To find the electric field at a point directly above the midpoint of the plastic wire, use the formula for the electric field due to a charged line. When the wire is bent into a circle, the electric field at a point directly above its center is zero.
Explanation:To find the electric field at a point directly above the midpoint of the plastic wire, you can use the formula for the electric field due to a charged line. The formula is given by E = (k * λ) / r, where E is the electric field, k is the Coulomb's constant (9 * 10^9 Nm²/C²), λ is the charge density, and r is the distance from the wire.
In this case, the charge density is 100nC/m and the distance is 6.0cm. Plug in the values to calculate the electric field. Since the wire is nonconducting, it does not affect the direction of the electric field, so the magnitude is the only value you need to find.
(b) When the wire is bent into a circle lying flat on the table, the electric field at a point directly above its center is zero. This is because the electric field due to each small segment of the wire cancels out by the symmetry of the circular shape.
Momentum is conserved in all collisions. If the initial motion of two trucks are known before a collision, the conservation of momentum can be used as long as what other information is known?
A) The masses of each car must be known.
B) The final motions of each car must be known.
C) The direction each car moves after collision must be known.
D) The distance each car traveled after the collision must be known.
Answer:
it's A
Explanation:
the masses of each car must be known
According to momentum conservation law, the total initial momentum of the colliding objects is equal to the total final momentum. To find the momentum, we have to know their masses. Hence, option A is correct.
What is momentum ?Momentum of an object is its ability to bring the applied force to make a maximum displacement. It is the product of mass and velocity of the object. Momentum is a vector quantity.
In all collisions momentum is conserved. Hence, total initial momentum is equal to the total final momentum of the colliding objects. Let m1 and m2 be the masses of the objects and u1 and u2 be their initial velocities then v1 and v2 are taken as their final velocities.
Then,
initial momentum = m1 u1 + m2 u2
final momentum =m1 u1 + m2 u2
conservation of momentum is written as:
m1 u1 + m2 u2 = m1 u1 + m2 u2.
Hence, by knowing the initial motion we can predict their final motion and momentum but we must know the masses of colliding objects. Hence, option A is correct.
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two mechanical devices typically used in laboratories to accurately measure small objects or distances are the _____ and _____. a)microscope b)meter stick c)micrometerd) calipers
Answer:
Option c and d
Explanation:
The two mechanical devices which are used in laboratories for accurate measurement of small distances or small objects are calipers and micrometer.
Micrometer is a mechanical device used in laboratories like that of a screw gauge. It is used for the measurement of thickness of objects, length and the depth of the small objects which can be measured by holding the object in between the spindle and anvil of the micrometer.
Calipers is another mechanical device like that of vernier calipers used in laboratories for measurement of small distances, usually the distance between the opposing faces of the object. The measurement is usually taken on a digital display, a dial or a scale that is ruled.
The distance is measured by adjusting the tips of the caliper holding the object on a ruler.
An electrical short cuts off all power to a submersible diving vehicle when it is a distance of 28 m below the surface of the ocean. The crew must push out a hatch of area 0.70 m2 and weight 200 N on the bottom to escape.If the pressure inside is 1.0 atm, what downward force must the crew exert on the hatch to open it?
To open the hatch of a submerged vehicle, the crew needs to overcome the difference in pressure applied on both sides of the hatch. By calculating the external pressure at the depth, the force outside is found. Subtracting the force inside (internal pressure plus hatch weight) gives the requisite force (~1.95 x 10⁵ N).
Explanation:The question is about determining the force the crew must exert to open the hatch of a submerged vehicle. In physics, this problem involves understanding the principles of fluid statics and the pressure difference between the interior and exterior of the hatch, contributing to the net force on the hatch.
Given that the vehicle is submerged 28 meters in saltwater, with an internal pressure of 1 atm, the external pressure at that depth will be greater than the internal pressure. The pressure a diver experiences under water increases by 1 ATA every 33 feet of salt water. Hence, at 28 meters (approximately 92 feet), the external pressure is about 1 (atmospheric pressure) + 92/33 (pressure due to ocean) = ~3.8 ATA. This is the pressure acting against the crew from outside.
The force acting on the hatch from outside (pressurized water) can be obtained by multiplying the pressure with the area (Force = Pressure x Area). So the Force outside = 3.8 atm x 0.70 m² = ~2.66 x 10⁵ N (considering that 1 atm is approximately 1.013 x 10⁵ N/m²).
Now, the force acting from inside consists of the atmospheric pressure (1 atm) and the weight of the hatch. Force inside = (1 atm x 0.70 m²) + weight of the hatch = 0.71 x 10⁵ N + 200 N = ~0.71 x 10⁵ N
Therefore, the resultant force (force required to open the hatch) is the difference between the outside and inside forces. That becomes ~2.66 x 10⁵ N - 0.71 x 10⁵ N = ~1.95 x 10⁵ N.
This is the downward force the crew must exert on the hatch to open it and escape.
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The downward force that the crew must exert on the hatch to open it at a depth of 28 m below the surface of the ocean is [tex]\( {199567.2 \, \text{N}} \)[/tex].
To find the downward force (F) that the crew must exert on the hatch to open it at a depth of 28 m below the surface of the ocean, we need to consider the hydrostatic pressure acting on the hatch.
