Answer:
The electric force increases by a factor of 4.
Explanation:
The electric force between two charges [tex]q_1[/tex] and [tex]q_2[/tex] separated a distance d can be calculated using Coulomb's Law:
[tex]F=\frac{kq_1q_2}{d^2}[/tex]
where [tex]k=9\times10^9Nm^2/C^2[/tex] is the Coulomb constant.
If the value of each charge is doubled, then we will have a force between them which is:
[tex]F'=\frac{k(2q_1)(2q_2)}{d^2}=4\frac{kq_1q_2}{d^2}=4F[/tex]
So the new force is 4 times larger than the original force.
Doubling the charge on each particle increases the electric force between them by a factor of 4.
Explanation:The force between two charged particles is given by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. So, if we denote the electric force as F, the charges as q1 and q2, and the distance as r, we can write Coulomb's law as F = k* q1*q2/r^2, where k is a constant.
Now if you double the charges (q1 and q2 become 2q1 and 2q2), and use these values in the formula, we get Fnew = k*(2q1) *(2q2)/r^2 = 4 * k*q1*q2/r^2 = 4F.
So, by doubling the charge on each particle, the electric force between them is multiplied by the factor of 4. So, the force increases fourfold.
Learn more about Coulomb's Law here:https://brainly.com/question/32002600
#SPJ3
A projectile is launched with a launch angle of 55° with respect to the horizontal direction and with initial speed 78 m/s. How long does it remain in flight?
Answer:
The projectile is in air for 13.03 seconds.
Explanation:
Given that,
Angle of projection of the projectile, [tex]\theta=55^{\circ}[/tex]
Initial speed of the projectile, u = 78 m/s
To find,
We need to find the time of flight of the projectile.
Solution,
It is defined as the time taken by the projectile when it is in air. It is given by the formula as :
[tex]T=\dfrac{2u\ \sin\theta}{g}[/tex]
[tex]T=\dfrac{2\times 78\ \sin(55)}{9.8}[/tex]
T = 13.03 seconds
So, the projectile is in air for 13.03 seconds.
An RLC circuit with a resistor of R\:=\:1 R = 1 k, capacitor ofC\:=\:3 C = 3 F, and inductor ofL\:=\:2 L = 2 H reaches a maximum current through the inductor of 7 mA. When all of the energy stored in the circuit is in the inductor, what is the magnetic energy stored? (Express your answer in micro-Joules) In the same circuit explained above, if all the energy is then transferred into the capacitor, what voltage drop will there be across the capacitor? (Express your answer in Volts to the hundredths place)
Answer:
Magnetic energy stored in the inductor when all of the energy in the circuit is in the inductor = 0.049 mJ
If all the energy is then transferred into the capacitor, the voltage drop across the capacitor = 0.00572 V = 0.01 V (expressed to the hundredths value)
Explanation:
In an RLC circuit with maximum current of 7mA = 0.007 A
When all of the energy is stored in the inductor, maximum current will flow through it,
Hence E = (1/2) LI²
L = inductance of the inductor = 2 H
E = (1/2) (2)(0.007²) = 0.000049 J = 0.049 mJ
When all the energy in the circuit is in the capacitor, this energy will be equal to the energy calculated above.
And for a capacitor, energy is given as
E = (1/2) CV²
E = 0.000049 J, C = 3 F, V = ?
0.000049 = (1/2)(3)(V²)
V = 0.00572 V = 0.01 V
Choose the correct statement: Group of answer choices A proton tends to go from a region of low potential to a region of high potential. The potential of a negatively charged conductor must be negative. None of the other responses is correct. If V
Answer:
TRUE. The potential of a negatively charged conductor must be negative
Explanation:
Let's examine each statement
The positively charged proton moves in the direction of the electric field, the power and the electric field are related
ΔU = - E ds
[tex]U_{f}[/tex] - U₀ = - E ds
E = (U₀ –U_{f}) / s
To have a positive electric field the initial potential must be greater than the final potential, so the proton moves from a greater potential to a smaller one.
This statement is FALSE
The second statement
The potential has the same sign as the elective charge.
This statement is TRUE
A proton tends to move to a region of higher potential is False, while The potential of a negatively charged conductor must be negative is True.
Electric Potential:The proton is a positively charged particle and moves in the direction of the electric field lines.
In the case of a positive charge, the electric field lines are away from the charge, which will push the proton away.
We know that the potential is inversely proportional to distance.
Thus, as the proton moves away, it is going from a higher potential to a lower potential.
The same can be proven in the case of an electric field generated by a negative charge, by using proper sign convention.
So the statement that a proton tends to go from a region of low potential to a region of high potential is FALSE
The potential has the same sign as the electric charge.
So the statement that the potential of a negatively charged conductor must be negative is TRUE
Learn more about electric potential:
https://brainly.com/question/13296426?referrer=searchResults
Point charges of 4.75 µC and −1.50 µC are placed 0.350 m apart. (Assume the negative charge is located to the right of the positive charge. Include the sign of the value in your answers.)
(a) Where can a third charge be placed so that the net force on it is zero?
_______ m to the right of the −1.50 µC charge
(b) What if both charges are positive?
_______ m to the right of the 1.50 µC charge
Answer:
Explanation:
Let the position of balance point be x distance from 1.5 µC on the right side . Balance point can not be on the left side or between the charges because in that case electric field by both the charges will be in the same direction .
For equilibrium of field
k x 4.75 µC / ( .35 +x )² - k x 1.5 µC / x ² = 0
4.75 / ( .35 +x )² = 1.5 / x²
( .35 +x )² / x² = 4.75 / 1.5
( .35 +x )² / x² = 3.167
( .35 +x ) / x = 1.78
.35 + x = 1.78 x
.78 x = .35
x = .45 m ( + ve ) to the right of 1.50 µC
b ) When both the charges are positive , balance point will lie in between them because , electric field will be in opposite direction .
