Answer:
Time taken for trip = 12.74 hour (Approx)
Explanation:
Given:
Distance of trip = 710-mi
Average speed for the trip = 55.7 mi/h
Find:
Time taken for trip = ?
Computation:
⇒ Time = Distance / Speed
⇒ Time taken for trip = Distance of trip / Average speed for the trip
⇒ Time taken for trip = 710-mi / 55.7 mi/h
⇒ Time taken for trip = 12.74 hour (Approx)
A spherical conductor with a 0.103 m radius is initially uncharged. How many electrons should be removed from the sphere in order for it to have an electrical potential of 4.50 kV at the surface
Answer:
Explanation:
capacitance of the capacitor C = 4πε₀R
= 1 x .103 / (9 x 10⁹ ) [ 1/ 4πε₀ = 9 x 10⁹ . ]
= .0114 x 10⁻⁹ F
potential V = 4.5 x 10³ v
charge Q = CV
.0114 x 10⁻⁹ x 4.5 x 10³
= .0515 x 10⁻⁶ C
charge on one electron = 1.6 x 10⁻¹⁹
no of electrons to be removed
= .0515 x 10⁻⁶ / 1.6 x 10⁻¹⁹
= .032 x 10¹³
3.2 x 10¹¹ electrons.
A spherical shell contains three charged objects. The first and second objects have a charge of −18.0 nC−18.0 nC and 38.0 nC38.0 nC , respectively. The total electric flux through the shell is −218 N⋅m2/C−218 N⋅m2/C . What is the charge on the third object?
Answer:
q3 = 21.9 nC
Explanation:
By the Gauss theorem you have that the electric flux in a Gaussian surface is given by:
[tex]\Phi_E=\frac{Q}{\epsilon_o}[/tex] (1)
ФE: electric flux = -218Nm^2/C
Q: net charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/(Nm^2)
You can consider the spherical shell as a Gaussian surface. Then, the net charge inside the surface is:
[tex]Q=-18.0nC+38.0nC+q_3[/tex] (2)
where charge q3 is unknown charge of the third object:
You replace the equation (2) into the equation (1), and you solve for q3:
[tex]\epsilon_0 \Phi_E=-18.0*10^{-9}C+38.0*10^{-9}C+q_3\\\\\epsilon_0 \Phi_E=20*10^{-9}C+q_3\\\\q_3=(8.85*10^{-12}C^2/(Nm^2))(-218Nm^2/C)-20*10^{-9}C\\\\q_3=2.19*10^{-9}C=21.9nC[/tex]
hence, the charge of the third object is 21.9 nC
5. Infer What might happen to sand in a
valley if more sand is deposited on top of it?
Answer:
It may turn into sedimentary rock when exposed to pressure.
Explanation:
When successive layer of sand is deposited on top of sand, the bottom layers go deeper and deeper. They are exposed to high pressures and temperatures and turn into hard rocks.
A 1kw electric heater is switched on for ten minutes. how much heat does it produce
H= P × t
1kW= 1000w
10 min = 600s
H= 1000×600=600,000J
=> 143.40kcal
A rock with mass m = 3.60 kg falls from rest in a viscous medium. The rock is acted on by a net constant downward force of F = 15.0 N (a combination of gravity and the buoyant force exerted by the medium) and by a fluid resistance force f=kv, where v is the speed in m/s and k = 2.68 N×s/m.(a) Find the initial acceleration a0 (b) Find the acceleration when the speed is 3.60 m/s.
Answer:
1
Explanation:
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A 80-kg base runner begins his slide into second base when he is moving at a speed of 3.7 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.A 45-kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.
Answer:
the vaulter has a height of 6.112 meters
Explanation:
Base runner: no question content.
__
Pole vaulter:
Initial kinetic energy* is ...
KE = (1/2)mv^2 = (1/2)(45 kg)(11 m/s)^2 = 2722.5 J
Above the bar, her kinetic energy is ...
