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.
What is the magnitude (in N/C) and direction of an electric field that exerts a 4.00 x 10^−5 N upward force on a −2.00 µC charge? magnitude N/C
direction: upward, downward, to the left, to the right
Answer:
The magnitude of electric field is [tex]1.25\times10^{14}\ N/C[/tex] in downward.
Explanation:
Given that,
Force [tex]F= 4.00\times10^{-5}\ N[/tex]
Charge q= -2.00 μC
We know that,
Charge is negative, then the electric field in the opposite direction of the exerted force.
We need to calculate the magnitude of electric field
Using formula of electric force
[tex]F = qE[/tex]
[tex]E = \dfrac{F}{q}[/tex]
[tex]E=\dfrac{4.00\times10^{-5}}{-2.00\times1.6\times10^{-19}}[/tex]
[tex]E=-1.25\times10^{14}\ N/C[/tex]
Negative sign shows the opposite direction of electric force.
Hence, The magnitude of electric field is [tex]1.25\times10^{14}\ N/C[/tex] in downward.
Two infinite planes of charge lie parallel to each other and to the yz plane. One is at x = -5 m and has a surface charge density of σ = -2.6 µC/m^2. The other is at x = 3 m and has a surface charge density of σ = 5.8 µC/m^2. Find the electric field for the following locations. (a) x < -5 m
(b) -5 m < x < 3 m
(c) x > 3 m
Answer:
a) -180.7 kN/C
b) -474.3 kN/C
c) 180.7 kN/C
Explanation:
For infinite planes the electric field is constant on each side, and has a value of:
E = σ / (2 * e0) (on each side of the plate the field points in a different direction, the fields point towards positive charges and away from negative charges)
The plate at -5 m produces a field of:
E1 = 2.6*10^-6 / (2 * 8.85*10^-12) = 146.8 kN/C into the plate
The plate at 3 m:
E2 = 5.8*10^-6 / (2 * 8.85*10^-12) = 327.5 kN/C away from the plate
At x < -5 m the point is at the left of both fields
The field would be E = 146.8 - 327.5 = -180.7 kN/C
At -5 m < x < 3 m, the point is between the plates
E = -146.8 - 327.5 = -474.3 kN/C
At x > 3 m, the point is at the right of both plates
E = -146.8 + 327.5 = 180.7 kN/C
The electric fields at the given locations are calculated by using the formula E=σ/2ε₀, and considering that the fields always point toward negative charges. Hence, they vary depending on the position of the point relative to the planes of charge.
Explanation:The field contribution due to each individual infinite plane is E=σ/2ε₀, where ε₀ is the permittivity of the free space (8.85 x10⁻ⁱ² C²/Nm²). Now, we'll calculate the electric field for each given location:
(a) For x < -5 m: Here, the point sits to the left of both the planes. Both fields point to the right (towards negative charge), so they add up.(b) -5 m < x < 3 m: The point is between the planes. The field from the negative plane points to the right, and that of the positive plane also points to the right, hence they subtract. E = -2.6 / 2ε₀ - 5.8 / 2ε₀ .(c) x > 3 m: Here, the point is to the right of both planes. The fields from both planes point left (towards the negative charge), add up. E = -2.6 / 2ε₀ + 5.8 / 2ε₀ . Learn more about Electric Field here:
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A car travels for 30 minutes at 30 m/s due north. The car stops for 10 ninutes, then turns and travels for 20 minutes due south at 20 m/s. What is the average velocity of the car? What is the average speed of the car?
Answer:
Average speed 21.67 m/s
Average velocity=8.33 m/s
Explanation:
In order to solve this, We have to know the difference between speed and velocity.
Velocity is a vector value, so it means we have to specify the magnitud and direction, we take the displacement instead of the distance.
speed is an scalar value, we only care about the magnitud, we take in count the total distance.
first we have to get the total amount of time
Total time= 30min+10min+20min=60min
Total time is= 3600 seconds
the distance on the first trame is:
[tex]X1=v*t\\\\X1=30m/s*(30min)(60sec/1min)\\X1=54km[/tex]
the distance on the second trame is:
[tex]X2=v*t\\\\X2=20m/s*(20min)(60sec/1min)\\X2=24km[/tex]
The total distance is the sum of both values
Td=78km
the Displacement is the vectorial sum of them, because the second trame is opposite the first, we have to substract the second distance
D=54-24=30km
[tex]Speed(avg)=\frac{distance}{time} \\\\Speed(avg)=\frac{78000m}{3600} \\\\Speed(avg)=21.67 m/s[/tex]
[tex]Velocity(avg)=\frac{displacement}{time} \\\\Velocity(avg)=\frac{30000m}{3600} \\\\Velocity(avg)=8.33 m/s[/tex]
A skier is gliding along at 4.2 m/s on horizontal, frictionless snow. He suddenly starts down a 10° incline. His speed at the bottom is 18 m/s . a) What is the length of the incline?
Express your answer with the appropriate units.
b) How long does it take him to reach the bottom?
Express your answer with the appropriate units.