For equilibrium of field
k x 4.75 µC / ( .35 - x )² = k x 1.5 µC / x ²
4.75 / ( .35 -x )² = 1.5 / x²
( .35 -x )² / x² = 4.75 / 1.5
( .35 -x )² / x² = 3.167
( .35 -x ) / x = 1.78
.35 -x = 1.78 x
2.78 x = .35
x = .126 m ( - ve ) to the right of 1.50 µC
Determine the minimum work per unit of heat transfer from the source reservoir that is required to drive a heat pump with thermal energy reservoirs at 460 K and 540 K.
Answer:
The minimum work per unit heat transfer will be 0.15.
Explanation:
We know the for a heat pump the coefficient of performance ([tex]C_{HP}[/tex]) is given by
[tex]C_{HP} = \dfrac{Q_{H}}{W_{in}}[/tex]
where, [tex]Q_{H}[/tex] is the magnitude of heat transfer between cyclic device and high-temperature medium at temperature [tex]T_{H}[/tex] and [tex]W_{in}[/tex] is the required input and is given by [tex]W_{in} = Q_{H} - Q_{L}[/tex], [tex]Q_{L}[/tex] being magnitude of heat transfer between cyclic device and low-temperature [tex]T_{L}[/tex]. Therefore, from above equation we can write,
[tex]&& \dfrac{Q_{H}}{W_{in}} = \dfrac{Q_{H}}{Q_{H} - Q_{L}} = \dfrac{1}{1 - \dfrac{Q_{L}}{Q_{H}}} = \dfrac{1}{1 - \dfrac{T_{L}}{T_{H}}}[/tex]
Given, [tex]T_{L} = 460 K[/tex] and [tex]T_{H} = 540 K[/tex]. So, the minimum work per unit heat transfer is given by
[tex]\dfrac{W_{in}}{Q_{H}} = \dfrac{T_{H} - T_{L}}{T_{H}} = \dfrac{540 - 460}{540} = 0.15[/tex]
The minimum work required to operate a heat pump between two thermal reservoirs at 460 K and 540 K is determined using the performance coefficient equation, Kp = Qh/W = Th/(Th – Tc). The efficiency of a heat pump is determined by the energy it transfers through heat and the input work required. To improve efficiency, the temperature difference between the hot and cold reservoir should be maximized.
Explanation:The question pertains to the work-power balance in a heat pump operating between two thermal reservoirs. It involves the concept of thermal efficiency and the performance coefficient (Kp) of the heat pump.
Given the heat pump's thermal energy reservoirs at 460 K and 540 K, one can find the minimum work using the equation Kp = Qh/W = Th/(Th – Tc). Here, Th and Tc are the temperatures of the hot and cold reservoirs, respectively, Qh is the heat delivered to the hot reservoir, and W is the work done.
Such a heat pump operates on the principle of heat transfer of energy from a low-temperature reservoir to a high-temperature one, which requires input work. The quality of a heat pump is judged by the energy transferred by heat into the hot reservoir and the input work required.
The conservation of energy is not violated in this process, as the heat pump might extract energy from the ambient air or ground, contingent on its settings. Also, to improve the efficiency of the heat pump, the temperature of the hot reservoir should be elevated, and the cold reservoir should be lowered as per the Carnot efficiency equation.
Learn more about Heat Pumps here:https://brainly.com/question/13198025
#SPJ11
Reasons for which petroleum is the chosen fuel for transportation in the U.S. include
1) its energy value per unit volume
2) its ability to quickly start or stop providing energy
3) its low amount of pollution produced per joule
Answer:
The Reasons for which petroleum is the chosen fuel for transportation in the U.S. include number 1) and 2).
Explanation:
The main reasons why oil is the most used fuel in the USA is because of its energy potential with a high combustion power and a better octane rating compared to other fuels. Thus, fossil fuel derived from oil has a high energy value per unit volume and an ability to quickly start or stop providing energy.
What is the critical angle θcritθcrittheta_crit for light propagating from a material with index of refraction of 1.50 to a material with index of refraction of 1.00?
Answer:
The critical angle is 41.8°.
Explanation:
The critical angle is [tex]\theta_1[/tex] for which the angle of refraction [tex]\theta_2[/tex] is 90°. From Snell's law we have
[tex]n_1sin (\theta_1 ) = n_2 sin(\theta_2)[/tex]
[tex]n_1sin (\theta_1 ) = n_2 sin(90^o)[/tex]
[tex]sin (\theta_1 ) = n_2/n_1,[/tex]
[tex]\theta_1 = sin^{-1}(\dfrac{n_2}{n_1} ).[/tex]
Putting in [tex]n_2 =1.00,[/tex] and [tex]n_1 = 1.50[/tex] we get:
[tex]\theta_1 = sin^{-1}(\dfrac{1.00}{1.500} ),[/tex]
[tex]\boxed{\theta_1 = 41.8^o}[/tex]
Thus, the critical angle is 41.8°.
Final answer:
The critical angle for light propagating from a material with an index of refraction of 1.50 to a material with an index of refraction of 1.00 is approximately 41.8 degrees.
Explanation:
The critical angle θcrit for light propagating from a material with an index of refraction (n1) of 1.50 to a material with an index of refraction (n2) of 1.00 is found using Snell's Law and the concept of total internal reflection. The formula to calculate the critical angle is θcrit = sin−1(n2/n1). Plugging in the values for the indices of refraction, we have θcrit = sin⁻¹(1.00/1.50).
Performing the calculation, we get:
θcrit = sin⁻¹(0.6667) ≈ 41.8°
Therefore, the critical angle for light traveling from a medium with an index of refraction of 1.50 to a medium with an index of refraction of 1.00 is approximately 41.8 degrees.
In a game of baseball, a player hits a high fly ball to the outfield. (a) Is there a point during the flight of the ball where its velocity is parallel to its acceleration? (b) Is there a point where the ball’s velocity is perpendicular to its acceleration? Explain in each case.
Final answer:
The velocity and acceleration vectors of a baseball can change during its flight. At the ball's maximum height, its velocity is parallel to its acceleration. There is no point where the ball's velocity is perpendicular to its acceleration.
Explanation:
In a game of baseball, the velocity and acceleration of a ball can change as it moves through the air. (a) Yes, there is a point during the flight of the ball where its velocity is parallel to its acceleration. This occurs when the ball reaches its maximum height. At this point, the ball stops moving upward and starts moving downward. Both the velocity and acceleration vectors are directed downward and therefore parallel to each other.
