Answer:
77362.56 J
163730.28571 J
Explanation:
A = Area = 25 mm²
l = Length = 300 mm
K = Constant = [tex]3.33\times 10^{-6}[/tex]
[tex]\eta[/tex] = Heat transfer factor = 0.75
[tex]f_m[/tex] = Melting factor = 0.63
T = Melting point of low carbon steel = 1760 K
Volume of the fillet would be
[tex]V=Al\\\Rightarrow V=25\times 300\\\Rightarrow V=7500\ mm^3=7500\times 10^{-9}\ m^3[/tex]
The unit energy for melting is given by
[tex]U_m=KT^2\\\Rightarrow U_m=3.33\times 10^{-6}\times 1760^2\\\Rightarrow U_m=10.315008\ J/mm^3[/tex]
Heat would be
[tex]Q=U_mV\\\Rightarrow Q=10.315008\times 7500\\\Rightarrow Q=77362.56\ J[/tex]
Heat required to weld is 77362.56 J
Amount of heat generation is given by
[tex]Q_g=\dfrac{Q}{\eta f_m}\\\Rightarrow Q_g=\dfrac{77362.56}{0.75\times 0.63}\\\Rightarrow Q_g=163730.28571\ J[/tex]
The heat generated at the welding source is 163730.28571 J
Without the specific heat capacity and melting point of low carbon steel, the exact heat required for the weld cannot be calculated. However, given the heat transfer factor and melting factor, the heat generated at the welding source is 163,700 Joules according to the student's provided answer.
The quantity of heat required for welding low carbon steel with a fillet weld having a cross-sectional area of 25.0 mm2 and length of 300 mm depends on the specific heat capacity of the steel and the temperature change required to melt it. However, the question does not provide specific values for the specific heat capacity or the melting point of low carbon steel, which are essential to calculate the heat quantity. Normally, such calculations would also require knowledge of the latent heat of fusion for the steel.
Given the lack of necessary details to calculate the quantity of heat, we can address the part (b) of the question which relates to the heat generated at the welding source. The heat generated at the source can be calculated by dividing the actual heat needed to make the weld (which is given by the student as an unknown, hence represented by the question mark '?') by the product of the heat transfer factor and the melting factor, which are 0.75 and 0.63 respectively.
If the heat required to perform the weld ('?') is found, then the heat generated at the source can be calculated as follows: Heat at source = Heat required / (Heat transfer factor × Melting factor). According to the answer provided by the student, the heat at the source is 163,700 Joules.
The Earth’s radius is 6378.1 kilometers. If you were standing at the equator, you are essentially undergoing uniform circular motion with the radius of your circular motion being equal to the radius of the Earth. You are an evil mad scientist and have come up with the simultaneously awesome and terrifying plan to increase the speed of the Earth’s rotation until people at the Earth’s equator experience a centripetal (radial) acceleration with a magnitude equal to g, (9.81 m/s2 ), effectively making them experience weightlessness. If you succeed in your dastardly plan, what would be the new period of the Earth’s rotation?
a. 2.7 minutes b. 84 minutes c. 48 minutes d. 76 minutes
Answer:
b. 84 minutes
Explanation:
[tex]a_c=g[/tex] = Centripetal acceleration = 9.81 m/s²
r = Radius of Earth = 6378.1 km
v = Velocity
Centripetal acceleration is given by
[tex]a_c=\dfrac{v^2}{r}\\\Rightarrow v=\sqrt{a_cr}\\\Rightarrow v=\sqrt{9.81\times 6378100}\\\Rightarrow v=7910.06706\ m/s[/tex]
Time period is given by
[tex]T=\dfrac{2\pi r}{v60}\\\Rightarrow T=\dfrac{2\pi 6378.1\times 10^3}{7910.06706\times 60}\\\Rightarrow T=84.43835\ minutes[/tex]
The time period of Earth’s rotation would be 84.43835 minutes
The new period of the Earth’s rotation is mathematically given as
T=84.43835 min
What would be the new period of the Earth’s rotation?Question Parameter(s):
The Earth’s radius is 6378.1 kilometers.
g= (9.81 m/s2 ),
Generally, the equation for the is mathematically given as
[tex]a_c=\dfrac{v^2}{r}[/tex]
Therefore
[tex]v=\sqrt{a_cr}\\\\v=\sqrt{9.81*6378100}[/tex]
v=7910.06706 m/s
In conclusion
[tex]T=\dfrac{2\pi r}{v60}[/tex]
Hence
[tex]T=\dfrac{2\pi 6378.1*10^3}{7910.06706*60}[/tex]
T=84.43835 min
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A certain simple pendulum has a period on the earth of 2.00 s .
1. What is its period on the surface of Mars, where the acceleration due to gravity is 3.71 m/s^2?
The period of the simple pendulum on the surface of Mars is approximately [tex]\( 3.25 \, \text{s} \)[/tex].
Step 1
The period T of a simple pendulum is given by the formula:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
where:
- T is the period of the pendulum,
- L is the length of the pendulum, and
- g is the acceleration due to gravity.
To find the period [tex]\( T_{\text{Mars}} \)[/tex] on the surface of Mars, we can use the formula with the acceleration due to gravity on Mars, [tex]\( g_{\text{Mars}} = 3.71 \, \text{m/s}^2 \)[/tex], and the same length L as on Earth.
We have the period [tex]\( T_{\text{Earth}} = 2.00 \, \text{s} \)[/tex] on Earth, and we need to find [tex]\( T_{\text{Mars}} \).[/tex]
Step 2
Using the formula, we get:
[tex]\[ T_{\text{Earth}} = 2\pi \sqrt{\frac{L}{g_{\text{Earth}}}} \][/tex]
Solving for L, we find:
[tex]\[ L = \left( \frac{T_{\text{Earth}}}{2\pi} \right)^2 g_{\text{Earth}} \][/tex]
Now, using this value of L, we can find [tex]\( T_{\text{Mars}} \)[/tex] using the acceleration due to gravity on Mars:
[tex]\[ T_{\text{Mars}} = 2\pi \sqrt{\frac{L}{g_{\text{Mars}}}} \][/tex]
Substituting the known values, we get:
[tex]\[ T_{\text{Mars}} = 2\pi \sqrt{\frac{\left( \frac{T_{\text{Earth}}}{2\pi} \right)^2 g_{\text{Earth}}}{g_{\text{Mars}}}} \][/tex]
Step 3
Simplifying:
[tex]\[ T_{\text{Mars}} = T_{\text{Earth}} \sqrt{\frac{g_{\text{Earth}}}{g_{\text{Mars}}}} \]Now, let's calculate \( T_{\text{Mars}} \):\[ T_{\text{Mars}} = 2.00 \times \sqrt{\frac{9.81}{3.71}} \]\[ T_{\text{Mars}} = 2.00 \times \sqrt{2.643} \]\[ T_{\text{Mars}} \approx 2.00 \times 1.626 \]\[ T_{\text{Mars}} \approx 3.25 \, \text{s} \][/tex]
So, the period of the simple pendulum on the surface of Mars is approximately [tex]\( 3.25 \, \text{s} \)[/tex].
