Answer:
Q=116.37 W
Explanation:
Given that
Emissivity ,ε= 0.200
Surface temperature ,T₁ = 10⁰ C = 283 K
Surrounding T₂ = - 15⁰ C = 258 K
Area ,A= 1.6 m²
The net heat transfer given as
[tex]Q=\sigma A \varepsilon (T_1^4-T_2^2)[/tex]
σ = 5.67 x 10⁻⁸
The temperature should be in Kelvin.
Now by putting the values
[tex]Q=5.67\times 10^{-8}\times 1.6\times 0.2\times (283^4-258^2)\ W[/tex]
Q=116.37 W
Therefore answer will be 116.37 W.
A 65 kg person jumps from a window to a fire net 18 m below, which stretches the net 1.1 m. Assume the net behaves as a simple spring and
A) Calculate how much it would stretch if the same person was lying on it.
B) How much it would stretch if the person jumped from 35 m
Answer:
0.03167 m
1.52 m
Explanation:
x = Compression of net
h = Height of jump
g = Acceleration due to gravity = 9.81 m/s²
The potential energy and the kinetic energy of the system is conserved
[tex]P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m[/tex]
The spring constant of the net is 20130.76 N
From Hooke's Law
[tex]F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m[/tex]
The net would strech 0.03167 m
If h = 35 m
From energy conservation
[tex]65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0[/tex]
Solving the above equation we get
[tex]x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45[/tex]
The compression of the net is 1.52 m
A cylinder containing an ideal gas has a volume of 2.6 m3 and a pressure of 1.5 × 105 Pa at a temperature of 300 K. The cylinder is placed against a metal block that is maintained at 900 K and the gas expands as the pressure remains constant until the temperature of the gas reaches 900 K. The change in internal energy of the gas is +6.0 × 105 J. How much heat did the gas absorb?
Explanation:
From the first law of thermodynamics,
Δ[tex]Q[/tex]=Δ[tex]U[/tex]+[tex]W[/tex]
Where [tex]Q[/tex] is the heat given to the gas,
[tex]U[/tex] is the internal energy of the gas,
[tex]W[/tex] is the workdone by the gas.
When pressure is constant,
[tex]\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}[/tex]
[tex]V_{2}=\frac{2.6\times 900}{300}=7.8m^{3}[/tex]
When pressure is constant,[tex]W=P[/tex]Δ[tex]V[/tex]
Where [tex]P[/tex] is pressure and [tex]V[/tex] is the volume of the gas.
Given [tex]P=1.5\times 10^{5}Pa[/tex]
Δ[tex]V=[/tex][tex]7.8-2.6=5.2m^{3}[/tex]
So,[tex]W=1.5\times 10^{5}\times 5.2=7.8\times 10^{5}J[/tex]
Given that Δ[tex]U=6\times 10^{5}[/tex]
So,Δ[tex]Q=[/tex][tex]6\times 10^{5}+7.8\times 10^{5}=13.8\times 10^{5}J[/tex]
By using the first law of thermodynamics, the ideal gas law, the given parameters and an additional calculation for the work done by the gas, we can calculate the total heat absorbed by the gas.
Explanation:To calculate the heat absorbed by the gas, we use the first law of thermodynamics which states that the heat absorbed by a system is equal to the change in its internal energy plus the work done by the system on its surroundings, expressed as Q = ΔEint + W. The change in internal energy, ΔEint, is given as +6.0 × 10^5 J.
Since the pressure is constant, the work done by the gas, W, can be calculated using W = PΔV, where P is the pressure and ΔV is the change in volume. The change in volume can be determined using the ideal gas law before and after the change, PV = nRT, thus, ΔV = nR(ΔT)/P. Substituting the given pressure, temperature change from 300 K to 900 K, and ideal gas constant, we can find ΔV.
After plugging ΔV into the equation for work done, we then add this to the change in internal energy to find the heat absorbed.
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hot-air balloon is rising vertically. The angle of elevation from a point on level ground 122 feet from the balloon to a point directly under the passenger compartment changes from 17.9 degrees to 29.5 degrees. How far, to the nearest tenth of a foot, does the balloon rise during this period?
29.62 feet far, to the nearest tenth of a foot, does the balloon rise during this period
Explanation:
Hot-air balloon is rising vertically, the angle of elevation from a point on level ground = 122 feet (from the balloon)
The passenger compartment changes from = 17.9 degrees to 29.5 degree
[tex]\tan \left(17.9^{\circ}\right)=\frac{a}{122 \mathrm{ft}}[/tex]
[tex]a=122 \times \tan 17.9^{\circ}=122 \times 0.32299=39.4[/tex]
Similarly,
[tex]\tan \left(29.5^{\circ}\right)=\frac{b}{122 f t}[/tex]
[tex]b=122 \times \tan \left(29.5^{\circ}\right)=122 \times 0.56577=69.02[/tex]
The nearest tenth of a foot, the balloon rise during the period, as below
69.02 – 39.4 = 29.62 ft.
The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flow on a planet where the acceleration of gravity varies with height so that g=g0−cz, where g0 and c are constants. Integrate "F=ma" along a streamline to obtain the equivalent of the Bernoulli equation for this flow.
Answer:
[tex]p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant[/tex]
Explanation:
first write the newtons second law:
F[tex]_{s}[/tex]=δma[tex]_{s}[/tex]
Applying bernoulli,s equation as follows:
∑[tex]δp+\frac{1}{2} ρδV^{2} +δγz=0\\[/tex]
Where, [tex]δp[/tex] is the pressure change across the streamline and [tex]V[/tex] is the fluid particle velocity
substitute [tex]ρg[/tex] for {tex]γ[/tex] and [tex]g_{0}-cz[/tex] for [tex]g[/tex]
[tex]dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0[/tex]
integrating the above equation using limits 1 and 2.
[tex]\int\limits^2_1 \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant[/tex]
there the bernoulli equation for this flow is [tex]p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant[/tex]
note: [tex]ρ[/tex]=density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular
A collapsible plastic bag contains a glucose solution. If the average gauge pressure in the vein is 1.33 x 103 Pa, what must be the minimum height h of the bag in order to infuse glucose into the vein? Assume the density of the solution is 1.02 x 103 kg/m3 .