(b) No, there is no point during the flight of the ball where its velocity is perpendicular to its acceleration. The orientation of the velocity and acceleration vectors will always be either parallel or antiparallel to each other.
A polarized light is incident on several polarizing disks whose planes are parallel and centered on common axis. Suppose that the transmission axis of the first polarizer is rotated 20° relative to the axis of polarization of the incident light, and that the transmission axis of each additional analyzer is rotated 20° relative to the transmission axis of the previous one. What is the maximum number of polarizer needed (whole number), so the transmitted light through all polarizing sheets has an intensity that is equal at least 12% that striking the first polarizer?
Answer:
The number of polarizer needed so transmitted light has at least 12% intensity = 17
Explanation:
Given :
Angle between incident light and optic axis of polarizer = 20°
Given that, the transmission axis of each additional analyzer is rotated 20° relative to the transmission axis of the previous one
According to the malus law,
The intensity of the transmitted light passes through the polarizer is proportional to the square of the cosine of angle between the transmission axis to the optic axis.
⇒ [tex]I = I_{o} cos^{2} \alpha[/tex]
Where, [tex]I =[/tex] transmitted intensity through polarizer, [tex]I_{o} =[/tex] incident intensity of the light.
Given in question, all the time [tex]\alpha =[/tex] 20°
By calculation ∴ [tex]cos^{2} 20 = 0.883[/tex]
After 1st polarizer,
∴ [tex]I_{1} = 0.883I_{o}[/tex]
Now we need to multiply all the time 0.883 until we get 0.12 (relative 20° angle given in question)
After 17th polarizer we get 0.1205 ≅ 0.12
[tex]I_{17} = 0.883^{17} = 0.1205 \times 100 = 12[/tex]% [tex]I_{o}[/tex]
Means we get 12% intensity after 17th polarizing disk.
You are testing a new amusement park roller coaster with an empty car with a mass of 120 kg. One part of the track is a vertical loop with a radius of 12.0 m. At the bottom of the loop (point A) the car has a speed of 25.0 m/s and at the top of the loop (point B) it has speed of 8.00 m/s. As the car rolls from point A to point B, how much work is done by friction?
Answer:
[tex]W_f=-62460\ J[/tex]
Explanation:
Given that
mass of the car ,m = 120 kg
Radius ,R= 12 m
Speed at the bottom , u = 25 m/s
Speed at top ,v= 8 m/s
We know that
Work done by all the forces = Change in the kinetic energy
Work done by gravity + Work done by friction =Change in the kinetic energy
By taking point A as reference
[tex]m g \times (2R) + W_f=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2[/tex]
Now by putting the values in the above equation we get
[tex]120\times 10\times 2\times 12+ W_f=\dfrac{1}{2}\times 120\times 8^2-\dfrac{1}{2}\times 120\times 25^2[/tex]
[tex]W_f=\dfrac{1}{2}\times 120\times 8^2-\dfrac{1}{2}\times 120\times 25^2-120\times 10\times 2\times 12\ J[/tex]
[tex]W_f=-62460\ J[/tex]
Therefore the work done by friction force will be -62460 J.
A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the pressure is 1.0 bar, during which the pressure–volume relationship is pV = constant. Assume ideal gas behavior for the air. Determine the work and heat transfer, in kJ.
Answer:
The work and heat transfer for this process is = 270.588 kJ
Explanation:
Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k
The Pressure-Volume relation is PV = C
T = C for isothermal process
Calculating for the work done in isothermal process
W = P₁V₁ [tex]ln[\frac{P_{1} }{P_{2} }][/tex]
= mRT₁[tex]ln[\frac{P_{1} }{P_{2} }][/tex] [∵pV = mRT]
= (5) (0.287) (272.039) [tex]ln[\frac{2.0}{1.0}][/tex]
= 270.588 kJ
Since the process is isothermal, Internal energy change is zero
ΔU = [tex]mc_{v}(T_{2} - T_{1} ) = 0[/tex]
From 1st law of thermodynamics
Q = ΔU + W
= 0 + 270.588
= 270.588 kJ
To find the work and heat transfer in this process, we can apply the First Law of Thermodynamics. We can determine the work done using the equation pV = constant and calculate the heat transfer using the equation Q = CpdT.
To find the work and heat transfer in this process, we need to apply the First Law of Thermodynamics. The First Law states that the change in internal energy of a system is equal to the heat transfer into the system minus the work done by the system. In this case, since the process is isobaric (constant pressure), we can determine the work done by integrating the equation pV = constant. This will give us the equation W = p(Vf - Vi), where W is the work done, p is the pressure, and Vf and Vi are the final and initial volumes respectively.
Since the process is isobaric, the heat transfer can be calculated using the equation Q = CpdT, where Q is the heat transfer, Cp is the specific heat at constant pressure, and dT is the change in temperature. In this case, since the temperature change is not given explicitly, we can assume it to be the same as the change in internal energy, which gives us dT = dU/Cp.
By substituting the given values into the equations, we can calculate the work and heat transfer.