KE = (1/2)mv^2 = (1/2)(45 kg)(1.1 m/s)^2 = 27.225 J
Then the amount of kinetic energy converted to potential energy is ...
2722.5 J -27.225 J = 2695.275 J
This corresponds to a change in height of ...
PE = mgh
2695.275 J = (45 kg)(9.8 m/s^2)h
h = 2695.275/(45·9.8) m = 6.112 m
Her altitude above the bar is 6.112 meters.
_____
* Here, we use the traditional equations for kinetic and potential energy, which use "m" to represent mass. When we fill in the numbers, we attach units to the numbers in which "m" represents "meters". We trust you can sort this without confusion.
A model rocket is fired straight up from the top of a 45-m-tall building. The rocket has only enough fuel to burn for 4.0 s. But while the rocket engine is burning fuel, it produces an upward acceleration of 55 m/s2. After the fuel supply is exhausted, the rocket is in free fall and just misses the edge of the building as it falls back to the ground. Ignoring air resistance, calculate (a) the height above the ground and the velocity of the rocket when its fuel runs out; (b) the maximum height of the rocket; (c) the time the rocket is in the air; and (d) the rocket's velocity the moment before it hits the ground.
Answer:
a)y = 485 m , v = 220 m / s , b) y = 2954.39 m , c) t_total = 51 s ,
d) v = 240.59 m / s
Explanation:
a) We can use vertical launch ratios for this exercise
the speed of the rocket the run out the fuel is
v = v₀ + a t
the rocket departs with initial velocity v₀ = 0
v = a t
v = 55 4
v = 220 m / s
the height at this point is
y = y₀ + v₀t + ½ a t²
y = y₀ + 1/2 a t²
y = 45 + ½ 55 4²
y = 485 m
b) the maximum height of the rocket is when its speed is zero
for this part we will use as the initial speed the speed at the end of the fuel (v₀´ = 220 m / s) and the height of y₀´ = 485 m
v² = v₀´² + 2 g (y-y₀´ )
0 = v₀´² + 2 g (y-y₀´ )
y = y₀´ + v₀´² / 2g
y = 485 + 220 2/2 9.8
y = 2954.39 m
c) the time that the rocket is in the air is the acceleration time t₁ = 4 s, plus the rise time (t₂) plus the time to reach the ground (t₃)
let's calculate the rise time
v = v₀´- g t
v = 0
t₂ = v₀´ / g
t₂ = 220 / 9.8
t₂ = 22.45 s
Now let's calculate the time it takes to get from this point (y₀´´ = 2954.39 m) to the floor
y = y₀´´ + v₀´´ t - ½ g t²
0 = y₀´´ - ½ g t²
t = √ (2 y₀´´ / g)
t = √ (2 2954.39 / 9.8)
t = 24.55 s
the total flight time is
t_total = t₁ + t₂ + t₃
t_total = 4 + 22.45 + 24.55
t_total = 51 s
d) the veloicda right now
v = vo + g t
v = 9.8 24.55
v = 240.59 m / s
Dock diving is a great form of athletic competition for dogs of all shapes and sizes. Sheba, the American Pit Bull Terrier, runs and jumps off the dock with an initial speed of 8.62 m/s at an angle of 28° with respect to the surface of the water. (Assume that the +x axis is in the direction of the run and the +y axis is up.) (a) If Sheba begins at a height of 0.85 m above the surface of the water, through what horizontal distance does she travel before hitting the surface of the water? m
Answer:
62.4m
Explanation:
a) Horizontal distance traveled = x = [tex]v_x[/tex] * t
where,
[tex]v_x[/tex] = horizontal velocity
and t = time in the air)
Time in the air t can be solved using the equation for y:
[tex]i_y=y_o+v_o_yt - 0.5gt^2[/tex]
where, [tex]y_o[/tex] = initial height i.e 0.85 m
[tex]v_o_y[/tex] = initial vertical velocity
and g = 9.8 m/s^2
When y = 0, the dog has hit the water.