PART 1/2
A neutron in a reactor makes an elastic headon collision with the nucleus of an atom initially at rest.
Assume: The mass of the atomic nucleus is
about 14.1 the mass of the neutron.
What fraction of the neutron’s kinetic energy is transferred to the atomic nucleus?
PART 2/2
If the initial kinetic energy of the neutron is
6.98 × 10−13 J, find its final kinetic energy.
Answer in units of J.
Answer:
0.247
5.25×10⁻¹³ J
Explanation:
Part 1/2
Elastic collision means both momentum and energy are conserved.
Momentum before = momentum after
m v = m v₁ + 14.1m v₂
v = v₁ + 14.1 v₂
Energy before = energy after
½ m v² = ½ m v₁² + ½ (14.1m) v₂²
v² = v₁² + 14.1 v₂²
We want to find the fraction of the neutron's kinetic energy is transferred to the atomic nucleus.
KE/KE = (½ (14.1m) v₂²) / (½ m v²)
KE/KE = 14.1 v₂² / v²
KE/KE = 14.1 (v₂ / v)²
We need to find the ratio v₂ / v. Solve for v₁ in the momentum equation and substitute into the energy equation.
v₁ = v − 14.1 v₂
v² = (v − 14.1 v₂)² + 14.1 v₂²
v² = v² − 28.2 v v₂ + 198.81 v₂² + 14.1 v₂²
0 = -28.2 v v₂ + 212.91 v₂²
0 = -28.2 v + 212.91 v₂
28.2 v = 212.91 v₂
v₂ / v = 28.2 / 212.91
v₂ / v = 0.132
Therefore, the fraction of the kinetic energy transferred is:
KE/KE = 14.1 (0.132)²
KE/KE = 0.247
Part 2/2
If a fraction of 0.247 of the initial kinetic energy is transferred to the atomic nucleus, the remaining 0.753 fraction must be in the neutron.
Therefore, the final kinetic energy is:
KE = 0.753 (6.98×10⁻¹³ J)
KE = 5.25×10⁻¹³ J
When a neutron collides elastically with the nucleus of an atom, a fraction of its kinetic energy is transferred to the nucleus. The fraction of kinetic energy transferred can be calculated using the principle of conservation of momentum and kinetic energy. For the given scenario, the fraction is 0.8636. To find the final kinetic energy of the neutron, multiply the fraction of kinetic energy transferred by the initial kinetic energy of the neutron.
Explanation:When a neutron in a reactor undergoes an elastic head-on collision with the nucleus of an atom initially at rest, kinetic energy is transferred from the neutron to the atomic nucleus. The fraction of the neutron's kinetic energy transferred to the nucleus can be calculated using the principle of conservation of momentum and kinetic energy. Since the mass of the atomic nucleus is about 14.1 times the mass of the neutron, the fraction of kinetic energy transferred can be calculated as:
Fraction of kinetic energy transferred = (14.1 - 1) / (14.1 + 1) = 0.8636
For PART 2/2, to find the final kinetic energy of the neutron, we can multiply the fraction of kinetic energy transferred to the nucleus by the initial kinetic energy of the neutron:
Final kinetic energy = Fraction of kinetic energy transferred x Initial kinetic energy = 0.8636 x 6.98 × 10-13 J
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A continuous and aligned fiber-reinforced composite having a cross-sectional area of 1130 mm2 is subjected to an external tensile load. If the stresses sustained by the fiber and matrix phases are 156 MPa and 2.75 MPa, respectively, the force sustained by the fiber phase is 74,000 N and the total longitudinal strain is 1.25 x 10-3, what is the value of the modulus of elasticity of the composite material in the longitudinal direction?
Answer:
Ec=53.7×10⁹N/m² =53.7Gpa
Explanation:
To calculate the modulus of elasticity in the longitudinal direction. This is possible realizing Ec=σ/ε where σ=(Fm+Ff)/Ac
[tex]Ec=Sigma/E\\Ec=\frac{(Fm+Ff)/E}{Ac}\\ Ec=\frac{1802+74,000}{(1.25*10^{-3})(1130)(1/1000)^{2} }\\ Ec=53.7*10^{9}N/m^{2}\\or\\Ec=53.7GPa[/tex]
Final answer:
The modulus of elasticity of the composite material in the longitudinal direction is 124,800 MPa.
Explanation:
To find the modulus of elasticity of the composite material in the longitudinal direction, we can use the formula:
E = (stress sustained by the fiber phase)/(longitudinal strain)
Given that the stress sustained by the fiber phase is 156 MPa and the total longitudinal strain is 1.25 x 10^-3, we can plug in these values to calculate the modulus of elasticity:
E = 156 MPa / (1.25 x 10^-3) = 124,800 MPa
Therefore, the modulus of elasticity of the composite material in the longitudinal direction is 124,800 MPa.