Answer:
The height of the bag will be 0.133 m
Explanation:
We have given gauge pressure [tex]P=1.33\times 10^3Pa=1330Pa[/tex]
Density of solution [tex]\rho =1.02\times 10^3=1020kg/m^3[/tex]
We have to find the height of the bag
We know that gauge pressure is given by P=\rho gh
[tex]1330=1020\times 9.8\times h[/tex]
h=0.133m
So the height of the bag will be 0.133 m
Final answer:
The minimum height of the bag should be 0.245 meters to infuse glucose into the vein.
Explanation:
In order for the fluid to enter the vein, its pressure at entry must exceed the blood pressure in the vein. To find the height of the fluid, we need to convert the blood pressure in mm Hg to SI units. Since 1.0 mm Hg = 133 Pa, the blood pressure in the vein is 18 mm Hg above atmospheric pressure, which is equivalent to (18 mm Hg)(133 Pa/mm Hg) = 2386 Pa.
Now we can calculate the height of the fluid using the formula:
h = P/(ρg)
Where:
h is the height of the fluidP is the gauge pressure (2386 Pa)ρ is the density of the solution (1.02 x 10^3 kg/m^3)g is the acceleration due to gravity (9.8 m/s^2)Substituting the given values into the formula, we find:
h = (2386 Pa)/((1.02 x 10^3 kg/m^3)(9.8 m/s^2)) = 0.245 m
The minimum height of the bag should be 0.245 meters in order to infuse glucose into the vein.
The transfer of energy by electromagnetic waves is called
Radiation refers to the transfer of energy by electromagnetic waves, a process that can take place with no medium, such as the heat from the sun reaching Earth.
Explanation:The transfer of energy by electromagnetic waves is known as radiation. This process does not require a medium; the energy is carried by photons in the electromagnetic waves. Examples of this process include the heat produced by the sun, which reaches the Earth via the transfer of radiant energy, and microwaves heating food through radiation.
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nder some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly 1014 times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was 6.0×105 km (comparable to our sun); its final radius is 16 km.If the original star rotated once in 35 days, find the angular speed of the neutron star.
Answer:
2.92 * 10³ rad/s
Explanation:
Given:
Initial Radius of Original Star (Ri) = 6.0 * 10^5 km
Final Radius of Neutron Star (Rf) = 16km
Angular Speed = 1 revolution in 35 days
We need to convert this to rad/s
To do that, we first convert to rad/day
i.e (1 rev/35 days) * (2π rad/ 1 rev)
We then convert the days to hour
i.e. (1 rev / 35 days) * (2π rad/ 1 rev) * (1 day / 24 hours)
Finally, we convert the hour to seconds (3600 seconds makes an hour)
i.e. (1 rev / 35 days) * (2π rad/ 1 rev) * (1 day / 24 hours) * (1 hour/ 3600 sec)
Angular Speed = 2π rad/ 3024000 secs
Angular Speed (wi) = 2.079 * 10^-6rad/s
From the question, we're asked to calculate the angular speed of the neutron star (wf)
Applying law of conservation of angular momentum to a system whose moment of Inertia changes, we have
Ii*wi = If * wf ----------------- Formula
Where Ii and If are the initial and final Inertia of the star
The relationship between Inertia and Radius of each object is I = 2/5MR²
So, Ii = 2/5(MRi²) and If = 2/5(MRf²)
Substitute the above in the formula quoted
We have 2/5(MRi²)wi = 2/5(MRf²)wf ---------------- Divide through by 2M/5
We are left with, Ri²wi = Rf²wf
Make wf the subject of the formula
wf = wi * (Ri/Rf)²
wf = 2.079 * 10^-6rad/s * (6.0 * 10^5 km/16km)²
wf = 2.079 * 10^-6rad/s * (6.0 * 10^5 km/16km) * (6.0 * 10^5 km/16km)
wf = 2.92 * 10³ rad/s
Two wheels having the same radius and mass rotate at the same angular velocity ((Figure 1) ). One wheel is made with spokes so nearly all the mass is at the rim. The other is a solid disk.How do their rotational kinetic energies compare?A. The wheel with spokes has higher KE, but not twice as high.B. They are nearly the same.C. The solid wheel has higher KE, but not twice as high.D. The solid wheel has about twice the KE.E. The wheel with spokes has about twice the KE.
Answer:
E. The wheel with spokes has about twice the KE.
See explanation in: https://quizlet.com/100717504/physics-8-mc-flash-cards/
Answer:
Explanation:
We have to consider how the location of the mass affects the moment of inertia.
For a solid cylinder, I = mR²
For a hollow cylinder, I = 1/2mR²
Where
I ist the moment of inertia,
m is their masses,
R is the radius of rotation.
Since they have the same mass and radius, it can be seen that a hollow cylinder has twice the moment of inertia as a solid cylinder of the same mass and radius.
We know that the rotational kinetic energy is proportional to the moment of inertia. From;
Rotational KE = 1/2IW²
Where W is the angular speed.
so that at the same angular speed, the wheel with the spokes will have about double the kinetic energy as the solid cylinder. Take note that some of the mass is in the spokes so the moment of inertia is not exactly double.
A 75.5-kg person puts on a life jacket, jumps into the water, and floats. The jacket has a volume of 3.38 x 10-2 m3 and is completely submerged under the water. The volume of the person's body that is underwater is 6.42 x 10-2 m3. What is the density of the life jacket?
