Answer:
T would be two times the temperature That it was before the double so if the initial temperature was 20 it would now be 40
Explanation:
If the car initially had twice the speed ΔT would be 4 times the original change in temperature of the breaks.
If we assume no heat loss to the surrounding. Then the loss in kinetic energy (KE) of the car after applying the breaks is completely converted into heat energy (Q) of the breaks.
Loss in KE = Gain in Heat Energy
ΔKE = ΔQ
Originally, ΔKE = [tex]\frac{1}{2}mv^{2}[/tex] = ΔQ
Now, ΔQ is directly proportional to change in temperature
[tex]\frac{1}{2}mv^{2}[/tex] = ΔQ ∝ ΔT
If the speed is doubled, v' = 2v, then
ΔKE = [tex]\frac{1}{2}mv^{'} ^{2}[/tex] = [tex]\frac{1}{2}m(4v^{2})[/tex] = 4 × [tex]\frac{1}{2}mv^{2}[/tex] = 4ΔQ ∝ 4ΔT
Hence, the temperature increases 4 times is speed is doubled.
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A radio-controlled car increases its kinetic energy from 3 J to 10 J over a distance of 1 m. What was the average net force on the car during this interval?
Answer:
Net force, F = 7 N
Explanation:
It is given that,
Initial kinetic energy of the car, [tex]K_i=3\ J[/tex]
Final kinetic energy of the car, [tex]K_f=10\ J[/tex]
Distance, d = 1 m
We need to find the average net force on the car during this interval. It is given by using the work energy theorem as :
[tex]W=\Delta K[/tex]
[tex]W=K_f-K_i[/tex]
Also, W = F.d d = distance and F = net force
[tex]F.d=K_f-K_i[/tex]
[tex]F=\dfrac{K_f-K_i}{d}[/tex]
[tex]F=\dfrac{10\ J-3\ J}{1\ m}[/tex]
F = 7 Newton
So, the average net force on the car during this interval is 7 newton. Hence, this is the required solution.
The average net force on a radio-controlled car that increased its kinetic energy from 3 J to 10 J over a distance of 1 m is calculated using the work-energy theorem and found to be 7 newtons.
The student's question asks for the calculation of the average net force on a radio-controlled car that increased its kinetic energy from 3 J to 10 J over a distance of 1 m. To find this, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.
The work done on the car, which equals the change in kinetic energy, is:
Work Done (W) = Final Kinetic Energy (KEfinal) - Initial Kinetic Energy (KEinitial)W = 10 J - 3 J = 7 JSince work is also defined as the force times the distance (W = F * d) and we know the work done (7 J) and the distance (1 m), we can solve for the force (F):
F = W / dF = 7 J / 1 m = 7 NTherefore, the average net force on the car during this interval was 7 newtons.
An object essentially at infinity is moved to a distance of 90 cm in front of a thin positive lens. In the process its image distance triples. Determine the focal length of the lens.
Answer:
67.5 cm
Explanation:
u = - 90 cm, v = 3 x u = 3 x 90 = 270 cm
let f be the focal length
Use lens equation
1 / f = 1 / v - 1 / u
1 / f = 1 / 270 + 1 / 90
1 / f = 4 / 270
f = 67.5 cm
To determine the focal length of the lens, we use the lens formula and set up an equation based on the given information. Solving for the image distance, we find that it is zero, indicating the image is formed at infinity. Therefore, the focal length of the lens is 90 cm.
Explanation:To determine the focal length of the lens, we can use the lens formula:
1/f = 1/v - 1/u
Where f is the focal length, v is the image distance, and u is the object distance.
Given that the image distance triples when the object is moved from infinity to 90 cm in front of the lens, we can set up the following equation:
1/f = 1/(3v) - 1/(90)
Multiplying through by 90*3v, we get:
90*3v/f = 270v - 90*3v
90*3v/f = 270v - 270v
90*3v/f = 0
Simplifying further, we find that: v = 0
When the image distance is zero, it means the image is formed at infinity, so the lens is focused at the focal point. Therefore, the focal length of the lens is 90 cm.
Using a rope that will snap if the tension in it exceeds 356 N, you need to lower a bundle of old roofing material weighing 478 N from a point 7.50 m above the ground. (a) What magnitude of the bundle's acceleration will put the rope on the verge of snapping? (b) At that acceleration, with what speed would the bundle hit the ground?
Answer:
a) 2.5 m/s²
b) 6.12 m/s
Explanation:
Tension of rope = T = 356N
Weight of material = W = 478 N
Distance from the ground = s = 7.5 m
Acceleration due to gravity = g = 9.81 m/s²
Mass of material = m = 478/9.81 = 48.72
Final velocity before the bundle hits the ground = v
Initial velocity = u = 0
Acceleration experienced by the material when being lowered = a
a) W-T = ma
⇒478-356 = 48.72×a
[tex]\Rightarrow \frac{122}{48.72} = a[/tex]
⇒a = 2.5 m/s²
∴ Acceleration achieved by the material is 2.5 m/s²
b) v²-u² = 2as
⇒v²-0 = 2×2.5×7.5
⇒v² = 37.5
⇒v = 6.12 m/s
∴ Velocity of the material before hitting the ground is 6.12 m/s
A fan blade is rotating with a constant angular acceleration of +12.8 rad/s2. At what point on the blade, as measured from the axis of rotation, does the magnitude of the tangential acceleration equal that of the acceleration due to gravity? (g = 9.80 m/s2)
Answer:
At r= 0.76 m measured from the axis of rotation, does the tangential acceleration equals the acceleration of the gravity.