Step 1: Calculate the pressure at the depth
The hydrostatic pressure [tex]\( P_{\text{water}} \)[/tex] at a depth h below the ocean surface is given by:
[tex]\[P_{\text{water}} = \rho g h\][/tex]
Calculate [tex]\( P_{\text{water}} \):[/tex]
[tex]\[P_{\text{water}} = 1025 \times 9.81 \times 28\][/tex]
[tex]\[P_{\text{water}} = 285396 \, \text{Pa}\][/tex]
Step 2: Calculate the total pressure inside the submersible
The total pressure inside the submersible is the sum of the atmospheric pressure and the hydrostatic pressure at depth:
[tex]\[P_{\text{total}} = P_{\text{inside}} + P_{\text{water}}\][/tex]
[tex]\[P_{\text{total}} = 1.0 \times 1.013 \times 10^5 + 285396\][/tex]
[tex]\[P_{\text{total}} = 1.013 \times 10^5 + 285396\][/tex]
[tex]\[P_{\text{total}} = 386396 \, \text{Pa}\][/tex]
Step 3: Calculate the force required to open the hatch
The force F required to open the hatch against the pressure difference is:
[tex]\[F = (P_{\text{total}} - P_{\text{atm}}) \times A\][/tex]
Substitute the values:
[tex]\[F = (386396 - 1.013 \times 10^5) \times 0.70\][/tex]
[tex]\[F = (386396 - 101300) \times 0.70\][/tex]
[tex]\[F = 285096 \times 0.70\][/tex]
[tex]\[F = 199567.2 \, \text{N}\][/tex]
The froghopper Philaenus spumarius is supposedly the best jumper in the animal kingdom. To start a jump, this insect can accelerate at 4.00 km/s2 over a distance of 2.0 mm as it straightens its specially designed "jumping legs."
(a) Find the upward velocity with which the insect takes off.
(b) In what time interval does it reach this velocity?
(c) How high would the insect jump if air resistance were negligible? The actual height it reaches is about 70 cm, so air resistance must be a noticeable force on the leaping froghopper.
Answer:
Part a)
[tex]v_f = 4 m/s[/tex]
Part b)
[tex]t = 0.001 s[/tex]
Part c)
[tex]d = 0.815 m[/tex]
Explanation:
Part a)
As we know that initially the grass hopper is at rest at the ground position
Now the acceleration is given as
[tex]a = 4000 m/s^2[/tex]
distance of the legs that it stretched is given as
[tex]s = 2.0 mm[/tex]
so we have
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]v_f^2 - 0 = 2(4000)(0.002)[/tex]
[tex]v_f = 4 m/s[/tex]
Part b)
time taken to reach this speed is given as
[tex]v_f - v_i = at[/tex]
[tex]4 - 0 = 4000 t[/tex]
[tex]t = 0.001 s[/tex]
Part c)
as the grass hopper reach the maximum height its final speed would be zero
so we will have
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - 4^2 = 2(-9.81) d[/tex]
[tex]d = 0.815 m[/tex]
Final answer:
The froghopper Philaenus spumarius takes off at an upward velocity of 4 m/s and reaches this velocity in 0.001 seconds. In the absence of air resistance, it can potentially jump to a height of 0.815 m, while air resistance reduces its actual jump to about 70 cm.
Explanation:
The froghopper Philaenus spumarius, known for its powerful jumping ability, can accelerate at 4.00 km/s² over a distance of 2.0 mm. To find the upward velocity (a) with which the insect takes off, the following kinematic equation can be used:v² = u² + 2as
where v is the final velocity, u is the initial velocity (0 m/s in this case), a is the acceleration, and s is the distance.
Firstly, convert the given units to make them consistent. The acceleration becomes 4000 m/s² and the distance becomes 0.002 m. Plugging the values into the equation gives:
v² = 0 + 2(4000)(0.002) = 16
The takeoff velocity, v, is √16, which equals 4 m/s.
For the time interval (b), we use the equation:
v = u + at
So, t = v / a = 4 / 4000 = 0.001 s.
To calculate how high the insect would jump if air resistance were negligible (c), we can use the equation:
s = ut + ½at²
Since initial velocity u is 0 when coming back down, and final velocity v is 0 at the peak of the jump, we set v = 0 and solve for s:
0 = (4)2 - 2(9.81)s
s = 16 / 19.62 = 0.815 m (or 81.5 cm).
Comparing this to the actual jump height of 70 cm indicates a significant effect of air resistance on the insect's jump.
The position of a particle moving along the x axis depends on the time according to the equation x = ct3 - bt6, where x is in meters and t in seconds. Let c and b have numerical values 2.9 m/s3 and 2.0 m/s6, respectively. From t = 0.0 s to t = 1.3 s, (a) what is the displacement of the particle? Find its velocity at times (b) 1.0 s, (c) 2.0 s, (d) 3.0 s, and (e) 4.0 s. Find its acceleration at (f) 1.0 s, (g) 2.0 s, (h) 3.0 s, and (i) 4.0 s.