So set y = 0 and solve for t.
[tex]v_o_y[/tex] = 8.62 m/s [tex]\times[/tex] (sin 28) = 4.04 m/s
0 = 0.85 m + (4.04 m/s)t - 0.5gt²
0.5t²-4.04t-0.85
Solve this quadratic formula for t: the solutions are t = 8.2 s and t = -0.2. Reject the negative solution, so t = 8.2 s.
How far does the dog travel horizontally in 9.2 s?
x = [tex]v_x[/tex] * t = (8.62 m/s) [tex]\times[/tex] (cos 28) [tex]\times[/tex] 8.2 s = 62.4 m
She travels the horizontal distance before hitting the surface of the water is 62.4m
What is projectile?When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion.
a) Horizontal distance traveled = x = V(x) * t
where, V(x) is the horizontal velocity and t = time in the air
Time in the air t can be solved using the equation for y:
hy =yo + Voy x t - 1/2gt²
where, yo= initial height i.e 0.85 m and Voy = initial vertical velocity
and g = 9.8 m/s^2
When y = 0, the dock has hit the water.
So set y = 0 and solve for t.
Voy = 8.62 m/s x (sin 28) = 4.04 m/s
Substitute the values, we get time as
0 = 0.85 m + (4.04 m/s)t - 0.5gt²
0.5t²-4.04t-0.85 =0
Solving the quadratic formula for t, we have
time t = 8.2 s and t = -0.2.
As time can't be negative, so, t = 8.2 s.
The horizontal distance is
x = Vo * t = (8.62 m/s) (cos 28) 8.2 s = 62.4 m
Thus, she travels 62.4 m.
Learn more about projectile.
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A 60 cm diameter potter's wheel with a mass of 30 kg is spinning at 180 rpm. Using her hands, a potter forms a 14 cm-diameter pot that is centered on and attached to the wheel. The pot's mass is negligible compared to that of the wheel. As the pot spins, the potter's hands apply a net frictional force of 1.3 N to the edge of the pot. If the power goes out, so that the wheels motor no longer provides any torque, how long will it take the wheel to come to a stop? You can assume that the wheel rotates on frictionless bearings and that the potter keeps her hands on the pot as it slows.
Answer:
It will take the wheel 278.9 s to come to a stop
Explanation:
Mass of the potter's wheel, M = 30 kg
Diameter of the potter's wheel, d₁ = 60 cm = 0.6 m
Radius, r₁ = d/2 = 0.6/2
r₁ = 0.3 m
The moment of inertia of the wheel, [tex]I = 0.5Mr_1^{2}[/tex]
[tex]I = 0.5*30*0.3^{2}\\I = 1.35 kg.m^2[/tex]
d₂ = 14 cm = 0.14 m
r₂ = 0.14/2 = 0.07 m
Angular velocity, [tex]\omega = 180 rpm[/tex]
[tex]\omega = \frac{180*2\pi }{60} \\\omega = 18.85 rad/s[/tex]
Frictional Force, F = 1.3 N
The torque generated:
[tex]\tau = F*r_{2}\\\tau = 1.3*0.07\tau = 0.091 Nm[/tex]
Torque can also be calculated as:
[tex]\tau = I \alpha\\\tau = I \frac{\omega }{t} \\0.091 = 1.35*\frac{18.8 }{t} \\t = (18.8*1.35)/0.091\\t = 278.9 s[/tex]
The time taken for the potter's wheel to come to a stop is 280 s.