Answer:
a) 90m
b) 8.1s
Explanation:
The shortest way to finding the length of the incline is to apply an energetic analysis to determine the height of the incline. At the beginning, there is potential and kinetic energy that will turn into kinetic energy only:
[tex]mgh+\frac{1}{2}mv_{0}^{2} =\frac{1}{2}mv_{f}^{2}[/tex]
Before we input any information, let's solve for h:
[tex]mgh+\frac{1}{2}mv_{0}^{2} =\frac{1}{2}mv_{f}^{2}}\\\\m(gh+\frac{v_{0}^{2}}{2})=\frac{1}{2}mv_{f}^{2}}\\\\gh+\frac{v_{0}^{2}}{2}=\frac{v_{f}^{2}}{2}\\\\gh=\frac{1}{2}(v_{0}^{2}-v_{f}^{2})\\\\h=\frac{1}{2g}(v_{0}^{2}-v_{f}^{2})=\frac{1}{2*9.8\frac{m}{s^{2}}}(4.2\frac{m}{s}^{2}-18\frac{m}{s}^{2})=15.63m[/tex]
Using a sine formula we can solve for [tex]l[/tex]:
[tex]Sin(10)=\frac{h}{l}\\\\l=\frac{h}{Sin(10)}=\frac{15.63m}{0.17}=90m[/tex]
In order to find the time we will use the distance formula and final velocity formula as a system of equations to solve for t:
[tex]X=V_{0}t+\frac{at^2}{2}\\\\V_f=V_0+at[/tex]
Since the acceleration and time are both variables we will solve for acceleration in the final velocity formula and replace in the distance formula:
[tex]V_f=V_0+at\\V_f-V_0=at\\\frac{V_f-V_0}{t}=a\\\\l=V_0t+\frac{(\frac{V_f-V_0}{t})t^2}{2}\\l=V_0t+\frac{(V_f-V_0)t}{2}\\l=t(V_0+\frac{V_f-V_0}{2})\\\\t=\frac{l}{V_0+\frac{V_f-V_0}{2}}=\frac{90m}{4.2\frac{m}{s}+\frac{18\frac{m}{s}-4.2\frac{m}{s}}{2}}=8.1s[/tex]
Final answer:
In this question, we calculate the length of the incline and the time taken by a skier to reach the bottom based on given velocity and angle.
Explanation:
For part a: The length of the incline can be found using the principles of physics. Using the given information about the skier's initial and final speeds and the angle of the incline, you can calculate that the length of the incline is approximately 67.7 meters.
For part b: To determine the time it takes for the skier to reach the bottom, you can use the kinematic equations of motion along the incline. The time taken for the skier to reach the bottom is about 9.64 seconds.
A boat moves 60 kilometers east from point A to point B. There, it reverses direction and travels another 45 kilometers toward point A. What are the total distance and total displacement of the boat?
A. The total distance is 105 kilometers and the total displacement is 45 kilometers east.
B. The total distance is 60 kilometers and the total displacement is 60 kilometers east.
C. The total distance is 105 kilometers and the total displacement is 15 kilometers east.
D. The total distance is 60 kilometers and the total displacement is 45 kilometers east.
Answer:
C. The total distance is 105 kilometers and the total displacement is 15 kilometers east.
Explanation:
The main difference between distance and displacement is that: distance is a measure of the total length traveled along the road, the displacement only takes into account the length between the initial (departure) and final (arrival) position.
So the boat traveled 105 km, but the displacement between the start and finish point is only 15 km.
:)
Three negative point charges q1 =-5 nC, q2 = -2 nC and q3 = -5 nC lie along a vertical line. The charge q2 lies exactly between charge q1 and q3 which are 16 cm apart. Find the magnitude and direction of the electric field this combination of charges produces at a point P at a distance of 6 cm from the q2
Answer:
Ep= 5450N/C in direction (-x)
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
Equivalence
1nC= 10⁻⁹C
1cm= 10⁻²m
Graphic attached
The attached graph shows the field due to the charges q₁, q₂ y q₃ in the P (x=6, y=0):
As the charge q₁ ,q₂ , and q₃ are negative the field enter the charges.
E₁: Electric Field due to charge q₁.
E₂: Electric Field due to charge q₂.
E₃: Electric Field due to charge q₃.
Field calculation due to q₁ and q₃
Because q₁ = q₃ and d₁ = d₃, then, the magnitude of E₁ is equal to the magnitude of E₃
d₁=d₃=d
[tex]d=\sqrt{6^{2}+8^{2} } =10cm=0.1m[/tex]
q₁=q₃= -5nC= 5*10⁻⁹C
E₁ = E₃= k*q/d² = 9*10⁹*5*10⁻⁹/0.1² = 4500 N/C
E₁x = E₃x= - 4500*(6/10)= -2700 N/C
E₁y = -4500*(8/10)= -3600 N/C
E₃y= +4500*(8/10)=+3600N/C
Field calculation due to q₂
E₂x= k*q₂/d₂²= -9*10⁹*2*10⁻⁹/0.6²= -50 N/C
Magnitude and direction of the electric field in the point P (Ep)
Epx= E₁x+ E₂x+E₃x = -2700 N/C-50 N/C-2700 N/C= -5450N/C
Epy= E₁y+ E₃y= -3600 N/C+3600N = 0
Ep= 5450N/C in direction (-x)
Consider an equilateral triangle with sides measuring 1.0 m in length. At each point of the triangle is a +2.0 μC charge. Calculate the Coulomb force on each charge. Recall that forces are vectors and thus your answer will require a magnitude and direction for each of the three forces.
Answer:
Th magnitude of each Force will be [tex]=62.35\times10^{-3}\ \rm N[/tex]
Explanation:
Given:
Length of each side of the equilateral triangle, L=1 mMagnitude of each point charge [tex]Q=2\ \rm \mu C[/tex]Since all the charges are identical and distance between them is same so magnitude of the force between each charge is equal.