Arrange the listed objects according to their angular speeds, from largest to smallest.
a tire of radius 0.381 m rotating at 12.2 rpm
a bowling ball of radius 12.4 cm rotating at 0.456 rad/s
a top with a diameter of 5.09 cm spinning at 18.7∘ per second
a rock on a string being swung in a circle of radius 0.587 m with
a centripetal acceleration of 4.53 m/s2 a square, with sides 0.123 m long, rotating about its center with corners moving at a tangential speed of 0.287 m / s
Answer:
Explanation:
1 ) tire of radius 0.381 m rotating at 12.2 rpm
12.2 rpm = 12.2 /60 rps
n = .20333 rps
angular speed
= 2πn
= 2 x 3.14 x .20333
= 1.277 rad / s
2 ) a bowling ball of radius 12.4 cm rotating at 0.456 rad/s
angular speed = .456 rad/s
3 ) a top with a diameter of 5.09 cm spinning at 18.7∘ per second
18.7° per second = (18.7 / 180) x 3.14 rad/s
= .326 rad/s
4 )
a rock on a string being swung in a circle of radius 0.587 m with
a centripetal acceleration of 4.53 m/s2
centripetal acceleration = ω²R
ω is angular velocity and R is radius
4.53 = ω² x .587
ω = 2.78 rad / s
5 )a square, with sides 0.123 m long, rotating about its center with corners moving at a tangential speed of 0.287 m / s
The radius of the circle in which corner is moving
= .123 x √2
=.174 m
angular velocity = linear velocity / radius
.287 / .174
1.649 rad / s
The perfect order is
4 ) > 5> 1 >2>3.
Explanation:
To arrange the listed objects according to their angular speeds from largest to smallest, we must first calculate the angular speed ω (in radians per second) for each object, as this unit provides a common ground for comparison. The angular speed can be found using the formula ω = v/r where v is the linear velocity and r is the radius.
A tire rotating at 12.2 rpm: To convert this to radians per second, we use the conversion factor 1 rpm = 2π/60 rad/s. ω = 12.2 * (2π/60) = 1.279 rad/s.A bowling ball rotating at 0.456 rad/s: This is already in the correct unit, so ω = 0.456 rad/s.A top spinning at 18.7° per second: Converting degrees to radians (ω = 18.7 * (π/180)) gives ω = 0.326 rad/s.A rock with a centripetal acceleration of 4.53 m/s²: Using the formula a = ω² * r, we can rearrange to find ω (angular speed) = √(a/r) = √(4.53/0.587) = 2.780 rad/s.A square with corners moving at a tangential speed of 0.287 m/s: The radius for a square rotating about its center is half the diagonal length. For a square of 0.123 m side, the diagonal (d) = √(2) * side = √(2) * 0.123 m, and the radius (r) = d/2. So, ω = v/r = 0.287/(√(2)*0.123/2) = 4.205 rad/s.Arranging these angular speeds from largest to smallest:
Angular speed of square: 4.205 rad/sAngular speed of rock: 2.780 rad/sAngular speed of tire: 1.279 rad/sAngular speed of bowling ball: 0.456 rad/sAngular speed of top: 0.326 rad/sThere are usually _________ collisions in a motor vehicle crash.
Answer:
3
Explanation:
During a crash 3 types of collisions can occur.
There are usually three types of collisions involved in a motor vehicle crash.
The first type of collision is the vehicle collision, which involves the physical impact between two or more vehicles. When vehicles collide, their structures deform, and the forces involved can cause severe damage to the involved vehicles. The severity of this collision depends on factors like the speed, mass, and angle of impact.
The second type of collision is the human collision, which occurs inside the vehicle. During an accident, passengers inside the vehicle can collide with each other or with interior components, such as the dashboard, steering wheel, or windows. These collisions can result in injuries like whiplash, head injuries, or broken bones.
The third type of collision is the internal collision, which involves the organs and tissues within the human body. When a collision occurs, the human body is subjected to rapid deceleration, causing organs to collide with each other or with the skeletal structure. These internal collisions can lead to internal bleeding, organ damage, and other life-threatening injuries.
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(a) A proton is confined to the nucleus of an atom. Assume the nucleus has diameter 5.5 x 10-15 m and that this distance is the uncertainty in the proton's position. What is the minimum uncertainty in the momentum of the proton? Dpmin = kg-m/s
(b) An electron is confined in an atom. Assume the atom has diameter 1 x 10-10 m and that this distance is the uncertainty in the electron's position. What is the minimum uncertainty in the momentum of the electron?
Answer:
[tex]1.91738\times 10^{-20}\ kgm/s[/tex]
[tex]1.05456\times 10^{-24}\ kgm/s[/tex]
Explanation:
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
[tex]\Delta x[/tex] = Uncertainty in the position
[tex]\Delta p[/tex] = Uncertainty in the momentum
From the uncertainty principle
[tex]\Delta x\Delta p=\dfrac{h}{2\pi}\\\Rightarrow \Delta p=\dfrac{h}{2\pi \Delta x}\\\Rightarrow \Delta p=\dfrac{h}{2\pi \Delta x}\\\Rightarrow \Delta p=\dfrac{6.626\times 10^{-34}}{2\pi\times 5.5\times 10^{-15}}\\\Rightarrow \Delta p=1.91738\times 10^{-20}\ kgm/s[/tex]
The minimum uncertainty in the momentum of the proton is [tex]1.91738\times 10^{-20}\ kgm/s[/tex]
[tex]\Delta x\Delta p=\dfrac{h}{2\pi}\\\Rightarrow \Delta p=\dfrac{h}{2\pi \Delta x}\\\Rightarrow \Delta p=\dfrac{h}{2\pi \Delta x}\\\Rightarrow \Delta p=\dfrac{6.626\times 10^{-34}}{2\pi\times 1\times 10^{-10}}\\\Rightarrow \Delta p=1.05456\times 10^{-24}\ kgm/s[/tex]
The minimum uncertainty in the momentum of the electron is [tex]1.05456\times 10^{-24}\ kgm/s[/tex]
Final answer:
The minimum uncertainty in the momentum of a proton confined to the nucleus of an atom is 9.6 × 10^-12 kg m/s. The minimum uncertainty in the momentum of an electron confined in an atom is 5.3 × 10^-24 kg m/s.
Explanation:
The Heisenberg Uncertainty Principle states that there is a limit to the precision with which we can know both the position and momentum of a particle. The minimum uncertainty in the proton's momentum is given by the formula: Dpmin = ħ/2Ax, where Ax is the uncertainty in the position of the proton. In this case, Ax is given as the diameter of the nucleus, so we have:
Dpmin = (1.055 × 10^(-34) kg m^2/s) / (2(5.5 × 10^(-15) m)) = 9.6 × 10^(-12) kg m/s.
Similarly, for the electron:
Dpmin = (1.055 × 10^(-34) kg m^2/s) / (2(1 × 10^(-10) m)) = 5.3 × 10^(-24) kg m/s.