Explanation:
at=g= 9.8 m/s²
α= 12.8 rad/s²
r= at/α
r= 0.76m
You push downward on a box at an angle 25° below the horizontal with a force of 750 N. If the box is on a flat horizontal surface for which the coefficient of static friction with the box is 0.66, what is the mass of the heaviest box you will be able to move?
Answer:
72.75 kg
Explanation:
[tex]F[/tex] = force applied on a box = 750 N
[tex]m[/tex] = mass of the box
[tex]N[/tex] = Normal force on the box
[tex]\mu _{s}[/tex] = Coefficient of static friction = 0.66
From the force diagram, force equation along the vertical direction is given as
[tex]N = F Sin25 + mg[/tex]
[tex]N = 750 Sin25 + mg[/tex] eq-1
Static frictional force is given as
[tex]f_{s} = \mu _{s} N[/tex]
using eq-1
[tex]f_{s} = \mu _{s} (750 Sin25 + mg)[/tex]
For the box to move,
[tex]F Cos25 = f_{s}[/tex]
[tex]750 Cos25 = \mu _{s} (750 Sin25 + mg)[/tex]
[tex]750 Cos25 = (0.66) (750 Sin25 + m (9.8))[/tex]
m = 72.75 kg
Three particles are fixed on an x axis. Particle 1 of charge q1 is at x = a, and particle 2 of charge q2 is at x = - a. If their net electrostatic force on particle 3 of charge + Q is to be zero, what must be the ratio q2/q1 when particle 3 is at x = 0.500a?
Answer:
9
Explanation:
As the net force on Q is zero, so the force on Q due to q1 is balanced by the force on Q due to q2.
Let the force on Q due to q1 is F1 and force on Q due to q2 is F2.
F1 = F2
K Q q1 / (0.5 a)^2 = K Q q2 / (1.5 a)^2
q1 / 0.25 = q2 / 2.25
q2 / q1 = 2.25 / 0.25
q2 / q1 = 9
Calculate the amount of charge travelling through a car battery when a current of 800.00 A is produced for 1.2 min.
The average current passing through a device is given by:
I = Q/Δt
I is the average current
Q is the amount of charge that has passed through the device
Δt is the amount of elapsed time
Given values:
I = 800.00A
Δt = 1.2min = 72s
Plug in the values and solve for Q:
800.00 = Q/72
Q = 57600C
A battery and a resistor are wired into a circuit. The resistor dissipates 0.50 W. Now two batteries, each identical to the original one, are connected in series with the resistor. Part A What power does it dissipate?
When a battery is connected in a series to a resistor, its power quadruples. So, the power that a resistor dissipating 0.5 W originally would dissipate when connected to two identical batteries in series would be 2.00 W.
Explanation:In the world of physics, power dissipated by a resistor in a circuit is governed by the equation P = I2R, where P is the power, I is the current, and R is the resistance. If the resistance stays unchanged and you double the voltage (by adding identical battery in series), the current through the circuit doubles as well. Thus, with the power quadrupling as a result of two times the current squared (since 22 = 4), the resistor would now dissipate 2.00 W of power.
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When two identical batteries are connected in series with a resistor originally dissipating 0.50 W, the power dissipation increases to 2.0 W. This is because the total voltage supplied to the circuit is doubled, leading to a quadratic increase in power dissipation.
To find the new power dissipation, we start by applying Ohm's law and the power formula. Initially, the power dissipated is given by P = IV. Since power can also be expressed as P = V²/R and the resistor dissipates 0.50 W:
P = V²/R = 0.50 W
Now, connecting two identical batteries in series doubles the voltage:
[tex]V_{new[/tex] = 2V
The new power dissipation is then:
[tex]P_{new[/tex] = ([tex]V_{new[/tex])² / R
[tex]P_{new[/tex] = (2V)² / R
[tex]P_{new[/tex] = 4V² / R
[tex]P_{new[/tex] = 4 * 0.50 W
[tex]P_{new[/tex] = 2.0 W
Therefore, the resistor now dissipates 2.0 W of power.
A ball is dropped 5 meters from rest. Using conservation of energy, what is the final velocity of the ball? a.) 5.39 m/s b.) 4.45 m/s c.) 9.90 m/s d.) 1.23 m/s e.) None of the above
Answer:
The final velocity of the ball, v = 9.9 m/s
Explanation:
It is given that,
A ball is dropped 5 meters from rest, h = 5 meters
We need to find the final velocity of the ball. It can be calculated using the conservation of energy as :
[tex]KE_i+KE_f=PE_i+PE_f[/tex]
Initial kinetic energy, [tex]KE_i=0\ (rest)[/tex]
Final kinetic energy, [tex]KE_f=\dfrac{1}{2}mv^2[/tex]
Initial potential energy, [tex]PE_i=mgh[/tex]
Final potential energy, [tex]PE_f=0[/tex] (at ground, h = 0)
[tex]\dfrac{1}{2}mv^2=mgh[/tex]
[tex]v=\sqrt{2gh}[/tex]
[tex]v=\sqrt{2\times 9.8\ m/s^2\times 5\ m}[/tex]
v = 9.89 m/s
or
v = 9.9 m/s
So, the final velocity of the ball is 9.9 m/s. Hence, this is the required solution.