Let F be the force between the particles. According to Coulombs Law we have
[tex]F=\dfrac{kQ^2}{L^2}\\=\dfrac{9\times10^9\times (2\times10^{-6})^2}{1^2}\\F=36\times10^{-3}\ \rm N[/tex]
Now the the force on any charge by other two charges will be F and the angle between the two force is [tex]60^\circ[/tex]
Let [tex]F_{resultant}[/tex] be the force on nay charge by other two
By using vector Law of addition we have
[tex]F_{resultant}=\sqrt{(F^2+F^2+2F\times F \times cos60^\circ)}\\=\sqrt{3}F\\=\sqrt{3}\times36\times10^{-3}\ \rm N\\=62.35\times10^{-3}\ \rm N[/tex]
The angle made by the resultant vector will be
[tex]\tan\beta=\dfrac{F\sin60^\circ}{F+F\cos60^\circ}\\\beta=30^\circ[/tex]
Keisha looks out the window from a tall building at her friend Monique standing on the ground, 8.3 m away from the side of the building, as shown. If Keisha's line of sight makes a 30° angle with the side of the building, what is Keisha's height above the ground? Assume Monique is 1.5 m tall. A. 14 m B. 15 m C. 16 m D. 17 m
Answer:
Explanation:
GIVEN DATA:
Distance between keisha and her friend 8.3 m
angle made by keisha toside building 30 degree
height of her friend monique is 1.5 m
from the figure
[tex]\Delta ACB[/tex]
[tex]tan 30 = \frac{8.3}{h}[/tex]
[tex]h= \frac{8.3}{tan 30} = 14.376 m[/tex]
therefore
height of keisha is [tex]= h + 1.5 m[/tex]
= 14.376 + 1.5
[tex]= 15.876 \simeq 16 m[/tex]
therefore option c is correct
Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular aluminum plates, each 21 cm in diameter, separated by 1.0 cm . Part A How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume that value of 3× 10 6 N/C of the field causes a spark. Express your answer with the appropriate units. q max q m a x = nothing nothing SubmitRequest Answer Provide Feedback Next
Answer:
0.92 μC
Explanation:
In a parallel-plate capacitor, the electric field formed is equal to the charge density divited by the vacuum permisivity e0, as there are no dielectric between the plates. e0 is equal to 8.85*10^-12 C^2/Nm^2. The charge density is the total charge of each individual plate divided by its area. Then, the maximum charge allowed will be equal to:
[tex]E = \frac{o}{e_0} = \frac{Q}{Ae_0} \\ Q = E*A*e_0 = 3*10^6 N/C * (0.25*\pi *(0.21m)^2)*8.85*10^{-12}C^2/Nm^2 = 9.196 *10^{-7} C[/tex]
or 0.92 μC
A mass attached to a horizontal spring is stretched by 10 cm and released. It takes 0.2 sec for the mass to reach the equilibrium position. The mass is then stretched to 30 cm and released. How long does it take the mass to reach the equilibrium position?
Answer:
No change
Explanation:
The time period of an oscillating body depends on the mass of the body attached and the spring constant of the spring.
The time taken by the oscillating body to complete one vibration is called the time period of teh body.
As the time period does not depend on the amplitude of the oscillations so the time period does not change as the amplitude changes.
Thus, the time taken by the mass to reach to the equilibrium position remains same.
A moving van travels 10km North, then 4 km east, drops off some furniture and then drives 8 km south. (a) Sketch the path of the moving van. (2) How far is the van from its starting point? (Hint: it is the magnitude of the displacement vector).
Answer:
4.47 km
Explanation:
If we draw the path of the van then we get a shape with two exposed points A and D. If we draw a line from point D perpendicular to BA we get point E. This gives us a right angled triangle ADE.
From Pythagoras theorem
AD² = AE² + ED²
[tex]AD=\sqrt{AE^2+ED^2}\\\Rightarrow AD=\sqrt{2^2+4^2}\\\Rightarrow AD=\sqrt{20}\\\Rightarrow AD=4.47\ km[/tex]
Hence, the van is 4.47 km from its initial point
The horizontal bar rises at a constant rate of three hundred mm/s causing peg P to ride in the quarter circular slot. When coordinate y reaches 150mm, find the v and a vectors for P. Hint: for a circle with origin shown x2+y2=r2. The radius of the slot is 300 mm.
Answer:
Explanation:
Given
Horizontal bar rises with 300 mm/s
Let us take the horizontal component of P be
[tex]P_x=rcos\theta [/tex]
[tex]P_y=rsin\theta [/tex]
where [tex]\theta [/tex]is angle made by horizontal bar with x axis
Velocity at y=150 mm
[tex]150=300sin\theta [/tex]
thus [tex]\theta =30^{\circ}[/tex]
position of[tex]P_x=rcos\theta =300\cdot cos30=300\times \frac{\sqrt{3}}{2}[/tex]
[tex]P_x=259.80 mm[/tex]
[tex]P=259.80\hat{i}+150\hat{j}[/tex]
Velocity at this instant
[tex]u_x=-rsin\theta =300\times sin30=-150 mm/s[/tex]
[tex]u_y=rcos\theta =300\times cos30=259.80 mm/s[/tex]
Consider the resistances of short and open circuits. Fill in the blanks: The voltage across a short circuit will always be ______________ The voltage across an open circuit will always be ______________
Answer:
a) Zero
b) power source
Explanation:
According to Ohm's law, the voltage dropped in a resistance is proportional to the current flow and the resistor opposing to it.
[tex]V=I*R[/tex]
For the case of a short circuit, the resistance tends to zero, so the voltage will tend to zero too.
In the case of the open circuit, the resitance will tend to infinity, because we cannot obtain an infite voltage, it will be limited by the power source.
The voltage across a short circuit is always zero, due to the negligible resistance allowing easy current flow, whereas in an open circuit, which has infinite resistance preventing current flow, the voltage will be equal to the supply voltage.
Explanation:The question concerns the behaviors of electrical circuits specifically relating to short and open circuits. When considering the resistances and the resulting voltages across these types of circuits, it's crucial to understand two fundamentals:
The voltage across a short circuit will always be zero.The voltage across an open circuit will always be equal to the supply voltage.In a short circuit, a pathway exists with negligible resistance, allowing current to flow easily and resulting in virtually no voltage drop across the circuit. In contrast, an open circuit has infinite resistance because the path is broken, preventing any current from flowing; thus, the whole supply voltage appears across the open circuit.
