Answer:
1) We will have excess of electrons
2) The number of electrons transferred equals [tex]1.343\times 10^{10}[/tex]
Explanation:
Part a)
Since we know that the charge transfer occurs by the transfer of electrons only as it is given that the carpet has acquired a positive charge it means that it has lost some of the electron's since electrons are negatively charged and if a neutral body looses negative charge it will become positively charged. The electron's that are lost by the carpet will be acquired by the feet of the human thus making us negatively charged.Hence we will gain electrons making us excess in electrons.
Part b)
From charge quantinization principle we have
[tex]Q=ne[/tex]
where
Q = charge of body
n = no of electrons
e = fundamental charge
Applying values in the above equation we get
[tex]2.15\times 10^{-9}C=n\times 1.6\times 10^{-19}C\\\\\therefore n=\frac{2.15\times 10^{-9}C}{1.6\times 10^{-19}C}=1.343\times 10^{10}[/tex]
onsider a 20-cm-thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held constant at 50°C, whereas the right face is exposed to a flow of 22°C air with a convection heat transfer coefficient of 15 W/m2·K. Neglecting heat transfer by radiation, find the right wall surface temperature and the heat flux through the wall.
Answer:
q=heat flux=216.83W/m^2
Ts=36.45C
Explanation:
What you should do is raise the heat transfer equation from the inside of the granite wall to the air.
You should keep in mind that in the numerator you place the temperature differences, while the denominator places the sum of the thermal resistances by conduction and convection.
When you have the heat value, all you have to do is use the convection equation q = h (ts-t) and solve for the surface temperature.
I attached procedure
Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 of 1110°C fresh lava into 36.2°C surroundings, assuming lava's emissivity is 1.
The rate of heat loss by radiation is equal to -207.5kW
Why?
To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:
[tex]HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )[/tex]
Where,
E, is the emissivity of the body.
A, is the area of the body.
T, are the temperatures.
S, is the Stefan-Boltzmann constant, which is equal to:
[tex]5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }[/tex]
Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so, we need to convert the given values of temperature from Celsius degrees to Kelvin.
We know that:
[tex]K=Celsius+273.15[/tex]
So, converting we have:
[tex]T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K[/tex]
Therefore, substituting the given information and calculating, we have:
[tex]HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )[/tex]
[tex]HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW[/tex]
Hence, we have that the rate of heat loss is equal to -207.5kW.
How much heat is necessary to change 350 g of ice at -20 degrees Celsius to water at 20 Celsius?
Final answer:
To change 350 g of ice at -20 degrees Celsius to water at 20 Celsius, you need to calculate the heat required for two processes: warming the ice and melting the ice. The heat needed to warm the ice can be calculated using the equation: Q = mcΔT, and the specific heat capacity of ice is 2.09 J/g°C. The heat needed to melt the ice can be calculated using the equation: Q = mL, where L is the heat of fusion, which is 334 J/g for ice.
Explanation:
To calculate the heat necessary to change 350 g of ice at -20 degrees Celsius to water at 20 degrees Celsius, we need to consider the energy required for two processes: warming the ice from -20°C to 0°C and then melting the ice at 0°C to form water at 0°C.
The heat needed to warm the ice can be calculated using the equation: Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. The specific heat capacity of ice is 2.09 J/g°C.
The heat needed to melt the ice can be calculated using the equation: Q = mL, where L is the heat of fusion, which is 334 J/g for ice.
Let's calculate the heat needed for each process and add them together:
Heat to warm the ice: Q = 350 g x 2.09 J/g°C x (0°C - (-20°C)) Heat to melt the ice: Q = 350 g x 334 J/gFinally, add both values to find the total heat required.
The angular velocity of a rigid body that undergoes plane motion depends on the reference line from which its angle of rotation is measured. True False
That statement is false.
Angular velocity is the rate at which that angle changes. As long as the reference line doesn't move, it doesn't matter where it is. The rate at which the angle changes is still the same, constant number.
The angular velocity of a rigid body that undergoes plane motion depends on the reference line from which its angle of rotation is measured is False.
What is velocity?When anything is moving, its velocity tells us how quickly that something's location is changing from a certain vantage point and as measured by a particular unit of time.
When a point travels down a path and completes a certain distance in a predetermined amount of time, its average speed over that time is equal to the distance covered divided by the travel time. A train traveling 100 kilometers in two hours, for instance, is doing it at an average speed of 50 km/h.
The pace at which that angle change is known as the angular velocity. It doesn't matter where the reference line is, as long as it stays put. The rate of change of the angle remains constant and the same.
Therefore, the angular velocity of a rigid body that undergoes plane motion depends on the reference line from which its angle of rotation is measured is False.
To know more about velocity:
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A tank holds 10.3 mol of an ideal gas at an absolute pressure of 519 kPa while at a temperature of 361.38°C. (a) Compute the container's volume (b) The gas is now heated to 651.6 °C. What is the pressure of the gas now?