An atomic nucleus with a rest mass energy of 7134 MeV is accelerated to 99.217% the speed of light relative to the lab frame. What is the Kinetic Energy in MeV of the atomic nucleus in the lab frame? Assume 4 significant figures.
Answer:
4.999 × 10⁴ MeV
Explanation:
Kinetic energy is related to the rest energy by the Lorentz transformation equation as shown below.
Kinetic energy in the lab frame K= ( γ -1 ) Eo
The Lorentz constant γ = [tex]1/\sqrt{1-v^2/c^2}[/tex]
So γ = [tex]1/\sqrt{1- (0.99217 c)^2/c^2}[/tex]= 8.0067
⇒ K = (8.0067-1)(7134) = 49986 MeV
=49,999 MeV ( 4 significant figures)
A collapsible plastic bag contains a glucose solution. The acceleration of gravity is 9.8 m/s2 . h Glucose solution If the average gauge pressure in the vein is 14800 Pa, what must be the minimum height of the bag in order to infuse glucose into the vein? Assume that the specific gravity of the solution is 1.03. Answer in units of m.
Answer:
The height of the bag is 1.466 m.
Explanation:
Given that,
Acceleration of gravity [tex]g=9.8\ m/s^2[/tex]
Pressure = 14800 Pa
Specific gravity = 1.03
We need to calculate the density
Using formula of specific gravity
[tex]\rho_{s}=\dfrac{\rho}{\rho_{w}}[/tex]
[tex]rho=\rho_{s}\times{\rho_{w}}[/tex]
Where, [tex]\rho[/tex] = density of solution
[tex]\rho_{w}[/tex] = density of water
Put the value in to the formula
[tex]\rho=1.03\times1000[/tex]
[tex]\rho=1030\ kg/m^3[/tex]
We need to calculate the height
Using formula of pressure
[tex]P=\rho gh[/tex]
[tex]h=\dfrac{P}{\rho g}[/tex]
Where, P = pressure
g = acceleration due to gravity
h = height
Put the value into the formula
[tex]h = \dfrac{14800}{1030\times9.8}[/tex]
[tex]h=1.466\ m[/tex]
Hence, The height of the bag is 1.466 m.
The minimum height of the bag of glucose solution must be 1.46 meters above the entry point into the vein for the fluid to just enter, calculated using the pressure equation P = hρg.
Explanation:In order to infuse the glucose into a vein, the pressure must be greater than the pressure in the vein. This can be achieved by finding the height of the fluid that corresponds to this greater pressure. Using the pressure equation P = hρg, where P is the pressure, h is the height of the fluid, ρ is the density of the fluid, and g is the acceleration due to gravity. In the given question we can solve for the height h:
h = P/(ρg)
Let's substitute the given values into the equation. The density ρ of the glucose solution is 1.03 times water's density since its specific gravity is given as 1.03. The density of water is 1000 kg/m³, so the density of the glucose solution is 1030 kg/m³. So, we get:
h = 14800 Pa / (1030 kg/m³ * 9.8 m/s²)
This calculates the minimum height of the collapsible plastic bag to be 1.46 meters (rounded to two decimal places) above the entry point into the vein for the fluid to just enter the vein.
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A combination of 0.250 kg of water at 20.0°C, 0.400 kg of aluminum at 26.0°C, and 0.100 kg of copper at 100°C is mixed in an insulated container and allowed to come to thermal equilibrium. Ignore any energy transfer to or from the container and determine the final temperature of the mixture.
Answer:
23.63 °C
Explanation:
[tex]m_{w}[/tex] = mass of water = 0.250 kg
[tex]T_{wi}[/tex] = initial temperature of water = 20.0 °C
[tex]c_{w}[/tex] = Specific heat of water = 4186 J/(kg °C)
[tex]m_{a}[/tex] = mass of aluminum = 0.400 kg
[tex]T_{ai}[/tex] = initial temperature of aluminum = 26.0 °C
[tex]c_{a}[/tex] = Specific heat of aluminum = 900 J/(kg °C)
[tex]m_{c}[/tex] = mass of copper = 0.100 kg
[tex]T_{ci}[/tex] = initial temperature of copper = 100 °C
[tex]c_{c}[/tex] = Specific heat of copper = 386 J/(kg °C)
[tex]T_{f}[/tex] = Final temperature of mixture
Using conservation of heat
[tex]m_{w}[/tex] [tex]c_{w}[/tex] ([tex]T_{f}[/tex] - [tex]c_{w}[/tex]) + [tex]m_{a}[/tex] [tex]c_{a}[/tex] ([tex]T_{f}[/tex] - [tex]T_{ai}[/tex] ) + [tex]m_{c}[/tex] [tex]c_{c}[/tex] ([tex]T_{f}[/tex] - [tex]T_{ci}[/tex] ) = 0
(0.250) (4186) ([tex]T_{f}[/tex] - 20) + (0.400) (900) ([tex]T_{f}[/tex] - 26) + (0.100) (386) ([tex]T_{f}[/tex] - 100) = 0
[tex]T_{f}[/tex] = 23.63 °C
You hang a 3 kg Halloween decoration 21 cm from the right end of a curtain rod. How much force does one of the curtain rod supports need to be able to hold? The supports are at the very ends of the rod, which has mass 2.1 kg and is 114 cm long. Answer in newtons.