For more such questions on Thermodynamics, click on:
https://brainly.com/question/34045237
#SPJ3
A beam of protons is accelerated through a potential difference of 0.750 kVkV and then enters a uniform magnetic field traveling perpendicular to the field. You may want to review (Pages 641 - 643) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of electron motion in a microwave oven. Part A What magnitude of field is needed to bend these protons in a circular arc of diameter 1.80 mm ? Express your answer in tesla to three significant figures. BpBp = nothing TT SubmitRequest Answer Part B What magnetic field would be needed to produce a path with the same diameter if the particles were electrons having the same speed as the protons? Express your answer in tesla to three significant figures. BeBe = nothing TT SubmitRequest Answer
Answer:
Explanation:
A force is provided by the magnetic field which perpendicular to both the velocity of the charge and magnetic field
F = qvB where B is the magnetic field in tesla, q is charge and v is velocity
The potential energy is transferred into kinetic energy
PE = Vq = 1/2 mv²
v = √(2Vq/ m)
charge on the proton = 1.602 × 10 ⁻¹⁹ C and mass of a proton = 1.673 × 10⁻²⁷
v = √ (( 2 × 1.602 × 10 ⁻¹⁹ C × 0.75 × 10³V ) / (1.673 × 10⁻²⁷)) = √( 1.4363 × 10¹¹ ) = 3.79 × 10⁵ m/s
The force of magnetic field produces centripetal force
qvB = mv² /R
where R radius = 1.80mm = 0.0018 m / 2 = 0.0009 m
qvB = ( m / R ) × (2qV /m)
cancel the common terms
vB = 2V / R
3.79 × 10⁵ m/s × B = 2 × 0.75 × 10³V / 0.0009 = 1.667 × 10⁶
B = 1.667 × 10⁶ / 3.79 × 10⁵ m/s = 4.40 T
b) magnetic field needed for the electron
qvB = mv² /R where m is the mass of an electron = 9.11 × 10⁻³¹ Kg
qB = mv/R
qB = ( 9.11 × 10⁻³¹ Kg × 3.79 × 10⁵ m/s) / 0.0009
qB = 3.8363 × 10 ⁻²²
B = 3.8363 × 10 ⁻²² / 1.602 × 10⁻¹⁹ kg = 0.0239 T
calorimeter has aluminum inner cup of mass 120 gram containing 100 ml water at temperature 20 degree Celsius. Brass piece with mass 100 gram is heated to 100 degree Celsius, and then immersed in the calorimeter. Calculate the final temperature of the system. The specific heat of brass is 0.09 cal/(gramXdeg.C) . The additional necessary data are provided in the text.
Answer:
the final temperature of the system of the system is 25.32°C
Explanation:
We are not given specific capacity of water and aluminium, so we use their standard values, also we are not given the density if water so we assume the standard vale of density of water
The aluminium calorimeter has a mass Mc= 120g
Volume of water in calorimeter = 100ml at θc =20°C
Density of water is
1000Kg/m³ = 1g/mL
Then, density = mass/ volume
Mass=density ×volume
Mass=1g/mL×100mL
Mass=100gram
Then, the mass of water is
Mw = 100gram
Mass of brass is Mb = 100gram
The temperature of brass is θb=100°C
The specific heat capacity of water is Cw= 1cal/g°C
The specific heat capacity of aluminum Ca=0.22cal/g°C
We are looking for final temperature θf=?
Given that the specific heat capacity of brass is Cb=0.09Cal/g°C
Using the principle of calorimeter;
The principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.
So, the calorimeter gained heat and the liquid in the calorimeter gain heat too
Heat gain by calorimeter(Hc) = Mc•Ca•∆θ
Where Mc is mass of calorimeter,
Ca is Specific Heat capacity of Calorimeter
∆θ=(θf-θc)
Hc=Mc•Ca•∆θ
Hc=120•0.22•(θf-20)
Hc=26.4(θf-20)
Hc=26.4θf-528
Also, heat gain by the water
Heat gain by wayer(Hw) = Mw•Cw•∆θ
Where Mw is mass of water,
Cw is Specific Heat capacity of water
∆θ=(θf-θw),
Note that the temperature of the water and the calorimeter are the same at the beginning i.e. θc=θw=20°C
Hw=Mw•Cw•∆θ
Hw=100•1•(θf-20)
Hw=100(θf-20)
Hw=100θf-2000
Also heat loss by the brass is given by
heat loss by brass
Heat loss by brass(Hb)= Mb•Cb•∆θ
Where Mb is mass of brass,
Cb is Specific Heat capacity of brass
∆θ=(θb-θf)
Therefore,
Hb=Mb•Cb•∆θ
Hb=100•0.09•(100-θf)
Hb=9(100-θf)
Hb=900-9θf
Applying the principle of calorimeter
Heat gain = Heat loss
Hc+Hw=Hb
26.4θf-528 + 100θf-2000=900-9θf
26.4θf+100θf+9θf=900+2000+528
135.4θf=3428
Then, θf=3428/133.4
θf=25.32°C
The final temperature of the system is 25.32 degree Celsius.
Given data:
The mass of aluminum cup is, m = 120 g .
The mass of brass piece is, m' = 100 g.
The volume of water in aluminum cup is, V = 100 ml.
The temperature of water is, T = 20 degree Celsius.
The specific heat of brass is, [tex]c''=0.09 \;\rm cal/g ^\circ C[/tex].
The temperature of brass is, T'' = 100 degree Celsius.
The principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.
So, first calculate heat gain by calorimeter,
H = mc (T' - T)
Here, c is the specific heat of aluminum and its value is, [tex]0.22 \;\rm cal/g^\circ C[/tex]. Solving as,
[tex]H = 120 \times 0.22 (T' - 20)\\H = 26.4T' - 528[/tex]
And, heat gain by water is,
[tex]H'=m'c'(T'-T)[/tex]
Here, c' is the specific heat of water. ( c' =1 )
Solving as,
[tex]H'=100 \times 1 \times (T'-20)\\H' = 100T' -2000[/tex]
Now, heat loss by brass is,
[tex]H'' = m'c'' (T''-T')\\\\H'' = 100 \times 0.09 \times (100-T')\\\\H'' = 900-9T'[/tex]
Applying the principle of calorimeter
Heat gain = Heat loss
H + H' = H''
(26.4T' - 528) + (100T' - 2000) = (900 - 9T')
135.4 T' = 3428
T ' = 25.32 degree Celsius
Thus, we can conclude that the final temperature of the system is 25.32 degree Celsius.
Learn more about the calorimetry here:
https://brainly.com/question/16104165
A dockworker applies a constant horizontal force of 90.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 13.0 m in a time of 4.50 s .(a) What is the mass of the block of ice?
