Answer:
The weight at a distance 2 RE from surface of earth is W/9
Explanation:
For the value of acceleration due to gravity (g), we have a formula, that is:
g = (G)(ME)/(RE)² ----- equation (1)
where,
G = Gravitational Constant
ME = Mass of Earth
RE = Radius of Earth
g = Acceleration due to gravity on surface of earth = 9.8 ms²
When the person goes 2RE, distance above earth's surface. Then the total distance from center of earth becomes: 2RE + RE = 3RE.
Therefore, equation (1) becomes:
gh = (G)(ME)/(3RE)²
where,
gh = acceleration due to gravity at height
gh = (G)(ME)/(RE)²9
using equation (1), we get:
gh = g/9
Now, he weight is given by formula:
W = mg ------- equation (2)
At height 2RE
Wh = (m)(gh)
where,
Wh = Weight at height = ?
m = mass of astronaut
Therefore, using vale of gh, we get:
Wh = mg/9
Using equation (2), we get:
Wh = W/9
A 4 kW vacuum cleaner is powered by an electric motor whose efficiency is 90%. (Note that the electric motor delivers 4 W of net mechanical power to the fan of the cleaner). What is rate at which this vacuum cleaner supplies energy to the room when running
Answer:[tex]3.6\ kW[/tex]
Explanation:
Given
Power Supplied [tex][tex]P_{input}=4\ kW[/tex][/tex]
Efficiency of the motor [tex]\neta =90\%[/tex]
and [tex]\neta =\dfrac{\text{Power output}}{\text{Power input}}[/tex]
[tex]\Rightarrow 0.9=\dfrac{P_{out}}{4}[/tex]
[tex]\Rightarrow P=0.9\times 4[/tex]
[tex]\Rightarrow P=3.6\ kW[/tex]
So, vacuum cleaner delivers a power of [tex]3.6\ kW[/tex]
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:
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
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.
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.
Beats are the result of
Answer:
The phenomenon of beats is the result of sound interference and sound diffraction in periodic vibrations. In periodic vibration, the beats of sounds produced by the interference and diffraction of the waves of different frequencies.
Explanation:
Beats. When two sound waves of different frequency approach your ear, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud - a phenomenon which is called "beating" or producing beats.
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 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
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]
Some animals, such as dolphins, use ________ to navigate and find food. This method uses sound waves.
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.
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Need help on this question I’d really appreciate it thanks!
Answer:radiation
Explanation:
radiation is the only one that makes sense
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.
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 heater rod (10 mm diameter, 100 mm length) of emissivity 0.75 is enclosed within a hollow cylindrical vacuum chamber (50 mm diameter, 100 mm length) of emissivity 0.25. The entire setup is insulated at the top and bottom ends by a low emissivity material, preventing any conductive heat dissipation from the ends. The heater rod is known to have a surface temperature of 1000 K, while the vacuum chamber is at a surface temperature of 300 K. How much heat is dissipated from the heater rod to the vacuum chamber (W)
Answer:
Explanation:
Given that:
Heater temperature ,T₁ = 1000K
Vaccum Chamber ,T₂ = 300K
emissivity of heater E₁ = 0.75
emissivity vaccum E₂ = 0.25
Heater diameter d₁ = 10 * 10⁻³mm
vaccum chamber d₂ = 50 * 10⁻³mm
When there is vaccum, then no air resistance will be there,
F₁₂ = 1
F₁₁ = 0
[tex]R_1= \frac{1-E_1}{E_1A_1} \\\\=\frac{1-0.75}{0.75*\pi * 10^-^2*L}[/tex]
[tex]R_2=\frac{1}{F_1_2 * A_1} \\\\=\frac{1}{1* \pi *10^-^2*L}[/tex]
[tex]R_3=\frac{1-0.25}{F_1_2 * A_1} \\\\=\frac{1}{0.25* \pi *5*10^-^2*L}[/tex]
Heat leaving from heater surface 1 to vaccum
[tex]Q_1_2 = \frac{L \pi \sigma (T_1^4- T_2^4)}{R_1+R_2+R_3}[/tex]
[tex]Q_1_2 = \frac{1000*10^-^3*\pi * 5.67*10^-^8(1000^4-300^4)}{\frac{0.25}{0.75*10^-2}+\frac{1}{10^-2} +\frac{0.75}{0.25*10^-^2*5} }[/tex]
[tex]Q_1_2 = \frac{1000*10^-^3*\pi * 5.67*10^-^8(1000^4-300^4)} {0.3333+1+0.6}\\\\Q_1_2= 91.39 \text {watt}[/tex]
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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A constant-velocity horizontal water jet from a stationary nozzle impinges normally on a vertical flat plate that rides on a nearly frictionless track. As the water jet hits the plate, it begins to move due to the water force. As a result, the acceleration will _____.