A cruise ship is moving at constant speed through the water. The vacationers on the ship are eager to arrive at their next destination. They decide to try to speed up the cruise ship by gathering at the bow (the front) and running together toward the stern (the back) of the ship.(a) While they are running toward the stern, what is the speed of the ship?1.higher than it was before they started running
2. unchanged from what it was before they started running
3. lower than it was before they started running
4. impossible to determine
The speed of the ship remains the same when vacationers run from the front to the back. This effect is explained by the conservation of momentum, which states that the total momentum of an isolated system, in this case the ship and the vacationers, remains constant. Thus, the internal movement of the vacationers does not affect the speed of the ship.
Explanation:This question is related to the principle of conservation of momentum in physics. According to this principle, if a system is isolated (no external forces acting on it), the total momentum remains constant. So, the spectators' running from the bow to the stern won't affect the speed of the ship. Therefore, the answer is 2. The speed of the ship is unchanged from what it was before they started running. It is important to note that the center of mass of the system (ship and vacationers) remains the same before and after the vacationers run from one point to another inside the ship. This is because their movement is internal to the system and has no effect on the system's center of mass or the ship's speed.
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The speed of the cruise ship remains unchanged even if the vacationers move from the bow to the stern. This is due conservation of linear momentum and Newton's third law. Even though the ship's center of mass is slightly affected, the total speed of the ship remains the same.
Explanation:Physics principles, particularly those related to Newton's third law and the conservation of linear momentum, apply in this scenario. As the vacationers from the bow run towards the stern, their forward motion will push the ship slightly backward due to Newton's third law, which states that every action has an equal and opposite reaction. However, the overall speed (or velocity) of the cruise ship relative to the water or the Earth remains unchanged, option 2, because the total linear momentum of the cruise ship system (which includes the ship and the passengers) is conserved.
It's important to note that the shift in passengers’ positions does slightly change the center of mass of the entire system, but it does not alter the fact that the total speed of the ship is conserved if no external force is applied.
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Which term below best describes the forces on an object with a a net force of zero?
A. Inertia
B. Balanced Forces
C. Acceleration
D. Unbalanced Forces
Answer:
B. Balanced Forces
Explanation:
The net force is defined as the sum of all the forces acting on an object. Therefore, if the forces are balanced, they will counteract each other, causing the net force to be zero, then the object will continue at rest or moving with constant velocity.
The term below best describes the forces on an object with a a net force of zero B. Balanced Force
Which term below best describes the forces on an object with a a net force of zero?Balanced Forces refer to the situation where the forces acting on an object are equal in magnitude and opposite in direction, resulting in a net force of zero. When balanced forces act on an object, they cancel each other out, leading to no change in the object's state of motion.
This concept is tied to Newton's First Law of Motion, which states that an object at rest remains at rest, and an object in motion continues to move with a constant velocity, unless acted upon by an unbalanced force.
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This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. The acceleration function (in m/s2) and the initial velocity v(0) are given for a particle moving along a line. a(t) = 2t + 4, v(0) = −32, 0 ≤ t ≤ 6
a) Find the velocity at time t.
b) Find the distance traveled during the given time interval
Answer:
(a) Velocity at time t will be [tex]v(t)=t^2+4t-32[/tex]
(B) Distance will be -48 m
Explanation:
We have given [tex]a(t)=2t+4[/tex]
And [tex]v(0)=-32[/tex]
(a) We know that [tex]v(t)=\int a(t)dt[/tex]
So [tex]v(t)=\int (2t+4)dt[/tex]
[tex]v(t)=t^2+4t+c[/tex]
As [tex]v(0)=-32[/tex]
So [tex]-32=0^2+4\times 0+c[/tex]
c = -32
So [tex]v(t)=t^2+4t-32[/tex]
(b) We have to find the distance traveled
So [tex]s(t)=\int_{0}^{6}v(t)dt[/tex]
[tex]s(t)=\int_{0}^{6}(t^2+4t-32)dt[/tex]
[tex]s(t)=\int_{0}^{6}(\frac{t^3}{3}+2t^2-32t)[/tex]
[tex]s=(\frac{6^3}{3}+2\times 6^2-32\times 6)-0=72+72-192=-48m[/tex]
Final answer:
To find the velocity at time t, integrate the acceleration function and add the initial velocity. To find the distance traveled, integrate the velocity function over the given time interval.
Explanation:
To find the velocity at time t, we can integrate the acceleration function over the given time interval and add the initial velocity. The integral of 2t + 4 is t^2 + 4t, so the velocity function is v(t) = t^2 + 4t - 32.
To find the distance traveled, we can integrate the velocity function over the given time interval. The integral of t^2 + 4t - 32 is (1/3)t^3 + 2t^2 - 32t. Evaluating this integral from 0 to 6 gives us the distance traveled during the given time interval.
Which of the following should be measured by vernier calipers?
A. Height of a bed
B. Length of a torch
C. Thickness of a candle
D. Length of a windowpane
E. Width of a bed
Answer:
option C.
Explanation:
The correct answer is option C.
Vernier Calipers is a precision instrument that is used to measure the internal or outer dimensions or the depth of the material.
Vernier Caliper consists of two jaws, the main scale, and the vernier scale.
Graduation is present on both the main scale and vernier.
To measure the width of any material it should be placed in between the jaws and reading is taken with the help of the main scale and vernier scale.
The thickness of the candle can be measured using vernier calipers precisely.
A singly charged positive ion has a mass of 3.46 × 10−26 kg. After being accelerated through a potential difference of 215 V the ion enters a magnetic field of 0.522 T in a direction perpendicular to the field. The charge on the ion is 1.602 × 10−19 C. Find the radius of the ion’s path in the field. Answer in units of cm.
Answer:
1.8 cm
Explanation:
[tex]m[/tex] = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg
[tex]q[/tex] = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C
[tex]\Delta V[/tex] =Potential difference through which the ion is accelerated = 215 V
[tex]v[/tex] = Speed of the ion
Using conservation of energy
Kinetic energy gained by ion = Electric potential energy lost
[tex](0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}[/tex]
[tex]r[/tex] = Radius of the path followed by ion
[tex]B[/tex] = Magnitude of magnetic field = 0.522 T
the magnetic force on the ion provides the necessary centripetal force, hence
[tex]qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm[/tex]
A proton initially traveling at 50,000 m/s is shot through a small hole in the negative plate of a parallal-plate capacitor. The electric field strength inside the capacitor is 1,500 V/m. How far does the proton travel above the negative plate before temporarily coming to rest and reversing course? Assume the proton reverses course before striking the positive plate.
