357.02259 N/C
690831.6619 N/C
Explanation:
x = Distance from the ring
R = Radius of ring = 10 cm
q = Charge = 78 μC
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
Electric field at a point x is from a ring given by
[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}[/tex]
For 1 cm
[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 78\times 10^{-6}\times 0.01}{(0.01^2+0.1^2)^{1.5}}\\\Rightarrow E=6908316.619\ N/C[/tex]
The electric field is 6908316.619 N/C
For 5 cm
[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 78\times 10^{-6}\times 0.05}{(0.05^2+0.1^2)^{1.5}}\\\Rightarrow E=25087609.3949\ N/C[/tex]
The electric field is 25087609.3949 N/C
For 30 cm
[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 78\times 10^{-6}\times 0.3}{(0.3^2+0.1^2)^{1.5}}\\\Rightarrow E=6652357.02259\ N/C[/tex]
The electric field is 6652357.02259 N/C
For 100 cm
[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 78\times 10^{-6}\times 1}{(1^2+0.1^2)^{1.5}}\\\Rightarrow E=690831.6619\ N/C[/tex]
The electric field is 690831.6619 N/C
A vertical, piston-cylinder device containing a gas is allowed to expand from 1 m3 to 3 m3. The heat added to the system during the constant pressure expansion was 200 kJ and the decrease in the energy of the system is 340 kJ. Calculate the gas pressure, in kPa.
Answer:
Explanation:
Given
Initial Volume [tex]v_1=1\ m^3[/tex]
final Volume [tex]v_2=3\ m^3[/tex]
Heat added at constant Pressure [tex]Q=200\ kJ[/tex]
Decrease in Energy of System [tex]\Delta U=-340\ kJ[/tex]
According to First law of thermodynamics
[tex]Q=\Delta U+W[/tex]
[tex]W=Q-\Delta U[/tex]
[tex]W=200-(-340)[/tex]
[tex]W=540\ kJ[/tex]
Work done in a constant Pressure Process is given by
[tex]W=P\Delta V[/tex]
where P is the constant Pressure
[tex]540=P\times (3-1)[/tex]
[tex]P=270\ kPa[/tex]
What will be the induced magnetic field strength 7.5 cm radially outward from the center of the plates?
Answer:
B = 9.867 * 10^-8 T
Explanation:
Given:
- Rate of accumulating charge I = 37.0 mC/s
- radial distance from center of the plate r = 7.5 cm
- magnetic constant u_o = 4*pi *10^-7 H/m
Find:
The induced magnetic field strength 7.5 cm radially outward from the center of the plates?
Solution:
- Apply Gaussian Law on the surface:
B.(dA) = u_o*I
- The surface integral is dA = (2*pi*r):
B.(2*pi*r) = u_o*I
B = u_o*I / (2*pi*r)
- Plug values in:
B = (4*pi *10^-7)*(37*10^-3) / (2*pi*0.075)
B = 9.867 * 10^-8 T
Answer:
answer is B on Plato if you have it
Explanation:
Substance A has a heat capacity that is much greater than that of substance B. If 10.0 g of substance A initially at 30.0 ∘C is brought into thermal contact with 10.0 g of B initially at 80.0 ∘C, what can you conclude about the final temperature of the two substances once the exchange of heat between the substances is complete?
When substances with different heat capacities are brought into thermal contact, heat transfers until they reach thermal equilibrium. In this case, the final temperature will be closer to the initial temperature of substance A.
When two substances with different heat capacities are brought into thermal contact, heat will transfer from the substance with a higher initial temperature to the substance with a lower initial temperature until they reach thermal equilibrium. In this case, since substance A has a much greater heat capacity than substance B, it can absorb and transfer a larger amount of heat. Therefore, the final temperature of the two substances will likely be closer to the initial temperature of substance A and lower than the initial temperature of substance B.
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A battery has emf E and internal resistance r = 1.50 Ω. A 14.0 Ω resistor is connected to the battery, and the resistor consumes electrical power at a rate of 92.0 J/s. What is the emf of the battery?
Answer:
E = 39.68 V
Explanation:
EMF ( Electromotive force) : This is defined as the magnitude of the potential difference of both the external circuit and the inside of the cell. The S.I unit is Volt.
The Expression for E.M.F is given as,
E = I(R+r) ................... Equation 1
Where E = EMF, I = current, R = External Resistance, r = internal resistance.