Answer:
Given the equation of the particle, we know that:
[tex]x=(2.9\frac{m}{s^{3} } )*t^{3} -(2\frac{m}{s^{6} } )*t^{6}[/tex]
a) In [tex]t=0.0s[/tex] we evaluate [tex]t[/tex] in the former equation:
[tex]x_{1} =(2.9\frac{m}{s^{3} } )*(0)^{3} -(2\frac{m}{s^{6} } )*(0)^{6}=0m[/tex]
In [tex]t=1.3s[/tex]
[tex]x_{2} =(2.9\frac{m}{s^{3} } )*(1.3s)^{3} -(2\frac{m}{s^{6} } )*(1.3s)^{6}=-3,28232m[/tex]
a) So, we know that the displacement of te particle is given by:
[tex]x=x_{2}-x_{1} =-3.28232m-0m=-3.28232m[/tex]
To find it's velocity, we need to derivate the equation of position by the formula:
[tex]v_{x} =\frac{dx}{dt} =3ct^{2}-6bt^{5} =(8.7\frac{m}{s^{3} })t^{2} -(12\frac{m}{s^{6} } )t^{5}[/tex]
And evaluate this expression at each specified t:
b) [tex]v(1s)_{x} =(8.7\frac{m}{s^{3} })(1s)^{2} -(12\frac{m}{s^{6} } )(1s)^{5}=-3.3\frac{m}{s}[/tex]
c) [tex]v(2s)_{x} =(8.7\frac{m}{s^{3} })(2s)^{2} -(12\frac{m}{s^{6} } )(2s)^{5}=-349.2\frac{m}{s}[/tex]
d) [tex]v(3s)_{x} =(8.7\frac{m}{s^{3} })(3s)^{2} -(12\frac{m}{s^{6} } )(3s)^{5}=-2837.7\frac{m}{s}[/tex]
e) [tex]v(4s)_{x} =(8.7\frac{m}{s^{3} })(4s)^{2} -(12\frac{m}{s^{6} } )(4s)^{5}=-12148.8\frac{m}{s}[/tex]
To find it's acceleration, we need to derivate the equation of velocity by the formula:
[tex]a_{x} =\frac{dv}{dt} =(17.4\frac{m}{s^{3} } )t-(60\frac{m}{s^{6} } )t^{4}[/tex]
And evaluate this expression at each specified t:
f) [tex]a(1s)_{x} =(17.4\frac{m}{s^{3} } )(1s)-(60\frac{m}{s^{6} } )(1s)^{4}=-42.6\frac{m}{s^{2} }[/tex]
g) [tex]a(2s)_{x} =(17.4\frac{m}{s^{3} } )(2s)-(60\frac{m}{s^{6} } )(2s)^{4}=-925.2\frac{m}{s^{2} }[/tex]
h) [tex]a(3s)_{x} =(17.4\frac{m}{s^{3} } )(3s)-(60\frac{m}{s^{6} } )(3s)^{4}=-4807.8\frac{m}{s^{2} }[/tex]
i) [tex]a(4s)_{x} =(17.4\frac{m}{s^{3} } )(4s)-(60\frac{m}{s^{6} } )(4s)^{4}=-15290.4\frac{m}{s^{2} }[/tex]
A child bounces a 52 g superball on the sidewalk. The velocity change of the superball is from 20 m/s downward to 14 m/s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk? Answer in units of N.
Answer:
0.982 N
Explanation:
mass of ball, m = 52 g = 0.052 kg
initial velocity, u = - 20 m/s (downward)
final velocity, v = 14 m/s (upward)
time of contact, t = 1.8 s
According to Newton's second law, the rate of change of momentum of the body is equal to the forced exerted on that body.
initial momentum , pi = mass x initial velocity = 0.052 x (-20) = - 1.04 kg m/s
final momentum, pf = mass x final velocity = 0.052 x 14 = 0.728 kg m/s
Change in momentum = final momentum - initial momentum
= 0.728 - (- 1.04)= 1.768 kg m/s
So, force = change in momentum / time
Force = 1.768 / 1.8 = 0.982 N
The magnitude of the average force exerted on the superball by the sidewalk is 1414.4 N. This is calculated by determining the impulse based on the change in velocity and mass of the superball, and dividing the impulse by the contact time.
Explanation:The question asks to calculate the magnitude of the average force exerted on a superball by the sidewalk during its bounce. The superball's mass is 52 g (0.052 kg), and it experiences a velocity change from 20 m/s downward to 14 m/s upward. The contact time with the sidewalk is given as 1/800 s. First, we must calculate the change in velocity (impulse) and then use it to find the average force.
To calculate the impulse, we add the magnitudes of both components of velocity together since the ball changes direction, the total change in velocity (Δv) is 20 m/s (downward) + 14 m/s (upward) = 34 m/s. The impulse (J) is then given by Δp = mΔv, where m is the mass of the ball. Therefore, J = 0.052 kg * 34 m/s = 1.768 kg·m/s.
The average force (Favg) can be found by dividing the impulse by the contact time (Δt): Favg = J / Δt. Thus, Favg = 1.768 kg·m/s / (1/800 s), which equals an average force of 1414.4 N.
Consider two cylinders of gas identical in all respects except that one contains O₂ and the other He. Both hold the same volume of gas at STP and are closed by a movable piston at one end. Both gases are now compressed adiabatically to one-third their original volume. Which gas will show the greater pressure increase?
A)It's impossible to tell from the information given.
B)the oxygen (O₂)
C)Neither; both will show the same increase.
D)the He
Answer:
D) He
Explanation:
For adiabatic change
[tex]PV^\gamma = constant[/tex]
\frac{P_2}{P_1} =(\frac{V_1}{V_2} )^\gamma
[tex]\frac{P_2}{P_1} =(\frac{3}{1} )^\gamma[/tex]
[tex]P_2 = P_1 3^\gamma[/tex]
As the value of \gamma increases, P_2 increases.
\gamma for He is 1.67 and \gamma for O₂ IS 1.4
Naturally The value of P_2 is greater for He. Increase in pressure for He is greater.
What is the entropy change of a 29.8 g ice cube that melts completely in a bucket of water whose temperature is just above the freezing point of water?
Answer:
[tex]\Delta S = 36.55 J/K[/tex]
Explanation:
As we know that entropy change for phase conversion is given as
[tex]\Delta S = \frac{\Delta Q}{T}[/tex]
Here we know that heat required to change the phase of the ice is given as
[tex]\Delta Q = mL[/tex]
here we have
m = 29.8 g = 0.0298 kg
L = 335000 J/kg
now we have
[tex]\Delta Q = 0.0298\times 335000[/tex]
[tex]\Delta Q = 9979.4 J[/tex]
also we know that temperature is approximately same as freezing temperature
so we have
[tex]T = 273 k[/tex]
so here we have
[tex]\Delta S = \frac{9979.4}{273}[/tex]
[tex]\Delta S = 36.55 J/K[/tex]
You observe a light ray move from one piece of glass to another (a different type of glass) and the light ray bends towards the glass-glass interface (away from the normal) when it enters the 2nd piece of glass. You deduce that:
Answer:
second medium is less optical denser than first medium
Explanation:
As per Snell's law we know that
[tex]\mu_1 sin\theta_1 = \mu_2 sin\theta_2[/tex]
so here we know that
[tex]\theta_1 [/tex] = angle of incidence
[tex]\theta_2[/tex] = angle of refraction
As we know that light ray goes away from the normal when it move from medium 1 to medium 2
so here angle of refraction will me more than the angle of incidence
so we have
[tex]\frac{sin\theta_1}{sin\theta_2} = \frac{\mu_2}{\mu_1}[/tex]
since we know that
[tex]\theta_1 < \theta_2[/tex]
so we have
[tex]\frac{sin\theta_1}{sin\theta_2} < 1[/tex]
[tex]\frac{\mu_2}{\mu_1} <1[/tex]
so we can say that
[tex]\mu_2 < \mu_1[/tex]
so we can say that second medium is less optical denser than first medium
An American manufacturer supplied a customer with refrigerators with electrical cords that were one yard long instead of 1 meter long. What was the difference in cord length (cm)?a. 0.856 cmb. 85.6 cmc. 8.56 cmd. 0.00856 cme. 856 cm
Answer:
C.