Answer:
x = 8.699 10⁻³ m
Explanation:
The proton feels an electric charge that is the opposite direction of speed, let's look for acceleration using Newton's second law
F = m a
F = q E
a = q E / m
a = 1.6 10⁻¹⁹ 1500 / 1.67 10⁻²⁷
a = 1,437 10¹¹ m / s²
Now we can use kinematic relationships
v² = v₀² - 2 a x
When at rest the speed is zero (v = 0)
x = v₀² / 2 a
Let's calculate
x = 50,000² / (2 1,437 10¹¹)
x = 8.699 10⁻³ m
The 94-lb force P is applied to the 220-lb crate, which is stationary before the force is applied. Determine the magnitude and direction of the friction force F exerted by the horizontal surface on the crate. The friction force is positive if to the right, negative if to the left.
Answer
given,
force = 94 lb
weight of crate = 220 lb
Assuming the static friction be equal = 0.47
kinetic friction = 0.36
Maximum force applied to move the object is when object is just start to move.
F = μ N
F = 0.47 x 220
F = 103.4 lb
As the frictional force is more than applied then the object will not move.
so, the friction force will be equal to the force applied on the object that is equal to 94 lb.
hence, the direction of force will left.
A metal bar is used to conduct heat. When the temperature at one end is 100°C and at the other is 20°C, heat is transferred at a rate of 16 J/s.
If the temperature of the hotter end is reduced to 80°C, what will be the rate of heat transfer?
a. 4 J/sb. 8 J/sc. 9 J/sd. 12 J/s
Answer:
a. 4 J/s
Explanation:
Fourier's law states for the case in which there is stationary heat flow in only one direction, that is, linearly, the heat transmitted per unit of time is proportional to the temperature difference:
[tex]\frac{Q}{t}\propto \Delta T[/tex]
When the temperature at one end is 100°C and at the other is 20°C, we have:
[tex]\Delta T_1=100^\circ C-20^\circ C\\\Delta T_1=80^\circ C[/tex]
If the temperature of the hotter end is [tex]80^\circ C[/tex], we have:
[tex]\Delta T_2=100^\circ C-80^\circ C\\\Delta T_2=20^\circ C[/tex]
So:
[tex]\Delta T_1=4\Delta T_2\\\Delta T_2=\frac{\Delta T_1}{4}[/tex]
Finally, we calculate the rate of heat transfer:
[tex]\frac{Q_2}{t_2}=\frac{\frac{Q_1}{t_1}}{4}\\\\\frac{Q_2}{t_2}=\frac{16\frac{J}{s}}{4}\\\frac{Q_2}{t_2}=4\frac{J}{s}[/tex]
First, design an experiment you could conduct that might measure how thermal equilibrium occurs. What materials would you use? What would you measure? What results would you expect? What if the results were different; what would that indicate?
Answer:
What material would you use: I would use a beaker, water, heated metal
What would you measure: Measure the changes in temperature before and after pouring the metal.
What results would you expect: Whether Thermal Equilibrium has occurred or not.
What if the results were different; what would that indicate : The Beaker might absorb some of the heat causing an error in the reading.
Explanation:
step 1: Take a well insulated beaker and pour water of known mass in it.
step 2: Record its initial temperature.
step 3: Place heated metal into the beaker and make sure that the beaker is tightly packed so that no heat escapes from the beaker.
step 4: Record temperature of the beaker at different intervals and after temperature has become constant ( No heat is being gained) , note the final temperature
In end we will check whether thermal equilibrium is established or not.
heat lost by the metal = heat gained by water + heat gained by the beaker
Possible Error:
There might be disparities in the values acquired as the beaker will absorb some heat.
Arace car accelerates uniformly at 11.3 m/s2. If the race car starts from rest how fast will it
be going after 6.7 seconds.
Answer:
75.71 m/s
Explanation:
From equation of motion, acceleration is given by
[tex]a=\frac {v-u}{t}[/tex]where v is the final velocity, u is the initial velocity and t is time taken.
Making v the subject of the above formula
v=at+u
Substituting 6.7 s for time, t and 11.3 for a and taking u as zero since it starts from rest
v=11.3*6.7=75.71 m/s
Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of 1012 W) pulses of electromagnetic waves that last for an extremely short time (a few nanoseconds). These short pulses scramble the interior of a cell without causing it to explode, as long pulses would do. We can model a typical such cell as a disk 4.6 μm in diameter, with the pulse lasting for 3.4 ns with an average power of 2.46×1012 W . We shall assume that the energy is spread uniformly over the faces of 100 cells for each pulse.
Part A
How much energy is given to the cell during the pulse?
Express your answer to two significant figures.
The energy given to each cell during the pulse can be calculated by multiplying the power of the pulse by its duration, and then dividing by the number of cells.
Explanation:The energy supplied to the cell during the pulse is determined by the power multiplied by the duration of the pulse. In this scenario, the power is 2.46×1012 W and the duration is 3.4 ns (which is 3.4x10-9 s when converted to seconds for mathematical calculations).
We use the formula:
E = P * t
Where,
E is the Energy
P is the Power
t is the time (duration of the pulse)
Substituting the given values into the formula:
E = 2.46x1012 W * 3.4x10-9 s
This gives the total energy supplied. We know the energy is spread uniformly over the faces of 100 cells, so each cell will get 1/100 of the total energy. Using these calculations, we can determine the amount of energy given to each cell during the pulse.
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Interference occurs with not only light waves but also all frequencies of electromagnetic waves and all other types of waves, such as sound and water waves. Suppose that your physics professor sets up two sound speakers in the front of your classroom and uses an electronic oscillator to produce sound waves of a single frequency. When she turns the oscillator on (take this to be its original setting), you and many students hear a loud tone while other students hear nothing.
The professor adjusts the oscillator to produce sound waves of twice the original frequency. What happens?
a. Some of the students who originally heard a loud tone again hear a loud tone, but others in that group now hear nothing.
b. The students who originally heard a loud tone again hear a loud tone, and the students who originally heard nothing still hear nothing.
c. Among the students who originally heard nothing, some still hear nothing but others now hear a loud tone.
d. The students who originally heard a loud tone now hear nothing, and the students who originally heard nothing now hear a loud tone.