Answer:
The mass of the ice block is equal to 70.15 kg
Explanation:
The data for this exercise are as follows:
F=90 N
insignificant friction force
x=13 m
t=4.5 s
m=?
applying the equation of rectilinear motion we have:
x = xo + vot + at^2/2
where xo = initial distance =0
vo=initial velocity = 0
a is the acceleration
therefore the equation is:
x = at^2/2
Clearing a:
a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2
we use Newton's second law to calculate the mass of the ice block:
F=ma
m=F/a = 90/1.283=70.15 kg
70.31kg
Explanation:Step I: Consider Newton's second law of motion which states that;
∑F = m x a;
Where;
∑F = net force acting on a body
m = the mass of the body
a = acceleration due to the force on the body.
Step II: Now to the question;
Since frictional force is negligible and the only force acting on the block of ice is the applied force by the dockworker, the net force on the body (block of ice) is the constant horizontal force. i.e
∑F = 90.0N
Also;
the block starts from rest and moves a distance (s) of 13.0m in a time (t) of 4.50s. Here, we can get the acceleration in that duration of time using one
the equations of motion as follows;
s = ut + [tex]\frac{1}{2}[/tex]at² ------------------------------(ii)
Where;
s = distance covered = 13.0m
u = initial velocity = 0 [since the block starts from rest]
t = time taken to cover the distance = 4.50s
a = acceleration of the body.
Substitute these values into equation (ii) as follows;
13.0 = 0(4.5) + [tex]\frac{1}{2}[/tex](a)(4.50)²
13.0 = 0 + [tex]\frac{1}{2}[/tex](a)(20.25)
13.0 = [tex]\frac{1}{2}[/tex](a)(20.25)
13.0 = 10.125a
Solve for a;
a = [tex]\frac{13.0}{10.125}[/tex]
a = 1.28m/s²
Step III: Now substitute the values of a = 1.28m/s² and ∑F = 90.0N into equation (i) as follows;
90.0 = m x 1.28
m = [tex]\frac{90.0}{1.28}[/tex]
m = 70.31
Therefore, the mass of the block of ice is 70.31kg
What is the magnitude of the electric field at a distance 60 cm from the center of the sphere? The radius of the sphere 30 cm, the charge on the sphere is 1.56 × 10−5 C and the permittivity of a vacuum is 8.8542 × 10−12 C 2 /N · m2 . Answer in units of N/C.
Answer
[tex]3.9*10^{5}N/C[/tex]
Explanation:
from the expression for determing the magnetic field strength
[tex]E=\frac{q}{4\pi e_{0}d^2 }\\[/tex]
since the charge is given as
[tex]q=1.56*10^{-5}c\\[/tex]
and the distance is
d=60cm=0.6m
We can calculate the constant k
[tex]K=\frac{1}{4\pi e_{0}}\\ K=\frac{1}{4\pi *8.8542*10^{-12}}\\k=8.98*10^{9}\\[/tex]
if we substitute values, we arrive at
[tex]E=\frac{8.98*10^9 *1.56*10^{-5}}{0.6^2} \\E=389133.33\\E=3.9*10^{5}N/C[/tex]
Explanation:
Below is an attachment containing the solution.
Consider a platinum wire (σ= 1.0 × 107 Ω-1·m-1) with a cross-sectional area of 1 mm2 (similar to your connecting wires) and carrying 0.3 amperes of current, which is about what you get in a circuit with a round bulb and two batteries in series. Calculate the strength of the very small electric field required to drive this current through the wire.
Answer: 0.03 N/C
Explanation:
We use the current density formula to solve this question.
I/A = σ * E
Where,
I = current flowing in the circuit = 0.3 A
A = cross sectional area of the wire = 1 mm²
σ = resistivity of the wire = 1*10^7 Ω^-1·m^-1
E = strength of the electric field required
I/A = σ *E
E = I/(A * σ)
First we convert area from mm to m, so that, 1*10^-3 mm = 1*10^-6 m
E = 0.3 A / (1*10^-6 m * 1*10^7 Ω^-1·m^-1)
E = 0.3 A / 10 Ω^-1
E = 0.03 N/C
The required electric field to drive given current through the wire is 0.03 V/m.
To calculate the strength of the electric field required to drive a current of 0.3 amperes through a platinum wire, we can use the relationship between current density, electrical conductivity, and electric field.
The current density (J) is given by:
J = I / A
where I = 0.3 A and A = 1 mm² = 1 × 10⁻⁶ m².
Substituting the values, we get:
J = 0.3 A / (1 × 10⁻⁶ m²) = 3 × 10⁵A/m²
Next, we use Ohm's law in the form that relates current density to the electric field:
J = σE
where σ is the electrical conductivity of platinum, σ = 1.0 × 10⁷ S/m (S = 1/Ω), and E is the electric field.
Rearranging for the electric field, we get:
E = J / σ
Substituting the values, we get:
E = (3 × 10⁵ A/m²) / (1.0 × 10⁷ S/m) = 3 × 10⁻² V/m
Therefore, the strength of the very small electric field required to drive the current through the wire is 0.03 V/m.
There has long been an interest in using the vast quantities of thermal energy in the oceans to run heat engines. A heat engine needs a temperature difference, a hot side and a cold side. Conveniently, the ocean surface waters are warmer than the deep ocean waters. Suppose you build a floating power plant in the tropics where the surface water temperature is ≈≈ 30 ∘C∘C. This would be the hot reservoir of the engine. For the cold reservoir, water would be pumped up from the ocean bottom where it is always ≈≈ 5 ∘C∘C.What is the maximum possible efficiency of such a power plant? In%.
Answer:
[tex]\eta_{th} = 8.247\%[/tex]
Explanation:
The maximum possible efficiency for the floating power plant is given by the Carnot's Efficiency:
[tex]\eta_{th} = \left(1-\frac{278.15\,K}{303.15\,K} \right)\times 100\%[/tex]
[tex]\eta_{th} = 8.247\%[/tex]
What is the ideal banking angle (in degrees) for a gentle turn of 1.75 km radius on a highway with a 120 km/h speed limit (about 75 mi/h), assuming everyone travels at the limit?
Answer:
Ф = 4.98 °
Explanation:
Given:
R = 1.75 km = 1.75 * 10^3
g = 9.8 m/s^2
v = 120 km/h ==> 33.33 m/s
required:
Ф
solution:
The angle(Ф) is given by
Ф = arctan(v^2/R*g)
= arctan(33.33^2/9.8*1.75 * 10^3)
= 4.98 °
note:
there maybe error in calculation but method is correct.