The given parameters;
diameter of the potter's wheel, d = 60 cm radius of the wheel, r = 30 cm = 0.3 mangular speed, ω = 180 rpmThe momentum of inertia of the potter's wheel is calculated ;
[tex]I = \frac{1}{2} Mr^2\\\\I = (0.5)(30)(0.3)^2\\\\I = 1.35 \ kgm^2[/tex]
The angular speed of the potter's wheel is calculated as follows;
[tex]\omega = 180 \ \frac{rev}{\min} \ \times \ \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s} \\\\\omega = 18.85 \ rad/s[/tex]
The time taken for the wheel to come to a stop is calculated as;
[tex]Fr = I \alpha \\\\Fr= I \times \frac{\omega}{t} \\\\t = \frac{I \omega }{Fr} \\\\[/tex]
d = 14 cm, r = 7 cm = 0.07 m
[tex]t = \frac{1.35 \times 18.85}{1.3 \times 0.07} \\\\t = 279.6 \ s\\\\t\approx 280 \ s[/tex]
Thus, the time taken for the potter's wheel to come to a stop is 280 s.
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The planet Earth travels in a circular orbit at constant speed around the Sun. What is the net work done on the Earth by the gravitational attraction between it and the Sun in one complete orbit
Answer:
WT = 3.32*10^34 J
Explanation:
The work done by the gravitational attraction between the Sun and the Earth in one complete orbit of the Earth can be calculated by using the following formula:
[tex]W_T=\int F_g dr[/tex] (1)
Fg: gravitational force between Sun and Earth
The gravitational force is given by:
[tex]F_g=G\frac{m_sm_e}{r^2}[/tex] (2)
G: Cavendish's constant = 6.674*10^-11 m^3 kg^-1 s^-2
ms: mass of the sun = 1.989*10^30 kg
me: mass of the Earth = 5.972 × 10^24 kg
r: distance between Earth and Sun, this value is a constant r = R = 149,597,870 km
You replace the formula (2) in (1):
[tex]W_T=\int G\frac{m_sm_e}{R^2}dr=G\frac{m_sm_e}{R^2}\int dr\\\\W_T=G\frac{m_sm_e}{R^2}(2\pi R)=2\pi G\frac{m_sm_e}{R}[/tex]
Next, you replace the values of all variables and solve obtain WT:
[tex]W_T=2\pi (6.674*10^{-11}m^3kg^{-1}s^{-2})\frac{(1.989*10^{30}kg)(5.972*10^{24}kg)}{(149597870*10^3 m)}\\\\W_T=3.32*10^{34}J[/tex]
hence, the work done on the Earth, in one orbit, is 3.32*10^34 J
Where is the potential energy equal to zero?
Answer:
im sure your already past this but it's E.
Explanation:
This is because in this case potential energy is linear to height, which means that the higher the more potential energy.
A substance is boiled repeatedly and stirred, but the solute never mixes with the solvent. Which best describes why? The temperature was not high enough to mix the solute and solvent. The chemical properties of the solute and solvent are different. The pressure was increased. The solution was saturated.
Answer:
b
Explanation:
If a substance is boiled repeatedly and stirred, but the solute never mixes with the solvent then the chemical properties of the solute and solvent are different, therefore the correct answer is option B.
What is a Chemical compound?A chemical compound is a combination of two or more either similar or dissimilar chemical elements.
For example, H₂O is a chemical compound made up of two oxygen atoms and a single hydrogen atom.
As given in the problem a substance is boiled repeatedly and stirred, but the solute never mixes with the solvent,
The correct response is option B because the chemical characteristics of the solute and solvent are different if the material is repeatedly boiled and stirred without the solute and solvent mixing.
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A leaf floating down from a tree is an example of an object in free fall.
A. True
B. False
Answer:
its false
Explanation:
because if an leaf floats down from a tree it is not considered an object for a free-fall
ASAP pls answer right I will mark brainiest . All I know is 4. Is A
Answer:
Q1: B.2 Q2: B.Waxing crescent Q3: A.Waxing Gibbous
Explanation:
PLZZZ HELP
Question 2 Multiple Choice Worth 2 points)
(04.05 LC)
Which of the following scenarios describe an object being acted on by a non-conservative force?
O Apuck hovers and glides across an air track at constant velocity.
A dishwasher was pulled up to a window by a rope using a pulley system.
O A feather falls from one end of a tube to the other inside a vacuum.
O A charged particle was pushed away by another charged particle.