Answer:
a = ½ ρ A/M v₁²
Explanation:
This is a problem of fluid mechanics, where the jet of water at constant speed collides with a paddle, in this collision the water remains at rest, we write the Bernoulli equation, we will use index 1 for the jet before the collision the index c2 for after the crash
P₁ + ½ ρ v₁² + ρ g h₁ = P₂ + ½ ρ v₂² + ρ g h₂
in this case the water remains at rest after the shock, so v₂ = 0, as well as it goes horizontally h₁ = h₂
P₁-P₂ = ½ ρ v₁²
ΔP = ½ ρ v₁²
let's use the definition of pressure as a force on the area
F / A = ½ ρ v₁²
F = 1/2 ρ A v₁²
the density is
ρ = m / V
the volume is
V = A l
F = ½ m / l v₁²
knowing the force we can focus on the acceleration of the mass palette M
F = M a
a = F / M
a = ½ m/M 1/l v₁²
as well it can be given depending on the density of the water
a = ½ ρ A/M v₁²
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.
Learning Goal: To understand the cause of constructive and destructive interference for the double-slit experiment, and to explain how the interference pattern depends on the parameters of the emitted waves. For this tutorial, use the PhET simulation Wave Interference. This simulation allows you to send waves through a variety of barriers and look at the resulting interference patterns.
Answer:
a) it is essential that the waves lurk coherently
b)the light passes through the slits, the relative phase between the two rays is due to the optical path difference of each one,
Explanation:
In double slit interference experiments it is essential that the waves lurk coherently, that is, that the relative phase of the waves that reaches each slit is maintained over time, this is achieved when point sources are used by passing light through a initial slit and if a laser is used it is already consistent at the output.
When the light passes through the slits, the relative phase between the two rays is due to the optical path difference of each one, when this path difference is equal to a whole number of donut lengths, it has constructive interference and if it is a number sowing we have destructive interference.
d sin θ = m λ constructive
d sin θ = (m + ½) λ destructive
where d is the distance between the two slits, lam is the wavelength and m is an integer called the interference order
The amplitudes and phase differences for four pairs of waves of equal wavelengths are (a) 2 mm, 6 mm, and π rad; (b) 3 mm, 5 mm, and π rad; (c) 7 mm, 9 mm, and π rad; (d) 2 mm, 2 mm, and 0 rad. Each pair travels in the same direction along the same string. Without written calculation, rank the four pairs according to the amplitude of their resultant wave, greatest first. (Hint: Construct phasor diagrams.)
Answer:
a = d > b = c
Explanation:
The information about amplitudes and phase differences for four pairs of waves of equal wavelengths are given below:
(a) 2 mm, 6 mm, and π rad
(b) 3 mm, 5 mm, and π rad
(c) 7 mm, 9 mm, and π rad
(d) 2 mm, 2 mm, and 0 rad
Whenever a wave has zero phase difference, its amplitude of the resultant wave will be twice the amplitude of any of the two waves. Nevertheless, let assume that the amplitude is a vector having angle Ø between them. The resultant vector will help us rank the four pairs according to the amplitude of their resultant wave by using phasor diagrams.
a.) 6 - 2 = 4mm
b.) 5 - 3 = 2mm
c.) 9 - 7 = 2mm
d.) 2 + 2 = 4mm
Therefore,
a = d and b = c
a = d > b = c
Please find the attached file for the phasor diagrams
What is the acceleration of an object that takes 20 sec to change from a speed of 200 m/s to 300 m/s ?
Given:
initial velocity, u = 200 m/sFinal velocity, v = 300 m/s Time taken, t = 20 secTo be calculated:
Calculate the acceleration of given object ?
Formula used:
Acceleration = v - u / t
Solution:
We know that,
Acceleration = v - u / t
☆ Substituting the values in the above formula,we get
Acceleration ⇒ 300 - 200 / 20
⇒ 100/20
⇒ 5 m/s²
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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What does it mean that an exoplanet exists in the "habitable zone"? Question 6 options: (a) It is located the same distance from its star that Earth is from the Sun. (b) It is located the right distance from its star to enable liquid water to exist on its surface. (c) It means that the exoplanet has an oxygen rich atmosphere. (d) It is located a distance from its star that makes extinction level asteroid impacts unlikely.
Answer:
(b) It is located the right distance from its star to enable liquid water to exist on its surface.
Explanation:
The habitable zone according to astronomy and astrobiology is a region around a star that the planets around the star can hold and support liquid water. For our solar system, the habitable zone coincides approximately with the distance from the sun to Earth.