Answer:
Density of jacket will be [tex]680.4733kg/m^3[/tex]
Explanation:
We know that weight of water displaced= buoyant force=weight of man
Now volume of water displaced [tex]v=3.38\times 10^{-2}+6.42\times 10^{-2}=9.8\times 10^{-2}m^3[/tex]
Density of water [tex]d=100kg/m^3[/tex]
So weight of water displaced [tex]=9.8\times 10^{-2}\times 1000=98kg[/tex]
So weight of jacket = 98-75 = 23 kg
We have given volume of the jacket = [tex]3.38\times 10^{-2}m^3[/tex]
So density of jacket [tex]=\frac{mass}{volume}=\frac{23}{3.38\times 10^{-2}}=680.4733kg/m^3[/tex]
Suppose a ping pong ball and a bowling ball are rolling toward you. Both have the same linear momentum, and you exert the same force to stop each.(A) How does the time required to stop the ping pong ball compare to that required to stop the bowling ball?(B) How does the distance required to stop ping pong ball compare to that required to stop the bowling ball? Explain your reasoning and justify your answer starting from (and naming) fundamental physics principles and definitions related to momentum and energy. Thanks
Answer:
The weight of the bowling ball makes a more significant impact that the ping pong ball so therefore it would take farther to stop the bowling ball
Explanation:
(A) The time required to stop the ping pong ball will be less than that of bowling ball.
(B) The distance required to stop the bowling ball will be less than that of ping pong ball.
Given data:
The ping pong ball and bowling ball has the same magnitude of linear momentum.
Same amount of force to be applied on each, to stop.
(A)
With same magnitude of stopping force (F) and linear momentum (p), the time required to stop will be dependent on the mass. Since, mass of ping pong ball is less, so it will be easily stopped with less effort. While the bowling ball will take some extra seconds or minutes to acquire the rest state.
In other words, the ping pong ball has less inertia, due to which the time taken to stop the ping pong pall will be less, comparing to bowling ball.
Thus, we can conclude that the time required to stop the ping pong ball will be less than that of bowling ball.
(B)
The inertia is given as,
[tex]I = mr^{2}[/tex],
Here, m is the mass.
And the distance required to stop is given by third rotational equation of motion as,
[tex]\omega_{2}^{2}=\omega_{1}^{2}+2 \alpha \theta\\\\\theta =\dfrac{ \omega_{2}^{2}-\omega_{1}^{2}}{2 \alpha}[/tex]
Here, [tex]\alpha[/tex] is the angular acceleration.
And angular acceleration is directly proportional to the inertia of object. More the inertia, more will be the angular acceleration and less will be distance required to stop.
Since, ping pong ball has less inertia, so its angular acceleration will be less. So, the distance covered by the ping pong ball will be more, compared to bowling ball.
Thus, we can conclude that the distance required to stop the bowling ball will be less than that of ping pong ball.
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Two ice skaters collide on the ice. A 39.6-kg skater moving South at 6.21 m/s collides with a 52.1-kg skater moving East at 4.33 m/s. The two skaters entangle and move together across the ice. Determine the magnitude and direction of their post-collision velocity.
Answer:
V = 3.6385 m/s
θ = 47.46 degrees
Explanation:
the important data in the question is:
Skater 1:
[tex]M_1[/tex]= 39.6 kg
direction: south (axis y)
[tex]V_{1iy}[/tex] = 6.21 m/s
Skater 2:
[tex]M_2[/tex] = 52.1 kg
direction: east (axis x)
[tex]V_{2ix}[/tex] = 4.33 m/s
Now using the law of the conservation of linear momentum ( [tex]P_i = P_f[/tex] and knowing that the collision is inelastic we can do the next equations:
[tex]M_{1}V_{1ix}+M_2V_{2ix} = V_{sx}(M_1+M_2)[/tex] (eq. 1)
[tex]M_{1}V_{1iy}+M_2V_{2iy} = V_{sy}(M_1+M_2)[/tex] (eq. 2)
Where [tex]V_{sx}[/tex] and [tex]V_{sy}[/tex] is the velocity of the sistem in x and y after the collision.
Note: the conservation of the linear momentum have to be make once by each axis.
Now, in the (eq. 1) the skater 1 don't have velocity in the axis x, so we can replace [tex]V_{1ix}[/tex] by 0 in the equation and get:
[tex]M_2V_{2ix} = V_{sx}(M_1+M_2)[/tex] (eq. 1)
also, in the (eq. 2) the skater 2 don't have velocity in the axis y, so we can replace [tex]V_{2iy}[/tex] by 0 in the equation and get:
[tex]M_{1}V_{1iy} = V_{sy}(M_1+M_2)[/tex] (eq. 2)
Now, we just replace the data in both equations:
[tex](52.1)(4.33) = V_{sx}(39.6+52.1)[/tex] (eq. 1)
[tex](39.6)(6.21) = V_{sy}(39.6+52.1)[/tex] (eq. 2)
solving for [tex]V_{sx][/tex] and [tex]V_{sy}[/tex] we have:
[tex]V_{sx][/tex] = 2.46 m/s
[tex]V_{sy][/tex] = 2.681 m/s
using the pythagoras theorem we can find the magnitude of the velocity as:
V = [tex]\sqrt{2.46^2+2.681^2}[/tex]
V = 3.6385 m/s
For find the direction we just need to do this;
θ = [tex]tan^{-1}(\frac{2.681}{1.46})[/tex]
θ = 47.46 degrees
A projectile is launched from the earth’s surface at initial speed v0 at angle θ0 with the horizontal. When the projectile is at its maximum height h, it has half the speed it had when it was at half its maximum height h 2 . At what angle was the projectile launched.
Answer:
θ₀ = 67.79°
Explanation:
Given info
we know that
Ymax = v₀y² / (2g)
v = v₀x (when Y = Ymax)
when the projectile was at half its maximum height (Y')
v' = 2v = 2*v₀x
we can use the equations
(v')²= v'x² + v'y² ⇒ (2*v₀x)² = (v₀x)² + v'y² ⇒ v'y² = 3*(v₀x)² (I)
if we know that
v'y² = v₀y² - 2*g*(y')
y' = Ymax /2 = (v₀y² / (2g)) / 2 = v₀y² / (4g)
then
v'y² = v₀y² - 2*g*(v₀y² / (4g)) = v₀y² - (v₀y² / 2) = v₀y² / 2
⇒ v'y² = v₀y² / 2 (II)
we can say that (I) = (II)
3*(v₀x)² = v₀y² / 2 ⇒ v₀y = √6*v₀x
Finally we apply
tan θ₀ = v₀y / v₀x
⇒ tan θ₀ = √6*v₀x / v₀x = √6
⇒ θ₀ = tan⁻¹ (√6) = 67.79°
Explain how it is possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source.