Explanation:
A meter is 8.56 centimeters longer than a yard. Something to keep in mind is that a meter is about 10% longer than a yard.
Hope this helps :)
Answer:
C. 8.56cm
Explanation:
Since we are to get the difference of cord length in centimeter, there fore we will have to convert the one yard and one metre length of wires to centimeters.
For 1m length of cord
1m = 100cm
For 1yard length of cord
1yard = 91.44cm
Difference in cord length = 100cm - 91.44cm
= 8.56cm
You drive a car 2.0 h at 40 km/h, then 2.0 h at 80 km/h. (a) What is your average velocity? km/h (b) Do you get the same answer if you drive 120 km at each of the two speeds above? no yes
Answer:
a)(2×40+2×80)/(2+2)=240/4=60
b)(240)/(120/40+120/80)=240/(3+1.5)
240/4.5=53.33333
no
To qualify to run in the 2005 Boston Marathon, a distance of 26.2 miles, an 18-year-old woman had to have completed another marathon in 3 hours and 40 minutes or less. a) To qualify, what must a woman's average speed have been in miles per hour?b) To qualify, what must a woman's average speed have been in meters per second? There are 1609 m in a mile.
Answer:
7,14545 mph and 3,1936 m/s
Explanation:
The average speed is calculated by dividing the displacement over time, then it is 26,2 miles/(3 2/3 hours), here 3 (2/3) hours is a mixed number, that represents 11/3 hours or 3,66 hours. Then the average speed is 7,14545 mph, now to turn this into meters per second, we notice as mentioned that 1 mile =1609 meters and 1 hour=3600 seconds. Then 7,14545 miles/hour* (1 hour/3600 seconds) * (1609 meters/1 mile)=3,1936 m/s
A light-year is the distance that light travels in one year. The speed of light is 3.00 × 108 m/s. How many miles are there in one light-year? (1 mi = 1609 m, 1 y = 365 d)
a 5.88 × 1012 mi
b 9.46 × 1015 mi
c 5.88 × 1015 mi
d 9.46 × 1012 mi
Answer:
(a) [tex]5.88\times 10^{12}mi[/tex]
Explanation:
We have given speed of light [tex]c=3\times 10^8m/sec[/tex]
Given time = 1 year =365 days
We know that in 1 minute = 60 sec
1 hour = 60×60 = 3600 sec
In one day = 24 hour = 24×60×60=86400 sec
So in 365 days = 365×86400[tex]=3.1536\times 10^7sec[/tex]
We know that distance = speed ×time[tex]=3.1536\times 10^7\times 3\times 10^8=9.46\times 10^{15}m[/tex]
We have given that 1 mi = 1609 m
So [tex]9.46\times 10^{15}m=\frac{9.46\times 10^{15}}{1609}=5.88\times 10^{12}mi[/tex]So option (a) is correct
Which elements exist naturally as diatomic molecules
Hydrogen, nitrogen, and oxygen are examples of diatomic molecules naturally.
What is an atom?The atom consists of matter that may be split without releasing electrical charges.
In the nucleus proton and the neutron is exists. The condition of the atom to be electrically neutral is that the number of the proton and electron should be the same.
Two or more two atoms combine to form the molecules. A diatomic molecule is one that is made up of atoms from two distinct elements. Hydrogen, nitrogen, and oxygen are examples of diatomic molecules.
Hence, hydrogen, nitrogen, and oxygen are examples of diatomic molecules naturally.
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A baseball pitcher throws a baseball with a speed of 42 m/s . In throwing the baseball, the pitcher accelerates it through a displacement of about 3.5 m, from behind the body to the point where it is released.
Answer:
a = 252 m / s²
Explanation:
The initial speed Vο = 0
The speed of baseball with witch it is thrown Vf = 42 m/s
Displacement covered = 3.5 m
Hence acceleration will be constant at this moment, so as per the equation of constant acceleration : ( Vf² - Vο² ) / 2d = a
a = 252 m / s²
Technician A says that a diesel engine’s higher compression ratio helps it to produce more power and use less fuel. Technician B says that higher compression ratios only help the diesel engine produce more torque. Who is correct?
Answer:
None of them is 100% correct.
Technician A is correct when he says that the higher compression rate helps the car to use less fuel. A higher compression ratio gives the engine a higher thermal efficiency which in turn translates to higher fuel efficiency but it does not produce more power than a gasoline engine.
Technician B is correct when he says the diesel engine produces more torque, but he also says that it is the only thing that is better with the diesel engine which is wrong because we already know that it also helps on fuel efficiency.
A boy is standing at a railroad crossing for a track that runs east and west. As he faces the track, east is to his right and west is to his left. Two trains on the track some distance apart are headed west, both at speeds of 8.40 m/s, and blowing their whistles (which have the same frequency). One train is approaching him from the east and the other is traveling away from him toward the west. (Assume the speed of sound is 343 m/s.) (a) If he hears a beat frequency of 6.50 Hz, determine the frequency (in Hz) emitted by the two whistles.