Also,
P = I²R
I = √(P/R) ..................... Equation 2
Where P = power, R = External resistance.
Given: P = 92 J/s, R = 14 Ω.
Substitute into equation 2
I = √(92/14)
I = √(6.57)
I = 2.56 A.
Also Given: r = 1.5 Ω.
Substitute into equation 1
E = 2.56(1.5+14)
E = 2.56(15.5)
E = 39.68 V.
A battery has emf E and internal resistance r = 2.00 Ω. A 11.5 Ω resistor is connected to the battery, and the resistor consumes electrical power at a rate of 96.0 J/s.
Part APart complete
What is the emf of the battery?
Express your answer with the appropriate units.
E =
39.0 V
Two 1.4 g spheres are charged equally and placed 1.6 cm apart. When released, they begin to accelerate at 110 m/s2 . Part A What is the magnitude of the charge on each sphere? Express your answer to two significant figures and include the appropriate units.
Answer:
[tex]q = 2.17\times 10^{-15}~{\rm C}[/tex]
Explanation:
The relation between the acceleration and the electrical force between the spheres can be found by Newton's Second Law.
[tex]\vec{F} = m\vec{a}\\\vec{F} = (1.4\times 10^{-3})(110) = 0.154~N[/tex]
This is equal to the Coulomb's Force.
[tex]F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{2q}{(1.6 \times 10^{-2})^2} = 0.154\\q = 2.17\times 10^{-15}~C[/tex]
a skydiver jumps out of a plane waering a suit and decleration due to air i s 0.45 m/s^2 for each 1 m/s, skydyvers initial value problems that models the skydiver velocity is v(t). temrinal speed assuming that accelration due to gravity is 9.8m/s
Answer:
[tex]v = 21.77\ m/s[/tex]
Explanation:
Given,
Air resistance, = 0.45 m/s² for each 1 m/s
Air resistance for velocity v = 0.45 v.
Terminal velocity = ?
acceleration due to gravity= g = 9.8 m/s²
now,
a = 0.45 v
we know,
[tex]a =\dfrac{dv}{dt}[/tex]
[tex]\dfrac{dv}{dt}=0.45 v[/tex]
[tex]\dfrac{dv}{v} = 0.45 dt[/tex]
integrating both side
[tex]\int \dfrac{dv}{v} = 0.45\int dt[/tex]
[tex] ln(v) = 0.45 t[/tex]
[tex]t = e^{0.45 t}[/tex]
When a body is moving at terminal velocity, Force due to gravity is equal to force due to air resistance.
m a = m g
0.45 v = 9.8
[tex]v = \dfrac{9.8}{0.45}[/tex]
[tex]v = 21.77\ m/s[/tex]
Hence, the terminal velocity of the skydiver is equal to 21.77 m/s.
Sam heaves a 16-lb shot straight up, giving it a constant upward acceleration from rest of 35.0 m/s2 for 64.0 cm. He releases it 2.20 m above the ground. Ignore air resistance. (a) What is the speed of the shot when Sam releases it? (b) How high above the ground does it go? (c) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 m above the ground?
Answer:
a. 6.69m/s
b. y=4.48m
c. t=1.43secs
Explanation:
Data given, acceleration,a=35m/s^2
distance covered,d=64cm=0.64m,
a. to determine the speed, we use the equation of motion
initial velocity,u=0m/s
if we substitute values we arrive at
[tex]v^{2}=u^{2}+2as\\v^{2}=0+2*35*0.64\\v^{2}=44.8m/s\\v=\sqrt{44.8}\\ v=6.69m/s\\[/tex]
b. After taking the shot,the acceleration value is due to gravity i.e a=9.81m/s^2
and the distance becomes (y-2.2) above the ground. When it reaches the maximum height, the final velocity becomes zero and the initial velocity becomes 6.69m/s.
Hence we can write the equation above again
[tex]v^{2}=u^{2}-2a(y-2.2)\\[/tex]
if we substitute values we have
[tex]v^{2}=u^{2}-2a(y-2.2)\\0=6.69^{2}-2*9.81(y-2.2)\\y-2.2=\frac{44.76}{19.62} \\y=2.28+2.2\\y=4.48m[/tex]
c. the time it takes to arrive at 1.83m is obtain by using the equation below
[tex]1.83-2.2=6.69t-\frac{1}{2} *9.81t^{2}\\4.9t^{2}-6.69t-0.37\\using \\t= \frac{-b±\sqrt{b^{2}-4ac} }{2a}\\ where \\a=4.9, b=-6.69, c=-0.37[/tex]
if we insert the values, we solve for t , hence t=1.43secs
(a) The speed of the shot when Sam releases it 6.75 m/s.