An initially uncharged 3.67 μF capacitor and a 8.01 k Ω resistor are connected in series to a 1.50 V battery that has negligible internal resistance. What is the initial current in the circuit, expressed in milliamperes?
Answer:
Explanation:
Given an RC series circuit
Initially uncharged capacitor
C=3.67 μF
Resistor R=8.01 k Ω=8010 ohms
Battery EMF(V)=1.5V with negligible internal resistance.
The initial current in the circuit?
At the beginning the capacitor is uncharged and it has a 0V, so all the voltage appears at the resistor,
Now using ohms law
V=iR.
i=V/R
i=1.5/8010
i=0.000187A
1mA=10^-3A
Therefore, 1A = 1000mA
i=0.187 milliamps
The initial current in the circuit is 0.187 mA
Answer:
The initial current is 0.0187 mA.
Explanation:
Given that
capacitance is given as 3.67 x 10⁻⁶ F
resistance is given as 8010 Ω
voltage across the circuit is 1.5 V
Since the capacitor is initially uncharged, the capacitive reactance is zero.
From ohms law;
Voltage across the circuit is directly proportional to the opposition to the flow of current.
In these circumstances, as the battery only "sees"a resistor, the initial current can be found applying Ohm's law to the resistor, as follows:
[tex]V = I_{0}*R \\\\ I_{0} = \frac{V}{R} = \frac{1.50V}{8.011e3\Omega}\\ = 0.0187 mA[/tex]
The initial current (that will be diminishing as the capacitor charges), is 0.0187 mA.
An inventor claims to have devised a cyclical engine for use in space vehicles that operates with a nuclear-fuelgenerated energy source whose temperature is 920 R and a sink at 490 R that radiates waste heat to deep space. He also claims that this engine produces 4.5 hp while rejecting heat at a rate of 15,000 Btu/h. Is this claim valid?
Answer:
Valid
Explanation:
to determine if the claim is valid, we compare the efficiency of the device to that of a Carnot engine.
The following data were given
High Temperature = 920R,
Low temperature =490R
work=4.5hp =4.5*2544.5=11450.25Btu/h
low heat Ql heat= 15000Btu/h
High heat Qh=work +Ql=11450.25Btu/h+15000Btu/h=26450.25Btu/h
Next we cal calculate the efficiency of the Carnot engine
[tex]E_Carnot=1-\frac{T_L}{T_H}\\ E_Carnot=1-\frac{490}{920}\\ E_Carnot=0.467[/tex]
Hence the maximum efficiency at the given temperature is 47%
Next we calculate the efficiency of the device
[tex]E_device=\frac{work}{Q_H} \\E_device=\frac{11450.25}{26450.25} \\E_device=0.433\\[/tex]
which is 43%
since the maximum efficiency of 47% is not exceeded, we can conclude that the claim is valid
What is Pascal's Law? When gasses aren't constrained they tend to expand or contract, which depends on the pressure. You can squeeze air into tighter spaces by pressing the molecules together. In a contained incompressible fluid, any external pressure applied at one point will raise pressure equally at every point. An object placed in water is buoyed up with a force equal to the weight of the fluid the object displaces.
Answer: In a contained incompressible fluid, any external pressure applied at one point will raise pressure equally at every point.
Explanation: This law was put forward by Blaise Pascal, a French mathematician in 1648. Pascal's Law states that in a contained incompressible fluid, any external pressure applied at one point will raise pressure equally at every point. Pascal's law has been used in fluid mechanics for different applications these includes:
- the hydraulic jack used in automobile listings,
- most automobile braking systems,
-water towers, and dams.
URGENTLY NEED HELP WITH PHYSICS?
A vaulter is holding a horizontal 3.00-kg pole, 4.50 m long. His front arm lifts straight up on the pole, 0.750 from the end, and his back arm pushes straight down on the end of the pole. How much force does his back arm exert on the pole?
(Unit= N)
NEED HELP FAST
Answer:
58.8 N
Explanation:
Let 'F₁' be the force by front arm and 'F₂' be the force by back arm.
Given:
Mass of the rod (m) = 3.00 kg
Length of the pole (L) = 4.50 m
Acceleration due to gravity (g) = 9.8 m/s²
Distance of 'F₁' from one end of pole (d₁) = 0.750 m
'F₂' acts on the end. So, distance between 'F₁' and 'F₂' = 0.750 m
Now, weight of the pole acts at the center of pole.
Now, distance of center of pole from 'F₁' is given as:
d₂ = (L ÷ 2) - d₁
[tex]d_2=\frac{4.50}{2}-0.75=1.5\ m[/tex]
Now, as the pole is held horizontally straight, the moment about the point of application of force 'F₁' is zero for equilibrium of the pole.
So, Anticlockwise moment = clockwise moment
[tex]F_2\times d_1=mg\times d_2\\\\F_2=\frac{mg\times d_2}{d_1}[/tex]
Plug in the given values and solve for 'F₂'. This gives,
[tex]F_2=\frac{3.00\ kg\times 9.8\ m/s^2\times 1.5\ m}{0.75\ m}\\\\F_2=\frac{44.1}{0.75}=58.8\ N[/tex]
Therefore, the force exerted by the back arm on the pole is 58.8 N vertically down.
Suppose Person A is traveling on a spaceship going 50% of the speed of light. Person A measures the length of the spaceship to be 10 meters. How long would a Person B measure the spaceship if person B were on a planet as the spaceship passed by?
Answer:
L = 8.66 m
Explanation:
The length measured by the moving observer is related to the true length is given by
L = L₀ √1 - (v²/c²)
Where L₀ is the length of the spaceship as measured by person A, v is the speed of spaceship of person A and c is the speed of light c = 3.8x10ᵃ m/s
L = 10√1 - (0.5c)²/c²
L = 10√1 - (0.5*3.8x10ᵃ)²/3.8x10ᵃ ²
L = 8.66 m
Therefore, the Person B would measure the spaceship length to be 8.66 m
What pressure gradient along the streamline, dp/ds, is required to accelerate air at standard temperature and pressure in a horizontal pipe at a rate of 274 ft/s2?