Answer:
[tex]f_o = 132.91 Hz[/tex]
Explanation:
Frequency of the train whistle which is approaching him is given as
[tex]f_1 = (\frac{v}{v - v_s}) f_o[/tex]
[tex]f_1 = (\frac{343}{343 - 8.40}) f_o[/tex]
now similarly the frequency of the train whistle which is moving away from him is given as
[tex]f_2 = (\frac{v}{v + v_s})f_o[/tex]
[tex]f_2 = (\frac{343}{343 + 8.40})f_o[/tex]
now we know that
[tex]f_{beat} = f_1 - f_2[/tex]
so we have
[tex]6.50 = (\frac{343}{343 - 8.40}) f_o - (\frac{343}{343 + 8.40}) f_o[/tex]
[tex]6.50 = 1.025 f_o - 0.976 f_o[/tex]
[tex]f_o = 132.91 Hz[/tex]
Final answer:
The Doppler effect can be used to determine the frequency emitted by the two train whistles. By applying the formula for the Doppler effect, the equation can be derived and solved to find the frequency of each whistle. In this case, the frequency emitted by the whistle of each train is approximately 155.46 Hz.
Explanation:
The situation described in the question is an example of the Doppler effect, which explains the change in frequency of a sound wave when the source of the sound is moving relative to the observer. In this case, the boy is standing at a railroad crossing and is hearing two trains approaching and moving away from him.
When a sound source approaches the observer, the frequency of the sound waves appears higher, resulting in a higher pitch. Conversely, when a sound source moves away, the frequency appears lower, resulting in a lower pitch. The difference between the frequencies heard when the two trains are approaching and moving away from the boy is called the beat frequency, which is given as 6.50 Hz in the question.
To determine the frequency emitted by the two whistles, we can use the formula for the Doppler effect:
f' = fs * (v + vo) / (v + vs)
Where:
f' is the observed frequency
fs is the source frequency
v is the speed of sound
vo is the velocity of the observer
vs is the velocity of the source
In this case, both trains have the same speed of 8.40 m/s, so the velocity of the observer is zero. The speed of sound is given as 343 m/s. Given these values, we can substitute them into the formula.
Let's assume that the frequency emitted by the whistle of the approaching train (Train A) is fA and the frequency emitted by the whistle of the receding train (Train B) is fB.
For the approaching train:
f' = fA * (v + vo) / (v + vs)
Since the observer is not moving, the equation simplifies to:
f' = fA * v / (v + vs) .......... (1)
And for the receding train:
f' = fB * (v + vo) / (v - vs)
f' = fB * v / (v - vs) .......... (2)
We are given that the beat frequency is 6.50 Hz. This means that the difference in frequencies of the two trains is 6.50 Hz.
Substituting the values into equations (1) and (2), we get:
fA * v / (v + vs) - fB * v / (v - vs) = 6.50 Hz
fA / (v + vs) - fB / (v - vs) = 6.50 / v Hz
Since both trains have the same source frequency (fA = fB), we can simplify the equation:
2fA * vs / (v^2 - vs^2) = 6.50 / v Hz
Solving for fA, we can rewrite the equation as:
fA = (6.50 / v) * ((v^2 - vs^2) / (2vs)) Hz
Substituting the given values, we have:
fA = (6.50 / 343) * ((343^2 - 8.4^2) / (2 * 8.4)) ≈ 155.46 Hz
Therefore, the frequency emitted by the whistle of each train is approximately 155.46 Hz.
What is the approximate mass of a column of air 1 cm2 in area that extends from sea level to the upper atmosphere? What is the weight of this amount of air? What is the pressure at the bottom?
Answer:
(a) the approximate mass of a column of air is: 1225 (Kg), (b) the weight of this amount of air is: 12017.25 (N) and (c) the pressure at the bottom is: 120172.5 (KPa)
Explanation:
We need to remember that the equation that related mass and volume is the density as:[tex]p=\frac{m}{V}[/tex], where p is the density, m is the mass and V is the volume and assuming that air density is 1.225 (Kg/m3) and knowing that exosphere is the upper atmosphere and its height is 10000 (Km). We now can get the volume as:[tex]Volume=A*h_{upper atmosphere} =0.0001*10000000=1000(metre^{3})[/tex], so that the approximate mass of a column of air is: [tex]m=p*V=1.225*1000=1225 (Kg)[/tex]. Then the weight of this amount of air is:[tex]W=m*g[/tex], where W is the weight, m is the mass and g the gravity acceleration, so:[tex]W=1225*9.81=12017.25 (N)[/tex]. Finally to find the pressure we know that gauge pressure equation [tex]P=p*g*h[/tex] where P is the pressure, p is the density of the column and h is the height of the column, so [tex]P=1.225*9.81*10000000=120172.5 (KPa)[/tex].
The approximate mass of the air column can be calculated using the density of air and the volume of the column. The weight of the air column can be calculated by multiplying the mass by the acceleration due to gravity. The pressure at the bottom can be calculated by dividing the weight by the area.
Explanation:The approximate mass of a column of air with an area of 1cm² that extends from sea level to the upper atmosphere can be calculated using the density of air. The density of air near the Earth's surface is about 1.29 kg/m³. Using this density, we can calculate the mass by multiplying the density with the volume of the column of air. The volume can be calculated by multiplying the area of 1 cm² with the height of the column, which is the distance from sea level to the upper atmosphere.
To calculate the weight of this amount of air, we need to multiply the mass of the air column by the acceleration due to gravity. The acceleration due to gravity is approximately 9.81 m/s². Multiplying the mass by the acceleration due to gravity will give us the weight of the air column in newtons.
The pressure at the bottom, or at sea level, can be calculated using the weight of the air column and the area of 1 cm². The pressure is equal to the weight divided by the area.