Answer:
34.3 N and 15.7 N
Explanation:
[tex]F_{left}[/tex] = force on the left end of the rod
[tex]F_{right} [/tex] = force on the right end of the rod
M = mass of Halloween decoration = 3 kg
[tex]F_{h}[/tex] = weight of the Halloween decoration = Mg = 3 x 9.8 = 29.4 N
m = mass of rod = 2.1 kg
[tex]F_{r}[/tex] = weight of the rod = mg = 2.1 x 9.8 = 20.6 N
From the force diagram, using equilibrium of torque about A
[tex]F_{h}[/tex] (AB) + [tex]F_{r}[/tex] (AC) = [tex]F_{right}[/tex] (AD)
(29.4) (21) + (20.6) (57) = [tex]F_{right}[/tex] (114)
[tex]F_{right}[/tex] = 15.7 N
Using equilibrium of force along the vertical direction
[tex]F_{right}[/tex] + [tex]F_{leftt}[/tex] = [tex]F_{h}[/tex] + [tex]F_{r}[/tex]
15.7 + [tex]F_{leftt}[/tex] = 29.4 + 20.6
[tex]F_{leftt}[/tex] = 34.3 N
To find the force a curtain rod support must hold, calculate the torques due to the rod's weight and the decoration's weight, considering their distances from the support. Then, balance these torques with the force at the other support. The calculated force for one support is approximately 15.7 newtons.
To calculate the force one of the curtain rod supports must be able to hold, we must consider the torques about one end of the rod and include the mass of the rod and the Halloween decoration. The rod and the decoration together exert a torque about the pivot point due to gravity. The support must exert an equal and opposite torque to keep the system in equilibrium. We do this by summing the torques and setting them to zero, since the system is not rotating.
The total torque ( au) about the right support due to the rod and the decoration can be calculated using au = r imes F where r is the distance from the pivot and F is the force due to gravity (weight). The weight of the rod acts at its center of mass, which is at the midpoint of the rod, while the weight of the decoration acts at 21 cm from the right end. So, the distances from the pivot (right end) to the centers of mass are:
Rod: 57 cm (half of the total length of 114 cm)
Decoration: 21 cm
The weights are:
Rod: 2.1 kg imes 9.8 m/s2 = 20.58 N
Decoration: 3 kg imes 9.8 m/s2 = 29.4 N
The torques about the right support are:
Rod: 57 cm imes 20.58 N = 1173.06 N ext{cm}
Decoration: 21 cm imes 29.4 N = 617.4 N ext{cm}
Since the rod is in equilibrium, the torques must balance, meaning the support at the left end must provide an upwards force resulting in a torque that equals the sum of the other two torques. Let F be the force at the left end support. It must satisfy:
F imes 114 cm = 1173.06 N ext{cm} + 617.4 N ext{cm}
F = (1173.06 + 617.4) / 114
F ≈ 15.7 N
Therefore, the force one of the curtain rod supports needs to be able to hold is approximately 15.7 newtons.
An electric heater is used to boil small amounts of water and consists of a 18-Ω coil that is immersed directly in the water. It operates from a 120-V socket. How much time is required for the heater to raise the temperature of 0.71 kg of water from 15 °C to the normal boiling point? Take the specific heat capacity of water to be 4186 J/(kg·C°).
Answer:
Is required 5.26 min to raise the temperature.
Explanation:
R= 18 ohms
V= 120 volts
m= 0.71 kg
C= 4186 J/kg °C
T1= 15°C
T2= 100 °C
Q= m * C * (T2-T1)
Q= 252.62 *10³ J
V/R= I
I= 6.66 A
P= I² * R
P= 800 W = 800 J/s
P= Q/t
t= Q/P
t= 315.77 s = 5.26 min
If the radius of the earth was suddenly tripled and its mass doubled, the surface gravitational acceleration would become: A) 9.8 m/s^2
B) 2.18 m/s^2
C) 7.35 m/s^2
D) 14.7 m/s^2
E) 13.1 m/s^2
Answer:
B) 2.18 m/s²
Explanation:
M = Original mass of earth
R = Original radius of earth
g = original acceleration due to gravity of earth = 9.8 m/s²
M' = New mass of earth = 2 M
R' = New radius of earth = 3 R
g = original acceleration due to gravity of earth = 9.8 m/s²
Original acceleration due to gravity of earth is given as
[tex]g=\frac{GM}{R^{2}}[/tex]
[tex]9.8 =\frac{GM}{R^{2}}[/tex] eq-1
g' = new acceleration due to gravity of earth
New acceleration due to gravity of earth is given as
[tex]g' =\frac{GM'}{R'^{2}}[/tex]
[tex]g' =\frac{G(2M)}{(3R)^{2}}[/tex]
[tex]g'=\left ( \frac{2}{9} \right )\frac{GM}{R^{2}}[/tex]
Using eq-1
[tex]g'=\left ( \frac{2}{9} \right )(9.8)[/tex]
g' = 2.18 m/s²
Final answer:
When the Earth's radius is tripled and its mass is doubled, the new surface gravitational acceleration would be calculated using the gravitational acceleration formula and would result in a new acceleration of B) 2.18 m/s².