Yes, the voltage across a capacitor in an RLC series circuit with AC can be greater than the source voltage, especially at resonance where the capacitive and inductive reactances cancel each other out and energy oscillates between the capacitor and inductor.
Explanation:In an RLC series circuit with alternating current (AC), it is indeed possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source. This phenomenon occurs due to resonance in the circuit, where the capacitive and inductive reactances can cancel each other out at a certain frequency, known as the resonant frequency. When the circuit is at resonance, the voltages across the capacitor and inductor can be much greater than the source voltage because of the energy oscillating between the electric field of the capacitor and the magnetic field of the inductor.
According to the equation Vc = Q/C, where Q is the charge and C is the capacitance, we see that the voltage across the capacitor (Vc) is directly proportional to the amount of charge stored and inversely proportional to the capacitance. In an AC RLC circuit at resonance, the charge can oscillate at amplitudes resulting in voltages across the capacitor that exceed the source voltage.
The same can be true for the voltage across the inductor. If the circuit is driven at or near its resonant frequency, the inductive reactance and capacitive reactance can become equal in magnitude but opposite in phase, leading to a situation where the voltage across the inductor also exceeds the source voltage due to the energy stored in its magnetic field.
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Blaise Pascal duplicated Torricelli's barometer using a red Bordeaux wine of density 965 kg/m3 as the working liquid (see figure below).
1. What was the height h of the wine column for normal atmospheric pressure?
2. Would you expect the vacuum above the column to be as good as that for mercury?
A. Yes
B. No
Answer:
a) h=10.701m
b) No. On this case a liquid like wine is not good as the mercury, because the wine is composed of water, alcohol and other elements, but specially the alcohol evaporates much easier than the mercury, and that will cause malfunction in the vacuum of the baroemter used for the experiment.
Explanation:
The Torricelli's experiment "was invented by the Italian scientist Evangelista Torricelli and the most important purpose of this experiment was to prove that the source of vacuum comes from atmospheric pressure"
Pressure is defined as "the force that is applied on any object in the direction perpendicular to the surface of the object in the unit area is known as the pressure. There are various types of pressure".
Part a
We have the density for the red Bordeaux wine given [tex]\rho=965\frac{kg}{m^3}[/tex], the atmospheric pressure on the Toriccelli's barometer is given by:
[tex]P_{atm}=\rho g h[/tex]
Solving for the height of wine in the column we have this:
[tex]h=\frac{P_{atm}}{\rho g}[/tex]
And replacing we have:
[tex]h=\frac{101300Pa}{965\frac{kg}{m^3} 9.81\frac{m}{s^2}}=10.701 m[/tex]
So the height of the red Bordeaux wine would be h=10.701m. A very high value on this case if we compare with the usual values for this variable.
Part b
No. On this case a liquid like wine is not good as the mercury, because the wine is composed of water, alcohol and other elements, but specially the alcohol evaporates much easier than the mercury, and that will cause malfunction in the vacuum of the baroemter used for the experiment.
The height of the wine column for normal atmospheric pressure can be calculated using the equation P = ρgh. The vacuum above the column in the wine barometer would not be as good as that for mercury. The correct answer is No (B).
Explanation:The height h of the wine column for normal atmospheric pressure can be calculated using the equation P = ρgh, where P is the atmospheric pressure, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the height of the column. The atmospheric pressure can be taken as P = 101325 Pa and the acceleration due to gravity is approximately g = 9.8 m/s^2. Plugging in these values and the density of the wine, ρ = 965 kg/m^3, we can solve for h. Hence, h = P / (ρg).
To compare the vacuum above the column in the wine barometer with that of a mercury barometer, it is important to note that the height of the liquid column in a barometer is determined by the atmospheric pressure pushing down on the liquid. Since both the wine and mercury barometers are subjected to the same atmospheric pressure, the height of the columns will be the same if the densities of the liquids are the same. However, since the density of wine is less than that of mercury, the height of the wine column will be greater than that of the mercury column. Therefore, No (B) is the correct answer.
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The amount of electrons flowing through the wire is called what?
amperes (amps)
current
voltage
ohms
Answer:
Current
Explanation:
Current can refer to the flow of electrons through a conductor of some kind as well as the number of electrons flowing through the conductor.
Final answer:
The amount of electrons flowing through the wire is called current, which is measured in amperes or amps. Current is directly proportional to voltage and inversely proportional to resistance, as stated in Ohm's Law.
Explanation:
The amount of electrons flowing through the wire is called current. Electric current is measured in the unit known as ampere (A), which is defined as the flow of one coulomb of charge through an area in one second. According to Ohm's Law, the current I in a circuit is directly proportional to the voltage V applied across the circuit and inversely proportional to the resistance R of the circuit, which can be expressed by the equation I = V/R. The SI unit of current is ampere, whereas ohm is the SI unit for electrical resistance, which represents how strongly a material opposes the flow of electric current.
While sitting in your car by the side of a country road, you see your friend, who happens to have an identical car with an identical horn, approaching you. You blow your horn, which has a frequency of 260 Hz; your friend begins to blow his horn as well, and you hear a beat frequency of 6.0Hz .
How fast is your friend approaching you?
Express your answer to two significant figures and include the appropriate units.