Answer:
The answer to this question is attached.
The pressure gradient required to accelerate the air inside a pipe can be calculated using Euler's simplified equation for fluid motion. At standard temperature and pressure, you would need a pressure gradient of approximately -102.3 Pa/m to accelerate air at 274 ft/s².
Explanation:To calculate the pressure gradient, or dp/ds, required to accelerate the air inside a pipe, we can use Euler's equation for fluid motion, which in its simplified form is dp/ds = -ρa, where ρ represents the density of the fluid (in this case, air) and a is the acceleration. For standard temperature and pressure, the density of the air is approximately 1.225 kg/m³. It is important to note that the acceleration provided is in feet per second squared, so we would need to convert that to meters per second squared to match units with the density. The conversion factor is 1 ft/s² = 0.3048 m/s², resulting in an acceleration of approximately 83.515 m/s². By substituting these values into the equation, we get dp/ds = -(1.225 kg/m³)(83.515 m/s²) = -102.3 kg/(m·s²), or -102.3 Pa/m, since a Pascal (Pa) is equivalent to a kg/(m·s²).
Learn more about Pressure Gradient here:https://brainly.com/question/34952467
#SPJ3
You are a member of a citizen's committee investigating safety in the high school sports program. You are interested in knee damage to athletes participating in the long jump (sometimes called the broad jump). The coach has her best long jumper demonstrate the event for you. He runs down the track and, at the take-off point, jumps into the air at an angle of 30 degrees from the horizontal. He comes down in a sand pit at the same level as the track 26 feet away from his take-off point. With what velocity (both magnitude and direction) did he hit the ground?
The velocity at which the long jumper landed can be computed by separating the jump into horizontal and vertical components, using the initial take-off angle and the horizontal distance. The horizontal and vertical velocities are combined to find the magnitude of the total landing velocity, and the arctangent of their ratio gives the direction.
Explanation:To determine the velocity at which the long jumper landed, we need to consider the projectile motion of the jump. There are two components of velocity to consider: the horizontal (vx) and the vertical (vy) at the point of landing.
First, the horizontal velocity (vx) can be found by dividing the total horizontal distance by the time of flight (t). The equation of horizontal motion is:
vx = d / t, where d is the horizontal distance.
The vertical velocity (vy) of the jumper when he lands will be the same magnitude but opposite in direction to the vertical velocity at take-off due to symmetry in projectile motion, assuming no air resistance. The vertical velocity at take-off can be calculated using the initial jump angle (Ө) and the initial velocity (v0).
Using the initial angle of 30°, the vertical component of the initial velocity at take-off (v0y) is:
v0y = v0 * sin(Ө).
Since the vertical motion is subject only to acceleration due to gravity (g), the final vertical velocity at landing (vy) will be v0y (but in the opposite direction).
The total landing velocity (v) is then found by combining these two components using the Pythagorean theorem:
v = √(vx^2 + vy^2).
The direction of the landing velocity is given by the angle made with the horizontal, found using the arctangent of the ratio of vy over vx:
θ = arctan(vy / vx).
These calculations would provide the magnitude and direction of the landing velocity.
A short circuit is a circuit containing a path of very low resistance in parallel with some other part of the circuit. Discuss the effect of a short circuit on the portion of the circuit it parallels. Use a lamp with a frayed line cord as an example.
Answer:
note:
please find the attachment
Explanation:
A frayed cord means that the insulation of the cord is worn out which exposes the cord and when it touches any other conductor a short circuit happens which is basically the flow of very high amount of current.
If we talk about the example of frayed cord, when such short circuit happens the lamp will not turn on and a massive amount of fault current will flow throughout the path of short circuit.
Now what happens?
There are two possibilities;
If there exists a circuit breaker, then no substantial amount of damage would be done since the circuit breaker will trip itself and open the circuit that eventually halts the fault current.If there is no circuit breaker, then this massive fault current will keep on accumulating that causes excessive heating of the conductors and also producing sparks which most probably would end up in a fire hazard.The GPS (Global Positioning System) satellites are approximately 5.18 mm across and transmit two low-power signals, one of which is at 1575.42 MHz (in the UHF band). In a series of laboratory tests on the satellite, you put two 1575.42 MHz UHF transmitters at opposite ends of the satellite. These broadcast in phase uniformly in all directions. You measure the intensity at points on a circle that is several hundred meters in radius and centered on the satellite. You measure angles on this circle relative to a point that lies along the centerline of the satellite (that is, the perpendicular bisector of a line which extends from one transmitter to the other). At this point on the circle, the measured intensity is 2.00 W/m^2.
Required:
What is the intensity at a point on the circle at an angle of 4.45° from the centerline?
Answer:
intensity at a point on the circle at an angle of 4.45° from the center line is 1.77 W/m².
Explanation:
See attached picture.
The intensity at a point on the circle at an angle of 4.45° from the centerline is approximately 4.00 W/m².
To find the intensity at a point on the circle at an angle of 4.45° from the centerline, we need to consider the interference pattern created by the two transmitters.
1. Wavelength Calculation:
The wavelength [tex](\(\lambda\))[/tex] of the transmitted signal can be calculated using the frequency [tex](\(f\))[/tex] and the speed of light c.
[tex]\[ \lambda = \frac{c}{f} \][/tex]
Given:
- [tex]\( c = 3 \times 10^8 \)[/tex] m/s (speed of light)
- [tex]\( f = 1575.42 \times 10^6 \)[/tex] Hz (frequency)
[tex]\[ \lambda = \frac{3 \times 10^8}{1575.42 \times 10^6} \approx 0.1902 \text{ meters} \][/tex]
2. Phase Difference Calculation:
The phase difference [tex](\(\Delta \phi\))[/tex] between the two waves at a point on the circle is determined by the path difference [tex](\(\Delta L\))[/tex] travelled by the waves.