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]
Suppose multiple stations are connected in wired network where CSMA/CD implemented. Among them, station A wants to communicate with B and station C wants to communicate with D. Previously, both of these sender stations have faced collisions at different timings and are still unsuccessful in their transmission. The value of k for station A has reached to 6, whereas the value of k for station B has reached to 4. The maximum attempts allowed to any station is set to 10 and transmission time for a frame is 0.5ms.
This time, Both of the sender stations (A & C) sensed the medium, both got the channel idle and started transmission. Sooner, the station A detected the collision.
i.What will happen now?
ii.How long station A has wait before re-trying
iii.Meanwhile what will happen at station C
Answer:
bro i think ur from the uni of arid agriculture rawalpindi and today is your networking paper
Explanation:
A positive test charge of 5.0 x 10^-4 C is in an electric field that exerts a force of 2.5 x 10^-4 N on it. What is the magnitude of the electric field at the location of the test charge?
Answer:
E = 0.5N/C
Explanation:
In order to calculate the magnitude of the electric field you use the following formula:
[tex]E=\frac{F}{q}[/tex]
q: charge = 5.0*10^-4 C
F: force on the charge = 2.5*10^-4N
You replace the values of q and F in the equation for E:
[tex]E=\frac{2.5*10^{-4}N}{5.0*10^{-4}C}=0.5\frac{N}{C}[/tex]
hence, the magnitude of the electric field at the position of the carge is 0.5N/C
At the location of the test charge, the magnitude will be:
"0.5 N/C".
Electric fieldWhenever charge seems to be available inside any form, an electric property has been linked among each spatial position. This same size and direction are represented either by the quantity of E.
According to the question,
Charge, q = 5.0 × 10⁻⁴ C
Force on the charge, F = 2.5 × 10⁻⁴ N
We know the relation of the magnitude of electric field be
→ E = [tex]\frac{F}{q}[/tex]
By substituting the values, we get
= [tex]\frac{2.5\times 10^{-4}}{5.0\times 10^{-4}}[/tex]
= 0.5 N/C
Thus the approach above is correct.
Find out more information about electric field here:
https://brainly.com/question/14372859
The displacement of the air molecules in a sound wave is modeled with the wave function s(x, t) = 3.00 nm cos(50.00 m−1x − 1.71 ✕ 104 s−1t). (a) What is the wave speed (in m/s) of the sound wave? 342 Correct: Your answer is correct. m/s (b) What is the maximum speed (in m/s) of the air molecules as they oscillate in simple harmonic motion? m/s (c) What is the magnitude of the maximum acceleration (in m/s2) of the air molecules as they oscillate in simple harmonic motion?
Answer:
a) 342 m/s
b) 51*10^-6 m/s
c) 0.87m/s^2
Explanation:
The following function describes the displacement of the molecules in a sound wave:
[tex]s(x,t)=3.00nm\ cos(50.00\ m^{-1}x-1.71*10^4s^{-1}t)[/tex] (1)
The general form of a function that describes the same situation is:
[tex]s(x,t)=Acos(kx-\omega t)[/tex] (2)
By comparing equations (1) and (2) you have:
k: wave number = 50.00 m^-1
w: angular frequency = 1.71*10^4 s^-1
A: amplitude of the oscillation = 3.00nm
a) The speed of the sound is obtained by using the formula:
[tex]v=\frac{\omega}{k}=\frac{1.71*10^4s^-1}{50.00m^{-1}}=342\frac{m}{s}[/tex]
b) The maximum speed of the molecules is the maximum value of the derivative of s(x,t), in time. Then, you first obtain the derivative:
[tex]\frac{ds}{st}=-\omega A sin(kx-\omega t)[/tex]
The max value is:
[tex]v_{max}=\omega A[/tex]
[tex]v_{max}=(1.71*10^4s^-1)(3.00nm)=51300\frac{nm}{s}=51\frac{\mu m}{s}[/tex] = 51*10^-6 m/s
c) The acceleration is the max value of the derivative of the speed, that is, the second derivative of the displacement s(x,t):
[tex]a=\frac{dv}{dt}=\frac{d^2s}{dt^2}=-\omega^2A cos(kx-\omega t)\\\\a_{max}=\omega^2 A[/tex]
Then, the maximum acceleration is:
[tex]a_{max}=(1.71*10^4s^{-1})^2(3.00nm)=0.87\frac{m}{s^2}[/tex]
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.
A group of college students eager to get to Florida on a spring break drove the 710-mi trip with only minimum stops. They computed their average speed for the trip to be 55.7 mi/h.How many hours did the trip take?
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)
Which of these is not a factor that influences animal growth? *
2 points
A. inherited diseases
B. gravitational force
C. genetic information
D. availability of food