Learn more about Mass, weight, and pressure of an air column here:https://brainly.com/qu estion/13340426
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A straight, nonconducting plastic wire 9.50 cm long carries a charge density of 130 nC/m distributed uniformly along its length. It is lying on a horizontal tabletop.A) Find the magnitude and direction of the electric field this wire produces at a point 4.50 cm directly above its midpoint.B)If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 4.50 cm directly above its center.
Answer:
A) E = 3.70*10^{4} N/C
B) E = 2.281*10^3 N/C
Explanation:
given data:
charge density [tex] \lambda = 130*10^{-9} C/m[/tex]
length of wire = 9.50 cm
a) at x = 4.5 m above midpoint, electric field is calculated as
[tex]E = \frac{1}{ 2\pi \epsilon} * \frac{ \lambda}{x\sqrt{(x^2/a^2)+1}}[/tex]
x = 4.5 cm
midpoint a = 4.5 cm = 0.0475 m
[tex]E =2{\frac{1}{ 8.99*10^9} * \frac{130*10^{-9} }{0.045\sqrt{(4.5^2/4.75^2)+1}}[/tex]
E = 3.70*10^{4} N/C
B) when wire is in circle form
[tex]Q = \lambda * L[/tex]
[tex]= 130*10^{-9} *9.5*10^{-2}[/tex]
= 1.235*10^{-8} C
Radius of circle
[tex]r = \frac{L}{2\pi}[/tex]
[tex]r = \frac{9.5*10^{-2}}{2\pi}[/tex]
r = 1.511*10^{-2} m
[tex]E = \frac{1}{ 2\pi \epsilon} * \frac{Qx}{(x^2+r^2)^{3/2}}[/tex]
[tex]E =8.99*10^{9} * \frac{1.23*10^{-8}*4.5*10^{-2}}{((4.5*10^{-2})^2+(1.511*10^{-2})^2)^{3/2}}[/tex]
E = 2.281*10^3 N/C
Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle orbiting in Saturn's D Ring, whose distance from Saturn is 69000 km. How many times larger is the velocity of the particle in the D Ring compared to the particle in the Encke Division?
Answer:
The particle in the D ring is 1399 times faster than the particle in the Encke Division.
Explanation:
The circular velocity is define as:
[tex]v = \frac{2 \pi r}{T}[/tex]
Where r is the radius of the trajectory and T is the orbital period
To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:
[tex]T^{2} = r^{3}[/tex]
Where T is orbital period and r is the radius of the trajectory.
Case for the particle in the Encke Division:
[tex]T^{2} = r^{3}[/tex]
[tex]T = \sqrt{(133370 Km)^{3}}[/tex]
[tex]T = \sqrt{(2.372x10^{15} Km)}[/tex]
[tex]T = 4.870x10^{7} Km[/tex]
It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( [tex]1.50x10^{8} Km[/tex] )
1 AU is defined as the distance between the earth and the sun.
[tex]\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU[/tex]
[tex]T = 0.324 AU[/tex]
But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.
[tex]T = \frac{0.324 AU}{1 AU} . 1 year[/tex]
[tex]T = 0.324 year[/tex]
That can be expressed in units of days
[tex]T = \frac{0.324 year}{1 year} . 365.25 days[/tex]
[tex]T = 118.60 days[/tex]
Circular velocity for the particle in the Encke Division:
[tex]v = \frac{2 \pi r}{T}[/tex]
[tex]v = \frac{2 \pi (133370 Km)}{(118.60 days)}[/tex]
For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.
[tex]118.60 days .\frac{86400 s}{1 day}[/tex] ⇒ 10247040 s
[tex]133370 Km .\frac{1000 m}{1 Km}[/tex] ⇒ 133370000 m
[tex]v = \frac{2 \pi (133370000 m)}{(10247040 s)}[/tex]
[tex]v = 81.778 m/s[/tex]
Case for the particle in the D Ring:
For the case of the particle in the D Ring, the same approach used above can be followed
[tex]T^{2} = r^{3}[/tex]
[tex]T = \sqrt{(69000 Km)^{3}}[/tex]
[tex]T = \sqrt{(3.285x10^{14} Km)}[/tex]
[tex]T = 1.812x10^{7} Km[/tex]
[tex]\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU[/tex]
[tex]T = 0.120 AU[/tex]
[tex]T = \frac{0.120 AU}{1 AU} . 1 year[/tex]
[tex]T = 0.120 year[/tex]
[tex]T = \frac{0.120 year}{1 year} . 365.25 days[/tex]
[tex]T = 43.83 days[/tex]
Circular velocity for the particle in D Ring:
[tex]v = \frac{2 \pi r}{T}[/tex]
[tex]v = \frac{2 \pi (69000 Km)}{(43.83 days)}[/tex]
For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.
[tex]43.83 days . \frac{86400 s}{1 day}[/tex] ⇒ 3786912 s
[tex]69000 Km . \frac{1000 m}{ 1 Km}[/tex] ⇒ 69000000 m
[tex]v = \frac{2 \pi (69000000 m)}{(3786912 s)}[/tex]
[tex]v = 114.483 m/s[/tex]
[tex]\frac{114.483 m/s}{81.778 m/s} = 1.399[/tex]
The particle in the D ring is 1399 times faster than the particle in the Encke Division.
A student driving home for the holidays starts at 8:00 am to make the 675-km trip, practically all of which is on nonurban interstate highway. If she wants to arrive home no later than 3:00 pm, what must be her minimum average speed? Will she have to exceed the speed limit?
8 am to 3 pm is 7 hours.
(675 km) / (7 hrs) = 96.4 km/hr .
Her average speed for the whole 7 hours has to be not less than 96.4 km/hr. Any less, she's not home by 3:00.
We don't technically know the speed limit at every point on her trip, because you technically haven't told us.
But in 7 hours, she MUST stop for gas, she MUST get some rest, she MUST make a pit stop, and she most likely encounters some traffic somewhere. So in order to average 96.4 for the whole trip, she MUST exceed it for PART of the trip ... possibly by a lot.