Answer:
C
Explanation:
A feather falls from one end of a tube to the other inside a vacuum.
Horizontal beam AB is 200 kg, 2.4 m long, and is welded at point A. The man is 80 kg and applies a tension of 300 N on the cable. Diameter of the pulley is 300 mm and BC = 300 mm. Determine:
(a) horizontal and vertical components of force at A,
(b) magnitude and direction of the moment supported at A.
Answer:
[tex]a)-3346.8\;N \\ b)-4937.04\;N-m[/tex]
Explanation:
a) - In Free-body diagram :
At point D, the free body diagram of a man :
[tex]D = Fn\\Fn=T+mg\\put\;values\;in\;it\\ .\;\;\;\;=300+(80)(9.81)=1084.8\;N[/tex]
[tex]Mg=200\times9.81=1962\;N[/tex]
[tex]\sum Fx=0\; where\; Ax=0[/tex]
[tex]\sum Fy=0\; where\; Ay-Mg-Fn-T=0[/tex]
Then, put the value in the equation.
[tex]Ay=3346.8\;N[/tex]
b)-
[tex]Ma=Mg(AE)+Fn(AD)+T=4937.04\;N-m[/tex]
A 30-kg skater moving at 3 m/s overtakes a 35-kg skater moving at 1 m/s in the same direction and collides with her. The two skaters stick together. It’s final speed is 1.9 m/s. How much kinetic energy is lost? Round all your answers to one decimal place.
Answer:
The lost in kinetic energy is [tex]KE_l = 125.5 \ J[/tex]
Explanation:
From the question we are told that
The mass of the first skater is [tex]m_1 = 30 \ kg[/tex]
The speed of the first skater is [tex]v_1 = 3 \ m/s[/tex]
The mass of the second skater is [tex]m_2 = 35 \ kg[/tex]
The speed of the second skater is [tex]v_2 = 1 \ m/ s[/tex]
The final speed of both skater are [tex]v_f = 1.9 m/s[/tex]
The initial kinetic energy of both skaters is mathematically represented as
[tex]KE_i = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2[/tex]
substituting values
[tex]KE_i = \frac{1}{2} * 30 * 3^2 + \frac{1}{2} * 35 * 1^2[/tex]
[tex]KE_i = 242.5 \ J[/tex]
The final kinetic energy of both skaters is mathematically represented as
[tex]KE_f = \frac{1}{2} * (m_1 + m_2 ) v_f^2[/tex]
substituting values
[tex]KE_f = \frac{1}{2} * (30 + 35 ) * 1.9^2[/tex]
[tex]KE_f = 117 \ J[/tex]
The lost in kinetic energy is
[tex]KE_l = 242.5 -117[/tex]
[tex]KE_l = 125.5 \ J[/tex]
4) (5 points) Given are the magnitudes and orientations (with respect to x-axis) of 3
vectors:
TĀ = 4.5N, 0,= 55
BI = 1.5N, 0,= 145
C = 6.00, 0 = 235
Determine the resultant vector Ř of these 3 vectors through the analytic method
involving components. You have state both the magnitude and the angle with respect to
x-axis.
Expand each vector into their component forms:
[tex]\vec A=(4.5\,\mathrm N)(\cos\theta_A\,\vec\imath+\sin\theta_A\,\vec\jmath)=(2.58\,\vec\imath+3.69\,\vec\jmath)\,\mathrm N[/tex]
Similarly,
[tex]\vec B=(-1.23\,\vec\imath+0.860\,\vec\jmath)\,\mathrm N[/tex]
[tex]\vec C=(-3.44\,\vec\imath-4.91\,\vec\jmath)\,\mathrm N[/tex]
Then assuming the resultant vector [tex]\vec R[/tex] is the sum of these three vectors, we have
[tex]\vec R=\vec A+\vec B+\vec C[/tex]
[tex]\vec R=(-2.09\,\vec\imath-0.368\,\vec\jmath)\,\mathrm N[/tex]
and so [tex]\vec R[/tex] has magnitude
[tex]\|\vec R\|=\sqrt{(-2.09)^2+(-0.368)^2}\,\mathrm N\approx2.12\,\mathrm N[/tex]
and direction [tex]\theta_R[/tex] such that
[tex]\tan\theta_R=\dfrac{-0.368}{-2.09}\implies\theta_R=-170^\circ=190^\circ[/tex]
For the different values given for the radius of curvature RRR and speed vvv, rank the magnitude of the force of the roller-coaster track on the cart at the bottom of each dip. Rank from largest to smallest. To rank items as equivalent, overlap them.