Answer:
(a) 62.57 L
(b) 801.94 kPa
Explanation:
Given:
[tex]n[/tex] = number of moles of gas = 10.3 mol
[tex]P_1[/tex] = initial pressure of the gas = [tex]519\ kPa = 5.19\times 10^5\ Pa[/tex]
[tex]T_1[/tex] = initial temperature of the gas = [tex]361.38^\circ C = (361.38+273)\K = 598.38\ K[/tex]
[tex]T_2[/tex] = final temperature of the gas = [tex]651.6^\circ C = (651.6+273)\K = 924.6\ K[/tex]
[tex]V[/tex] = volume of the tank
R = universal gas constant = [tex]8.314 J/mol K[/tex]
Part (a):
Using Ideal gas equation, we have
[tex]PV=nRT\\\Rightarrow V = \dfrac{nRT}{P}\\\Rightarrow V = \dfrac{10.3\times 8.314\times 598.38}{5.19\times 10^{5}}\\\Rightarrow V = 6.2567\times10^{-2}\ m^3\\\Rightarrow V = 62.567\ L[/tex]
Hence, the volume of the container is 62.567 L.
Part (b):
As the volume of the container remains constant.
Again using ideal gas equation,
[tex]PV=nRT\\\because V,\ n, R\ are\ constant\\\therefore \dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}\\\Rightarrow P_2 = \dfrac{T_2}{T_1}P_1\\\Rightarrow P_2 = \dfrac{924.6}{598.38}\times 519\\\Rightarrow P_2 = 801.94\ kPa[/tex]
Hence, the final pressure of the gas is now 801.94 kPa.
Two large, flat, horizontally oriented plates are parallel to each other, a distance d apart. Half way between the two plates the electric field has magnitude E. If the separation of the plates is reduced to d/2 what is the magnitude of the electric field half way between the plates? Assume the charge of plates is constant.
Answer:
E the electric field remains unchanged.
Explanation:
potential difference between the plates of a capacitor
V = Q / C Where Q is charge on the capacitor and C is capacity of capacitor
Here Q is unchanged.
C the capacity has value equal to ε A / d
Here d is distance between plates. when it is halved, capacitance C becomes double.
V = Q /C , When C becomes double V becomes half
E = V/d , E is electric field between plates having separation of d.
When V becomes half and d also becomes half , there is no change in the value of E.
Hence E the electric field remains unchanged.
The magnitude of the electric field between two charged parallel plates remains constant when the plate separation is reduced from d to d/2, provided the charge density on the plates stays the same. Thus, the magnitude of the electric field halfway between the plates would still be E.
Explanation:The student asked what the magnitude of the electric field would be if the separation between two large, flat, parallel, charged plates is reduced from d to d/2, assuming the charge on the plates remains constant.
According to Essential Knowledge 2.C.5, the electric field (E) between two oppositely charged parallel plates with uniformly distributed electric charge is constant in magnitude and direction at points far from the edges of the plates.
The magnitude of the electric field E between parallel plates is given by the equation E = σ/ε₀, where σ is the surface charge density and ε₀ is the permittivity of free space. Since the surface charge density (σ = Q/A, charge per area) does not change when the separation between the plates is altered and the permittivity of free space (ε₀) is a constant, the magnitude of the electric field between the plates remains the same even if the separation is reduced to d/2.
Therefore, the magnitude of the electric field halfway between the plates when the separation is reduced to d/2 would still be E.
Determine the velocity and position as a function of time for the time force F(t)=F Cos^2(WT). Generate plots for the resulting equation.
Answer with Explanation:
We know from newton's second law that acceleration produced by a force 'F' in a body of mass 'm' is given by
[tex]a=\frac{Force}{Mass}=\farc{F}{m}[/tex]
In the given case the acceleration equals
[tex]a=\frac{F_{o}cos^{2}(\omega t)}{m}[/tex]
Now by definition of acceleration we have
[tex]a=\frac{dv}{dt}\\\\\int dv=\int adt\\\\v=\int adt\\\\v=\int \frac{F_{o}cos^{2}(\omega t)}{m}\cdot dt\\\\v=\frac{F_{o}}{m}\int cos^{2}(\omega t)dt\\\\v=\frac{F_{o}}{\omega m}\cdot (\frac{\omega t}{2}+\frac{sin(2\omega t)}{4}+c)[/tex]
Similarly by definition of position we have
[tex]v=\frac{dx}{dt}\\\\\int dx=\int vdt\\\\x=\int vdt[/tex]
Upon further solving we get
[tex]x=\int [\frac{F_{o}}{\omega m}\cdot (\frac{\omega t}{2}+\frac{sin(2\omega t)}{4}+c)]dt\\\\x=\frac{F_{o}}{\omega m}\cdot ((\int \frac{\omega t}{2}+\frac{sin(2\omega t)}{4}+c)dt)\\\\x(t)=\frac{F_{o}}{\omega m}\cdot (\frac{\omega t^{2}}{4}-\frac{cos(2\omega t)}{8\omega }+ct+d)[/tex]
The plots can be obtained depending upon the values of the constants.
A nervous squirrel gets startled and runs 5.0\,\text m5.0m5, point, 0, space, m leftward to a nearby tree. The squirrel runs for 0.50\,\text s0.50s0, point, 50, space, s with constant acceleration, and its final speed is 15.0\,\dfrac{\text m}{\text s}15.0 s m 15, point, 0, space, start fraction, m, divided by, s, end fraction. What was the squirrel's initial velocity before getting startled
Answer:
−5.0
Explanation:
Answer:
The squirrel's initial velocity is 5 m/s.