Whatever the speed limit may be, I think it's likely that she'll exceed it SOMEwhere, at least for SOME time.
The student must maintain a minimum average speed of 96.43 km/h.
Whether this exceeds the speed limit depends on the highway's speed limit.
To determine the minimum average speed the student needs to maintain to complete the 675-km trip by 3:00 pm, starting at 8:00 am,
We first calculate the total travel time allowed, the difference between 3:00 pm and 8:00 am is 7 hours.
Therefore, to find the minimum average speed, we divide the total distance by the total time:
Average Speed = Total Distance / Total Time
Average Speed = 675 km / 7 hours
= 96.43 km/h
The student must maintain a minimum average speed of 96.43 km/h to reach the destination on time.
Determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.(a) 1s22s22p63s23p5(b) 1s22s22p63s23p63d74s2(c) 1s22s22p63s23p63d104s24p6(d) 1s22s22p63s23p64s1(e) 1s22s22p63s23p63d104s24p64d55s2(f) 1s22s22p63s2
The electron configurations correspond to a halogen, transition metals, an inert gas, an alkali metal, and an alkaline earth metal, identified by their completed electron orbitals and the presence of electrons in specific orbitals such as d or p.
Explanation:To determine whether an electron configuration corresponds to an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal, you must look at the position that the electron configuration would occupy on the periodic table.(a) 1s2 2s2 2p6 3s2 3p5: This is a halogen because it is one electron short of a noble gas configuration, indicating it is in group 17.(b) 1s2 2s2 2p6 3s2 3p6 3d7 4s2: This is a transition metal due to the partially filled d orbital.(c) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6: This is an inert gas because all orbitals are filled, indicating that it is in group 18.(d) 1s2 2s2 2p6 3s2 3p6 4s1: This is an alkali metal, in group 1, with a single electron in the outermost s orbital.(e) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d5 5s2: This configuration belongs to a transition metal because of the partially filled d orbital (4d5).(f) 1s2 2s2 2p6 3s2: This is an alkaline earth metal with two electrons in the outermost s orbital, indicating it is in group 2.James suffers from atherosclerosis, w condition that causes artery walls to harden and thicken. Atherosclerosis restricts blood flow to organs and tissues. The restricted blood flow increases blood pressure in which system does atherosclerosis develop
Answer:
Artherosclerosis develops in the nervous system.
Explanation:
Artherosclerosis is a condition that develops when plaque or fatty deposits block the arteries. This restricts the blood flow through the arteries partially or completely. Artherosclerosis can develop in the arteries of heart, brain or legs.
When the arteries are blocked by the thickening of the blood vessels the blood flow to the corresponding organs reduces. Thus the supply of oxygen and other vital nutrients to these parts reduces.
Based on the location of development of plaque and the type of artery affected various conditions arise.
Examples are coronary heart disease, angina, carotid artery disease, peripheral artery disease, chronic kidney disease etc.
Answer:
Cardiovascular system
Explanation:
This is because Atherosclerosis is the build up of fats and cholesterol in your Artery walls, this can restrict blood flow in your body.Your Artery walls are apart of your Cardiovascular system.
A 230.-mL sample of a 0.240 M solution is left on a hot plate overnight; the following morning, the solution is 1.75 M. What volume of water evaporated from the 0.24 M solution?
Answer:
The volume of water evaporated is 199mL
Explanation:
Concentration is calculated with the following formula
[tex]C=\frac{n}{V}[/tex]
where n is the number of moles of solute and V is the volume of the solution (in this case is the same as the solvent volume) in liters.
So we isolate the variable n to know the amount of moles, using the volume given in liters
[tex]230mL=0.23L[/tex]
[tex]n=C*V=0.240 M*0.23L=0.055 mol[/tex]
Now, we isolate the variable V to know the new volume with the new concentration given.
[tex]V=\frac{n}{C} =0.055mol/1.75M=0.031L=31mL[/tex]
Finally, the volume of water evaporated is the difference between initial and final volume.
[tex]V_{ev}= V_{i} -V_{f} =230mL-31mL=199mL[/tex]
Two of the types of infrared light, IR-C and IR-A, are both components of sunlight. Their wavelengths range from 3000 to 1,000,000 nm for IR-C and from 700 to 1400 nm for IR-A. Compare the energy of microwaves, IR-C, and IR-A.
Explanation:
3000 to 1,000,000 nm for IR-C
700 to 1400 nm for IR-A
Now, we are working with light, so we can use that the energy of a photon with frequency f is : E(f) = h*f
where h is the plank constant h = 6.62607015×10−34 j*s
and f is the frequency that f= c/λ, where λ is the wavelenght.
so, for IR-C the energy lies between h*c/3000 and h*c/1000000
where c is the light speed.
For IR-A the energy lies between h*c/700 and h*c/1400.
so here you can se that IR-A has a lot more energy than IR-C, you can se that the minimal energy of IR-A is h*c/1400 and the maximal of IR-C is h*c/3000 so the minimal energy of IR-A is almost twice times the maximum energy of IR-C
A solid sample of a metal is submerged in a large graduated cylinder. The water level in the graduate was 55.25 mL and the final level was 61.00 mL. If the mass of the object was determined to be 23.795 g, what is the density of the metal?
Answer:
ρm=4.14g/ml :metal density
Explanation:
Conceptual analysis
We apply the formula to calculate the density:
ρ= m /V Formula (1)
Where:
ρ:density in g/mL
m: mass in g (grams)
V= volume in ml (milliliters)
Known data
[tex]V_{il}[/tex]: final volume of liquid=55.25 ml
[tex]V_{fl}[/tex]: final volume of liquid=61.00 ml
ms: metal mass=23.795 g
Problem development
The volume of the metal (vm) is equal to the change in volume of the liquid.