This event is a result of interference, a physics phenomenon where waves combine to create a new wave. Changing the frequency altered the phase difference between the waves at various places in the room, leading to some students now hearing a tone because they are at points of constructive interference.
Explanation:The correct answer is c. Among the students who originally heard nothing, some still hear nothing but others now hear a loud tone. This situation is described by the phenomenon of interference. Interference occurs when two waves combine to form a resultant wave. When two identical sound waves from the speakers meet at a point in space, they can either constructively or destructively interfere depending on the phase difference.
If the phase difference is such that the waves reinforce each other, it results in a loud sound (constructive interference). However, if the phase difference is such that one wave cancels the other, no sound is heard (destructive interference). By doubling the frequency, the professor effectively changes the phase difference between the waves at the various points in the room. Therefore, some students' locations might now be at points of constructive interference, allowing them to hear the sound.
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The correct answer is option a. When the oscillator's frequency is doubled, the interference pattern shifts, causing some students who originally heard a loud tone to hear nothing, while others still hear a loud tone.
The student asked what happens when the professor adjusts the oscillator to produce sound waves of twice the original frequency. The correct answer is a) Some of the students who originally heard a loud tone again hear a loud tone, but others in that group now hear nothing.
The phenomenon described in the question is due to the interference of sound waves. Interference occurs when sound waves from the two speakers overlap, creating areas of constructive interference (where waves are in phase and amplify the sound) and destructive interference (where waves are out of phase and cancel each other out).
When the frequency is changed, the interference pattern shifts, causing some students who previously experienced a loud sound to now be in a zone of destructive interference, and vice versa. This adjustment results in a new distribution of loud and quiet spots within the classroom.
A solid cylinder attached to a horizontal spring (???? = 3.00 N/m) rolls without slipping along a horizontal surface. If the system is released from rest when the spring is stretched by 0.250 m, find (a) the translational kinetic energy and (b) the rotational kinetic energy of the cylinder as it passes through the equilibrium position. (c) Show that under these conditions the cylinder’s center of mass executes simple harmonic motion with period
Answer:
a. Ek = 62.5 x 10⁻³ J
b. Ek = 31.25 x 10 ⁻³ J
Explanation:
E = [ Us + (M * v²) / 2 ] + [ (I * ω² ) / 2 ]
T = 2π * √ 3M / 2 k , K = 3.0 N / m , d = 0.25m
a.
Ek = ¹/₂ * m * v² = ¹/₃ * k * d²
Ek = ¹/₃ * k * d² = ¹/₃ * 3.0 N / m * 0.25²m
Ek = 62.5 x 10⁻³ J
b.
Ek = ¹/₂ * I * ω² = ¹/₄ * M * v²
Ek = ¹/₆ * k * Xm² = ¹/₆ * 3.0 N / m * 0.25²m
Ek = 31.25 x 10 ⁻³ J
c.
d Emech / dt = d / dt * [ 3 m * v² / 4 + k * x² / 2 ]
acm = - ( 2 k / 3M)
ω = √ 2k / 3m ⇒ T = 2π * √ K / m
To understand the formula representing a traveling electromagnetic wave.Light, radiant heat (infrared radiation), X rays, and radio waves are all examples of traveling electromagnetic waves. Electromagnetic waves comprise combinations of electric and magnetic fields that are mutually compatible in the sense that the changes in one generate the other.The simplest form of a traveling electromagnetic wave is a plane wave. For a wave traveling in the x direction whose electric field is in the y direction, the electric and magnetic fields are given byE? =E0sin(kx??t)j^,B? =B0sin(kx??t)k^.This wave is linearly polarized in the y direction.1.a. In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E0 and B0are the __________ of the electric and magnetic fields.Choose the best answer to fill in the blank.1. maxima2. amplitudes3. wavelengths4. velocitiesb. The variable ? is called the __________ of the wave.Choose the best answer to fill in the blank.1. velocity2. angular frequency3. wavelengthc. The variable k is called the __________ of the wave.1. wavenumber
2. wavelength
3. velocity
4. frequency
Answer:
1) Eo and Bo. They are maximum amplitudes. Answer 1 and 2
2) .w is angular frequency. Answer 2
3) k is wave number. Answer 1
Explanation:
The electromagnetic wave is given by
[tex]E_{y}[/tex] = E₀ sin (kx –wt)
This is the equation of a traveling wave on the x axis with the elective field oscillating on the y axis
The terms represent E₀ the maximum amplitude of the electric field,
The wave vector
k = 2π /λ
Angular velocity
w = 2π f
To answer the questions let's use the previous definitions
1) Eo and Bo. They are maximum amplitudes. Answer 1 and 2
2) .w is angular frequency. Answer 2
3) k is wave number. Answer 1
The inductance in the drawing has a value of L = 9.4 mH. What is the resonant frequency f0 of this circuit?
Answer:
The resonant frequency of this circuit is 1190.91 Hz.
Explanation:
Given that,
Inductance, [tex]L=9.4\ mH=9.4\times 10^{-3}\ H[/tex]
Resistance, R = 150 ohms
Capacitance, [tex]C=1.9\ \mu F=1.9\times 10^{-6}\ C[/tex]
At resonance, the capacitive reactance is equal to the inductive reactance such that,
[tex]X_C=X_L[/tex]
[tex]2\pi f_o L=\dfrac{1}{2\pi f_oC}[/tex]
f is the resonant frequency of this circuit
[tex]f_o=\dfrac{1}{2\pi \sqrt{LC}}[/tex]
[tex]f_o=\dfrac{1}{2\pi \sqrt{9.4\times 10^{-3}\times 1.9\times 10^{-6}}}[/tex]
[tex]f_o=1190.91\ Hz[/tex]
So, the resonant frequency of this circuit is 1190.91 Hz. Hence, this is the required solution.
In Bob Shaw's short story, "The Light of Other Days," he describes something called slow glass. In the story, a married couple buys a 4-foot-wide window of slow glass that has been out on a beautiful hillside in Ireland, collecting light for 10 years. The idea is that the light takes 10 years to pass through the glass, so if you mount the window in your house it will give a view of the Irish landscape for the next 10 years, slowly unveiling everything that happened there. You can read the full short story via the link below, if you are interested.
Link to Bob Shaw's short story: The Light of Other Days.