Explanation:
To find the new surface gravitational acceleration (g') when the radius of the Earth is tripled and its mass is doubled, we use the formula for gravitational acceleration:
g' = (G × new mass) / (new radius)²
Where:
G is the gravitational constantThe new mass is twice Earth's mass (2M)The new radius is three times Earth's radius (3R)Substituting the new mass and radius into the formula, we get:
g' = (G × 2M) / (3R)²
g' = (2G × M) / (9R²)
g' = (2/9) × (G × M / R²)
Since G × M / R² is the original gravitational acceleration (g) of Earth, which is 9.8 m/s², we can now substitute:
g' = (2/9) × 9.8 m/s²
g' = 2.18 m/s²
So the new surface gravitational acceleration would be 2.18 m/s², which matches option B.
A current density of 7.00 10-13 A/m2 exists in the atmosphere at a location where the electric field is 185 V/m. Calculate the electrical conductivity of the Earth's atmosphere in this region.
Answer:
electrical conductivity is given as
[tex]\sigma = 3.8 \times 10^{-15} [/tex]
Explanation:
For a current carrying conductor we will have relation between current density and electric field inside the conductor as
[tex]j = \sigma E[/tex]
here we have
current density = j
electric field = E
given that
[tex]j = 7.00 \times 10^{-13}[/tex]
[tex]E = 185 V/m[/tex]
now we will have
[tex]\sigma = \frac{j}{E}[/tex]
[tex]\sigma = \frac{7.00 \times 10^{-13}}{185}[/tex]
[tex]\sigma = 3.8 \times 10^{-15} [/tex]
A specimen of copper having a rectangular cross section 15.2 mm X19.1 mm (0.6 in. X 0.75 in.) is pulled in tension with 44500 N(10000 lbf) force, producing only elastic deformation. Calculate the resulting strain.
Answer:
The elastic deformation is 0.00131.
Explanation:
Given that,
Force F = 44500 N
Cross section [tex]A =15.2mm\times19.1\ mm[/tex]
We Calculate the stress
Using formula of stress
[tex]\sigma=\dfrac{F}{A}[/tex]
Where, F = force
A = area of cross section
Put the value into the formula
[tex]\sigma=\dfrac{44500}{15.2\times10^{-3}\times19.1\times10^{-3}}[/tex]
[tex]\sigma=153.27\times10^{6}\ N/m^2[/tex]
We need to calculate the strain
Using formula of strain
[tex]Y=\dfrac{\sigma}{\epsilon}[/tex]
[tex]epsilon=\dfrac{\sigma}{Y}[/tex]
Where,
[tex]\sigma[/tex]=stress
Y = young modulus of copper
Put the value into the formula
[tex]\epsilon=\dfrac{153.27\times10^{6}}{117\times10^{9}}[/tex]
[tex]\epsilon =0.00131[/tex]
Hence, The elastic deformation is 0.00131.
Two small plastic spheres between them has magnitude 0.22 N. What is the charge on each sphere is one the other? Explain whether it can be determined from the information give, whether the charges are positive or negative.
Answer:
Explanation:
To find the force between the two charges, we required the charges on each object and the distance between the two objects.
In the question, the distance between two charges is not given so we cannot find the charge on each sphere.
If the force is attractive, then the charges are opposite in nature and if the force is repulsive then the charges are same in nature.
Suppose 150W of heat flows by conduction from the blood capillaries beneath the skin to the body's surface area of 1.5 m^2. If the temperature difference is 0.50 ºC, estimate the average distance of capillaries below the skin surface. Use 'mm' as your units.
Answer:
[tex]l =1mm[/tex]
Explanation:
Given:
Rate of heat transfer, P = 150 W
Body surface Area, A = 1.5 m²
Temperature difference, ΔT = 0.50°C
Also,
The rate of heat transfer, P is given as:
[tex]P = \frac{kA\Delta T}{l}[/tex]
Where,
l =length of material (or here it isaverage distance of the capillaries below the skin surface)
k = Thermal conductivity
Here the transfer of heat is through the skin. Thus, k for human tissue is given as 0.2
substituting the values in the above equation, we get
[tex]150 = \frac{0.2\times 1.5\times 0.50}{l}[/tex]
or
[tex]l = \frac{0.2\times 1.5\times 0.50}{150}[/tex]
or
[tex]l = 1\times 10^{-3}m=1mm[/tex]
Using the formula for thermal conduction, it is estimated that the capillaries lie about 3 mm below the skin surface. This is a simplification, actual distances can vary based on specific factors.
Explanation:To estimate the average distance between the capillaries and skin surface, we can use the formula for thermal conduction, which states that heat flow equals the thermal conductivity constant (k) times the surface area of the skin (A) times the temperature difference (ΔT), divided by the thickness of the skin (d), or Q = k*A*ΔT/d.
Assuming that human skin has a thermal conductivity similar to water (k~0.6 W/mK), we can rearrange the formula to solve for d: d = k*A*ΔT/Q. Plugging in the given values, we get d = (0.6 W/mK * 1.5 m^2 * 0.50 ºC) / 150 W, which simplifies to approximately 0.003 m, or 3 mm.