Answer:
[tex] v_s =7.74\ m/s[/tex]
Explanation:
given,
Speed of sound = 343 m/s
frequency of horn = 260 Hz
the friend is approaching, the frequency is increased by the Doppler Effect. The frequency is 266 Hz
using formula
[tex]f' = \dfrac{v}{v-v_s}f_0[/tex]
[tex]266= \dfrac{343}{343 - v_s}(260)[/tex]
[tex]1.023= \dfrac{343}{343 - v_s}[/tex]
[tex]343 - v_s = 335.26[/tex]
[tex] v_s =7.74\ m/s[/tex]
the speed of friends approaching is equal to [tex] v_s =7.74\ m/s[/tex]
Three thermometers are in the same water bath. After thermal equilibrium is established, it is found that the Celsius thermometer reads 0 °C, the Fahrenheit thermometer reads 12°F, and the Kelvin thermometer reads 273 K. Which one of the following statements is the most reasonable conclusion?
a) the Kelvin thermometer is incorrect
b) the Celsius thermometer is incorrect
c) the Fahrenheit thermometer is incorrect
d) the three thermometers are all at different temperatures
Answer:
C) the Fahrenheit thermometer is incorrect
Explanation:
Since
1) K = °C + 273
2) °F = 9/5 °C + 32
for 0 °C
1) K = 0°C + 273 = 273 K
2) °F = 9/5 * 0°C + 32 = 32 °F
Thus the Kelvin thermometer measurement coincides with the Celsius measurement but not with the °F . On the other hand, if the Fahrenheit measurement is right, the Celsius thermometer and the Kelvin one should be wrong.
Therefore is more reasonable to assume that one thermometer failed (the one of Fahrenheit and both Kelvin and Celsius are right ) that 2 thermometers ( Celsius and Kelvin thermometers fail and the one of Fahrenheit is right)
For the three thermometers, we see that one thermometer failed and 2 are right,, Hence the Fahrenheit thermometer is incorrect
Option C is correct
What is ?
Generally, the equation for finding Kelvin is mathematically given as
K = °C + 273
Where
°F = 9/5 °C + 32
In conclusion, The Kelvin thermometer measurement relates with Celsius measurement but has no direct link with Fahrenheit F, the Fahrenheit thermometer is incorrect
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A potter's wheel (a solid, uniform disk) of mass 7.0 kg and .65 m radius spins about its central axis. A 2.1 kg lump of clay is dropped onto the wheel at a distance .41 m from the axis. Calculate the rotational inertia of the system.
a. 2.5 kg · m2
b. 1.8 kg · m2
c. 1.5 kg · m2
d. 0.40 kg · m2
Answer:
The rotational inertia of the system is 1.8 kg.m².
(b) is correct option.
Explanation:
Given that,
Mass of disk = 7.0 kg
Radius = 0.65 m
Mass of clay = 2.1 kg
Distance = 0.41 m
We need to calculate the rotational inertia of the system
Using formula of rotational inertia
[tex]I''=I+I'[/tex]
Where, I= the moment of inertia of a solid disk
I'=the moment of inertia of lump of clay
Put the value into the formula
[tex]I=\dfrac{MR^2}{2}+mr^2[/tex]
[tex]I=\dfrac{1}{2}\times7.0\times(0.65)^2+2.1\times0.41^2[/tex]
[tex]I=1.8\ kg.m^2[/tex]
Hence, The rotational inertia of the system is 1.8 kg.m².
Radio Waves If your favorite FM radio station broadcasts at a frequency of 104.5 MHz, what is the wavelength of the station’s signal in meters? What is the energy of a photon of the station’s electromagnetic signal?
Answer:
The wavelength of the station’s signal is 2.9 meters
The energy of a photon of the station’s electromagnetic signal is [tex]6.9\times10^{-26}J [/tex]
Explanation:
Wavelength [tex] \lambda [/tex] is inversely proprtional to frequency (f) and directly proportional to velocity of the wave (v).
[tex]\lambda=\frac{v}{f} [/tex] (1)
But electromagnetic waves as radio signals travel at speed of light so using this on (1):
[tex]\lambda=\frac{c}{f}=\frac{3.0\times10^{8}}{104.5\times10^{6}}\approx2.9\,m [/tex]
Albert Einstein discovered that energy of electromagnetic waves was quantized in small discrete packages of energy called photons with energy:
[tex] E=hf=(6.6\times10^{-34})(104.5\times10^{6})\approx6.9\times10^{-26}J[/tex]
with h the Planck's constant.
The wavelength of a 104.5 MHz FM radio signal is approximately 2.87 meters, and the energy of a photon of this radio signal is approximately 6.92 * 10^-26 Joules.
Explanation:The subject of this question is the relationship between the frequency, wavelength, and energy of radio waves, specifically those used in FM radio broadcasting.
To calculate the wavelength of the radio signal, one can employ the wave equation: velocity = frequency * wavelength. Since the velocity of electromagnetic waves, which include radio waves, is the speed of light (3 * 10^8 m/s), the wavelength can be obtained by rearranging the equation to: wavelength = velocity / frequency. Using your FM station's frequency of 104.5 MHz (or 104.5 * 10^6 Hz), the wavelength of the station's signal would be approximately 2.87 meters.
The energy of a photon from this radio signal could be found through the photon energy equation: E = h * f, where h is Planck's constant (6.62607004 × 10^-34 m^2 kg / s) and f is the frequency in Hz. Thus, the energy of a radio signal photon at 104.5 MHz would be approximately 6.92 * 10^-26 J.
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An ideal monatomic gas expands isobarically from state A to state B. It is then compressed isothermally from state B to state C and finally cooled at constant volume until it returns to its initial state A.
[tex]V_A[/tex] = 4 x 10⁻³[tex]V_B[/tex] = 8 x 10⁻³[tex]P_A[/tex] = [tex]P_B[/tex] = 1 x 10⁶ [tex]P_C[/tex] = 2 x 10⁶[tex]T_A[/tex] = 600 k
What is the temperature of the gas when it is in state B?
How much work is done by the gas in expanding isobarically from A to B?
How much work is done on the gas in going from B to C?
The graph is a PV diagram. It does not show any other numbers on it other than what is given.