Explanation:
It is given that,
Distance covered by the squirrel, d = 5 m
Time taken, t = 0.5 s
Final speed of the squirrel, v = 15 m/s
To find,
The squirrel's initial velocity.
Solution,
Let a is the acceleration of the squirrel. Using the first equation of motion to find it :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{15-u}{0.5}[/tex]..............(1)
Let u is the initial speed of the squirrel. Using again third equation of kinematics to find it :
[tex]v^2-u^2=2ad[/tex]
Use equation (1) to put value of a
[tex](15)^2-u^2=2\times \dfrac{15-u}{0.5}\times 5[/tex]
[tex]225-u^2=20(15-u)[/tex]
On solving the above quadratic equation, we get the value of u as :
u = 5 m/s
Therefore, the squirrel's initial velocity before getting started is 5 m/s.
During a hard sneeze, your eyes might shut for 0.32 s. If you are driving a car at 76 km/h during such a sneeze, how far does the car move during that time?
Answer:
Distance, d = 6.75 meters
Explanation:
Given that,
Time taken by the eyes to shut during a hard sneeze, t = 0.32 s
Speed of the car, v = 76 km/h = 21.11 m/s
We need to find the distance covered by the car during this time. The product of speed and the time taken by the car is called the distance covered by the it. Mathematically,
[tex]d=v\times t[/tex]
[tex]d=21.11\times 0.32[/tex]
d = 6.75 meters
So, the car will cover 6.75 meters during that time. Hence, this is the required solution.
Three faces of a cube of side length 2 cm each have a uniform electric field of magnitude 500 N/C pointing directly inward. The other three faces each have a uniform electric field of magnitude 200 N/C pointing directly outward. What is the magnitude and sign of the net charge enclosed in the cube?
Answer:
[tex]Q=-8.85*10^{-12}*0.36 C =-3.186*10^{-12} C[/tex]
(negative charge)
Explanation:
Hi!
To solve this problem we use Gauss Law for electric fields, which relates the flux of electric field E through a closed surface S with the charge Q enclosed by that surface:
[tex]\int_{S} \vec{E}\cdot \vec{n} \;dA = \frac{Q}{\epsilon_0}[/tex]
[tex]\vec{n} = \text{outwards surface normal}\\\epsilon_0 = 8.85*10^{-12}\frac{C^2}{Nm^2}[/tex]
In this problem S is a cube of side length 2cm. The integral is easy because the electric field is uniform in each face, and normal to the face. The total integral is the sum of the integrals in each of the six faces.
[tex]\int_{S} \vec{E}\cdot \vec{n} \;dA = \frac{Q}{\epsilon_0} = 3(-500\frac{N}{C} (2cm)^2) + 3(200\frac{N}{C} (2cm)^2)[/tex]
[tex](-1500*4*10^{-4}\frac{Nm^2}{C} + 600*4*10^{-4}\frac{Nm^2}{C} )=-0.36\frac{Nm^2}{C}[/tex]
[tex]Q=-8.85*10^{-12}*0.36 C =-3.186*10^{-12} C[/tex]
From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.40 m/s and angle of 15.0° below the horizontal. It strikes the ground 5.00 s later. (a) If the base of the building is taken to be the origin of the coordinates, with upward the positive y-direction, what are the initial coordinates of the ball? (Use the following as necessary: y0. Assume SI units. Do not substitute numerical values; use variables only.)
Answer:
[tex]y_{o}=v_{o}*sin(\alpha )*t + 1/2*gt^{2}[/tex]
Explanation:
Kinematics equation, with gravity as deceleration
[tex]y =y_{o}-v_{o}*sin(\alpha )*t - 1/2*gt^{2}[/tex]
If the base of the building is taken to be the origin of the coordinates, then :
[tex]y_{o} [/tex] is the initial coordinates of the ball
for t=5.00, the ball strikes the ground ⇒ y=0m
then:
[tex]y_{o}=v_{o}*sin(\alpha )*t + 1/2*gt^{2}[/tex]
Final answer:
The initial coordinates of the ball below the horizontal are (0, y0), with y0 representing the height from which the ball was tossed and 0 representing the horizontal origin.
Explanation:
The student's question asks for the initial coordinates of a ball tossed from a window with a given initial velocity and angle relative to the horizontal. Since we're dealing with projectile motion, we need to resolve the initial velocity into its horizontal (x-direction) and vertical (y-direction) components.
The initial horizontal velocity (vx0) can be found using trigonometry: vx0 = v0 × cos(θ), where v0 is the initial velocity of 8.40 m/s and θ is the launch angle of 15.0°. However, since the angle is below the horizontal, the vertical component of the initial velocity (vy0) will be negative: vy0 = v0 × sin(θ), pointing downward.
Given the vertical direction is positive upwards, and taking the base of the building to be at the origin (0,0), the initial coordinates of the ball will be (0, y0), where y0 represents the height of the window from which the ball was tossed.
Two men pushing a stalled car generate a net force of +690Nfor
6.6 sec. What is the final momentum of the car?