[tex]V_{m} = V_{fl}-V_{il} =61.00 ml - 55.25 ml =5.75ml[/tex]
We replace the data in formula (1 )to calculate the density of the metal(ρm):
ρm= mm /Vm=23.795 g/5.75ml
ρm=4.14g/ml
A plane flying from Townsville, Australia, has an air speed of 264 m/s in a direction 5.0° south of west. It is in the jet stream, which is blowing at 37 m/s in a direction 15° south of east. In this problem you are going to be asked about the velocity of the airplane relative to the Earth.
Answer:
Speed of plane is 300.5 m/s at angle of 6.22 degree South of West
Explanation:
Air speed of the plane is given as
v = 264 m/s in direction 5 degree South of West
So we have
[tex]v_1 = 264 cos5 \hat i + 264 sin5 \hat j[/tex]
[tex]v_1 = 263 \hat i + 23 \hat j[/tex]
Also we have speed of air is given as
v = 37 m/s at 15 degree South of West
so it is
[tex]v_2 = 37 cos15\hat i + 37 sin15 \hat j[/tex]
[tex]v_2 = 35.74 \hat i + 9.58 \hat j[/tex]
So the net speed of plane with respect to ground is given as
[tex]v_p = v_1 + v_2[/tex]
[tex]v_p = (263 \hat i + 23 \hat j) + (35.74 \hat i + 9.58 \hat j)[/tex]
[tex]v_p = 298.74 \hat i + 32.58\hat j[/tex]
so it is
[tex]v_p = \sqrt{298.74^2 + 32.58^2}[/tex]
[tex]v_p = 300.5 m/s[/tex]
direction is given as
[tex]\theta =tan^{-1} \frac{v_y}{v_x}[/tex]
[tex]\theta = tan^{-1} \frac{32.58}{298.74}[/tex]
[tex]\theta = 6.22 degree[/tex]
Use vectors to decide whether the triangle with vertices P(1, −2, −1), Q(2, 1, −3), and R(6, −1, −4) is right-angled. Yes, it is right-angled. No, it is not right-angled.
Answer:
Yes, it is right-angled
Explanation:
Two vectors are orthogonal if the scalar product between them is zero. Then, we will match each pair of vertices with a vector, wich is formed with the following formula:
Given two points A and B, the vector going from A to B is
[tex]AB=B-A=(B_{x}-A_{x},B_{y}-A_{y},B_{z} -A_{z})[/tex]
So, we calculate each component separately.
[tex]PQ=Q-P=(2-1,1-(-2),-3-(-1))=(1,3,-2)[/tex]
[tex]QR=R-Q=(6-2,-1-1,-4-(-3))=(4,-2,-1)[/tex]
[tex]RP=P-R=(1-6,-2-(-1),-1-(-4))=(-5,-1,3)[/tex]
Finally, using the scalar product formula
[tex]A*B=A_{x}* B_{x}+ A_{y}* B_{y}+ A_{z}* B_{z}[/tex]
we see if the products is zero
[tex]PQ*QR=1*4+3(-2)+(-2)*(-1)=0[/tex]
In this case we don't even have to calculate the other products as we've found that PQ and QR form a right angle.
Final answer:
By calculating the dot product of vectors representing the sides of the triangle, it is found that the dot product of PQ and QR is zero, indicating that these vectors are orthogonal. Therefore, the triangle PQR is right-angled.
Explanation:
To determine whether the triangle with vertices P(1, -2, -1), Q(2, 1, -3), and R(6, -1, -4) is right-angled, we can use the concept of the dot product between two vectors. A triangle is right-angled if one of the pairs of vectors corresponding to its sides is orthogonal, meaning their dot product is zero.
First, we find the vectors representing the sides of the triangle:
⇨ PQ = Q - P = (2, 1, -3) - (1, -2, -1) = (1, 3, -2)
⇨ PR = R - P = (6, -1, -4) - (1, -2, -1) = (5, 1, -3)
⇨ QR = R - Q = (6, -1, -4) - (2, 1, -3) = (4, -2, -1)
Next, calculate the dot products of each pair of vectors:
PQ ⋅ PR = (1, 3, -2) ⋅ (5, 1, -3) = 1*5 + 3*1 - 2*(-3) = 5 + 3 + 6 = 14
PQ ⋅ QR = (1, 3, -2) ⋅ (4, -2, -1) = 1*4 + 3*(-2) - 2*(-1) = 4 - 6 + 2 = 0
PR ⋅ QR = (5, 1, -3) ⋅ (4, -2, -1) = 5*4 + 1*(-2) - 3*(-1) = 20 - 2 + 3 = 21
As we see, the dot product of PQ and QR is zero, which means these two vectors are orthogonal, and hence the triangle PQR is right-angled.
Think of an example that you could use to convice a friend that an object at rest has an "internal hold back property" that is different from a frictional force or the force of the earth's gravity.
Answer:
Explanation:
There are various examples around us that states " internal hold back property " and is different from frictional force or the force of Earth's gravity.
The plaster which is the mixture of cement and water has an adhesive nature and molecules inside them has a internal hold back property. They stick to the walls and ceiling and do not fall.
Though gravity is applied on it still its molecules tends to hold the ceiling firmly and do not fall upon.
A race car driver on the Flats first heads north for 6.71 km, then makes a sharp turn and heads southwest for 1.33 km, then makes another turn and heads east for 3.67 km. How far is she from where she started?
Answer:
D=7.385km
Explanation:
First we calculate how much it moves vertically (north and south), we will call this distance Y,.
taking into account that the car moves to the southwest and that it has a slope of 45 degrees, we can say that the component that moved south can be subtracted from the distance that moved north.
Y=6.71-1.33sen45=5.77km
then the distance that moves in the east we will call it X
we can add the component of the distance that moves to the southwest with the distance that moves to the east
X=3.67+1.33cos45=4.61Km
then we use the distance equation
[tex]D=\sqrt{x^{2} +y^{2} }[/tex]
[tex]D=\sqrt{4.61^{2}+5.77^{2} }[/tex]
D=7.385km