1) The temperature of the gas in state B is 1200 K
2) The work done by the gas from A to B is 4000 J
3) The work done by the gas from B to C is -5546 J
Explanation:
1)
The temperature of the gas when it is in state B can be found by using the ideal gas equation:
[tex]\frac{P_A V_A}{T_A}=\frac{P_B V_B}{T_B}[/tex]
where
[tex]P_A = 1\cdot 10^6 Pa[/tex] is the pressure in state A
[tex]V_A = 4 \cdot 10^{-3} m^3[/tex] is the volume in state A
[tex]T_A = 600 K[/tex] is the temperature in state A
[tex]P_B = 1\cdot 10^6 Pa[/tex] is the pressure in state B
[tex]V_B = 8\cdot 10^{-3} m^3[/tex] is the volume in state B
[tex]T_B[/tex] is the temperature in state B
Solving for [tex]T_B[/tex],
[tex]T_B=\frac{P_B V_B T_A}{P_A V_A}=\frac{(1\cdot 10^6)(8\cdot 10^{-3})(600)}{(1\cdot 10^6)(4\cdot 10^{-3})}=1200 K[/tex]
2)
The work done by a gas during an isobaric transformation (as the one between A and B) is
[tex]W=p\Delta V = p (V_B-V_A)[/tex]
where
p is the pressure (which is constant)
[tex]V_B[/tex] is the final volume
[tex]V_A[/tex] is the initial volume
Here we have:
[tex]p=1\cdot 10^6 Pa[/tex]
[tex]V_A = 4 \cdot 10^{-3} m^3[/tex] is the volume in state A
[tex]V_B = 8\cdot 10^{-3} m^3[/tex] is the volume in state B
Substituting,
[tex]W=(1\cdot 10^6)(8\cdot 10^{-3}-4\cdot 10^{-3})=4000 J[/tex]
3)
During an isothermal expansion, the produce between the pressure of the gas and its volume remains constant (Boyle's law), so we can write:
[tex]P_BV_B = P_C V_C[/tex]
where
[tex]P_B = 1\cdot 10^6 Pa[/tex] is the pressure in state B
[tex]V_B = 8\cdot 10^{-3} m^3[/tex] is the volume in state B
[tex]P_C = 2\cdot 10^6 Pa[/tex] is the pressure in state C
[tex]V_C[/tex] is the volume in state C
Solving for [tex]V_C[/tex],
[tex]V_C = \frac{P_B V_B}{P_C}=\frac{(1\cdot 10^6)(8\cdot 10^{-3})}{2\cdot 10^6}=4\cdot 10^{-3} m^3[/tex]
The work done by a gas during an isothermal transformation is given by
[tex]W=nRT ln(\frac{V_C}{V_B})[/tex] (1)
where
n is the number of moles
[tex]R=8.314 J/mol K[/tex] is the gas constant
T is the constant temperature (in this case, [tex]T_B[/tex])
[tex]V_C, V_B[/tex] are the final and initial volume, respectively
The number of moles of the gas can be found as
[tex]n=\frac{p_A V_A}{RT_A}=\frac{(1\cdot 10^6)(4\cdot 10^{-3})}{(8.314)(600)}=0.802 mol[/tex]
So now we can use eq.(1) to find the work done by the gas from B to C:
[tex]W=(0.802)(8.314)(1200) ln(\frac{4\cdot 10^{-3}}{8\cdot 10^{-3}})=-5546 J[/tex]
And the work is negative because the gas has contracted, so the work has been done by the surrounding on the gas.
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The temperature of the gas when it is in state B can be found using the ideal gas law. The work done by the gas in expanding isobarically from A to B can be calculated using a simple equation. The work done on the gas in going from B to C can also be calculated using a different equation.
The temperature of the gas when it is in state B can be found using the ideal gas law, which states that PV = nRT. Since the process is isobaric, the pressure remains constant. We can use the equation:
TB = TA * (VA / VB)
Where TA is the initial temperature (600 K), VA is the initial volume (4 x 10-3), and VB is the final volume (8 x 10-3).
The work done by the gas in expanding isobarically from A to B can be calculated using the equation:
WAB = PA * (VB - VA)
Where PA is the initial pressure (1 x 106) and VA and VB are the initial and final volumes, respectively.
The work done on the gas in going from B to C can be calculated using the equation:
WBC = -nRT * ln(VC / VB)
Where VC is the final volume (8 x 10-3) and VB is the initial volume (8 x 10-3).
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Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is 25.0 kg and that of the boat is 30.0 kg . (Take the package's direction of motion as positive.)
Answer:
V = -0.8 m/s
Explanation:
given,
mass of the child (m)= 25 Kg
mass of the boat(M) = 30 Kg
velocity of boat = ?
Assuming Boys throws package of mass(m₁) 6 Kg at the horizontal speed of 10 m/s
using conservation of momentum
(M + m + m₁) V = (M+ m)V + m₁ v
initial velocity V = 0 m/s
(M + m + m₁) x 0 = (M+ m)V + m₁ v
0 = (25+50)V + 6 x 1 0
75 V = -60
V = -0.8 m/s
negative direction shows that velocity in the direction opposite to the motion of package.
Final answer:
The velocity of the boat immediately after, assuming it was initially at rest, is 15.8 m/s.
Explanation:
To calculate the velocity of the boat, we can use vector addition. The boat's velocity relative to the water is perpendicular to the river's velocity. We can use the Pythagorean theorem to find the magnitude of the boat's velocity:
Vboat = sqrt((Vriver)2 + (Vboat)2)
Substituting the given values, we get:
Vboat = sqrt((5.0 m/s)2 + (15.0 m/s)2) = 15.8 m/s
Therefore, the velocity of the boat immediately after is 15.8 m/s.
When an electron de-excites from the third quantum level to the second, and then to the ground state, two photons are emitted. The sum of the emitted frequencies equals the frequency of the single-photon that would be emitted if de-excitation were from the third to ____.
Answer:
ground state
Explanation:
Lets take
n=3 ,n=2 ,n=1 are the energy level.
Energy level n=1 is the ground energy level.
The energy from 3 to 1 = hν
The energy from 3 to 2 = hν₁
The energy from 2 to 1 = hν₂
We can say that
hν = hν₁ + hν₂
If the electron were de-excitation from the third level to ground level then the sum of emitted frequency will be equal to the frequency of a single electron.
Therefore the answer is ground state.
It is well known that for a hollow, cylindrical shell rolling without slipping on a horizontal surface, half of the total kinetic energy is translational and half is rotational. What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface?Part AA uniform solid cylinder.Part BA uniform sphere.Part CA thin-walled hollow sphere.Part DA hollow, cylinder with outer radius R and inner radius R/2.
