(i) 625 V/m
(ii) 2876.25 x 10⁻¹² C
(iii) 0.000575.25 x 10⁻⁶ F
Explanation:(i) The electric field (E) between the plates of a parallel plate capacitor is related to the potential difference (V) between the plates and the distance (d) of separation between the plates as follows;
E = V / d ----------------(i)
From the question;
V = 5.0V
d = 0.80cm = 0.008m
Substitute these values into equation (i) as follows;
E = 5.0 / 0.008
Solve for E;
E = 625 V/m
Therefore, the electric field in the region between the two plates is 625 V/m.
(ii) To make things easier, let's calculate the capacitance of the parallel plates first.
The capacitance (C) of a parallel plate capacitor is given as;
C = A x ε₀ / d --------------------------(ii)
Where;
A = Area of either of the plates of the capacitor = 0.52m²
ε₀ = permittivity of free space = 8.85 x 10⁻¹² F/m
d = distance between the plates = 0.8cm = 0.008m
Substitute these values into equation (ii) as follows;
C = 0.52 x 8.85 x 10⁻¹² / 0.008
Solve for C;
C = 575.25 x 10⁻¹² F
The capacitance (C) is related to potential difference (V) and charge (Q) on the plates as follows;
Q = C x V -------------------------(iii)
Where;
C = 575.25 x 10⁻¹² F
V = 5.0V
Substitute these values into equation (iii)
Q = 575.25 x 10⁻¹² x 5
Q = 2876.25 x 10⁻¹² C
Therefore, the charge on the plates is 2876.25 x 10⁻¹² C
(iii) The capacitance (C) of the parallel plates has been calculated in (ii) above.
Its value is 575.25 x 10⁻¹² F = 0.000575.25 x 10⁻⁶F
The electric field in the region between two plates is 625 V/m. The capacitance of the plates is 0.0572 nanoFarads and the accumulated charge Q on each plate is 2.86 x 10^-10 Coulombs.
Explanation:The electric field between two plates can be calculated using the formula E = V/d, where V represents voltage and d represents the distance between plates. So, the electric field E would be 5.0 V / 0.008 m = 625 N/C or 625 V/m. The charge Q on the plates can be determined using the formula Q = CV, where C is the capacitance and V is the voltage. To find the charge, we first need to calculate the capacitance. The capacitance of the parallel plates can be calculated using the formula C = ε0*(A/d), where ε0 is the permittivity of free space (8.85 x 10^-12 F/m), A is the area and d is the distance between the plates. The capacitance then would be 8.85 x 10^-12 F/m * 0.52 m^2 / 0.008 m = 5.72 x 10^-11 F or 0.0572 nF (nanoFarads). So, after finding the capacitance, we can now calculate the charge Q, which is Q = CV = 5.72 x 10^-11 F * 5.0 V = 2.86 x 10^-10 C (Coulombs).
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A rock is thrown straight up into the air with an initial speed of 55 m/s at time t = 0. Ignore air resistance in this problem. At what times does it move with a speed of 36 m/s? Note: There are two answers to this problem.
Answer:
After 1.938 sec velocity of rock will be 36 m/sec
Explanation:
We have given initial velocity at which rock is thrown u = 55 m/sec
Final velocity v = 36 m/sec
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
According to first equation of motion we know that [tex]v=u+gt[/tex], here v is final velocity, u is initial velocity, g is acceleration due to gravity and t is time
So [tex]36=55-9.8t[/tex] ( Negative sign is due to rock is thrown upward )
So [tex]9.8t=19[/tex]
t = 1.938 sec
So after 1.938 sec velocity of rock will be 36 m/sec
The falling object in Example 2 satisfies the initial value problem dv/dt =9.8−(v/5), v(0) =0. (a) Find the time that must elapse for the object to reach 98% of its limiting velocity. (b) How far does the object fall in the time found in part (a)?
Answer:
a. [tex]t=19.56 s[/tex]
b.[tex]d=718.34[/tex]
Explanation:
The solution to the differential equation
[tex]\dfrac{dv}{dt}=9.8-\dfrac{v}{5}[/tex]
is the exponential function
[tex]v(t)=ce^{-0.2t}+49[/tex]
and we find [tex]c[/tex] from the initial condition [tex]v(0)=0:[/tex]
[tex]0=ce^{-0.2*0}+49\\\\0=c+49\\\\c=-49[/tex]
Therefore, we have
[tex]v(t)=-49e^{-0.2t}+49[/tex]
[tex]\boxed{ v(t)=49(1-e^{-0.2t})}[/tex]
Part A:
The maximum velocity that the object can reach is 49 (which the maximum value [tex]v(t)[/tex] can have).
Now, 98% of 49 is 48.02; therefore,
[tex]48.02=49(1-e^{-0.2t})[/tex]
[tex]0.98=1-e^{-0.2t}[/tex]
[tex]e^{-0.2t}=0.02[/tex]
[tex]\boxed{t=19.56 s}[/tex]
Part B:
The distance traveled is the integral of the speed:
[tex]d=\int_0^{19.56}v(t)*dt[/tex]
[tex]d=\int^{19.56}_0 {49(1-e^{-0.2t})} \, dt[/tex]
[tex]d=49[t+5e^{-0.2t}]_0^{19.56}[/tex]
[tex]\boxed{d=718.34}[/tex]
To find the time that must elapse for the object to reach 98% of its limiting velocity, we need to solve the differential equation. We can then find the distance the object falls by integrating the velocity function with respect to time.
Explanation:(a) Finding the time to reach 98% of the limiting velocityTo find the time it takes for the object to reach 98% of its limiting velocity, we need to solve the differential equation. First, we separate the variables by writing it as:
dv / (9.8 - (v/5)) = dt
Next, we integrate both sides:
∫ (1 / (9.8 - (v/5))) dv = ∫ dt
After evaluating the integrals, we can solve for v:
v = 49 - 49e^(-t/5)
Substituting v with 0.98 times the limiting velocity (which is 49), we can solve for t:
49 - 49e^(-t/5) = 0.98 * 49
Solving this equation will give us the time it takes for the object to reach 98% of its limiting velocity.