Cosmic rays are highly energetic particles which can travel at great speeds through outer space. A typical speed for a cosmic ray is 1.29 x 10^8 m/s. What is this speed converted to km/hr (kilometers per hour)? A. 1.29 x 10^5 km/hr B. 35.8 km/hr C. 7.74 x 10^6 km/hr D. 4.64 x 10^8 km/hr
Answer:
Speed of the cosmic rays, [tex]v=4.64\times 10^8\ km/hr[/tex]
Explanation:
It is given that, Cosmic rays are highly energetic particles which can travel at great speeds through outer space.
The speed of the cosmic rays, [tex]v=1.29\times 10^8\ m/s[/tex]
We need to convert the speed of cosmic rays to kilometers per hour. We know that :
1 kilometers = 1000 meters
1 hour = 3600 seconds
[tex]v=1.29\times 10^8\ m/s=\dfrac{1.29\times 10^8\times (1/1000\ km)}{(1/3600\ h)}[/tex]
On solving the above expression,
[tex]v=464400000\ km/h[/tex]
or
[tex]v=4.64\times 10^8\ km/hr[/tex]
So, the speed of the cosmic rays is [tex]4.64\times 10^8\ km/hr[/tex]. Hence, this is the required solution.
Consider the following statements about Newton's 2nd law in general. Select all of the statements that are true. Note, there may be more than one. Pay attention to synonyms: velocity and change in position are the same thing.
1. A net force causes velocity to change
2. If an object has a velocity, then we can conclude that there is a net force on the object
3. Accelerations are caused by the presence of a net force
Answer:
1. True
2. False
3. True
Explanation:
Newton's 2nd law states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:
[tex]\sum F = ma[/tex] (1)
where
[tex]\sum F[/tex] is the net force on the object
m is its mass
a is the acceleration
Furthermore, we know that acceleration is defined as the rate of change of velocity:
[tex]a = \frac{dv}{dt}[/tex]
So let's now analyize the three statements:
1. A net force causes velocity to change: TRUE. Net force (means non-zero) causes a non-zero acceleration, which means that the velocity of the object must change.
2. If an object has a velocity, then we can conclude that there is a net force on the object: FALSE. The fact that the object has a velocity does not imply anything about its acceleration: in fact, if its velocity is constant, then its acceleration is zero, which would mean that the net force on the object is zero. So this statement is not necessarly true.
3. Accelerations are caused by the presence of a net force: TRUE. This is directly implied by eq.(1): the presence of the net force results in the object having a non-zero acceleration.
Light travels faster in warmer air. On a sunny day, the sun can heat a road and create a layer of hot air above it. Let's model this layer as a uniform one with a sharp boundary separating it from the cooler air above. Use this model to explain the formation of a mirage appearing like the shiny surface of a pool of water.
Answer:
Explanation:
If we assume there is a sharp boundary between the two masses of air, there will be a refraction. The refractive index of each medium will depend on the relative speeds of light.
n = c / v
If light travels faster in warmer air, it will have a lower refractive index
nh < nc
Snell's law of refraction relates angles of incidence and refracted with the indexes of refraction:
n1 * sin(θ1) = n2 * sin(θ2)
sin(θ2) = sin(θ1) * n1/n2
If blue light from the sky passing through the hot air will cross to the cold air, then
n1 = nh
n2 = nc
Then:
n1 < n2
So:
n1/n2 < 1
The refracted light will come into the cold air at angle θ2 wich will be smaller than θ1, so the light is bent upwards, creating the appearance of water in the distance, which is actually a mirror image of the sky.
The brakes are applied to a car traveling on a dry, level highway. A typical value for the magnitude of the car's acceleration is 4.60 m/s2. If the car's initial speed is 31.8 m/s, how long does it take to stop and how far does it travel, starting from the moment the brakes are applied?
Final answer:
The car takes approximately 6.91 seconds to stop and travels approximately 110.7 meters before coming to a stop.
Explanation:
The car's initial speed is 31.8 m/s and it decelerates at a rate of 4.60 m/s2. To find the time it takes to stop, we can use the equation:
Final velocity = Initial velocity + (Acceleration x Time)
Since the final velocity is 0 m/s when the car stops, we can rearrange the equation to solve for time:
Time = (Final velocity - Initial velocity) / Acceleration
Plugging in the values, we get:
Time = (0 m/s - 31.8 m/s) / -4.60 m/s2
Simplifying the equation, we find that it takes approximately 6.91 seconds for the car to stop.
To find the distance traveled, we can use the equation:
Distance = Initial velocity x Time + (0.5 x Acceleration x Time2)
Plugging in the values, we get:
Distance = 31.8 m/s x 6.91 s + (0.5 x -4.60 m/s2 x (6.91 s)2)
Simplifying the equation, we find that the car travels approximately 110.7 meters before it comes to a stop.
Three point charges are arranged along the x-axis. Charge q1 = +3.00 μC is at the origin, and charge q2 = -5.00 μC is at x = 0.200 m. Charge q3 = -8.00 μC. Where is q3 located if the net force on q1 is 7.00 N in the -x-direction?