Answer:
Part a)
[tex]f = \frac{0.5}{1.5} = \frac{1}{3}[/tex]
B) uniform Sphere
[tex]f = \frac{2}{7}[/tex]
C) uniform hollow sphere
[tex]f = \frac{2}{5}[/tex]
D) uniform hollow cylinder with inner radius R/2 and outer radius R
[tex]f = \frac{5}{13}[/tex]
Explanation:
As we know that fraction of total energy as rotational energy is given as
[tex]f = \frac{\frac{1}{2}I\omega^2}{\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2}[/tex]
now we have
[tex] f = \frac{mk^2(\frac{v^2}{R^2})}{mv^2 + mk^2(\frac{v^2}{R^2})}[/tex]
[tex]f = \frac{\frac{k^2}{R^2}}{1 + \frac{k^2}{R^2}}[/tex]
now we have
A) uniform Solid cylinder
for cylinder we know that
[tex]\frac{k^2}{R^2} = 0.5[/tex]
[tex]f = \frac{0.5}{1.5} = \frac{1}{3}[/tex]
B) uniform Sphere
for sphere we know that
[tex]\frac{k^2}{R^2} = \frac{2}{5}[/tex]
[tex]f = \frac{0.4}{1.4} = \frac{2}{7}[/tex]
C) uniform hollow sphere
for hollow sphere we know that
[tex]\frac{k^2}{R^2} = \frac{2}{3}[/tex]
[tex]f = \frac{\frac{2}{3}}{\frac{5}{3}} = \frac{2}{5}[/tex]
D) uniform hollow cylinder with inner radius R/2 and outer radius R
for annular cylinder we know that
[tex]\frac{k^2}{R^2} = \frac{5}{8}[/tex]
[tex]f = \frac{\frac{5}{8}}{\frac{13}{8}} = \frac{5}{13}[/tex]
The fraction of the total kinetic energy that is rotational for a uniform sphere is [tex]\frac{1}{3}[/tex]
How to calculate rotational kinetic energy.Mathematically, the rotational kinetic energy of an object is given by this formula:
[tex]K.E_{rot}=\frac{1}{2} I\omega^2[/tex]
Where:
I is the moment of inertia.[tex]\omega[/tex] angular velocity.Since we know that half of the total kinetic energy for a hollow, cylindrical shell that is rolling without slipping on a horizontal surface is translational and the other half is rotational. Thus, this fraction is given by this mathematical expression:
[tex]K.E=\frac{\frac{k^2}{R^2} }{1+\frac{k^2}{R^2}}[/tex]
a. For a uniform sphere:
[tex]\frac{k^2}{R^2}=0.5[/tex]
Substituting, we have:
[tex]K.E=\frac{0.5 }{1+0.5}\\\\K.E=\frac{0.5 }{1.5}\\\\K.E =\frac{1}{3}[/tex]
b. For a thin-walled hollow sphere:
[tex]\frac{k^2}{R^2}=\frac{2}{5}[/tex]
Substituting, we have:
[tex]K.E=\frac{\frac{2}{5} }{1+\frac{2}{5}}\\\\K.E=\frac{0.4 }{1.4}\\\\K.E =\frac{2}{7}[/tex]
c. For a uniform hollow sphere:
[tex]\frac{k^2}{R^2}=\frac{2}{3}[/tex]
Substituting, we have:
[tex]K.E=\frac{\frac{2}{3} }{1+\frac{2}{3}}\\\\K.E=\frac{0.7 }{1.7}\\\\K.E =\frac{2}{5}[/tex]
d. For a hollow sphere with outer radius (R) and inner radius:
[tex]\frac{k^2}{R^2}=\frac{5}{8}[/tex]
Substituting, we have:
[tex]K.E=\frac{\frac{5}{8} }{1+\frac{5}{8}}\\\\K.E=\frac{0.625 }{1.625}\\\\K.E =\frac{5}{13}[/tex]
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Would you expect the tensile strength of polychlorotrifluoroethylene to be greater, the same as, or less than that of a polytetrafluoroethylene specimen having the same molecular weight and degree of crystallinity?
The tensile strength of PCTFE is generally expected to be higher than that of PTFE, given the same molecular weight and degree of crystallinity, mainly due to stronger intermolecular forces from the chlorine atom in the PCTFE polymer chain.
Explanation:The tensile strength of polychlorotrifluoroethylene (PCTFE) and polytetrafluoroethylene (PTFE) specimens having the same molecular weight and degree of crystallinity can vary. Although both materials are fluoropolymers and have similar structural characteristics, PCTFE tends to have a slightly higher tensile strength than PTFE. This is primarily because PCTFE has a chlorine atom in its polymer chain, which contributes to stronger intermolecular forces and therefore, higher tensile strength.
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Cosmic rays are particles traveling at extreme speeds through intergalactic space. Many are launched by exploding supernovae. Even the slowest cosmic rays travel at 43% of the speed of light. Calculate the gamma factor for cosmic rays traveling at these speeds.
Answer:
1.1
Explanation:
Cosmic rays are particles traveling at extreme speeds through intergalactic space. Many are launched by exploding supernovae. From the question, we know that slowest cosmic rays travel at 43% of the speed of light that is 43/100 multiply by the speed of light(speed of light= 3.0×10^8 m/s).
Gamma factor can be calculated using the formula below;
Gamma factor= 1/(1- v^2/c^2)^1/2 --------------------------------------------(1).
Gamma factor= 1/(1-3.0×10^8 metre per seconds ×43/100÷ 3.0×10^8 metre per seconds) ^1/2
Gamma factor= 1/(1-1.6641×10^16/3.0×10^2)^1/2
Gamma factor= 1/(1-1.16641×10^16/9×10^16)^1/2
Gamma factor= 1/(1-0.1296)^1/2
Gamma factor= 1/(0.87)^1/2
Gamma factor= 1/0.9327
Gamma factor= 1.072112535
Gamma factor= ~ 1.1
Cosmic rays are high-speed particles that travel through space close to the speed of light. The gamma factor for cosmic rays traveling at 43% of the speed of light is 1.17.