This estimate suggests that, on average, the capillaries lie about 3 mm below the skin surface. However, this is a simplification and actual distances can vary based on factors such as the specific region of the body, individual physiology, and more. Thermal conduction is just one mechanism of heat transfer in the body, along with convection and radiation.
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An electron with charge −e and mass m moves in a circular orbit of radius r around a nucleus of charge Ze, where Z is the atomic number of the nucleus. Ignore the gravitational force between the electron and the nucleus. Find an expression in terms of these quantities for the speed of the electron in this orbit. (Use any variable or symbol stated above along with the following as necessary: k for Coulomb's constant.)
Answer:
[tex]v=\sqrt{\frac{kZe^2}{mr}}[/tex]
Explanation:
The electrostatic attraction between the nucleus and the electron is given by:
[tex]F=k\frac{(e)(Ze)}{r^2}=k\frac{Ze^2}{r^2}[/tex] (1)
where
k is the Coulomb's constant
Ze is the charge of the nucleus
e is the charge of the electron
r is the distance between the electron and the nucleus
This electrostatic attraction provides the centripetal force that keeps the electron in circular motion, which is given by:
[tex]F=m\frac{v^2}{r}[/tex] (2)
where
m is the mass of the electron
v is the speed of the electron
Combining the two equations (1) and (2), we find
[tex]k\frac{Ze^2}{r^2}=m\frac{v^2}{r}[/tex]
And solving for v, we find an expression for the speed of the electron:
[tex]v=\sqrt{\frac{kZe^2}{mr}}[/tex]
The speed of an electron in a circular orbit around a nucleus is determined by equating the Coulomb force and the centripetal force, resulting in the expression v = sqrt(kZe^2/mr).
Explanation:The speed v of an electron in a circular orbit around a nucleus can be found by equating the electrostatic force to the centripetal force required for circular motion. The electrostatic force, due to the Coulomb's interaction, between the electron and the nucleus is given by F = k(Ze)(-e)/r^2, where k is Coulomb's constant, Z is the atomic number, e is the magnitude of the charge of an electron, and r is the radius of the orbit. On the other hand, the centripetal force needed to keep the electron in circular motion is F = mv^2/r where m is the mass of the electron and v is its speed.
Setting the two expressions equal gives the equation for the electron's speed v:
k(Ze)(-e)/r^2 = mv^2/r
Solving for v results in the expression:
v = sqrt(kZe^2/mr)
This equation shows that the electron's speed in its orbit depends on the atomic number Z, Coulomb's constant k, the electron's mass m, and the orbit radius r.
A metallic sheet has a large number of slits, 5.0 mm wide and 20 cm apart, and is used as a diffraction grating for microwaves. A wide parallel beam of microwaves is incident normally on the sheet. What is the smallest microwave frequency for which only the central maximum occurs? (c = 3.00 × 108 m/s)
Answer:
The smallest microwave frequency is 1.5 GHz.
Explanation:
Given that,
Distance d= 20 cm
We need to calculate the wavelength
Using formula for maximum diffraction
[tex]d\sin\theta=\lambda[/tex]
For maximum, [tex]\sin\thea=1[/tex]
[tex]d=\lambda[/tex]
[tex]\lambda=20 cm=0.2 m[/tex]
We need to calculate the frequency
[tex]f =\dfrac{c}{\lambda}[/tex]
[tex]f=\dfrac{3\times10^{8}}{0.2}[/tex]
[tex]f=1.5\times10^{9}\ Hz[/tex]
[tex]f=1.5 GHz[/tex]
Hence, The smallest microwave frequency is 1.5 GHz.
A voltaic cell with an aqueous electrolyte is based on the reaction between Cd2+(aq) and Mg(s), producing Cd(s) and Mg2+(aq). Write half-reactions for the anode and cathode and then write a balanced cell reaction. Please include the states of matter in the equations.
Answer : The balanced two-half reactions will be,
Oxidation half reaction (anode) : [tex]Mg(s)\rightarrow Mg^{2+}(aq)+2e^-[/tex]
Reduction half reaction (cathode) : [tex]Cd^{2+}(aq)+2e^-\rightarrow Cd(s)[/tex]
Thus the overall reaction will be,
[tex]Mg(s)+Cd^{2+}(aq)\rightarrow Mg^{2+}(aq)+Cd(s)[/tex]
Explanation :
Voltaic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the galvanic cell or electrochemical cell.
The given redox reaction occurs between the magnesium and cadmium.
In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.
The balanced two-half reactions will be,
Oxidation half reaction (anode) : [tex]Mg(s)\rightarrow Mg^{2+}(aq)+2e^-[/tex]
Reduction half reaction (cathode) : [tex]Cd^{2+}(aq)+2e^-\rightarrow Cd(s)[/tex]
Thus the overall reaction will be,
[tex]Mg(s)+Cd^{2+}(aq)\rightarrow Mg^{2+}(aq)+Cd(s)[/tex]
half-reactions
cathode : Cd²⁺ (aq) + 2e⁻ ---> Cd (s)
anode : Mg (s) → Mg²⁺ (aq) + 2e−
a balanced cell reaction
Cd²⁺(aq) + Mg(s)→ Cd(s) + Mg²⁺ (aq)
Further explanationCell potential (E °) is the potential difference between the two electrodes in an electrochemical cell.