Final answer:
The final momentum of the car pushed by two men with a net force of 690 N for 6.6 seconds is 4554 N·s. This is calculated using the formula for impulse, which is the product of the net force and the time over which the force is applied.
Explanation:
The question is asking to calculate the final momentum of a car after two men have pushed it with a net force of 690 N for 6.6 seconds. According to Newton's second law of motion, the change in momentum (also known as impulse) is equal to the net force multiplied by the time over which the force is applied. The formula for impulse is:
Impulse (J) = Force (F) × Time (t)
By applying this formula, we can find the final momentum:
Force (F) = 690 NTime (t) = 6.6 secondsTherefore, the impulse is:
J = 690 N × 6.6 s = 4554 N·s
Since the car was initially at rest and assuming there is no external resistance, the final momentum of the car will be equal to the impulse given to it:
Final momentum = 4554 N·s
The final momentum of the car, with an applied force of +690 N for 6.6 seconds, is 4554 Ns. The initial momentum is zero since the car starts from rest.
To determine the final momentum of the car, we will utilize the relationship between force, time, and momentum. According to the impulse-momentum theorem, the impulse applied to an object is equal to the change in its momentum.
Impulse is given by the formula:
Impulse (J) = Force (F) × Time (t)
Given:
Force (F) = +690 NTime (t) = 6.6 sNow, calculate the impulse:
J = 690 N × 6.6 s = 4554 Ns
In this scenario, the car starts from rest, which means its initial momentum (p_initial) is 0. Hence, the final momentum ([tex]p_{final}[/tex]) after 6.6 seconds is equal to the impulse:
[tex]p_{final}[/tex] = 4554 Ns
The final momentum of the car is 4554 Ns.
A team exerts a force of 10 N towards the south in a tug of war. The other, opposite team, exerts a force of 17 N towards the north. The net force is A. 27 N east. B. 7 N south C. 7 N north D. 27 N north
Final answer:
In a tug-of-war where forces are exerted in opposite directions (10 N south and 17 N north), the net force is calculated by subtracting the smaller force from the larger, resulting in a 7 N force towards the north. The correct answer is D.
Explanation:
The net force in a tug of war scenario is the vector sum of all the forces acting on the object. Since force is a vector quantity, it has both magnitude and direction. In this case, the force exerted by one team (17 N north) and the force exerted by the other team (10 N south) are in opposite directions along the same line, so we subtract the smaller force from the larger one to find the net force.
|17 N| - |10 N| = 17 N - 10 N = 7 N north.
A ball is traveling at 60m/s, then increases to 90m/s in 3 second, what is the average velocity?
Answer:
75 m /s
Explanation:
Acceleration of ball a = change in velocity / time
=( 90 - 60)/ 3 = 10 m s⁻²
Initial velocity u = 60 m/s
a = 10 m s⁻²
time t = 3
distance traveled s = ut + 1/2 at²
= 60 x 3 + .5 x 10 x 3 x 3
= 225 m
Average velocity = total distance traveled /time
= 225 / 3 = 75 m/s
Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two charges of magnitude 2*Q and 2*Q when held at a distance r/2.?
Answer:197.504 N
Explanation:
Given
Two Charges with magnitude Q experience a force of 12.344 N
at distance r
and we know Electrostatic force is given
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
[tex]F=\frac{kQ\cdot Q}{r^2}[/tex]
[tex]F=\frac{kQ^2}{r^2}[/tex]
Now the magnitude of charge is 2Q and is at a distance of [tex]\frac{r}{2}[/tex]
[tex]F'=\frac{k2Q\cdot 2Q}{\frac{r^2}{2^2}}[/tex]
F'=16F
F'=197.504 N
A student walks (2.9±0.1)m, stops and then walks another (3.9 ±0.2)m in the same direction. What is the largest distance the student could possibly be from the starting point?
Answer:
7.1 m
Explanation:
Given:
Distance traveled by the student in the first attempt = [tex](2.9 \pm 0.1)\ m[/tex]
Distance traveled by the student in the second attempt = [tex](3.9 \pm 0.2)\ m[/tex]
So, the maximum distance that the student could travel in this attempt = [tex](2.9+0.1)\ m = 3.0\ m[/tex]
So, the maximum distance that the student could travel in this attempt = [tex](3.9+0.2)\ m = 4.1\ m[/tex]
Since the student first moves straight in a particular direction, rests for a while and then moves some distance in the same direction.
So, the largest distance that the student could possibly be from the starting point would be the largest distance of the final position of the student from the starting point.
And this distance is equal to the sum of the maximum distance possible in the first attempt and the second attempt of walking which is 7.1 m.
Hence, the largest distance that the student could possibly be from the starting point is 7.1 m.
In levelling, the following staff readings were observed involving an inverted staff, A = 2.915 and B = -2.028. What is the rise or fall from A to B (only enter numeric answer, no alpha)?
Answer:
The rise from A to B is 0.887
Solution:
As per the question:
The following reading of an inverted staff is given as:
A = 2.915
B = -2.028
Here, for inverted staff, the greater reading shows greater elevation and lesser reading shows lower elevation.
Thus
The rise from A to B is given as:
A - B = 2.915 - 2.028 = 0.887
A myopic person (assume no astigmatism) is diagnosed with a far point of 160 cm. What corrective prescription should they be supplied to enable the person to see objects in the distance?