(a) In the short story, the couple buys a window that is one-quarter-inch thick, and takes light 10 years to pass through. Let's say that you were able to locate a supplier of slow glass, and you bought some glass that was 5.00 mm thick, with the light taking 7.00 years to pass through. Taking one year to be 365.24 days, calculate the index of refraction of your piece of slow glass.
In 1999, Lene Hau, a physicist at Harvard University, received quite a bit of attention for getting light to travel at bicycle speed (later, she was able to temporarily stop light completely). The speed of a bicycle is a lot faster than light travels through the slow glass from the story, but it is still orders of magnitude less than the speed at which light travels through vacuum. If you're interested, you can follow this link to learn more about Lene Hau.
(b) Lene Hau used something called a Bose-Einstein condensate to slow down light. If the light is traveling at a speed of 40.0 km/hr through the Bose-Einstein condensate, what is the effective index of refraction of the condensate?
Answer:
Consider the following calculations
Explanation:
a ) velocity of the glass is v = distance / time
= 5 X 10-3 / 7 X 365.24 X 24 X 60 X 60
v = 2.263 X 10-11 m/sec
the speed of the light in vaccum is C
C = 3 X 108 m/sec
n = C / v
n = 3 X 108 / 2.263 X 10-11
n = 1.32567 X 1019
b ) given is 40 km/hr
= 40 X 103 / 60 X 60
= 11.11 m/sec
n = C / v
n = 3 X 108 / 11.11
n = 27002700.27
The index of refraction for the 5.00 mm thick slow glass taking light 7.00 years to pass through is approximately 1.33 x 10^19. For a Bose-Einstein condensate where light travels at 40.0 km/hr, the effective index of refraction is about 2.70 x 10^7.
In Bob Shaw's short story 'The Light of Other Days', a fictional material called slow glass is described, which delays the passage of light. To calculate the index of refraction of a 5.00 mm thick piece of slow glass where light takes 7.00 years to pass through, we can use the formula n = c/v, where c is the speed of light in a vacuum (3.00 x 108 m/s), and v is the speed of light through the material.
To find v, we can calculate the total distance light travels in 7.00 years and divide it by the time it takes to travel through the slow glass. Since the slow glass is 5.00 mm thick, which is equivalent to 5.00 x 10-3 m, and one year is 365.24 days, we calculate the speed as follows:
v = distance/time = 5.00 x 10-3 m / (7.00 years x 365.24 days/year x 24 hours/day x 3600 seconds/hour) = 5.00 x 10-3 m / 220,937,280 seconds ≈ 2.263 x 10-11 m/s.
Then, the index of refraction, n, can be calculated as n = c/v ≈ 3.00 x 108 m/s / 2.263 x 10-11 m/s ≈ 1.33 x 1019.
For Lene Hau's experiment using a Bose-Einstein condensate with a light speed of 40.0 km/hr, the index of refraction can also be calculated using n = c/v. Converting 40.0 km/hr to m/s:
v = 40.0 km/hr x (1000 m/km) / (3600 s/hr) = 11.1 m/s.
Using this value for v, we calculate n as n = c/v ≈ 3.00 x 108 m/s / 11.1 m/s ≈ 2.70 x 107.
A box of mass 17.6 kg with an initial velocity of 2.25 m/s slides down a plane, inclined at 19◦ with respect to the horizontal. The coefficient of kinetic friction is 0.48. The box stops after sliding a distance x. 17.6 kg µk = 0.48 2.25 m/s 19◦
How far does the box slide? The acceleration due to gravity is 9.8 m/s 2 . The positive x-direction is down the plane. Answer in units of m.
Final answer:
The sliding distance of the box can be calculated using the principles of conservation of energy and the work-energy theorem, considering the initial kinetic energy and the work done against the force of kinetic friction.
Explanation:
Calculating the Sliding Distance of a Box on an Inclined Plane
To determine how far the box slides, we can use the principles of conservation of energy and the work-energy theorem. The initial kinetic energy of the box is transformed into work done against friction. The work done by friction is equal to the force of friction times the distance the box slides. We start with calculating the force of kinetic friction, which is μ_k (coefficient of kinetic friction) times the normal force. The normal force is the component of the box's weight perpendicular to the inclined plane, calculated as m*g*cos(θ), where m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of the incline.
With given values: m = 17.6 kg, μ_k = 0.48, θ = 19°, βi = 2.25 m/s, and g = 9.8 m/s², we can calculate the force of kinetic friction (ƒ_k).
The component of gravity along the incline is m*g*sin(θ), and we know that the box stops when its initial kinetic energy is equal to the work done by friction. So, from the equation:
Kinetic Energy_initial = Work_friction,
½*m*βi^2 = ƒ_k * distance,
½*17.6 kg*(2.25 m/s)^2 = (0.48*17.6 kg*9.8 m/s²*cos(19°)) * distance,
We can then solve for the distance the box slides. After calculation, we obtain the sliding distance x.
The girl makes a microscope with a 3.0 cm focal length objective and a 5.0 cm eyepiece. The microscope tube length is 10 cm. Use the simple formula to find the expected magnification of this microscope.
(A) 12
(B) 14
(C) 17
(D) 20
(E) 24
To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".
The overall magnification of microscope is
[tex]M = \frac{Nl}{f_ef_0}[/tex]
Where
N = Near point
l = distance between the object lens and eye lens
[tex]f_0[/tex]= Focal length
[tex]f_e[/tex]= Focal of eyepiece
Given that the minimum distance at which the eye is able to focus is about 25cm we have that N = 25cm
Replacing,
[tex]M = \frac{25*10}{3*5}[/tex]
[tex]M = 16.67\approx 17\\[/tex]
Therefore the correct answer is C.
A student solving for the acceleration of an object has applied appropriate physics principles and obtained the expression a=a1 + F/m where a1 = 3.00 meter/ second2 F=12.0 kilogram.meter/second2 and m=7.00 kilogram. First which of the following is the correct step for obtaining a common denominator for the two fractions in the expression in solving for a?a. (m/m times a1/1) + (1/1 times F/m)b. (1/m times a1/1) + (1/m times F/m)c. (m/m times a1/1) + (F/F times F/m)d. (m/m times a1/1) +(m/m times F/m )
The correct step for obtaining a common denominator in the expression 'a = a1 + F/m' is to multiply each term by m/m, which results in '(m/m * a1/1) + (m/m * F/m)'. This leads to the acceleration, a, equalling 4.71 m/s2 when substituting the given values into the expression.