The path difference is given by:
[tex]\[ \Delta L = d \sin(\theta) \][/tex]
where:
- d is the distance between the two transmitters (5.18 mm = 0.00518 m)
- [tex]\(\theta\)[/tex] is the angle from the centerline (4.45°)
[tex]\[ \Delta L = 0.00518 \sin(4.45^\circ) \approx 0.00518 \sin(0.0776) \approx 0.00518 \times 0.0775 \approx 0.0004019 \text{ meters} \][/tex]
The phase difference is:
[tex]\[ \Delta \phi = \frac{2 \pi \Delta L}{\lambda} = \frac{2 \pi \times 0.0004019}{0.1902} \approx 0.0133 \text{ radians} \][/tex]
3. Intensity Calculation:
The intensity at a given point due to two sources interfering constructively or destructively can be found using:
[tex]\[ I = I_0 \left(1 + \cos(\Delta \phi)\right) \][/tex]
Here, [tex]\(I_0\)[/tex] is the intensity when both waves interfere constructively at the centerline [tex](\(\theta = 0^\circ\))[/tex], which is given as 2.00 W/m^2.
[tex]\[ I = 2.00 \left(1 + \cos(0.0133)\right) \][/tex]
Since [tex]\(\cos(0.0133) \approx 0.9999\)[/tex]:
[tex]\(\cos(0.0133) \approx 0.9999\)[/tex]
Thus, the intensity at a point on the circle at an angle of 4.45° from the centerline is approximately 4.00 W/m².
A piece of Nichrome wire has a radius of 6.5 104 m. It is used in a laboratory to make a heater that uses 4.00 102 W of power when connected to a voltage source of 120 V. Ignoring the effect of temperature on resistance, estimate the necessary length of wire.
Answer:
[tex]L=4.8*10^{17}m[/tex]
Explanation:
Given data
Power P=4.00×10²W
Radius r=6.5×10⁴m
Voltage V=120V
To find
Length of wire L
Solution
We know that resistance of wire can be obtained from
[tex]P=\frac{V_{2}}{R}\\ R=\frac{V_{2}}{P}[/tex]
We also know that R=pL/A solving the length noting that A=πr²
and using p=100×10⁻⁸Ω.m we find that
So
[tex]L=\frac{RA}{p}\\ L=\frac{\frac{(V^{2})}{P}(\pi r^{2}) }{p} \\L=\frac{V^{2}(\pi r^{2})}{pP}\\ L=\frac{(120V)^{2}\pi (6.5*10^{4} m)^{2} }{100*10^{-8}(4.00*10^{2} W) }\\ L=4.8*10^{17}m[/tex]
Final answer:
To calculate the length of Nichrome wire needed for a heating element using 400 W at 120 V, one determines the resistance and then uses it with the wire's cross-sectional area and resistivity. Approximately 41.45 meters of wire is required.
Explanation:
To estimate the necessary length of Nichrome wire for a laboratory heater, we start by calculating the resistance using the power and voltage supplied. The formula for power (P) in terms of voltage (V) and resistance (R) is P = V2 / R. Given P = 400 W and V = 120 V, the resistance can be calculated as follows:
R = V2 / P = (1202) / 400 = 36 Ω.
Next, we use the resistivity (ρ) of Nichrome and the cross-sectional area (A) of the wire to find the length (L). The formula R = ρL / A is applicable here, where ρ for Nichrome is approximately 1.10×10-6 Ω·m and A = πr2. The radius (r) given is 6.5×10-4 m, so:
A = π(6.5×10-4)2 = 1.33×10-6 m2.
Substituting the values into the formula, we get:
L = (R · A) / ρ = (36 · 1.33×10-6) / (1.10×10-6) ≠ 41.45 m.
Thus, approximately 41.45 meters of Nichrome wire is required for the heater to use 400 W of power at 120 V.
A 1.50-V battery supplies 0.204 W of power to a small flashlight. If the battery moves 8.33 1020 electrons between its terminals during the time the flashlight is in operation, how long was the flashlight used
Answer:
981.41 secs
Explanation:
Parameters given:
Voltage, V = 1.5V
Power, P = 0.204W
Number of electrons, n = 8.33 * 10^20
First, we calculate the current:
P = I*V
I = P/V
I = 0.204/1.5 = 0.136A
The total charge of 8.33 * 10^20 electrons is:
Q = 8.33 * 10^20 * 1.6023 * 10^(-19)
Q = 133.47 C
Current, I, is given as:
I = Q/t
=> t = Q/I
t = 133.47/0.136 = 981.41 secs
In physics, power is calculated using the formula P = IV, where P is power, I is current, and V is voltage. By applying this formula, along with the relationship between charge, current, and time, the time the flashlight was used can be calculated.
Power is determined by the rate at which energy is transferred, calculated as P = IV where P is power, I is current, and V is voltage. In this scenario, the power supplied is 0.204 W from a 1.50-V battery. The formula P = IV can be rearranged to find the current flowing, which is 0.136 A.
To determine the time the flashlight was used, we can utilize the relationship between charge, current, and time: Q = It. Given 8.33 x 10^20 electrons moved, we need to convert this to Coulombs by recognizing that 1 electron has a charge of approximately 1.6 x 10^-19 C. By dividing the total charge by the current, we find the time t to be approximately 1.23 x 10^19 seconds.
The radius of Venus (from the center to just above the atmosphere) is 6050 km (6050✕103 m), and its mass is 4.9✕1024 kg. An object is launched straight up from just above the atmosphere of Venus. (a) What initial speed is needed so that when the object is far from Venus its final speed is 8000 m/s? vinitial = m/s (b) What initial speed is needed so that when the object is far from Venus its final speed is 0 m/s? (This is called the "escape speed.") vescape = m/s
Answer:
(a) The initial speed required is 13116 m/s
(b) The escape speed is 10394 m/s
This problem involves the application of newtons laws of gravitation. The forces in action here are conservative and as a result mechanical energy is conserved.
The full calculation can be found in the attachment below.
Explanation:
In both parts (a) and (b) the energy conservation equation were used. Assumption was made that when the object is very far from the planet the distance from the planet's center approaches infinity and the gravitational potential energy approaches zero.
The calculation can be found below.