(b) Finding the distance the object fallsTo find the distance the object falls, we need to integrate the velocity function, v, with respect to time:
∫ v dt
By evaluating the integral, we can calculate the distance the object falls in the time found in part (a).
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In an experiment to measure the acceleration due to gravity g, two independent equally reliable measurements gave 9.67 m/s2 and 9.88 m/s2. Determine (i) the percent difference of the measurements (ii) the percent error of their mean. [Take the theoretical value of g to be 9.81 m/s
Answer:
i. +/- 1.43% and +/- 0.71% ii. +/- 0.33%
Explanation:
[tex]% Error = \frac{Error}{Measurement}* 100%\\[/tex]
If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision?
Using the physics equation of motion and the given initial velocity, reaction time, and deceleration, one can determine whether a truck can stop in time to avoid a collision.
Explanation:The question focuses on stopping distance and acceleration required to avoid a collision, indicating its base in Physics. If we have a truck moving at a constant velocity and it brakes at a certain distance from an obstacle, the minimum acceleration needed to avoid a collision can be calculated using the equation of motion v^2 = u^2 + 2as. Here, 'v' is the final velocity (0 m/s as the truck needs to stop), 'u' is the initial velocity, 'a' is the acceleration, and 's' is the distance over which the truck needs to stop.
To determine if the truck will hit the child, we must account for the driver's reaction time as well. During this reaction time, the truck continues to travel at its initial speed. After the reaction time, the truck will begin decelerating until it comes to a stop. The total stopping distance is the distance covered during the reaction time plus the distance covered during deceleration. The latter can be found using the deceleration rate and the formula mentioned above.
For the given scenario of the truck with an initial velocity of 10 m/s, a braking distance of 50 m, reaction time of 0.5 seconds, and deceleration of -1.25 m/s^2, we can calculate whether or not the truck will be able to stop in time to avoid hitting the child.
If Earth were completely blanketed with clouds and we couldn’t see the sky, could we learn about the realm beyond the clouds? What forms of radiation might penetrate the clouds and reach the ground?
The definition of waves that propagate through electric fields is called electromagnetic waves. The earth, despite being covered with clouds, can be 'affected' because waves such as sunlight or the moon have the ability to penetrate and be visible to the inhabitants of the earth. Microwaves and radio waves would be less affected by the clouds that cover the Earth.
Through these waves, you can know that there is beyond the clouds.
Ultraviolet light, microwaves and radio waves are the radiations that penetrate through the clouds and reach the Earth's surface.
Therefore, the answer is Yes, ultraviolet light, microwaves and radio waves are the forms of radiation that penetrate and reach the ground.
It is indeed possible to learn about the universe beyond the clouds due to other non-visual forms of radiation, mainly radio waves and gamma rays, which can penetrate through the clouds and reach the earth's surface.
Explanation:Yes, even if Earth were completely blanketed with clouds and we could not see the sky, we could still learn about the universe beyond the clouds. This is because, in addition to visible light which would be blocked by the clouds, the universe also emits various other forms of radiation that can penetrate the clouds and reach the ground.
Two major types of radiation that could penetrate the dense clouds are radio waves and gamma rays. Radio waves are a form of electromagnetic radiation used in many areas of science and technology, while gamma rays are highly energetic forms of radiation and are used in fields such as astronomy to get valuable information about distant celestial bodies.
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Suppose electrons in a TV tube are accelerated through a potential difference of 2.00 104 V from the heated cathode (negative electrode), where they are produced, toward the screen, which also serves as the anode (positive electrode), 25.0 cm away.At what speed would the electrons impact the phosphors on the screen? Assume they accelerate from rest, and ignore relativistic effects?
Answer:
83816746.4254 m/s
Explanation:
m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
V = Voltage = [tex]2\times 10^4\ V[/tex]
The kinetic energy of the electron is
[tex]K=\dfrac{1}{2}mv^2[/tex]
Energy is given by
[tex]E=qV[/tex]
Balancing the energy
[tex]qV=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2qV}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 2\times 10^4}{9.11\times 10^{-31}}}\\\Rightarrow v=83816746.4254\ m/s[/tex]
The velocity of the electrons is 83816746.4254 m/s
Henrietta is jogging on the side-walk at 3.05 m/s on the way to her physics class. Bruce realizes that she forgot her bag of bagels, so he runs to the window, which is 38.0 m above the street level and directly above the sidewalk, to throw the bag to her. He throws it horizontally 9.00 s after she has passed below the window, and she catches it on the run. Ignore air resistance. (a) With what initial speed must Bruce throw the bagels so that Henrietta can catch the bag just before it hits the ground? (b) Where is Henrietta when she catches the bagels?
Answer:
12.9121148614 m/s
35.9393048982 m
Explanation:
t = Time taken
u = Initial velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = a
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 38=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{38\times 2}{9.81}}\\\Rightarrow t=2.78337865516\ s[/tex]
Time taken for the bag to fall is 2.78337865516 seconds
Time she has been jogging for
9+2.78337865516 = 11.78337865516 seconds
Total distance traveled by her
[tex]s=vt\\\Rightarrow s=3.05\times 11.78337865516=35.9393048982\ m[/tex]
Henrietta is 35.9393048982 m away
Velocity of throwing
[tex]\dfrac{35.9393048982}{2.78337865516}=12.9121148614\ m/s[/tex]
The velocity of throwing is 12.9121148614 m/s
Final answer:
Bruce must throw the bagels at an initial speed of 12.92 m/s for Henrietta to catch them, and Henrietta will be 35.93 m from the point directly below Bruce's window when she catches the bagels.
Explanation:
Projectile Motion and Kinematics Problem
To find the initial speed Bruce must throw the bagels, we need to consider two aspects of projectile motion: the horizontal motion, which is constant because air resistance is neglected, and the vertical motion, which is influenced by gravity.