Answer:
The person should use a convex lens of power -0.625 D.
Explanation:
Given that, the far point of the person, with myopic eye, is 160 cm.
It means that the image of an object, which is at infinity, is formed at this point 160 cm distance away from the person's eye.
If the object is infinity, then the object distance, u = -[tex]\infty[/tex].
Image distance, v = -160 cm.
u and v are taken to be negative because the object and the image of the object are on the side of the eye from where the light is coming.
Let the focal length of the lens which is to be used for the correction of this myopic eye be f, then using lens equation,
[tex]\rm \dfrac 1f = \dfrac 1v-\dfrac 1u=\dfrac 1{-160 }-\dfrac1{-\infty}=\dfrac 1{-160}-0=\dfrac 1{-160}.\\\Rightarrow f=-160\ cm.[/tex]
The negative focal length indicates that the lens should be convex.
The power of the lens is given by
[tex]\rm P=\dfrac{1}{f}=\dfrac{1}{-160\ cm }=\dfrac{1}{-1.6\ m}=-0.625\ D.[/tex]
So, the person should use a convex lens of power -0.625 D.
Answer:
- 1.428 D
Explanation:
A myopic person is able to see the nearby objects clearly but cannot see the far off objects very clearly. It can be cured by using concave lens of suitable power.
far point = 70 cm
v = - 70 cm (position of image from the lens)
u = ∞ (position of object from lens)
Let f be the focal length of the lens
Use lens equation
[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]
[tex]\frac{1}{f}=\frac{1}{-70}-\frac{1}{∞}[/tex]
f = - 70 cm
[tex]P=\frac{1}{f}[/tex]
Where, P is the power of lens
P = - 1.428 Dioptre
Thus, the power of lens used is - 1.428 D.
An astronomer is trying to estimate the surface temperature of a star with a radius of 5.0×108m by modeling it as an ideal blackbody. The astronomer has measured the intensity of radiation due to the star at a distance of 2.5×1013m and found it to be equal to 0.055W/m2. Given this information, what is the temperature of the surface of the star?
Final answer:
To estimate the surface temperature of a star modeled as an ideal blackbody, we can use the Stefan-Boltzmann law. The law states that the power radiated by a blackbody is proportional to its surface area and temperature to the fourth power. We can rearrange the equation to solve for temperature: T = (I/(σ*A))^1/4.
Explanation:
To estimate the surface temperature of a star modeled as an ideal blackbody, we can use the Stefan-Boltzmann law. The law states that the power radiated by a blackbody is proportional to its surface area and temperature to the fourth power. We can rearrange the equation to solve for temperature: T = (I/(σ*A))^1/4, where T is the temperature, I is the intensity of radiation, σ is the Stefan-Boltzmann constant, and A is the surface area.
Plugging in the given values, we have T = (0.055W/m² / (5.670 x 10^-8 W/(m²K^4) * 4π(5.0 x 10^8m)^2))^1/4.
If a body travels half it’s total path in the last 1.10s if it’s fall from rest, find the total time of its fall (in seconds)
Answer:
3.75s
Explanation:
We can use the equations for constant acceleration motion. Let's call x, the total length of the path, then x/2 will be half of path. After falling from rest and reaching the half of its total path, the velocity of the body will be:
[tex]v_f^2 =v_0^2 + 2a(x/2)[/tex]
vf is the final velocity, v0 is the initial velocity, 0m/s because the body starts from rest. a is the acceleration, gravity = 9.81m/s^2 in this case. Now, clearing vf we get:
[tex]v_f=\sqrt{(0m/s)^2 + 2g(x/2)}\\v_f = \sqrt{g*x}
In the second half:
[tex]x/2 = \frac{1}{2}gtx^{2} + v_ot[/tex]
[tex]x/2 = \frac{1}{2}g*(t)^2 + \sqrt{g*x}*(t)[/tex]
[tex](\frac{1}{2}\frac{(x-gt^2)}{\sqrt{g}t})^2 = x\\\\\frac{1}{4gt^2}(x^2 - 2xgt^2 + g^2t^4) = x\\\\\frac{1}{4gt^2}x^2 - (\frac{2gt^2}{4gt^2}+1)x + \frac{g^2t^4}{4gt^2} = 0\\ \frac{1}{4gt^2}x^2 - \frac{3}{2}x + \frac{gt^2}{4} = 0\\[/tex]
[tex]0.0211 x^2 - 1.5x + 2.97 = 0\\[/tex]
Solving for x, you get that x is equal to 69.2 m or 2.03m. The total time of the fall would be:
[tex]x = \frac{1}{2}gt^2\\t=\sqrt{(2x/g)}[/tex]
Trying both possible values of x:
[tex]t_1 = 3.75 s\\t_2 = 0.64 s[/tex]
t2 is lower than 1.1s, therefore is not a real solution.