Electric current moves from a high potential pole to a low potential, so the cell potential is the difference between an electrode that has a high electrode potential (cathode) and an electrode that has a low electrode potential (anode)
[tex] \large {\boxed {\bold {E ^ osel = E ^ ocatode -E ^ oanode}}} [/tex]
or:
E ° cell = E ° reduction-E ° oxidation
(At the cathode the reduction reaction occurs, the anode oxidation reaction occurs)
The value of E cells uses a reference electrode which is used as a comparison called the Standard Electrode which is the hydrogen-platinum electrode
In reaction:
Cd²⁺ + Mg → Cd + Mg²⁺
half-reactions
at the cathode (reduction reaction) Cd²⁺ (aq) + 2e⁻ ---> Cd (s) at the anode (oxidation reaction) Mg (s) → Mg²⁺ (aq) + 2e−a balanced cell reaction
Cd²⁺(aq) + Mg(s)→ Cd(s) + Mg²⁺ (aq)
Learn moreThe standard cell potential
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At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 µT northward, and the electric field is about 100 N/C downward in fair weather. Find the gravitational, electric, and magnetic forces on an electron in this environment, assuming the electron has an instantaneous velocity of 5.90 ✕ 106 m/s directed to the east in this environment.
Answer:
[tex]F_g = 8.9 \times 10^{-30} N[/tex]
[tex]F_e = 1.6 \times 10^{-17} N[/tex]
[tex]F_m = 4.72 \times 10^{-17} N[/tex]
Explanation:
For gravitational force we know that
F = mg
now we have
[tex]m = 9.1 \times 10^{-31} kg[/tex]
[tex]F_g = (9.1 \times 10^{-31})(9.8)[/tex]
[tex]F_g = 8.9 \times 10^{-30} N[/tex]
Now electrostatic force
[tex]F = qE[/tex]
here we have
[tex]F = (1.6 \times 10^{-19})(100)[/tex]
[tex]F = 1.6 \times 10^{-17} N[/tex]
Now magnetic force on it is given by
[tex]F_{m} = qvB[/tex]
[tex]F_m = (1.6 \times 10^{-19})(5.90 \times 10^6)(50 \times 10^{-6})[/tex]
[tex]F_m = 4.72 \times 10^{-17} N[/tex]
The gravitational force on the electron is 8.90 x 10^-31 N downward, the electric force on the electron is 1.60 x 10^-17 N upward, and the magnetic force on the electron is 5.90 x 10^-25 N northward.
Explanation:The gravitational force on an electron near the surface of the Earth can be calculated using the formula Fg = m * g, where m is the mass of the electron and g is the acceleration due to gravity. The electric force on the electron can be calculated using the formula Fe = q * E, where q is the charge of the electron and E is the magnitude of the electric field. The magnetic force on the electron can be calculated using the formula Fm = q * v * B, where v is the velocity of the electron and B is the magnitude of the magnetic field.
Given that the electron has a mass of 9.11 x 10^-31 kg, a charge of -1.60 x 10^-19 C, and a velocity of 5.90 x 10^6 m/s, the gravitational force on the electron is approximately 8.90 x 10^-31 N downward. The electric force on the electron is approximately 1.60 x 10^-17 N upward. The magnetic force on the electron is approximately 5.90 x 10^-25 N northward.
Fresh water flows through a horizontal tapered pipe. At the wide end its speed is 8 m/s. The difference in pressure between the two ends is 5338 Pa. What is the speed (in m/s) of the water at the narrow end? Round your answer to the nearest tenth.
Answer:
8.64 m/s
Explanation:
v1 = 8 m/s, v2 = ?, P2 - P1 = 5338 Pa, density of water, d = 1000 kg/m^3
By the use of Bernoulli's theorem
P 1 + 1/2 x d x v1^2 = P2 + 1/2 x d x v2^2
P2 - P1 = 1/2 x d x (v2^2 - v1^2)
5338 = 0.5 x 1000 x (v2^2 - 64)
10.676 = v2^2 - 64
v2^2 = 74.676
v2 = 8.64 m/s
A football player with a mass of 88 kg and a speed of 2.0 m/s collides head-on with a player from the opposing team whose mass is 120 kg. The players stick together and are at rest after the collision. Find the speed of the second player, assuming the speed of light is 3.0 m/s.
Answer:
Speed of another player, v₂ = 1.47 m/s
Explanation:
It is given that,
Mass of football player, m₁ = 88 kg
Speed of player, v₁ = 2 m/s
Mass of player of opposing team, m₂ = 120 kg
The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :
[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]
V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.
[tex]v_2=-\dfrac{m_1v_1}{m_2}[/tex]
[tex]v_2=-\dfrac{88\ kg\times 2\ m/s}{120\ kg}[/tex]
[tex]v_2=-1.47\ m/s[/tex]
So, the speed of another player is 1.47 m/s. Hence, this is the required solution.