Firstly, we need to calculate the time it takes for the bagels to fall from the window to the ground. Using the equation for free fall h =
1/2 g t², where h is the height (38.0 m), and g is the acceleration due to gravity (9.81 m/s²), we can solve for t, the time to fall:
38.0 m = 1/2 * 9.81 m/s² * t²
t = sqrt(2 * 38.0 m / 9.81 m/s²) = sqrt(7.74) ≈ 2.78 s
Bruce throws the bagels 9.00 s after Henrietta has passed below the window. In this time, Henrietta has jogged a distance of d = speed * time = 3.05 m/s * 9.00 s = 27.45 m horizontally.
Since Henrietta is already past the point directly below the window, we need to add the distance she will cover in the time it takes for the bagels to fall. This distance is additional distance = jogging speed * fall time = 3.05 m/s * 2.78 s ≈ 8.48 m.
Overall, Henrietta will be approximately 27.45 m + 8.48 m = 35.93 m from the point directly below the window when she catches the bagels.
To find the initial speed with which Bruce throws the bagels, we use the horizontal motion formula initial speed = distance / time, which gives us an initial speed of approximately 35.93 m / 2.78 s ≈ 12.92 m/s.
Bruce must throw the bagels horizontally at an initial speed of approximately 12.92 m/s for Henrietta to catch them just before they hit the ground, at a distance of approximately 35.93 m from the point directly below Bruce's window.
Arctic sea ice has declined over the past few decades causing water levels to increase. This is an interaction of which two spheres?
Biosphere and geosphere
Cryosphere and hydrosphere
Geosphere and atmosphere H
ydrosphere and geosphere
Answer:
Option (2)
Explanation:
The Cryosphere refers to the frozen water bodies on earth. This includes the glaciers, icebergs, ice sheets and the frozen water surrounding the Arctic as well as Antarctica.
The Hydrosphere refers to all the water bodies on earth including the rivers, streams, lakes, and ponds.
The given condition is based on the interaction between the cryosphere and the hydrosphere.
The frozen ice in the Antarctic and Arctic is melting rapidly due to the increase in the global warming effect. This declining ice in the polar region results in the rise in the global sea level. This can be catastrophic as many of the big cities will be flooded because of this increasing height of sea level.
Thus, the correct answer is option (2).
The decline of Arctic sea ice and its impact on water levels is an interaction between two Earth system spheres: the cryosphere and hydrosphere.
The interaction of Arctic sea ice decline and increasing water levels involves the cryosphere and hydrosphere spheres. The cryosphere refers to the frozen components of the Earth system, including ice caps, glaciers, and sea ice. The hydrosphere encompasses all the water on Earth, including oceans, lakes, and rivers.
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A sly 1.5-kg monkey and a jungle veterinarian with a blow-gun loaded with a tranquilizer dart are 25 m above the ground in trees 70 m apart. Just as the veterinarian shoots horizontally at the monkey, the monkey drops from the tree in a vain attempt to escape being hit. What must the minimum muzzle velocity of the dart be for the dart to hit the monkey before the monkey reaches the ground?
Answer:
31 m/s
Explanation:
As both the monkey and the darts are subjected to constant gravitational acceleration g = 9.8 m/s2 and both start from rest (vertically speaking). Their vertical position will always be the same. For the dart to hit the monkey, its horizontal position must be the same as the monkey's, which is unchanged before reaching the ground. Therefore, the time it takes for the dart to travel across 70 m must be less than the time it takes for the monkey to drop 25m to the ground. We can find it out using the following equation of motion
[tex]s_m = gt_m^2/2[/tex]
[tex]25 = 9.8t_m^2/2[/tex]
[tex]t_m^2 = 50/9.8 = 5.1[/tex]
[tex]t_m = \sqrt{5.1} = 2.26 s[/tex]
For the dart to takes less that 2.26 s to travel 70m, its horizontal speed must at least be 70 / 2.26 = 31 m/s
You are observing a spacecraft moving in a circular orbit of radius 100,000 km around a distant planet. You happen to be located in the plane of the spacecraft’s orbit. You find that the spacecraft’s radio signal varies periodically in wavelength between 2.99964 m and 3.00036 m. Assuming that the radio is broadcasting at a constant wavelength, what is the mass of the planet?
To solve this problem we will apply the concepts related to centripetal acceleration, which will be the same - by balance - to the force of gravity on the body. To find this acceleration we must first find the orbital velocity through the Doppler formulas for the given periodic signals. In this way:
[tex]v_{o} = c (\frac{\lambda_{max}-\bar{\lambda}}{\bar{\lambda}}})[/tex]
Here,
[tex]v_{o} =[/tex] Orbital Velocity
[tex]\lambda_{max} =[/tex] Maximal Wavelength
[tex]\bar{\lambda}} =[/tex] Average Wavelength
c = Speed of light
Replacing with our values we have that,
[tex]v_{o} = (3*10^5) (\frac{3.00036-3}{3})[/tex]
Note that the average signal is 3.000000m
[tex]v_o = 36 km/s[/tex]
Now using the definition about centripetal acceleration we have,
[tex]a_c = \frac{v^2}{r}[/tex]
Here,
v = Orbit Velocity
r = Radius of Orbit
Replacing with our values,
[tex]a = \frac{(36km/s)^2}{100000km}[/tex]
[tex]a= 0.01296km/s^2[/tex]
[tex]a = 12.96m/s^2[/tex]
Applying Newton's equation for acceleration due to gravity,
[tex]a =\frac{GM}{r^2}[/tex]
Here,
G = Universal gravitational constant
M = Mass of the planet
r = Orbit
The acceleration due to gravity is the same as the previous centripetal acceleration by equilibrium, then rearranging to find the mass we have,
[tex]M = \frac{ar^2}{G}[/tex]
[tex]M = \frac{(12.96)(100000000)^2}{ 6.67*10^{-11}}[/tex]
[tex]M = 1.943028*10^{27}kg[/tex]
Therefore the mass of the planet is [tex]1.943028*10^{27}kg[/tex]
A straight wire 0.280 m in length carries a current of 3.40 A. What are the two angles between the direction of the current and the direction of a uniform 0.0400 T magnetic field for which the magnetic force on the wire has magnitude 0.0250 N?