Therefore, the path traveled will be 69.2m and the total time 3.75s
600-nm light is incident on a diffraction grating with a ruling separation of 1.7 10 m. The second order line occurs at a diffraction angle of: 42 45° 21° 10°
Answer:
45°
Explanation:
λ = 600 nm = 600 x 10^-9 m
d = 1.7 x 10^-6 m
For second order, N = 2
Use the equation
d Sinθ = N λ
1.7 x 10^-6 x Sinθ = 2 x 600 x 10^-9
Sinθ = 0.706
θ = 45°
Thus, the angle of diffraction is 45°.
The diffraction angle of the second-order line is approximately 44.22°.
Explanation:The diffraction angle of the second order line can be found using the formula:
dsinθ = mλ
Where d is the ruling separation of the diffraction grating, θ is the diffraction angle, m is the order of the line, and λ is the wavelength of light. Plugging in the values given in the question:
1.7 × 10-6 m × sinθ = 2 × 600 × 10-9 m
Simplifying the equation:
sinθ = 2 × 600 × 10-9 m / (1.7 × 10-6 m)
Calculating the value of sinθ, we find:
sinθ ≈ 0.7059
Finally, taking the inverse sine of 0.7059, we find:
θ ≈ 44.22°
The second-order line occurs at a diffraction angle of approximately 44.22°.
Learn more about Diffraction here:https://brainly.com/question/12290582
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A particle with mass m and charge e is accelerated through a potential difference (V). What is the wavelength of the particle?
Answer:
The wavelength of the particle is [tex]\lambda=\frac{h}{\sqrt{2V\cdot e\cdot m}}[/tex]
Explanation:
We know that after accelerating across a potential 'V' the potential energy of the charge is converted into kinetic energy as
[tex]\frac{1}{2}mv^{2}=V\cdot e\\\\\therefore v=\sqrt{\frac{2V\cdot e}{m}}[/tex]
now according to De-Broglie theory the wavelength associated with a particle of mass 'm' moving with a speed of 'v' is given by
[tex]\lambda =\frac{h}{mv}[/tex]
where
'h' is planck's constant
'm' is the mass of particle
'v' is the velocity of the particle
Applying the values in the above equation we get
[tex]\lambda =\frac{h}{m\cdot \sqrt{\frac{2V\cdot e}{m}}}[/tex]
Thus
[tex]\lambda=\frac{h}{\sqrt{2V\cdot e\cdot m}}[/tex]
Answer:
Wavelength of the particle is [tex]\frac{h}{\sqrt{2meV}}[/tex]
Solution:
As per the question:
The particle with mass, m and charge, e accelerates through V (potential difference).
The momentum of the particle, [tex]p_{p}[/tex] if it travels with velocity, [tex]v_{p}[/tex]:
[tex]p_{p} = mv_{p}[/tex]
Now, squaring both sides and dividing by 2.
[tex]\frac{1}{2}p_{p}^{2} = \frac{1}{2}m^{2}v_{p}^{2}[/tex]
[tex]K.E =\frac{1}{2m}p_{p}^{2} = \frac{1}{2}mv_{p}^{2}[/tex]
[tex]K.E = \frac{1}{2}mv_{p}^{2}[/tex]
Also,
[tex]p_{p} = \sqrt{2mK.E}[/tex] (1)
Now, we know that Kinetic energy of particle accelerated through V:
K.E = eV (2)
where
e = electronic charge = [tex]1.6\times 10^{- 19} C[/tex]
From eqn (1) and (2):
[tex]p_{p} = \sqrt{2mK.E}[/tex] (3)
From eqn (2) and (3):
[tex]p_{p} = \sqrt{2meV}[/tex]
From the de-Broglie relation:
[tex]\lambda_{p} = \frac{h}{p_{p}}[/tex] (4)
where
[tex]\lambda_{p}[/tex] = wavelength of particle
h = Planck's constant
From eqn (3) and (4):
[tex]\lambda_{p} = \frac{h}{\sqrt{2meV}}[/tex]
A 34.9-kg child starting from rest slides down a water slide with a vertical height of 16.0 m. (Neglect friction.)
(a) What is the child's speed halfway down the slide's vertical distance?
(b) What is the child's speed three-fourths of the way down?
Answer:
a) 12.528 m/s
b) 15.344 m/s
Explanation:
Given:
Mass of the child, m = 34.9 kg
Height of the water slide, h = 16.0 m
Now,
a) By the conservation of energy,
loss in potential energy = gain in kinetic energy
mgh = [tex]\frac{\textup{1}}{\textup{2}}\textup{m}\times\textup{v}^2[/tex]
where,
g is the acceleration due to the gravity
v is the velocity of the child
thus,
at halfway down, h = [tex]\frac{\textup{16}}{\textup{2}}[/tex]= 8 m
therefore,
34.9 × 9.81 × 8 = [tex]\frac{\textup{1}}{\textup{2}}\textup{34.9}\times\textup{v}^2[/tex]
or
v = 12.528 m/s
b)
at three-fourth way down
height = [tex]\frac{\textup{3}}{\textup{4}}\times16[/tex] = 12 m
thus,
loss in potential energy = gain in kinetic energy
34.9 × 9.81 × 12 = [tex]\frac{\textup{1}}{\textup{2}}\textup{34.9}\times\textup{v}^2[/tex]
or
v = 15.344 m/s
Answer:
(a) 12.52 m/s
(b) 15.34 m/s
Explanation:
mass, m = 34.9 kg
h = 16 m
(a) Initial velocity, u = 0
height = h / 2 = 16 / 2 = - 8 m (downward)
let the speed of child is v.