The student is trying to solve for the acceleration (a) of an object using the equation a = a1 + F/m, where a1 is given as 3.00 meters/second2, F as 12.0 kilogram.meter/second², and m as 7.00 kilograms. To obtain a common denominator for the two terms a1 and F/m, we look for an expression that allows us to combine these two fractions.
The correct step for obtaining a common denominator is option d. This is written as:
(m/m * a1/1) + (m/m * F/m)
This is because multiplying by m/m is equivalent to multiplying by 1, which does not change the value of the expression. For the term a1, since there is no denominator, multiplying by m/m effectively gives it a common denominator with F/m. The calculation becomes:
a = (m * a1 + F) / m
Substituting the given values:
a = (7.00 kg * 3.00 m/s2 + 12.0 kg*m/s2) / 7.00 kg
= (21.00 + 12.00) kg*m/s2 / 7.00 kg
= 33.00 kg*m/s2 / 7.00 kg
= 4.71 m/s2
Thus, the newton's second law, a = F/m, can be used to calculate the acceleration of the object.
When you walk at an average speed (constant speed, no acceleration) of 24 m/s in 94.1 sec
you will cover a distance of__?
Answer:
2258.4 m
Explanation:
Distance covered is a product of speed and time hence
s=vt where s is the displacement/distance covered, v is the speed and t is the time taken
s=24*94.1=2258.4 m
Therefore, the distance covered is 2258.4 m
A block of mass m = 0.775 kg is fastened to an unstrained horizontal spring whose spring constant is k = 83.6 N/m. The block is given a displacement of +0.113 m, where the + sign indicates that the displacement is along the +x axis, and then released from rest. What is the force (magnitude) that the spring exerts on the block just before the block is released?
Answer:
F= 9.45 N
Explanation:
If the mass is fastened to an unstrained horizontal spring, this means that at this position, the spring doesn't exert any force, because it keeps his equilibrium length.
If then the block is given a displacement of +0.113m, this means that the spring has been stretched in the same length.
According to Hooke's Law, the spring exerts a restoring force (trying to return to his equilibrium state) that opposes to the displacement, and which is proportional (in magnitude) to it, being the proportionality constant, a quantity called spring constant, which depends on the type of spring.
We can write the Hooke's Law as follows:
F = - k * Δx
Just before the block is released, we can get the value of F as follows:
⇒ F = 83.6 N/m* 0.113 m = 9.45 N (in magnitude)
Calculate the rotational inertia of a meter stick, with mass 0.71 kg, about an axis perpendicular to the stick and located at the 18 cm mark. (Treat the stick as a thin rod.)
To solve this problem we will use the parallel axis theorem for which the inertia of a point of an object can be found through the mathematical relation:
[tex]I = I_{cm} +mx^2[/tex]
Where
[tex]I_{cm}[/tex] = Inertia at center of mass
m = mass
x = Displacement of axis.
Our mass is given as 0.71kg,
m = 0.71kg
Para a Stick with length (L) the Moment of Inertia of the stick about and axis passing through the center and perpendicular to stick is
[tex]I_{cm} = \frac{1}{12} mL^2[/tex]
[tex]I_{cm} = \frac{1}{12} (0.71)(1)^2[/tex]
[tex]I_{cm} = 0.05916Kg\cdot m^2[/tex]
The distance between center of mass to the specific location is
[tex]x = 50cm - 18cm[/tex]
[tex]x = 38cm = 0.38m[/tex]
So, from parallel axis theorem ,
[tex]I = I_{cm} + mx^2[/tex]
[tex]I =0.05916Kg\cdot m^2+ (0.71kg)(0.38m)^2[/tex]
[tex]I = 0.161684Kg\cdot m^2[/tex]
Therefore the rotational inertia is [tex]0.161684Kg\cdot m^2[/tex]
An explosion occurs at the end of a pier. The sound reaches the other end of the pier by traveling through three media: air, fresh water, and a slender metal handrail. The speeds of sound in air, water, and the handrail are 339, 1480, and 5060 m/s, respectively. The sound travels a distance of 141 m in each medium.
(a) After the first sound arrives, how much later does the second sound arrive?
(b) After the first sound arrives, how much later does the third sound arrive?
Answer:
a) 0.0674s
b) 0.3880s
Explanation:
Based on the speed of sound in each medium ([tex]s_{air}=339m/s, s_{water}=1480m/s, s_{metal}=5060m/s[/tex]) the sound will first arrive via the metal handrail, then the water and lastly the air.
[tex]t=\frac{d}{s}[/tex] , where 't' is time, 'd' is distance and 's' is speed
[tex]t_{metal}=\frac{d}{s_{metal}}[/tex]
[tex]t_{metal}=\frac{141}{5060}[/tex]
[tex]t_{metal}=0.0279s[/tex]
[tex]t_{water}=\frac{d}{s_{water}}[/tex]
[tex]t_{water}=\frac{141}{1480}[/tex]
[tex]t_{water}=0.0953s[/tex]
[tex]t_{air}=\frac{d}{s_{air}}[/tex]
[tex]t_{air}=\frac{141}{339}[/tex]
[tex]t_{air}=0.4159s[/tex]
a) Time difference between the first and second sound
[tex]t_{water}-t_{metal}[/tex]
[tex]0.0953-0.0279[/tex]
[tex]0.0674s[/tex]
b) Time difference between the first and third sound
[tex]t_{air}-t_{metal}[/tex]
[tex]0.4159-0.0279[/tex]
[tex]0.3880s[/tex]
The second sound arrives 0.321 seconds after the first, and the third sound arrives 0.388 seconds after the first sound, demonstrating how sound speed varies in different media.
The question involves calculating the time differences for sound to travel through different media (air, water, and a metal handrail) over the same distance. To find out how much later one sound arrives compared to another, we use the formula time = distance / speed. The distances for all three media are the same, 141 m, but the speeds vary: 339 m/s for air, 1480 m/s for water, and 5060 m/s for the metal handrail.
Air: time = 141 m / 339 m/s = 0.416 s
Water: time = 141 m / 1480 m/s = 0.095 s
Metal handrail: time = 141 m / 5060 m/s = 0.028 s
(a) The second sound (through water) arrives 0.416 s - 0.095 s = 0.321 s after the first sound (through air).
(b) The third sound (through the metal handrail) arrives 0.416 s - 0.028 s = 0.388 s after the first sound (through air).