Answer:
The angle θ between the direction of the current and the direction of the uniform magnetic field is 41° or 139°.
Explanation:
The force on a current carrying wire is given by the following equation:
[tex]\vec{F} = I\vec{L}\times \vec{B}[/tex]
The cross-product can be written with a sine term:
[tex]F = ILB\sin(\theta)\\0.025 = (3.4)(0.28)(0.04)\sin(\theta)\\\sin(\theta) = 0.6565[/tex]
Therefore, the angle θ is 41.03° and 138.97°
The angles can be calculated using the formula sin(θ) = F / (I L B), giving two symmetrical values about 90° in the first and second quadrants because the sine function is periodic.
The question is asking for the angles at which the force on a current-carrying wire in a magnetic field is a specific magnitude. The magnitude of the force exerted on a current-carrying wire placed in a magnetic field is given by the formula F = I L B sin(θ), where F is the force, I is the current, L is the length of the wire, B is the magnetic field strength, and θ is the angle between the direction of the current and the direction of the magnetic field.
By rearranging for θ, we get the equation sin(θ) = F / (I L B). Plugging in the values from the question, we find sin(θ) = 0.0250 N / (3.40 A imes 0.280 m imes 0.0400 T). This gives us θ values that correspond to the sine of this ratio.
There are two angles that will produce the same sine value because sine is a periodic function, which are θ and 180°-θ. Therefore, the two angles between the direction of the current and the direction of the uniform magnetic field for which the force on the wire has a magnitude of 0.0250 N will be symmetrical about 90° in the first and second quadrants.
What is a constellation as astronomers define it today? What does it mean when an astronomer says, "I saw a comet in Capricorn last night?"
A constellation, in astronomy, is a conventional grouping of stars, whose position in the night sky is apparently invariable. The peoples, generally of ancient civilizations, decided to link them through imaginary strokes, thus creating virtual silhouettes on the celestial sphere. From 1928, the International Astronomical Union (UAI) decided to officially regroup the celestial sphere into 88 constellations with precise limits, such that every point in the sky would be within the limits of a figure. When an astronomer says he saw a comet in Capricorn last night, it means that he saw a comet in the direction of the constellation of Capricorn.
How many kWh of energy does a 550-W toaster use in the morning if it is in operation for a total of 5.0 min? At a cost of 9.0 cents/k Wh, estimate how much this would add to your monthly electric energy bill if you made toast four mornings per week.
Answer:
0.0458 kWh
6.5736 cents
Explanation:
The formula for electric energy is given as
E = Pt................. Equation 1
Where E = Electric energy, P = Electric power, t = time.
Given; P = 550 W, t = 5 min = (5/60) h = 0.083 h.
Substituting into equation 1
E = 550(0.083)
E = 45.83 Wh
E = (45.83/1000) kWh
E = 0.0458 kWh.
Hence the kWh = 0.0458 kWh.
If the makes a toast four morning per week, and the are Four weeks in a month.
Total number days he makes toast in a month = 4×4 = 16 days.
t = 16×0.083 h = 1.328 h.
Total energy used in a month = 550(1.328)
E = 730.4 Wh
E = 0.7304 kWh.
If the cost of energy is 9.0 cents per kWh,
Then for 0.7304 kWh the cost will be 9.0(0.7304) = 6.5736 cents.
Hence this would add 6.5736 cents to his monthly electric bill
A 550-W toaster in operation for 5.0 minutes uses 2.75 kWh of energy. If you make toast four mornings per week, it would add an estimated cost of $4.30 to your monthly electric energy bill.
Explanation:To calculate the energy used by the toaster, we can use the formula E = Pt, where P is the power and t is the time. In this case, the power of the toaster is 550 watts and the time it is in operation is 5.0 minutes. Plugging these values into the formula, we get E = (550 W)(5.0 min) = 2750 W.min. To convert this to kilowatt-hours (kWh), we need to divide by 1000, so the energy used by the toaster is 2.75 kWh.
To estimate how much this would add to your monthly electric energy bill, we need to know how many times you use the toaster in a month. If you use it four mornings per week, that would be 4 days x 52 weeks / 12 months = 17.33 days per month. Multiplying the energy used by the toaster (2.75 kWh) by the number of days in a month (17.33), we get an estimate of 47.75 kWh per month. Finally, to find the cost, we multiply the energy (47.75 kWh) by the cost per kilowatt-hour (9.0 cents/kWh) and convert it to dollars, giving us an estimated cost of $4.30 per month.
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An orange loses 1.2 kJ of heat as it cools per °C drop in its temperature. What is the amount of heat loss from the orange per °F drop in its temperature?
To solve this problem we will apply the conversion rate between Celcius and Fahrenheit degrees. We will use the direct relationship clearly and not the added degrees of scale conversion. We know from the statement that the orange loses heat at the rate of
[tex]Q = 1.2kJ/\°C[/tex]
We have the conversion to °F is given as
[tex]T (\°F) = 1.8T+32[/tex]
Calculate the amount of heat loss from orange per °F
[tex]Q = \frac{1.2}{1.8}[/tex]
[tex]Q = 0.667kJ/\°F[/tex]
Therefore the amount of heat loss from the orange per °F drop in its temperature is 0.667kJ/°F
The heat loss from an orange per °F drop is 0.67 kJ, calculated by taking 1.2 kJ per °C drop and dividing it by 1.8 to convert it to Fahrenheit,
Explanation:The heat loss from the orange per °F drop in its temperature can be found by converting 1.2 kJ lost per 1 °C drop in temperature to kJ lost per 1 °F drop. This can be achieved using the formula that 1 °C equals 1.8 °F.
Therefore, the heat loss per degree Fahrenheit will be less than the heat loss per degree Celsius. We calculate this as follows:
(1.2 kJ / °C) / 1.8 = 0.67 kJ per °F.
So for every degree Fahrenheit that the orange cools, it will lose 0.67 kilojoules of heat.