(b) The height risen by the shot above the ground is 4.52 m.
(c) The time taken for the shot to return to 1.8 m above the ground is 1.44 s.
The given parameters;
constant acceleration, a = 35 m/s²height above the ground, h₀ = 64 cm = 2.2 mheight traveled, Δh = 64 cm = 0.64 mThe speed of the shot when Sam releases it is calculated as;
[tex]v^2 = u^2 + 2as\\\\v^2 = 0 + 2a(\Delta h)\\\\v = \sqrt{2a(\Delta h)} \\\\v = \sqrt{2\times 35(0.65)} \\\\v = 6.75 \ m/s[/tex]
The height risen by the shot is calculated as follows;
[tex]v^2 = u^2 + 2gh\\\\at \ maximum \ height , v = 0\\\\0 = (6.75)^2 + 2(-9.8)h\\\\19.6h = 45.56 \\\\h = \frac{45.56}{19.6} \\\\h =2.32 \ m[/tex]
The total height above the ground = 2.20 m + 2.32 m = 4.52 m.
The time taken for the shot to return to 1.8 m above the ground is calculated as follows;
the time taken to reach the maximum height is calculated as;
[tex]h = vt - \frac{1}{2} gt^2\\\\2.32 = 6.75t - (0.5\times 9.8)t^2\\\\2.32 = 6.75t -4.9t^2\\\\4.9t^2 -6.75t + 2.32=0\\\\a = 4.9, \ b = -6.75, \ c = 2.32\\\\t = \frac{-b \ \ + /- \ \ \sqrt{b^2-4ac} }{2a} \\\\t = \frac{-(-6.75) \ +/- \ \ \sqrt{(-6.75)^2 - 4(4.9\times 2.32)} }{2(4.9)} \\\\t = 0.72 \ s\ \ or \ \ 0.66 \ s[/tex]
[tex]t \approx 0.7 \ s[/tex]
height traveled downwards from the maximum height reached = 4.52 m - 1.8 m = 2.72 m
[tex]h = vt + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2(2.72)}{9.8} } \\\\t = 0.74 \ s[/tex]
The total time spent in air;
[tex]t = 0.7 \ s \ + \ 0.74 \ s\\\\t = 1.44 \ s[/tex]
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The mass of the Sun is 2x1030 kg, and the mass of Mars is 6.4x1023 kg. The distance from the Sun to Mars is 2.3X1011 m. Calculate the magnitude of the gravitational force exerted by the Sun on Mars. N Calculate the magnitude of the gravitational force exerted by Mars on the Sun. N
The magnitude of the gravitational force exerted by the Sun on Mars is approximately 1.49x10^22 N, while the magnitude of the gravitational force exerted by Mars on the Sun is approximately 5.92x10^15 N.
To calculate the magnitude of the gravitational force exerted by the Sun on Mars, we can use Newton's law of gravitation. The formula is given by F = G * (m1 * m2) / r^2, where F is the magnitude of the force, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between them.
Plugging in the values for the Sun's mass (2x10^30 kg), Mars' mass (6.4x10^23 kg), and the distance between them (2.3x10^11 m), we get
F = (6.673x10^-11 N·m²/kg²) * ((2x10^30 kg) * (6.4x10^23 kg)) / (2.3x10^11 m)^2
Simplifying the equation and performing the calculations, the magnitude of the gravitational force exerted by the Sun on Mars is approximately 1.49x10^22 N.
Similarly, to calculate the magnitude of the gravitational force exerted by Mars on the Sun, we can use the same formula with the masses and distance reversed. Plugging in the values, we get
F = (6.673x10^-11 N·m²/kg²) * ((6.4x10^23 kg) * (2x10^30 kg)) / (2.3x10^11 m)^2
Simplifying and calculating the equation, the magnitude of the gravitational force exerted by Mars on the Sun is approximately 5.92x10^15 N.
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One of the primary visible emissions from a distant planet occurs at 425 nm. Calculate the energy of a mole of photons of this emission.]