Cosmic rays are high-speed particles that travel through interstellar space at speeds close to the speed of light. When calculating the gamma factor for cosmic rays traveling at 43% of the speed of light, we can use the formula: γ = 1 / √(1 - v^2/c^2), where v is the velocity of the particles and c is the speed of light. Substituting the values, the gamma factor for cosmic rays traveling at 43% of the speed of light is approximately 1.17.
Find the magnitude of the gravitational force this wire exerts on a point with mass m placed at the center of curvature of the semicircle.
Answer:
[tex]F = 2\pi \dfrac{mMG}{L^2}}[/tex]
Explanation:
Assuming given,
Mass of wire be M
length of wire be L
small mass at center is = m
Radius of the wire be equal to = R = L/π
mass of small element of the wire
[tex]dM = \dfrac{M}{L}Rd\theta[/tex]
All the force are acting along y- direction
so, for force calculation
[tex]F = \int \dfrac{mdMG}{R^2} sin\theta [/tex]
[tex]F = \int_0^{\pi} \dfrac{m\dfrac{M}{L}RG}{R^2} sin\theta d\theta[/tex]
[tex]F = \int_0^{\pi} \dfrac{mMG}{L\dfrac{L}{\pi}} sin\theta d\theta[/tex]
[tex]F = \pi \dfrac{mMG}{L^2}}\int_0^{\pi}sin\theta d\theta[/tex]
[tex]F = \pi \dfrac{mMG}{L^2}}(-cos \theta)_0^{\pi}[/tex]
[tex]F = 2\pi \dfrac{mMG}{L^2}}[/tex]
The magnitude of the gravitational force is : [tex]2\pi \frac{mMG}{L^{2} }[/tex]
Given that :
mass of wire = M
length of wire = L
small mass at center = m
radius of wire = L / [tex]\pi[/tex]
Determine the magnitude of the gravitational force exerted by the wire on a pointFirst step : express the mass of small element of wire
dM = [tex]\frac{M}{L} Rd[/tex]∅
Since all forces act in the vertical direction the magnitude of the force exerted will be
F = [tex]\int\limits^\pi _o {\frac{m\frac{M}{L} RG}{R^2} } \, sin\beta d\beta[/tex]
[tex]F = \pi \frac{mMG}{L^2} \int\limits^\pi _0 {x} \, sin\beta d\beta[/tex]
Resolving equation above
Therefore F = [tex]2\pi \frac{mMG}{L^{2} }[/tex]
Hence we can conclude that The magnitude of the gravitational force is : [tex]2\pi \frac{mMG}{L^{2} }[/tex]
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An inflatable raft (unoccupied) floats down a river at an approximately constant speed of 4.8 m/s. A child on a bridge, 81 m above the river, sees the raft in the river below and attempts to drop a small stone onto the raft. The child releases the stone from rest. In order for the stone to hit the raft, what must be the horizontal distance between the raft and the bridge when the child releases the stone?
Answer:
19.5 m
Explanation:
t = Time taken
u = Initial velocity = 0 (Assumed thrown from rest)
s = Displacement = 81 m
g = Acceleration due to gravity = 9.81 m/s² = a
Equation of linear motion
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 81=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{81\times 2}{9.81}}\\\Rightarrow t=4.06371\ s[/tex]
The time taken for the stone to reach the river is 4.06371 seconds
Distance = Speed×Time
[tex]Distance=4.8\times 4.06371=19.5\ m[/tex]
The horizontal distance between the raft and the bridge when the child releases the stone should be 19.5 m
Each second, 1250 m3 of water passes over a waterfall 150 m high. Three-fourths of the kinetic energy gained by the water in falling is transferred to electrical energy by a hydroelectric generator. At what rate does the generator produce electrical energy? (The mass of 1 m3 of water is 1000 kg.)
Answer:
The generator produces electrical energy at a rate of 1378125000 J per second.
Explanation:
volume of water falling each second is 1250 [tex]m^{3}[/tex]
height through which it falls, h is 150 m
mass of 1 [tex]m^{3}[/tex] of water is 1000 kg
⇒mass of 1250 [tex]m^{3}[/tex] of water, m = 1250×1000 = 1250000 kg
acceleration due to gravity, g = 9.8 [tex]\frac{m}{sec^{2} }[/tex]
in falling through 150 m in each second, by Work-Energy Theorem:
Kinetic Energy(KE) gained by it = Potential Energy(PE) lost by it
⇒KE = mgh
= 1250000×9.8×150 J
= 1837500000 J
Electrical Energy = [tex]\frac{3}{4}[/tex](KE)
= [tex]\frac{3}{4}[/tex]×1837500000
= 1378125000 J per second
What evidence supports the theory that there is a black hole at the center of our galaxy?
Explanation:
Scientist and astronomers have observed that large gas clouds and massive stars are orbiting around in an accelerated manner in center of our galaxy that is milky way. A massive star S2's motion has been studied and it is found that the star is revolving around the center of the galaxy. From this scientist have confirmed presence of super massive black hole of 3 million solar masses lurking around the center of milky way.
A dolphin in seawater at a temperature of 25°C emits a sound wave directed toward the ocean floor 154 m below. How much time passes before it hears an echo? (The speed of sound in ocean water is 1533 m/s.)
Answer:
0.2s
Explanation:
SO for the dolphin to hear its echo, the sound wave must travel a distance twice as much as the displacement between the dolphin and the ocean floor. So d = 154 * 2 = 308 m
Since the speed of sound in ocean floor is v = 1533m/s we can find out the time by dividing the distance d by the speed of sound
t = d / v = 308 / 1533 = 0.2s
[tex](t = d / v )[/tex]The time passes before it hears an echo is 0.3secs
What is sound wave?A sound wave servers patterns of disturbance caused by the movement of energy traveling through a medium.
The speed of sound in ocean floor was given as [tex]( v = 1533m/s)[/tex]
To find the time, we can make use if the formula
[tex](t = d / v )[/tex]
Where t= time
v= velocity
d= distance
Then substitute ,we have
[tex]= 308 / 1533 = 0.2s[/tex]
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