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By standard convention, both the electric potential and the the electric potential energy between two charges is taken to be zero in what configuration?
Answer: at when distance r = infinity.
Explanation: The formulae for the electric potential of an electric charge to an arbitrary point is given by the formulae below
V = q/4πεr
V = electric potential (volts)
q = magnitude of electric charge
ε = permittivity of free space
r = distance between arbitrary point and charge.
In the equation above, it can be seen that only electric potential (v) and distance (r) is a variable, and there is an inverse relationship between them (an increase in one leads to a decrease in the other)
Thus to have zero value of electric potential (v= 0) we have to have the largest value of r ( r = infinity).
Same goes for electric potential energy between two charges, the formulae is given below as
W = q1 *q2/4πεr
W= electric potential energy
q1 = magnitude of first charge.
q2 = magnitude of second charge
ε = permittivity of free space
r = distance between arbitrary point and charge.
Also, all values are constant aside from electric potential energy (w) and distance (r) which have an inverse relationship.
Thus to have zero value of electric potential energy (w =0), we have to get an infinite value of distance ( r =infinity)
A ball is tossed with a velocity of 10 m/s directed vertically upward from a window located 20 m above the ground. Determine the following: (a) The velocity v and elevation y of the ball above the ground at any time t. (b) The highest elevation reached by the ball and its corresponding time t. (c) The time when the ball will hit the ground and the impact velocity.
Answer:
Explanation:
Given
Initial velocity of ball [tex]u=10\ m/s[/tex]
height of window [tex]h=20\ m[/tex]
Using Equation of motion
[tex]y=ut+\frac{1}{2}at^2[/tex]
where u=initial velocity
t=time
a=acceleration
As ball is already is at a height of 20 m so
[tex]Y=ut+\frac{1}{2}at^2+20[/tex]
[tex]Y=10\times t+0.5\times (-9.8)t^2+20[/tex]
[tex]Y=-4.9t^2+10t+20[/tex]
(b)highest point is obtained at v=0
[tex]v^2-u^2=2as[/tex]
where
v=final velocity
u=initial velocity
a=acceleration
s=displacement
[tex](0)-10^2=2\times (-9.8)\times s[/tex]
[tex]s=\frac{100}{19.6}[/tex]
[tex]s=5.102\ m[/tex]
Highest Point will be [tex]s+20=25.102\ m[/tex]
(c)Time taken when the ball hit the ground i.e. at Y=0
[tex]-4.9t^2+10t+20=0[/tex]
[tex]t=3.28\ s[/tex]
impact velocity [tex]v=\sqrt{2\times 9.8\times 25.102}[/tex]
[tex]v=22.181\ m/s[/tex]
(a) The equation be "Y = -4.9t² + 10t + 20".
(b) The highest point be "25.102 m".
(c) The impact velocity be "22.181 m/s"
Equation of motionAccording to the question,
Ball's initial velocity, u = 10 m/s
Window's height, h = 20 m
(a) By using equation of motion,
Y = ut + [tex]\frac{1}{2}[/tex]at²
By substituting the values,
= ut + [tex]\frac{1}{2}[/tex]at² + 20
= 10 × t + 0.5 × (9.8)t² + 20
= -4.9t² + 10t + 20
(b) We know that,
→ v² - u² = 2as
here, Final velocity, v = 0
0 - (10)² = 2 × (-9.8) × s
s = [tex]\frac{100}{19.6}[/tex]
= 5.102 m
(c) Time taken will be:
→ -4.9t² + 10t + 20 = 0
t = 3.28 s
hence,
The impact velocity,
v = [tex]\sqrt{2\times 9.8\times 25.102}[/tex]
= 22.181 m/s
Thus the above response is correct.
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Calculate the molecular weight of a polyethylene molecule with n=750. Express your answer to three significant figures.
To calculate the molecular weight of a polyethylene molecule with n=750, multiply the molecular weight of the ethylene unit by n.
Explanation:To calculate the molecular weight of a polyethylene molecule with n=750, we need to know the molecular formula and the atomic weights of the elements present in the molecule. Polyethylene is made up of repeating ethylene (C2H4) units, so we can calculate the molecular weight of the polyethylene molecule by multiplying the molecular weight of the ethylene unit (28.05 g/mol) by the value of n (750).
Calculation:
Molecular weight of polyethylene = Molecular weight of ethylene unit × n = 28.05 g/mol × 750 = 21,037.5 g/mol
Therefore, the molecular weight of the polyethylene molecule with n=750 is 21,037.5 g/mol, rounded to three significant figures.
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A load consists of a 70-Ω resistor in parallel with a 90-μF capacitor. If the load is connected to a voltage source vs(t) = 160cos 2000t, find the average power delivered to the load.
Answer:
Power delivered by the source will be 182.912 watt
Explanation:
We have given a load is consist of a resistor of 70 ohm in parallel with [tex]90\mu F[/tex] capacitance
Voltage source is given [tex]v_s(t)=160cos(2000t)[/tex]
So maximum value of voltage source is 160 volt
So rms value [tex]v_{r}=\frac{v_m}{\sqrt{2}}=\frac{160}{1.414}=113.154volt[/tex]
We know that average power delivered by the source will be equal to average power absorbed by the resistor
So power absorbed by the resistor [tex]P=\frac{v_r^2}{R}=\frac{113.154^2}{70}=182.912watt[/tex]
So power delivered by the source will be 182.912 watt
The front of an aircraft hanger is being built in the shape of a parabola, which is 48 ft. wide, and has a maximum height of 18 ft., AND must have a rectangular doorway that is 8 ft. tall. What is the maximum width of the doorway? (Round your answer to one decimal place.)