Answer:
Explanation:
Given
Wavelength of incoming light [tex]\lambda =425\ nm[/tex]
We know
[tex]speed\ of\ wave=frequency\times wavelength[/tex]
[tex]frequency=\frac{speed}{wavelength}[/tex]
[tex]\mu =\frac{3\times 10^8}{425\times 10^{-9}}[/tex]
[tex]\mu =7.058\times 10^{14}\ Hz[/tex]
Energy associated with this frequency
[tex]E=h\mu [/tex]
where h=Planck's constant
[tex]E=6.626\times 10^{-34}\times 7.058\times 10^{14}[/tex]
[tex]E=46.76\times 10^{-20}\ Hz[/tex]
Energy of one mole of Photon[tex]=N_a\times E[/tex]
[tex]=6.022\times 10^{23}\times 46.76\times 10^{-20}[/tex]
[tex]=281.58\times 10^{3}[/tex]
[tex]=281.58\ kJ[/tex]
To calculate the energy of a mole of photons of the emission at 425 nm, use the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Convert the wavelength to meters, substitute the values into the equation, and calculate to find the energy of a single photon. Multiply this by Avogadro's number to find the energy of a mole of photons.
Explanation:To calculate the energy of a mole of photons of the emission at 425 nm, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength (in meters).
Converting the wavelength to meters, we have 425 nm = 425 x 10^-9 m.
Substituting the values into the equation, we get E = (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m). Calculating this gives us the energy of a single photon of this emission. To find the energy of a mole of photons, we can multiply this value by Avogadro's number (6.02 x 10^23 photons/mol).
Therefore, the energy of a mole of photons of this emission is (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m) x (6.02 x 10^23 photons/mol).
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In the absence of air resistance, a ball is thrown vertically upward with a certain initial KE. When air resistane is a factor affecting the ball, does it return to its original level with the same, less or more KE? Does your answer contradict the law of energy of conservation?
Answer:
Explanation:
A ball is thrown vertically upward with a certain Kinetic Energy in the absence of air resistance and while returning it experiences air resistance.
Air resistance causes the ball to lose its kinetic energy as it provides resistance which will convert some of its kinetic energy to heat energy.
So in a way total energy is conserved but not kinetic energy as some portion of it is lost in the form of heat.
Final answer:
A ball thrown upward with air resistance will return with less kinetic energy due to the negative work done by air resistance, which transforms some of its kinetic energy into heat. This does not violate the conservation of energy as the energy is still conserved but in different forms. Mechanical energy decreases but is compensated by the increase in thermal energy in the air.
Explanation:
When a ball is thrown vertically upward in the absence of air resistance, it will return to its original level with the same kinetic energy (KE) because energy is conserved in a system where no external forces do work. However, when air resistance is a factor, it will return with less kinetic energy. This is because air resistance does negative work on the ball, converting some of its Kinetic Energy into thermal energy as it dissipates heat into the air.
The conservation of energy principle is not contradicted by this scenario. The total energy (kinetic plus potential plus any energy converted into heat due to air resistance) is still conserved. However, the mechanical energy of the ball (the sum of its kinetic and potential energy) decreases due to the work done by air resistance. This reduction in mechanical energy is exactly balanced by the increase in thermal energy of the air.
If an object like a feather is thrown upward, it will experience significant air resistance, and it will definitely return with less kinetic energy than it had when it was thrown up. Again, this doesn't violate the conservation of energy; instead, the energy is simply transformed into different forms, primarily heat, due to interactions with the air.
A 40-kg rock is dropped from an elevation of 10 m. What is the velocity of the rock when it is 5-m from the ground?
Answer:
Explanation:
Given
mass of rock [tex]m=40\ kg[/tex]
Elevation of Rock [tex]h=10\ m[/tex]
Distance traveled by rock with time
[tex]h=ut+\frac{1}{2}at^2[/tex]
where, u=initial velocity
t=time
a=acceleration
here initial velocity is zero
when rock is 5 m from ground then it has traveled a distance of 5 m from top because total height is 10 m
[tex]5=0\times t+\frac{1}{2}(9.8)(t^2)[/tex]
[tex]t^2=\frac{10}{9.8}[/tex]
[tex]t=1.004\approx 1\ s[/tex]
velocity at this time
[tex]v=u+at[/tex]
[tex]v=0+9.8\times 1.004[/tex]
[tex]v=9.83\ m/s[/tex]
Final answer:
To calculate the velocity of a 40-kg rock when it is 5 meters from the ground, after being dropped from 10 meters, we use the kinematic equation, leading to a final velocity of approximately 9.9 m/s.