acceleration, a = - 9.8 m/s^2 (downward)
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]v^{2}=0^{2}+2\times 9.8\times 8[/tex]
v = 12.52 m/s
(b) Initial velocity, u = 0
height = 3 h / 4 = 12 m = - 12 m (downward)
let the speed of child is v.
acceleration, a = - 9.8 m/s^2 (downward)
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]v^{2}=0^{2}+2\times 9.8\times 12[/tex]
v = 15.34 m/s
Two point charges are located on the y axis as follows: charge q1 = -2.30 nC at y1 = -0.600 m , and charge q2 = 2.80 nC at the origin (y = 0). What is the magnitude of the net force exerted by these two charges on a third charge q3 = 7.50 nC located at y3 = -0.300 m ?
Answer:
3.825*10^-6 N
Explanation:
As particle 1 and particle 3 has opposite types of charge, particle 3 will be attracted to particle 1. And as particle 2 and 3 has the same sign, they will repel each other. Due to the position of the particles, both the Force that 1 exerts on 3 and the force that 2 exerts on 3, will have the same direction. Now, you need the magnitude. You can use the following expression:
[tex]F_e = K\frac{q_1*q_3}{r^2}[/tex]
K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q1 and q2 is the charge of the particles, and r is the distance.
Force 1 on 3 is equal to:
[tex]F_e = K\frac{q_1*q_3}{r^2} = 9*10^9 \frac{Nm^2}{C^2} * \frac{2.3*10^{-9}C * 7.5*10^{-9}C}{(-0.6m - (-0.3m))^2} = 1.725 * 10^{-6} N[/tex]
Force 2 on 3:
[tex]F_e = K\frac{q_2*q_3}{r^2} = 9*10^9 \frac{Nm^2}{C^2} * \frac{2.8*10^{-9}C * 7.5*10^{-9}C}{(0m - (-0.3m))^2} = 2.1 * 10^{-6} N[/tex]
The magnitude of the resultant force is the addition of both forces:
[tex]F_e = 1.725*10^{-6} N + 2.1*10^{-6} N= 3.825 * 10^{-6} N[/tex]
The magnitude of the net force exerted by the two charges on the third charge is 1.38 N.
Explanation:The magnitude of the net force exerted by q1 and q2 on q3 can be found using Coulomb's Law. Coulomb's Law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is:
F = k(q1 * q2) / r^2
Where F is the force, k is Coulomb's constant (k = 9 x 10^9 N * m^2 / C^2), q1 and q2 are the charges, and r is the distance between them.
In this case, q1 = -2.30 nC, q2 = 2.80 nC, q3 = 7.50 nC, y1 = -0.600 m, y3 = -0.300 m, and y2 = 0 m.
Using the formula and substituting the values, we can calculate the force:
F = (9 x 10^9 N * m^2 / C^2) * (q1 * q3) / ((y3 - y1)^2)
F = (9 x 10^9 N * m^2 / C^2) * (-2.30 nC * 7.50 nC) / ((-0.300 m + 0.600 m)^2)
F = -1.38 N
Therefore, the magnitude of the net force exerted by the two charges on the third charge is 1.38 N.
The so-called Lyman-α photon is the lowest energy photon in the Lyman series of hydrogen and results from an electron transitioning from the n = 2 to the n = 1 energy level. Determine the energy in eV and the wavelength in nm of a Lyman-α photon. (a) the energy in eV
(b)the wavelength in nm
Answer:
a) 10.2 eV
b) 122 nm
Explanation:
a) First we must obtain the energy for each of the states, which is given by the following formula:
[tex]E_{n}=\frac{-13.6 eV}{n^2}[/tex]
So, we have:
[tex]E_{1}=\frac{-13.6 eV}{1^2}=-13.6 eV\\E_{2}=\frac{-13.6 eV}{2^2}=-3.4 eV[/tex]
Now we find the energy that the electron loses when it falls from state 2 to state 1, this is the energy carried away by the emitted photon.
[tex]E_{2}-E_{1}=-3.4eV-(-13.6eV)=10.2eV[/tex]
b) Using the Planck – Einstein relation, we can calculate the wavelength of the photon:
[tex]E=h\nu[/tex]
Where E is the photon energy, h the Planck constant and [tex]\nu[/tex] the frequency.
Recall that [tex]\nu=\frac{c}{\lambda}[/tex], Rewriting for [tex]\lambda[/tex]:
[tex]E=\frac{hc}{\lambda}\\\lambda=\frac{hc}{E}\\\lambda=\frac{(4.13*10^{-15}eV)(3*10^8\frac{m}{s})}{10.2eV}=1.22*10^{-7}m[/tex]
Recall that [tex]1 m=10^9nm[/tex], So:
[tex]1.22*10^{-7}m*\frac{10^9nm}{1m}=122nm[/tex]
what is the approximate radius of the n = 1orbit of gold ( Z
=19 )?