Answer:
q₃ is located at X= - 0.144m
Explanation:
Coulomb law:
Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
F=K*q₁*q₂/d² Formula (1)
F: Electric force in Newtons (N)
K : Coulomb constant in N*m²/C²
q₁, q₂: Charges in Coulombs (C)
d: distance between the charges in meters(m)
Equivalence
1uC= 10⁻⁶C
Data
F=0.66 N
K=8.99x10⁹N*m²/C²
q₁ = +3.00 μC = +3.00 *10⁻⁶C
q₂= -5.00 μC = -5.00 *10⁻⁶C
q₃= -8.00 μC= -8.00 *10⁻⁶C
d₁₂= 0.2m: distance between q₁ and q₂
Fn=7 N: Net force on q₁
Calculation of the magnitude of the forces exerted on q₁
F₂₁ = K*q₁*q₂/d₁₂² = (8.99 * 10⁹ * 3.00 * 10⁻⁶ * 5.00 * 10⁻⁶)/(0.2)² = 3.37 N
F₃₁ = K*q₁*q₃/d₁₃² = (8.99 * 10⁹ * 3.00 * 10⁻⁶ * 8.00 * 10⁻⁶)/(d₁₃)² = 0.21576/(d₁₃)² N
Calculation of the distance d₁₃ between q₁ and q₃
In order for the net force to be negative on the x axis, the charge q₃ must be located on the x axis to the left of the charge q₁
F₂₁: It is a force of attraction and goes to the right(+)
F₃₁: It is a force of attraction and goes to the left(-)
-Fn₁ = F₂₁ - F₃₁
-7 = 3.37 - 0.21576/(d₁₃)²
3.37 + 7 = 0.21576/(d₁₃)²
10.37*(d₁₃)²=0.21576
(d₁₃)² = (0.215769 )/ (10.37)
(d₁₃)² = 20.8 * 10⁻³
[tex]d_{13} = \sqrt{20.8 * 10^{-3}} = 0.144 m[/tex]
d₁₃=0.144m
q₃ is located at X= - 0.144m
If the gravitational force between objects of equal mass is 2.30 x 10^‐8 N when the objects are 10.0 m apart, what is the mass of each object?
Answer:
Mass of each objects, m = 185.69 kg
Explanation:
The gravitational force between two equal masses is, [tex]F=2.3\times 10^{-8}\ N[/tex]
Separation between the masses, d = 10 m
Let m is the mass of each object. The gravitational force is given by :
[tex]F=G\dfrac{m^2}{d^2}[/tex]
[tex]m=\sqrt{\dfrac{F.d^2}{G}}[/tex]
[tex]m=\sqrt{\dfrac{2.3\times 10^{-8}\times (10)^2}{6.67\times 10^{-11}}}[/tex]
m = 185.69 kg
So, the mass of each objects is 185.69 kg. Hence, this is the required solution.
To find the mass of each object, we can use Newton's law of gravitation and solve an equation.
Explanation:To find the mass of each object, we can use Newton's law of gravitation, which states that the gravitational force between two objects is given by the equation: F = G * M₁ * M₂ / R². Rearranging the equation to solve for the mass, we have: M = F * R² / (G * M₂), where M is the mass of one object.
Using the given values, the gravitational force is 2.30 x 10^‐8 N and the distance between the objects is 10.0 m. Plugging these values into the equation, we have: M = (2.30 x 10^‐8 N) * (10.0 m)² / (6.67 × 10-¹¹ Nm²/kg² * M₂).
Since the masses of the two objects are equal, we can assume M₁ = M₂. Therefore, the mass of each object would be the same and can be calculated by solving the equation.
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You've been called in to investigate a construction accidentin
which the cable broke while a crane was lifting a 5300 kg
container. The steel cable is 2.0 cm indiameter and has a safety
rating of 50,000 N. The crane is designednot to exceed speeds of
3.0 m/s or accelerations of 1.0m/s2, and your tests find
that the crane is notdefective. What is the maximum tension the
cable?
Answer:
57300 N
Explanation:
The container has a mass of 5300 kg, the weight of the container is:
f = m * a
w = m * g
w = 5300 * 9.81 = 52000 N
However this container was moving with more acceleration, so dynamic loads appear.
w' = m * (g + a)
w' = 5300 * (9.81 + 1) = 57300 N
The rating for the cable was 50000 N
The maximum load was exceeded by:
57300 / 50000 - 1 = 14.6%
The skateboarder in the drawing starts down the left side of the ramp with an initial speed of 7.03 m/s. Neglect nonconservative forces, such as friction and air resistance, and find the height h of the highest point reached by the skateboarder on the right side of the ramp. (g = 9.80 m/s2)
Answer:
h=5.04m
Explanation:
At the bigining he has only kinetic energy because he is at the lowest point. When he is at the highest point, he is no longer moving because he will start moving downwards: all his kinetic energy transformed into potential gravitational energy:
[tex]E_k=E_p[/tex]
[tex]\frac{mv^2}{2}=mgh\\ h= \frac{v^2}{g}[/tex]
h=5.04m
Answer:
The height of the highest point reached by the skateboarder on the right side of the ramp is [tex]h=2.52m[/tex]
Explanation:
In this problem, we use the Energy Conservation Principle, as there are not nonconservative forces, the energy is constant and we will call it E. Then, we choose two different moments that will be advantageous for our analysis.
The first moment is the moment where the skateboarder is situated at the height 0 meters (meaning that the potential energy is zero), i.e, the energy is totally kinetic. We will call it K.
The second moment is the moment where the skateboarder is situated at the highest point reached, this means that all energy will be potential, because for an instant, the velocity of the skateboarder is zero (and in the following instant he will go back the way he came due to the pull of gravity). We will call it V. This is the point with the height that we want to calculate.
Therefore, we can equalize both energies, as they are constant and equal to E. We write
[tex]E=K=V=\frac{1}{2}mv^2 =mgh\Leftrightarrow \frac{1}{2}v^2=gh\Leftrightarrow h=\frac{v^2}{2g}[/tex]
Now, we can replace the given data, to obtain
[tex]h=\frac{7.03^2}{2*9.80}m=2.52m[/tex]
which is the outcome.
A 6.0 kg ball is dropped from a 10-m height. Its kinetic energy, just before it hits the ground is : 588 J
392 J
280 J
140 J
882 J
Answer:
588 J
Explanation:
mass of ball, m = 6 kg
Height from it dropped, h = 10 m
initial velocity, u = 0
acceleration due to gravity, g = 9.8 m/s^2
Let it hits the ground with velocity v.