A) R = 60m v = 16 m/s
B) R = 15m v = 8m/s
C) R = 30m v = 4m/s
D) R = 45m v = 4m/s
E) 30m v = 16 m/s
F) R = 15m v =12 m/s
Explanation:
The force of the roller-coaster track on the cart at the bottom is given by :
[tex]F=\dfrac{mv^2}{R}[/tex], m is mass of roller coaster
Case 1.
R = 60 m v = 16 m/s
[tex]F=\dfrac{(16)^2m}{60}=4.26m\ N[/tex]
Case 2.
R = 15 m v = 8 m/s
[tex]F=\dfrac{(8)^2m}{15}=4.26m\ N[/tex]
Case 3.
R = 30 m v = 4 m/s
[tex]F=\dfrac{(4)^2m}{30}=0.54m\ N[/tex]
Case 4.
R = 45 m v = 4 m/s
[tex]F=\dfrac{(4)^2m}{45}=0.36m\ N[/tex]
Case 5.
R = 30 m v = 16 m/s
[tex]F=\dfrac{(16)^2m}{30}=8.54m\ N[/tex]
Case 6.
R = 15 m v =12 m/s
[tex]F=\dfrac{(12)^2m}{15}=9.6m\ N[/tex]
Ranking from largest to smallest is given by :
F>E>A=B>C>D
Which best describes what is made of matter?
all living things and objects
all atoms
all solid objects and atoms
all living things
A. All living things and objects
Answer:
A: All living things and objects. I Hope This Helped!
Explanation:
Describes Matter.
Describe how heat transfer occurs when you
place your hand in a sink full of hot water.
Answer:
If one places ones's hands into a sink full of hot water, heat is transferred from the hot water to the hands. Heat only flows when there is a temperature gradient, i.e when there is a difference in temperature between a hot and a cold body. Heat is always transferred from a hot body to a cold body and never the reverse in a normal system.
At the boundary of the hand and water, heat transfer is by conduction between the hot water molecules and the molecules of the hands. This heat is then further transferred to the internal body organs and body fluid through either convection or conduction. This heat transfer is maintained until both the water and the hands are at the same temperature, assuming the owner is able to withstand the heat.
What is potential energy
Answer:
Potential Energy is when the energy is kept in place the object isnt moving so it isnt creating Kinetic energy.
Answer:
[tex]\boxed{\mathrm{view \ explanation}}[/tex]
Explanation:
Potential energy is the energy stored in a body due to the body’s positioning or state.
What is the difference between reflection and refraction
Answer:
Reflection is when light bounces off an object, while refraction is when light bends while passing through an object.
Explanation:
I just learned about this 2 weeks ago actually.
Where does the energy released in nuclear reactions come from
Answer:
Nuclear energy comes from tiny mass changes in nuclei as radioactive processes occur. In fission, large nuclei break apart and release energy; in fusion, small nuclei merge together and release energy.
I hoep that was helpful! Let me know:)
Explanation:
What is kinetic energy
Answer:
Kinetic energy is the energy of motion
Answer:
Energy which a body possesses by virtue of being in motion
Explanation:
A learner driver travels at a speed of 3km/h for an hour and a half. What distance does it travel?
Answer:
distance covered is 4.5km or 1.25m
Explanation:
GIVEN DATA:
SPEED=v=3km/h
TIME=t=1.5h
TO FIND:
DISTANCE COVERED=d=?