Explanation:
The question concerns calculating the velocity of a 40-kg rock when it is 5 meters from the ground, after being dropped from an elevation of 10 meters. This problem is fundamentally based on the principles of kinematic equations that describe the motion of objects under the influence of gravity. Assuming the acceleration due to gravity (g) is 9.8 m/s2, and ignoring air resistance, we can use the equation of motion v² = u² + 2as, where v is the final velocity, u is the initial velocity (0 m/s, since the rock is dropped), a is the acceleration due to gravity, and s is the distance covered.
To find out how fast the rock is moving when it is 5 meters from the ground, we first calculate the distance it has fallen, which is 10 meters - 5 meters = 5 meters. Plugging the values into the equation, we have v² = 0 + 2(9.8)(5) = 98. Therefore, v = √98 = approximately 9.9 m/s.
A particle moves so that its position (in meters) as a function of time (in seconds) is . Write expressions (in unit vector notation) for (a) its velocity and (b) its acceleration as functions of time.
Answer:
a.Velocity=[tex]\vec{v}=(6t)\hat{j}+8\hat{k}[/tex]
b.[tex]\vec{a}=6\hat{j}[/tex]
Explanation:
We are given that
A particle moves so that its position (in m) as function of time is
[tex]\vec{r}=2\hat{i}+(3t^2)\hat{j}+(8t)\hat{k}[/tex]
a.We have to find its velocity
We know that
Velocity,[tex]v=\frac{dr}{dt}[/tex]
Using the formula
Velocity,v=[tex]\frac{d(2i+3t^2j+8tk)}{dt}[/tex]
Velocity=[tex]\vec{v}=(6t)\hat{j}+8\hat{k}[/tex]
b.Acceleration[tex]=\vec{a}=\frac{d\vec{v}}{dt}[/tex]
[tex]\vec{a}=\frac{d((6t)\hat{j}+8\hat{k})}{dt}[/tex]
[tex]\vec{a}=6\hat{j}[/tex]
The velocity and acceleration of a particle can be calculated from its position function using differentiation in respect to time. The first derivative gives the velocity, while the second derivative gives the acceleration. Unit vector notation is used to display the results.
Explanation:The question involves the calculations of a particle's velocity and acceleration over time, represented in unit vector notation. This scenario falls within the realm of physics, specifically kinematics. Given a particle's motion described by a position function, we can calculate its velocity and acceleration as functions of time.
V(t), the velocity of the particle, is the first derivative of the position function. Hence, to find the velocity, we simply differentiate the position function with respect to time. Now, to calculate A(t), the acceleration of the particle, we take the first derivative of the velocity function, or equivalently, the second derivative of the position function. This gives us the rate of change of velocity, which represents acceleration.
For example, suppose the particle's position function r(t) is given by r(t) = 3t^2 + 2 squares. Differentiating the position function once gives us v(t) = 6t + 2, the velocity function. Differentiating v(t) gives us a(t) = 6, the acceleration function. Both velocity and acceleration are given in unit vector notation.
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(5 pts) A sewing machine needle moves up and down in simple harmonic motion with an amplitude of 0.0127 m and a frequency of 2.55 Hz. How far doesthe needlemove in one period?A.0.0127 mB.0.0254 mC.0.0508 m
Answer:
B. 0.0254m
Explanation:
A is the amplitude of the oscillation, i.e. the
maximum displacement of the object from
equilibrium, either in the positive or negative x-direction. Simple harmonic motion is repetitive.
The period T is the time it takes the object tocomplete one oscillation and return to the startingposition.
d = 2A = 2×0.0127
The specific heat capacity of lead is 0.13 J/g-°C. How much heat (in J) is required to raise the temperature of of 91 g of lead from 22°C to 37°C?
Answer:
Q = 177J
Explanation:
Specific heat capacity of lead=0.13J/gc
Q=MCΔT
ΔT=T2-T1,where T1=22degrees Celsius and T2=37degree Celsius.
ΔT=37 - 22 = 15
Q = Change in energy
M = mass of substance
C= Specific heat capacity
Q = (91g) * (0.13J/gc) * (15c)= 177.45J
Approximately, Q = 177J