Answer:
maximum width of the doorway = 35.77ft
Explanation:
The detailed calculation and derivation from first principle is as shown in the attachment
Answer:
the maximum width is x= 4√2 ft = 5.656 ft
Explanation:
for the parabola
y= a*x² + b*x + c
where y= height and x= width
an aircraft hangar should be symmetric with respect to the y axis , then
y(-x)=y(x) → a*x² + b*x + c = a*x² - b*x + c →-2*b*x =0 → b=0
it also should be pointing downwards → a is negative
, then the parabola would be
y= c- a*x²
since c= maximum height = 18 ft
then for y=0 , x= 48 ft/2 = 24 ft → 0 = 18 ft - a*(24 ft)² → a= 1/32 ft⁻¹
then
y= 18 ft- 1/32 ft⁻¹ *x²
since the doorway cannot go beyond the parabola , the maximum possible doorway is obtained when the doorway touches the parabola.
then for a height y= 8 ft
8 ft = 18 ft- 1/32 ft⁻¹ *x²
x= 4√2 ft = 5.656 ft
A dripping water faucet steadily releases drops 1.0 s apart. As these drops fall, does the distance between them increase, decrease, or remain the same? Prove your answer.
Answer:
Distance between them increase
Explanation:
The position S of the water droplet can be determined using equation of motion
[tex]S=ut+\frac{1}{2} at^2[/tex]
where [tex]u[/tex] is the initial velocity which is zero here
[tex]t[/tex] is time taken, [tex]a[/tex] is acceleration due to gravity
the position of first drop after time [tex]t[/tex] is given by
[tex]S_{1} =0 \times t+ \frac{1}{2} at^2=\frac{1}{2} at^2............(1)[/tex]
the position of next drop at same time is
[tex]S_{2} =\frac{1}{2} a(t-1)^2 = \frac{1}{2} a(t^2+1-2t)............(2)[/tex]
distance between them is [tex]S_{1} -S_{2}[/tex] is [tex]a(t-1)[/tex]
from the above the difference will increase with the time
As the water drops fall, their velocity increases due to the force of gravity, which causes the distance between each subsequent drop to increase.
Explanation:The response to the student's question deals with the notion of acceleration due to gravity. As the water drops fall, they are accelerated by gravity, which means their velocity (speed) increases over time. If we consider two subsequent droplets, the second drop begins its descent 1.0 seconds after the first. Therefore, when the second drop begins to fall, the first drop has already accelerated for 1.0 seconds. This causes the distance between the two drops to increase as they fall.
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A projectile is fired with an initial speed of 40 m/s at an angle of elevation of 30∘. Find the following: (Assume air resistance is negligible. Your answer should contain the gravitational constant ????.)
a. The time at which the maximum height is achieved is functionsequation editor s.
b. The maximum height achieved by the projectile is functionsequation editor m.
c. The time when the projectile hits the ground is functionsequation editor s.
d. The range of the projectile is functionsequation editor m.
e. The speed of the projectile on impact with the ground is functionsequation editor m/s.
Answer:
a. 2.0secs
b. 20.4m
c. 4.0secs
d. 141.2m
e. 40m/s, ∅= -30°
Explanation:
The following Data are giving
Initial speed U=40m/s
angle of elevation,∅=30°
a. the expression for the time to attain the maximum height is expressed as
[tex]t=\frac{usin\alpha }{g}[/tex]
where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at
[tex]t=40sin30/9.81\\t=2.0secs[/tex]
b. the expression for the maximum height is expressed as
[tex]H=\frac{u^{2}sin^{2}\alpha }{2g} \\H=\frac{40^{2}0.25 }{2*9.81} \\H=20.4m[/tex]
c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,
Hence T=2t
T=2*2.0
T=4.0secs
d. The range of the projectile is expressed as
[tex]R=\frac{U^{2}sin2\alpha}{g}\\R=\frac{40^{2}sin60}{9.81}\\R=141.2m[/tex]
e. The landing speed is the same as the initial projected speed but in opposite direction
Hence the landing speed is 40m/s at angle of -30°
A prism-shaped closed surface is in a constant, uniform electric field E, filling all space, pointing right.The 3 rectangular faces of the prism are labeled A, B, and C. Face A is perpendicular to the E-field. The bottom face C is parallel to E. Face B is the leaning face. (The two triangular side faces are not labeled.)Which face has the largest magnitude electric flux through it?
a) A b) B c) C d) A and B have the same magnitude flux
Answer:
The correct answer is:
d) A and B have the same magnitude flux
Explanation:
Electric flux is the property of electric field that measures the electric field lines, passing through a surface and electric flux is also directly proportional to the number of electric field lines passing through a surface.
The formula of electric flux is:
Φ = E A Cos θ
(where E is the electric field, A is the area of face and θ is the angle between the face and the electric field).
Since, faces A and B are perpendicular to the electric field and the electric field lines passing through face A also passes through face B therefore, both of these faces have larger and same magnitude of electric flux.
Since, face C is parallel to the electric field so, the electric flux is smaller at face C, because the magnitude of Cos 180 (when face is parallel) is smaller than the magnitude of Cos 90 (when face is perpendicular).
Face A, which is perpendicular to the uniform electric field, has the largest magnitude electric flux through it because the angle between the field lines and the normal to the surface is zero, maximizing the electric flux.
Explanation:The question revolves around calculating the electric flux through different faces of a prism in a uniform electric field. Electric flux (Φ) is given by the equation Φ = E ⋅ A ⋅ cos(θ), where E is the magnitude of the electric field, A is the area through which the field lines pass, and θ is the angle between the field lines and the normal (perpendicular) to the surface.
Face A is perpendicular to the electric field, which means the angle θ is 0 degrees and cos(θ) is 1. Thus the flux through Face A is maximum. For Face B, the leaning face, θ is greater than 0 degrees but less than 90 degrees, thus cos(θ) will be less than 1. Hence, flux through Face B will be less than through Face A. Face C, being parallel to the electric field, has θ as 90 degrees, and cos(90) is 0, so the flux through Face C is zero. Therefore, in comparison, Face A has the largest magnitude electric flux through it.
A certain carbon monoxide molecule consists of a carbon atom mc = 12 u and an oxygen atom mo = 17 u that are separated by a distance of d = 128 pm, where "u" is an atomic unit of mass.