SOLUTION:
As we know that
speed=distance covered /time taken
here we have to find distance
speed×time taken=distance covered
3km/h×1.5h=distance covered
4.5km=distance covered or in international system of unit the answer is 1.25m
An earth observer sees an alien ship pass overhead at 0.3c. The ion gun of the alien ship shoots ions straight ahead of the ship at a speed 0.4c relative to the ship. What is the speed of the ions relative to the earth observer?
Answer:
The velocity of the ions relative to the earth observer is [tex]v = 0.63 c[/tex]
Explanation:
From the question we are told that
The speed of the alien ship is [tex]v_a = 0.3c[/tex]
The speed of the iron gun is [tex]v_g = 0.4c[/tex]
The motion of the ions relative to the earths observer can be mathematically represented as
[tex]v = \frac{v_a + v_g}{[\frac{1+ v_av_g}{c^2} ]}[/tex]
substituting values
[tex]v = \frac{ 0.3c + 0.4c}{[\frac{1+ 0.3c * 0.4c}{c^2} ]}[/tex]
=> [tex]v = 0.63 c[/tex]
(a) At an air show a jet flies directly toward the stands at a speed of 1220 km/h, emitting a frequency of 3270 Hz, on a day when the speed of sound is 342 m/s. What frequency (in Hz) is received by the observers?
Answer:
f' = 358442.3 Hz
Explanation:
This is a problem about the Doppler's effect. To find the perceived frequency of the observers you use the following formula:
[tex]f'=f(\frac{v}{v\pm v_s})[/tex]
v: speed of sound = 342 m/s
vs : speed of the source (jet) = 1220 km/h (1h/3600s)*(1000 m/ 1km) = 338.88 m/s (it is convenient to convert the units of vs to m/s)
f: frequency emitted by the source = 3270 Hz
f': perceived frequency
Due to the jet is getting closer to the observers, the sing of the denominator in equation (1) is minus (-). Then, you replace the values of f, vs and v:
[tex]f'=(3270s^{-1})(\frac{342m/s}{342m/s-338.88m/s})=358442.3 Hz[/tex]
Hence, the perceived frequency by the observers is 358442.3 Hz
Answer:
[tex]f_o=359466.42Hz[/tex]
Explanation:
You can solve this problem using doppler effect formula. The Doppler effect is the phenomenon by which the frequency of the waves perceived by an observer varies when the emitting focus or the observer itself moves relative to each other
Doppler effect general case is given by:
[tex]f_o=f\frac{v\pm v_o}{v \mp v_s} \\\\Where:\\\\f=Actual\hspace{3}frequency\\f_o=Observed\hspace{3}frequency\\v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves\\v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer\\v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source[/tex]
Now, you can use the following facts:
[tex]+v_o[/tex] Is used when the observer moves towards the source
[tex]-v_o[/tex] Is used when the observer moves away from the source
[tex]-v_s[/tex] Is used when the source moves towards the observer
[tex]+v_s[/tex] Is used when the source moves away from the observer
Since the source is alone in motion towards the observer, the formula is given by:
[tex]f_o=f\frac{v}{v-v_s} \\\\Where:\\\\v_s=1220km/h\approx 338.8888889m/s\\v=342m/s\\f=3270Hz[/tex]
Therefore:
[tex]f_o=(3270) \frac{342}{342-338.8888889} = 359466.42Hz[/tex]
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An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99537 c relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 40.0 km .
As measured by the scientist, how much time does it take the particle to travel the 40.0 km to the surface of the earth?
Answer:
Explanation:
Velocity relative to earth = .99537 c
= .99537 x 3 x 10⁸ m /s
= 2.98611 x 10⁸ m /s
distance relative to earth = 40 km = 40 x 10³ m .
Time measured by scientist
= 40 x 10³ / 2.98611 x 10⁸
= 13.395 x 10⁻⁵ s
= 134 μs.