Use third equation of motion
[tex]v^{2} = u^{2}+2as[/tex]
[tex]v^{2}=0^{2}+2\times 9.8\times 10[/tex]
v = 14 m/s
The formula for the kinetic energy is given by
[tex]K=\frac{1}{2}mv^{2}[/tex]
where, m is the mass of the ball and v be the velocity of the ball as it hits the ground
K = 0.5 x 6 x 14 x 14
K = 588 J
1.48 Assume that the air volume in a small automobile tire is constant and equal to the volume between two concentric cylinders 13 cm high with diameters of 33 cm and 52 cm. The air in the tire is initially at 25 °C and 202 kPa. Immediately after air is pumped into the tire, the temperature is 30 °C and the pressure is 303 kPa. What mass of air was added to the tire? What would be the air pressure after the air has cooled to a temperature of 0 °C?
Answer:
a) 0.018 kg
b) 262 kPa
Explanation:
The volume of the concentric cylinders would be:
V = π/4 * h * (D^2 - d^2)
V = π/4 * 13 * (52^2 - 33^2) = 16500 cm^3 = 0.0165 m^3
The state equation of gases:
p * V = m * R * T
Rearranging:
m = (p * V) / (R * T)
R is 287 J/(kg * K) for air
25 C = 298 K
m0 = 202000 * 0.0165 / (287 * 298) = 0.039 kg
After pumping more air the volume remains about the same, but temperature and pressure change.
30 C = 303 K
m1 = 303000 * 0.0165 / (287 * 303) = 0.057 kg
The mass that was added is
m1 - m0 = 0.057 - 0.039 = 0.018 kg
If that air is cooled to 0 C
0 C is 273 K
p = m * R * T / V
p = 0.057 * 278 * 273 / 0.0165 = 262000 Pa = 262 kPa
A cube with sides of area 48 cm^2 contains a 28.7 nanoCoulomb charge. Find the flux of the electric field through the surface of the cube in unis of Nm^2/C.
Please conceptually explain this question answer to me! Thanks!!
Answer:
The flux of the electric field through the surface is 3.24\times10^{3}\ Nm^/C[/tex].
Explanation:
Given that,
Area of cube = 48 cm²
Charge = 28.7 nC
We need to calculate the flux of the electric field through the surface
Using formula Gauss's law
The electric flux through any closed surface,
[tex]\phi =\dfrac{q}{\epsilon_{0}}[/tex]
Where, q = charge
Put the value into the formula
[tex]\phi=\dfrac{28.7\times10^{-9}}{8.85\times10^{-12}}[/tex]
[tex]\phi =3.24\times10^{3}\ Nm^/C[/tex]
Hence, The flux of the electric field through the surface is 3.24\times10^{3}\ Nm^/C[/tex].
Electric fields are vector quantities whose magnitudes are measured in units of volts/meter (V/m). Find the resultant electric field when there are two fields, E1and E2, where E1 is directed vertically upward and has magnitude 100 V/m and E2 is directed 45 degrees to the left of E1 and has magnitude 150 V/m. Use a graph to show vector drawing!
The resultant electric field (Er) is approximately 231.76 V/m, directed at an angle of about 62.76° to the left of the vertical E1 direction.
Find the resultant electric field when there are two fields, E1 and E2:
Step 1: Resolve the electric fields into their x and y components.
E1:
Ex1 = 0 V/m (because E1 is directed vertically upward)
Ey1 = 100 V/m
E2:
Since E2 is directed 45 degrees to the left of E1, we can use trigonometry to find its x and y components.
Ex2 = 150 V/m * sin(45°) = 106.07 V/m (directed to the left, so negative)
Ey2 = 150 V/m * cos(45°) = 106.07 V/m
Step 2: Add the x and y components of each electric field separately.
Σ Ex = Ex1 + Ex2 = 0 V/m - 106.07 V/m = -106.07 V/m
Σ Ey = Ey1 + Ey2 = 100 V/m + 106.07 V/m = 206.07 V/m
Step 3: Find the magnitude of the resultant electric field (Er) using the Pythagorean theorem.
Er = √(Σ Ex)² + (Σ Ey)²
Er = √(-106.07 V/m)² + (206.07 V/m)²
Er ≈ 231.76 V/m
Step 4: Find the direction of the resultant electric field using arctangent.
tan(θ) = Σ Ey / Σ Ex
tan(θ) = 206.07 V/m / -106.07 V/m
θ ≈ arctan(-1.94) ≈ -117.24°
Since the arctangent function only outputs values between -90° and 90°, we need to add 180° to get the angle within the range of 0° to 180°.
θ = -117.24° + 180° ≈ 62.76°
Therefore, the resultant electric field has a magnitude of approximately 231.76 V/m and is directed approximately 62.76° to the left of E1.
A solid steel ball is thrown directly downward, with an initial speed of 7.95 m/s, from the top of a building at a height of 29.8 m. How much time (in s) does it take before striking the ground?
Answer:
1.78 s
Explanation:
Initial speed of the ball = u = 7.95 m/s and is vertically downwards.
Acceleration due to gravity = g = 9.8 m/s/s , vertically downwards.
Height of the building h = 29.8 m (traversed downwards by the steel ball).
h = u t + 1/2 g t²
29.8 = 7.95 t + 0.5 (9.8) t²
⇒ 4.9 t² +7.95 t - 29.8 = 0
Using the quadratic formula , solve for t.
t= [tex]= \frac{-b\pm \sqrt{b^2-4\times a \times c}}{2\times a}[/tex]
t = [tex]\frac{-7.95 \pm \sqrt{7.95^2-4\times 4.9 \times (-29.8)}}{2\times 9.8}[/tex] = 1.78 s, -3.4 s
Since time does not have a negative value, time taken by the stone to reach the ground = t = 1.78 s
Final answer:
To calculate the time taken for a solid steel ball to strike the ground when thrown downward from a building, use the equation of motion and plug in the given values to find the time, which is about 1.7 seconds.
Explanation:
The time taken for the ball to strike the ground can be calculated using the equation of motion:
h = (1/2) * g * t2
Where h is the initial height of the ball, g is the acceleration due to gravity, and t is the time taken. Substituting the given values, we find that it takes approximately 1.7 seconds for the ball to strike the ground.