Answer:
[tex]r=6.72\times 10^{-13}\ m[/tex]
Explanation:
Let r is the radius of the n = 1 orbit of the gold. According to Bohr's model, the radius of orbit is given by :
[tex]r=\dfrac{n^2h^2\epsilon_o}{Z\pi me^2}[/tex]
Where
n = number of orbit
h = Planck's constant
Z = atomic number (for gold, Z = 79)
m = mass of electron
e = charge on electron
[tex]r=\dfrac{(6.63\times 10^{-34})^2 \times 8.85\times 10^{-12}}{79\pi \times 9.1\times 10^{-31}\times (1.6\times 10^{-19})^2}[/tex]
[tex]r=6.72\times 10^{-13}\ m[/tex]
So, the radius of the n = 1 orbit of gold is [tex]6.72\times 10^{-13}\ m[/tex]. Hence, this is the required solution.
Write down an equation describing a sinusoidal traveling wave (in 1-D). Tell us (words and/or equations) what in your equation tells us the speed and direction of the wave? [Hint: you can google this if you do not know the answer. Be sure you understand it though!]
Answer:
Explanation:
The standard equation of the sinusoidal wave in one dimension is given by
[tex]y = A Sin\left ( \frac{2\pi }{\lambda }\left ( vt-x \right )+\phi \right )[/tex]
Here, A be the amplitude of the wave
λ be the wavelength of the wave
v be the velocity of the wave
Φ be the phase angle
x be the position of the wave
t be the time
this wave is travelling along positive direction of X axis
The frequency of wave is f which relates with velocity and wavelength as given below
v = f x λ
The relation between the time period and the frequency is
f = 1 / T.
What is the net electrostatic force (magnitude and direction) on a particle with charge +5μC situated at the apex of an equili!ateral triangle if each of the other corners contain identical charges of-6 μC and the length of a side of the triangle is 0.10 m?
Answer:
net electrostatic force = 46.76 N
and direction is vertical downward
Explanation:
given data
charge = +5μC
corners identical charges = -6 μC
length of side = 0.10 m
to find out
What is the net electrostatic force
solution
here apex is the vertex where the two sides of equal length meet
so upper point A have charge +5μC and lower both point B and C have charge -6 μC
so
force between +5μC and -6 μC is express as
force = [tex]k \frac{q1q2}{r^2}[/tex] ..............1
put here electrostatic constant k = 9 × [tex]10^{9}[/tex] Nm²/C² and q1 q2 is charge given and r is distance 0.10 m
so
force = [tex]9*10^{9} \frac{5*6*10^(-12)}{0.10^2}[/tex]
force = 27 N
so net force is vector addition of both force
force = [tex]\sqrt{x^{2}+x^{2}+2x^{2}cos60}[/tex]
here x is force 27 N
force = [tex]\sqrt{27^{2}+27^{2}+2(27)^{2}cos60}[/tex]
net electrostatic force = 46.76 N
and direction is vertical downward
A +17 nC point charge is placed at the origin, and a +8 nC charge is placed on the x axis at x=5m. At what position on the x axis is the net electric field zero? (Be careful to keep track of the direction of the electric field of each particle.) (Please explain how to do the problem.)
Answer:
2.97m
Explanation:
Hi!
To solve this problem we must consider the directions of the electric fields, since both charges are positive each field will point outwards form the sources meaning that the field will point yo the -x direction formpoints along the x-axis to the left of the charges and the field will point to the +x direction for points along the x-axis to the rigth of the sources, having said this, the only place where the fields point to oposite ditections is between them:
In the following diagram X is the 17nC charge and O the 8nC charge, the arrows represent the direction of their respective electric fields and d is an arbitrary distance measured from the origin.
|-------d------|-----5-d------|
X-----------> | <------------O
|---------------5--------------|
At d the electric fields of the charges are:
[tex]E_{X}=k\frac{17nC}{d^{2}} \\E_{O}=-k\frac{8nC}{(5-d)^{2}}[/tex]
And the total field is the sum of both. Since we are looking for a position d at which the total or net electric field is zero we must solve the following equation:
[tex]0=k(\frac{17}{d^{2}} - \frac{8}{(5-d)^2} )[/tex]
Reorganizing terms:
17*(5-d)^2-8*d^2=0
17*25-170d+9d^2=0
Solving for d, we find two solutions
d1 = 2.9655 d2=15.923
The second solution implies a distance outside the region between the charges for which the equations are no longer valid, since the direction of the 8nC field would be wrong.
So the solution is
d=2.97 m
A centrifuge in a medical laboratory rotates at an
angularspeed of 3800 rev/min. When switched off, it rotates through
46.0revolutions before coming to rest. Find the constant
angularacceleration of the centrifuge.
Answer:
- 273.77 rad/s^2
Explanation:
fo = 3800 rev/min = 3800 / 60 rps = 63.33 rps
f = 0
ωo = 2 π fo = 2 x 3.14 x 63.33 = 397.71 rad/s
ω = 2 π f = 0
θ = 46 revolutions = 46 x 2π radian = 288.88 radian
Let α be the angular acceleration of the centrifuge
Use third equation of motion for rotational motion
[tex]\omega^{2}=\omega _{0}^{2}+2\alpha \theta[/tex]
[tex]0^{2}=397.71^{2}+2 \times \alpha \times 288.88[/tex]
α = - 273.77 rad/s^2