Part (a) write a symbolic equation for the location of the center of mass of the carbon monoxide molecule relative to the position of the oxygen atom. This expression should be in terms of the masses of the atoms and the distance between them. 50%
Part (b) Calculate the numeric value for the center of mass of carbon monoxide in units of pm. Grade Summary Deductions Potential 0% 100%
Answer:
a) x_{cm} = m₂/ (m₁ + m₂) d , b) x_{cm} = 52.97 pm
Explanation:
The expression for the center of mass is
[tex]x_{cm}[/tex] = 1 / M ∑ [tex]x_{i}[/tex] [tex]m_{i}[/tex]
Where M is the total masses, mI and xi are the mass and position of each element of the system.
Let's fix our reference system on the oxygen atom and the molecule aligned on the x-axis, let's use index 1 for oxygen and index 2 for carbon
x_{cm} = 1 / (m₁ + m₂) (0+ m₂ x₂)
Let's reduce the magnitudes to the SI system
m₁ = 17 u = 17 1,661 10⁻²⁷ kg = 28,237 10⁻²⁷ kg
m₂ = 12 u = 12 1,661 10⁻²⁷ kg = 19,932 10⁻²⁷ kg
d = 128 pm = 128 10⁻¹² m
The equation for the center of mass is
x_{cm} = m₂/ (m₁ + m₂) d
b) let's calculate the value
x_{cm} = 19.932 10⁻²⁷ /(19.932+ 28.237) 10⁻²⁷ 128 10-12
x_{cm} = 52.97 10⁻¹² m
x_{cm} = 52.97 pm
(a) The expression for the center mass of these two atoms relative to oxygen atom is [tex]X_{cm} = \frac{m_1 d_0 \ +\ m_2d}{m_1 + m_2}[/tex]
(b) The numeric value for the center of mass of carbon monoxide is 53 pm.
The given parameters;
mass of the carbon atom = 12umass of the oxygen atom, = 17 udistance between the atoms, = 128 pmThe center mass of these two atoms relative to oxygen atom is calculated as follows;
[tex]X_{cm} = \frac{m_1 d_0 \ +\ m_2d}{m_1 + m_2}[/tex]
where;
[tex]d_0[/tex] is distance of the atom in the fixed reference point (oxygen atom)(b)
The numeric value for the center of mass of carbon monoxide in units of pm is calculated as follows;
[tex]X_{cm} = \frac{17u(0) \ +\ 12u(128 \ pm)}{(12u + 17u)}\\\\X_{cm} = \frac{(12 \times 128u) \ pm}{29u} \\\\X_{cm} = 53 \ pm[/tex]
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A used car is pushed off an 87-ft-high sheer seaside cliff with a speed of 8 ft/s. Find the speed at which the car hits the water.
Final Answer:
The speed at which the car hits the water is approximately 75.2 feet per second.
Explanation:
To find the speed at which the car hits the water, we can use one of the kinematic equations that relates the initial velocity, acceleration due to gravity, the height it fell from, and the final velocity. The kinematic equation that we need is:
[tex]\[ v^2 = u^2 + 2gh \][/tex]
Where:
- v is the final velocity,
- u is the initial velocity,
- g is the acceleration due to gravity (which we will use [tex]\( 32.174 \, \text{ft/s}^2 \)[/tex] for since we are dealing with feet),
- h is the height.
Here, we are given:
- [tex]\( u = 8 \, \text{ft/s} \)[/tex] (initial velocity)
- [tex]\( h = 87 \, \text{ft} \)[/tex] (height)
- [tex]\( g = 32.174 \, \text{ft/s}^2 \)[/tex] (acceleration due to gravity)
Let's find the final velocity \( v \) using these values.
[tex]\[ v^2 = u^2 + 2gh \][/tex]
[tex]\[ v^2 = (8 \, \text{ft/s})^2 + 2 \cdot 32.174 \, \text{ft/s}^2 \cdot 87 \, \text{ft} \][/tex]
[tex]\[ v^2 = 64 \, \text{ft}^2/\text{s}^2 + 2 \cdot 32.174 \, \text{ft/s}^2 \cdot 87 \, \text{ft} \][/tex]
[tex]\[ v^2 = 64 \, \text{ft}^2/\text{s}^2 + 5591.148 \, \text{ft}^2/\text{s}^2 \][/tex]
[tex]\[ v^2 = 5655.148 \, \text{ft}^2/\text{s}^2 \][/tex]
Now we take the square root of both sides to solve for the final velocity \( v \):
[tex]\[ v = \sqrt{5655.148} \, \text{ft/s} \][/tex]
Performing the square root calculation, we get:
[tex]\[ v \approx 75.2 \, \text{ft/s} \][/tex]
So, the speed at which the car hits the water is approximately 75.2 feet per second.
If instead the distance between the moon and the planet were 7 times as large (no change in mass), what would the magnitude of the force be?
Answer:
Reduced by 49 times
Explanation:
We have Newton formula for attraction force between 2 objects with mass and a distance between them:
[tex]F_G = G\frac{M_1M_2}{R^2}[/tex]
where G is the gravitational constant. [tex]M = M_1 = M_2[/tex] are the masses of the 2 objects. and R is the distance between them.
Since R squared is in the denominator of the formula, if we make it 7 times as large with no change in mass, gravitational force would be dropped by 7*7 = 49 times
To solve the problem we should know about Newton's Law of gravity.
What is Newton's Law of gravity?
According to Newton's law of gravity, there is an attractive force between any two-particle carrying mass, such that the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
[tex]F \propto m_1m_2\\\\F \propto \dfrac{1}{R^2}[/tex]
[tex]F = G\dfrac{m_1m_2}{R^2}[/tex]
Where G is the proportionality constant and the value of G is 6.67 x 10-11 N m² / kg².
The force between the two will be [tex]\dfrac{1}{49}[/tex] time of the force before.
Given to us,
Mass of the planet = [tex]m_1[/tex]Mass of the earth = [tex]m_2[/tex]distance between the moon and the planet is 7 timesAssumption
Let's assume that the distance between the moon and the planet is d.