A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true anomaly of 150. If the speed of the meteoroid at that time is 2.23 km/s, calculate
(a) the eccentricity of the trajectory;
(b) the altitude at closest approach; and
(c) the speed at the closest approach.
Explanation:
Given that,
Distance = 402000 km
Speed = 2.23 m/s
Angle = 150
(a). We need to calculate the eccentricity of the trajectory
Using formula of eccentricity
[tex]\epsilon=\dfrac{v^2}{2}-\dfrac{\mu}{r}[/tex]
Put the value into the formula
[tex]\epsilon=\dfrac{2.23^2}{2}-\dfrac{398600}{402000}[/tex]
[tex]\epsilon=1.4949\ km^2/s^2[/tex]
We need to calculate the angular momentum
Using formula of the angular momentum
[tex]h^2=-\dfrac{1}{2}\dfrac{\mu^2}{\epsilon}(1-e^2)[/tex]
[tex]h^2=-\dfrac{1}{2}\dfrac{(398600)^2}{1.4949}(1-e^2)[/tex]
[tex]h^2=-5.3141\times10^{10}(1-e^2)[/tex]...(I)
The orbit equation is
[tex]h^2=\mu r(1+e\cos\theta)[/tex]
[tex]h^2=398600\times402000(1-+\cos150)[/tex]
[tex]h^2=16.02372\times10^{10}(1-e0.8660)[/tex]
[tex]h^2=16.02372\times10^{10}-e13.877\times10^{10}[/tex]....(II)
Equating the value of h²
[tex]-5.3141\times10^{10}(1-e^2)=16.02372\times10^{10}-e13.877\times10^{10}[/tex]
[tex]-5.3141+5.3141e^2=13.877e+16.02372[/tex]
[tex]5.3141e^2+13.877e-21.33782=0[/tex]
[tex]e = 0, 1.086[/tex]
(b). We need to calculate the altitude at closest approach
Put the value of e in equation (I)
[tex]h^2=16.02372\times10^{10}-1.086\times13.877\times10^{10}[/tex]
[tex]h^2=9.53298\times10^{9}\ km^4/s^2[/tex]
Now, using the formula of the altitude at closest
[tex]r_{perigee}=\dfrac{h^2}{\mu}\dfrac{1}{1+e}[/tex]
[tex]r_{perigee}=\dfrac{9.53298\times10^{9}}{398600}\dfrac{1}{1+1.086}[/tex]
[tex]r_{perigee}=11465\ km[/tex]
So, The altitude is
[tex]z_{perigee}=r_{perigee}-r_{earth}[/tex]
[tex]z_{perigee}=11465-6378[/tex]
[tex]z_{perigee}=5087\ km[/tex]
(c). We need to calculate the speed at the closest approach.
Using formula of speed
[tex]v_{perigee}=\dfrac{h}{r_{perigee}}[/tex]
[tex]v_{perigee}=\dfrac{\sqrt{9.53298\times10^{9}}}{11465}[/tex]
[tex]v_{perigee}=8.516\ km/s[/tex]
Hence, This is the required solution.
(a) The eccentricity of the meteoroid's trajectory is approximately -0.226. This value indicates the deviation of the orbit from a perfect circle, with negative eccentricity suggesting an elliptical orbit.
(b) At its closest approach, or perigee, the meteoroid is at an altitude of about 595,261 kilometers from the Earth's center. This is the point in its trajectory where it is closest to Earth.
(c) The speed of the meteoroid at its closest approach is approximately 3.69 kilometers per second. This velocity is characteristic of its orbital motion when it is nearest to Earth and is determined by the specific orbital energy and gravitational influences.
To calculate the eccentricity (e), altitude at closest approach (perigee), and speed at the closest approach, we can use the vis-viva equation and orbital mechanics. The vis-viva equation relates the specific orbital energy (ε) to the semi-major axis (a), eccentricity (e), and velocity (v) of an object in orbit:
ε = v²/2 - μ/a
Where:
ε = specific orbital energy
v = velocity of the object
μ = standard gravitational parameter of Earth (approximately 3.986 x 10^5 km³/s²)
a = semi-major axis of the trajectory
We are given the speed of the meteoroid at a distance of 402,000 km from the Earth's center, which we'll use as v. We'll also use μ for Earth's gravitational parameter.
First, let's find the semi-major axis (a):
ε = v²/2 - μ/a
Rearrange to solve for a:
a = μ / (2μ/v² - 1/v²)
a = (3.986 x 10^5 km³/s²) / (2 * (3.986 x 10^5 km³/s²) / (2.23 km/s)² - (1 / (2.23 km/s)²))
a ≈ 491,453 km
(a) The semi-major axis (a) is approximately 491,453 km.
Next, let's calculate the eccentricity (e) using the true anomaly (ν) and the semi-major axis:
e = cos(ν) - r/a
e = cos(150°) - (402,000 km) / (491,453 km)
e ≈ -0.226
(b) The eccentricity (e) is approximately -0.226.
Now, to find the altitude at closest approach (perigee), we need to calculate the perigee distance (rp) using the semi-major axis and eccentricity:
rp = a(1 - e)
rp = (491,453 km) * (1 - (-0.226))
rp ≈ 595,261 km
(c) The perigee distance is approximately 595,261 km.
To find the speed at closest approach (closest approach velocity), we can use the vis-viva equation again:
v = sqrt(μ * (2/r - 1/a))
Where r is the distance from the center of the Earth (rp at perigee).
v = sqrt((3.986 x 10^5 km³/s²) * (2 / (595,261 km) - 1 / (491,453 km)))
v ≈ 3.69 km/s
(c) The speed at closest approach is approximately 3.69 km/s.
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A secret agent skis off a slope inclined at θ = 30.2 degrees below horizontal at a speed of v0 = 20.4 m/s. He must clear a gorge, and the slope on the other side of the gorge is h = 11.7 m below the edge of the upper slope. Does he make it?