A football player pushes a sled on a level surface with a weight force of 760 N that acts perpendicular to the surface. The coefficient of kinetic friction between the sled and grass is 0.78 How much work does the football player perform if he pushes the sled and displaces it 10 m. Assume ) the average force performed is equal to the force required to keep the sled in motion and 2) the applied force is paraîlel to the sleds displacement. (3 points) a. b. How much power did the football player produce if he performed the sled push in 3.21 seconds? (2 point)
Answer:
Part a)
W = 5928 J
Part b)
P = 592.8 Watt
Explanation:
As we know that force of friction on the sled is given by
[tex]F_f = \mu F_n[/tex]
here we know that
[tex]\mu = 0.78[/tex]
also we know that normal force on the sled is counter balanced by the weight of the object
[tex]F_n = mg = 760 N[/tex]
now we have
[tex]F_f = (0.78)(760) = 592.8 N[/tex]
Now the applied force must be equal to this friction force so that it will start sliding on the grass
now if we push it by 10 m distance then work done to slide it given by
[tex]W = F.d[/tex]
[tex]W = (592.8)(10) = 5928 J[/tex]
Part B)
Power required to push the sled in 10 s
[tex]Power = \frac{work}{time}[/tex]
[tex]Power = \frac{5928}{10} = 592.8 Watt[/tex]
The top of the ramp is h1 = 1.17 m above the ground; the bottom of the ramp is h2 = 0.298 m above the ground. The block leaves the ramp moving horizontally, and lands a horizontal distance d away. Calculate the distance d.
Answer:
d = 1.02 m
Explanation:
By energy conservation we can find the speed by which ball will leave the ramp
[tex]U_i + KE_i = U_f + KE_f[/tex]
here we know that
[tex]mgh_1 + 0 = mgh_2 + \frac{1}{2}mv^2[/tex]
here we have
[tex]h_1 = 1.17 m[/tex]
[tex]h_2 = 0.298 m[/tex]
so we have
[tex](9.8)(1.17) = (9.8)(0.298) + \frac{1}{2} v^2[/tex]
[tex]v = 4.13 m/s[/tex]
now the time taken by the block to reach the ground is given by
[tex]h_2 = \frac{1}{2}gt^2[/tex]
[tex]t = \sqrt{\frac{2h_2}{g}}[/tex]
[tex]t = \sqrt{\frac{2(0.298)}{9.8}}[/tex]
[tex]t = 0.25 s[/tex]
now the distance covered by it is given as
[tex]d = 0.25 \times 4.13[/tex]
[tex]d = 1.02 m[/tex]
Pendulum A has a bob of mass m hung from the string of length L; pendulum B is identical to A except its bob has the length 2L. Compare the frequencies of small oscillations of the two pendulums.
Answer:
[tex]f_{B}: f_{A} = \sqrt{\frac{2}{1}}[/tex]
Explanation:
For pendulum A: Length = L and gravity = g
The frequency of pendulum A is given by
[tex]f = \frac{1}{2\pi }\sqrt{\frac{g}{L}}[/tex]
Here, f is the frequency, L be the length
[tex]f_{A} = \frac{1}{2\pi }\sqrt{\frac{g}{L}}[/tex] ... (1)
For pendulum B: Length = 2L, gravity = g
The frequency of pendulum B is given by
[tex]f_{B} = \frac{1}{2\pi }\sqrt{\frac{g}{2L}}[/tex] .... (2)
Divide equation (1) by (2)
[tex]f_{B}: f_{A} = \sqrt{\frac{2}{1}}[/tex]
How large must the coefficient of static friction be between the tires and road, if a car rounds a level curve of radius 85 m at a speed of 23.5 m/s?
Answer:
0.66
Explanation:
By using the formula
u = v^2 / r g
Where u is coefficent of friction
u = 23.5 × 23.5 / (85 × 9.8)
u = 0.66
A force of 18 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching it from its natural length to 11 in. beyond its natural length? W = ft-lb
Answer:
11.54 ft - lb
Explanation:
F = 18 lb, x = 8 in = 8 / 12 = 0.66 ft
F = k x
k = F / x = 18 / 0.66 = 27.27 lb/ft
y = 11 in = 11 / 12 ft = 0.92 ft
Work done = 1 /2 x k x y^2
W = 0.5 x 27.27 x 0.92 x 0.92 = 11.54 ft - lb
Work done of spring is the product of average force and the displacement.
The work required in stretching spring from its natural length to 11 in. beyond its natural length is 11.54 ft-Ib.
How to calculate the work done required stretching the spring?Work done of spring is the product of average force and the displacement. It can be given as,
[tex]W=\dfrac{1}{2}kx^2[/tex]
Here, [tex]k[/tex] is the spring constant. The spring constant can be given as,
[tex]k=\dfrac{F}{x}[/tex]
Here, [tex]F[/tex] is the force and [tex]x[/tex] is the displacement of spring.
Given information-
The value of the force is 18 Ib.
The length of spring stretched is 8 in beyond its natural length.
Change the length in feet as,
[tex]x=\dfrac{8}{12} \\x=0.66\rm ft[/tex]
Put the value in the above formula as,
[tex]k=\dfrac{18}{0.66}\\k=27.27 \rm Ib/ft[/tex]
Work W is required in stretching it from its natural length to 11 in. beyond its natural length, Change this length in feet as,
[tex]y=\dfrac{11}{12} \\y=0.94\rm ft[/tex]
Put the values in the formula of work done as,
[tex]W=\dfrac{1}{2}\times27.27\times (0.92)^2\\W=11.54 \rm ft-Ib[/tex]
Thus the work required in stretching spring from its natural length to 11 in. beyond its natural length is 11.54 ft-Ib.
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