Values
As it is given that there is no change in the mass of the moon or the planet, therefore,
Mass of the planet = [tex]m_1[/tex]Mass of the earth = [tex]m_2[/tex]Also, it is given that the distance between them changes to 7 times, therefore,
distance between the moon and the planet =7dNewton's Law of gravitySubstitute the value Newton's Law of gravity,
[tex]F = G\dfrac{m_1m_2}{(7d)^2}\\\\\\F = G\dfrac{m_1m_2}{49d^2}[/tex]
Thus, the force between the two will be [tex]\dfrac{1}{49}[/tex] time of the force before.
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A phonograph record has an initial angular speed of 37 rev/min. The record slows to 14 rev/min in 1.6 s. What is the record’s average angular acceleration during this time interval? Answer in units of rad/s 2
Answer:
Acceleration will be [tex]\alpha =-1.50rad/sec^2[/tex]
Explanation:
We have given initial angular velocity [tex]\omega _i=37rpm[/tex]
In radian/sec initial angular velocity will be [tex]\omega _i=\frac{2\times \pi 37}{60}=3.873rad/sec[/tex]
Angular velocity after 1.6 sec is 14 rpm
So final angular velocity [tex]\omega _f=\frac{2\times \pi\times 14}{60}=1.465rad/sec[/tex]
Time t = 1.6 sec
We have to find the angular angular acceleration
From first equation of motion we know that
[tex]\omega _f=\omega _+\alpha t[/tex]
[tex]1.465=3.873+\alpha \times 1.6[/tex]
[tex]\alpha =-1.50rad/sec^2[/tex] here negative sign indicates that motion is deaccelerative in nature
A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help." When she has fallen for 3.0 s, she hears the echo of her shout from the valley floor below. The speed of sound is 340 m/s. (a) How tall is the cliff? (b) If we ignore air resistance, how fast will she be moving just before she hits the ground? (Her actual speed will be less than this, due to air resistance.)
Answer:
532.0725 m
102.17270893 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = g
H = Height of cliff
Distance traveled in 3 seconds
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 3^2\\\Rightarrow s=44.145\ m[/tex]
Distance traveled by sound = 2H-44.145 m
[tex]2H-44.145=ut+\dfrac{1}{2}at^2\\\Rightarrow 2H-44.145=340\times 3\\\Rightarrow H=\dfrac{340\times 3+44.145}{2}\\\Rightarrow H=532.0725\ m[/tex]
The height of the cliff is 532.0725 m
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 532.0725+0^2}\\\Rightarrow v=102.17270893\ m/s[/tex]
Her speed just before she hits the ground is 102.17270893 m/s
Asteroids, meteoroids, and comets are remnants of the early solar system. (T/F)
Answer: Asteroids, meteoroids, and comets are remnants of the early solar system. This Statement is TRUE.
Explanation:
METEOROID: these are small rocky or metallic objects found in outer space.
ASTEROIDS: these are also known as minor planets of the inner solar system. They are irregularly shaped object in space that orbits the Sun.
COMETS: these are dusty chunk of ice, that moves in a highly elliptical orbit about the sun.
Asteroids, meteoroids, and comets as remnants of the early solar system was further proved in nebular hypothesis
initially proposed in the eighteenth century by German philosopher Immanuel Kant and French mathematician Pierre-Simon Laplace. (The word nebula means a gaseous cloud.) According to the modern version of the theory, about 4.5 to 5 billion years ago the solar system developed out of a huge cloud of gases and dust floating through space. These materials were at first very thin and highly dispersed.
What length of tube would be required to produce a second tone under the same experimental conditions? Explain your answer.
To produce a second tone or the first overtone in a tube closed at one end, the length of the tube required is three times the length used for the fundamental frequency, resulting in a length of 1.008 m.
To understand the length required to produce a second tone or the first overtone in a tube closed at one end, it's essential to grasp the concept of harmonics in sound resonance. In such a tube, the resonant frequencies occur in odd multiples of the fundamental frequency. The first resonance the students observed, with the fundamental frequency of 256 Hz at a length of 0.336 m, corresponds to a quarter wavelength of the sound wave in the tube.
For the first overtone (second resonance), the air column in the tube must accommodate three-quarters of a wavelength, meaning the effective length will be three times larger than that of the fundamental. Thus, if the fundamental resonance occurs at a length of 0.336 m, the length for the second resonance will be:
0.336 m x 3 = 1.008 m.This calculation is based on the understanding that the second tone or first overtone in a closed tube happens at three times the length necessary for the fundamental frequency, leading to the described increase in the length of the air column.
To find the length of tube for the second resonance, halve the initial length where the first resonance occurred at a fundamental frequency of 256 Hz.
Explanation:The length required to produce a second tone under the same experimental conditions can be calculated based on the concept of resonance in a closed tube.
To find the length for the second resonance (first overtone), we know that the first resonance occurs at 0.336m for a fundamental frequency of 256 Hz. The second resonance, in this case, would occur at half the wavelength of the fundamental frequency, so the length would be half of the initial length: 0.168m.
Which of the following statement(s) about energy and phase is/are correct? Select all that apply. Choose one or more: A. While only one phase is present, adding or removing energy changes PE but not KE. B. While only one phase is present, adding or removing energy changes KE but not PE. C. During a phase change, adding or removing energy changes KE but not PE. D. During a phase change, adding or removing energy changes PE but not KE.
In a single phase, the addition or removal of energy changes Kinetic Energy not Potential Energy. However, during a phase change, this energy addition or subtraction results in a change in Potential Energy, not Kinetic Energy.
Explanation:The subject of this question is energy and phase, particularly in the context of Potential Energy (PE) and Kinetic Energy (KE). When only one phase is present, adding or removing energy will mainly change the KE, not the PE. This is because the energy is utilized to speed up or slow down the particles, thus changing their kinetic energy. However, during a phase change, adding or removing energy changes PE but not KE as it alters the state rather than the speed of the particles. Statement B is the one that is accurate while only one phase is present, whereas the correct option for the phase change scenario is option D.
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