Answer:
He will make it if the gorge is no wider than 14.4 m
Explanation:
The secret agent follows a parabolic motion. We have the following data:
[tex]v_0 = 20.4 m/s[/tex] is the initial speed
[tex]\theta=30.2^{\circ}[/tex] below the horizontal is the initial angle
h = 11.7 m is the vertical distance covered by the agent before landing on the other side
Let's start by analyzing the vertical motion. The initial vertical velocity is
[tex]u_y = v_0 sin \theta = (20.4) (sin 30.2^{\circ})=10.3 m/s[/tex]
Where we have chosen downward as positive direction. Now we use the following equation:
[tex]h=u_y t + \frac{1}{2}gt^2[/tex]
where [tex]g=9.8 m/s^2[/tex] (acceleration of gravity) to find the time t at which the agent lands. Substituting the numbers:
[tex]11.7 = 10.3 t + 4.9 t^2\\4.9t^2 + 10.3t -11.7 = 0[/tex]
Which has two solutions: t = -2.92 s and t = 0.82 s. Since the negative solution is meaningless, we discard it, so the agent reaches the other side of the gorge after 0.82 s.
Now we want to find what is the maximum width of the gorge that allows the agent to safely land on the other side. For that, we need to calculate the horizontal velocity of the agent, which is constant during the motion:
[tex]u_x = u_0 cos \theta = (20.4)(cos 30.2^{\circ})=17.6 m/s[/tex]
So, the horizontal distance covered by the agent is
[tex]d = u_x t = (17.6)(0.82)=14.4 m[/tex]
So, the agent will land safely if the gorge is at most 14.4 m wide.
The secret agent will make it if the width of the gorge is less than 14.4 m.
The given parameters;
inclination of the slope, θ = 30.2⁰initial velocity, v = 20.4 m/sheight of the gorge, h = 11.7 mThe time to travel the vertical distance is calculated as follows;
[tex]h = v_0_yt + \frac{1}{2} gt^2\\\\11.7 = (20.4\times sin(30.4))t + (0.5\times 9.8)t^2\\\\11.7 = 10.32t + 4.9t^2\\\\4.9t^2 + 10.32t - 11.7 = 0\\\\solve \ the \ quadratic \ equation, \ using \ formula \ method;\\\\a = 4.9, \ b = 10.32, \ c = - 11.7\\\\t = \frac{-b \ \ + /- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = 0.82 \ s[/tex]
The horizontal distance traveled by the secret agent is calculated as;
[tex]X = V_0_x t\\\\X = (20.4 \times cos (30.4) \times 0.82\\\\X = 14.4 \ m[/tex]
Thus, we can conclude that the secret agent will make it if the width of the gorge is less than 14.4 m.
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A person starts to run around a square track with sides of 50 m starting in the bottom right corner and running counter-clockwise. Running at an average speed of 5 m/s, the jogger runs for 53.80 seconds. What is the difference between the magnitude of the person's average velocity and average speed in m/s? (please provide detailed explanation)
Answer:3.71 m/s
Explanation:
Given
square track with sides 50 m
average speed is 5 m/s
Total running time=53.8 s
Total distance traveled in this time[tex]=53.8\times 5=269 m[/tex]
i.e. Person has completed square track one time and another 69 m in second round
So displacement is 269-200=69 m
average velocity[tex]=\frac{Displacement}{time}[/tex]
[tex]=\frac{69}{53.8}=1.28 m/s[/tex]
Difference between average velocity and average speed is
5-1.28=3.71 m/s
Answer:
The difference in average speed and average velocity in terms of magnitude is 3.993 m/s
Solution:
As per the question:
The side of a square track, l = 50 m
Average speed of the runner, [tex]v_{avg} = 5 m/s[/tex]
Time taken, t = 53.80 s
Now,
The distance covered by the runner in this time:
s = [tex]v_{avg}t[/tex]
s = [tex]5\times 53.80[/tex]
s = 269 m
After covering a distance of 269 m, the person is at point A:
[tex]AQ^{2} = AR^{2} + QR^{2}[/tex]
where
AR = 19 m
QR = 50 m
Refer to fig 1.
As the runner starts from the bottom right, i.e., at Q and traveled 269 m.
After completion of 250 m , he will be at point R after one complete round and thus travels 19 m more to point A to cover 269 m.
Thus
[tex]AQ = \sqrt{19^{2} + 50^{2}} = 53.48 m[/tex]
where
AQ is the displacement
Hence,
Average velocity, v' = [tex]\frac{AQ}{t}[/tex]
v' = [tex]\frac{53.48}{53.80} = 1.007 m/s[/tex]
The difference in average speed and average velocity is:
[tex]v_{avg} - v' = 5 - 1.007 = 3.993 m/s[/tex]
There are two vectors à and b, with an angle between them. Then the dot product of them is A. a. b = ab cos B. a b = (a + b) cos C. a. b = ab sino D.. b = (a + b) sino
Answer:
[tex]a.b=|a||b|\ cos\theta[/tex]
Explanation:
Let a and b are two vectors such that [tex]\theta[/tex] is the angle between them. Dot product is also known as scalar product. It is used to find the angle between two vectors such that,
[tex]a.b=|a||b|\ cos\theta[/tex]
[tex]\theta[/tex] is the angle between a and b. It can be calculated as :
[tex]\theta=cos^{-1}(\dfrac{a.b}{|a||b|})[/tex]
[tex]|a|\ and\ |b|[/tex] are the magnitude of vectors a and b such that :
[tex]|a|=\sqrt{x^2+y^2+z^2}[/tex] if a = xi +yj +zk
and
[tex]|b|=\sqrt{p^2+q^2+r^2}[/tex] if a = pi +qj +rk
So, the correct option is (a). Hence, this is the required solution.