Answer:
Explanation:
The formula holding the relationship between wavelength and frequency of a wave can be expressed as:
[tex]\lambda = \dfrac{v}{f}[/tex]
where;
[tex]\lambda =[/tex] wavelength
v = speed
f = frequency
Given that:
v = 192 m/s and f = 240Hz
Then;
[tex]\lambda = \dfrac{192 \ m/s}{240 \ Hz}[/tex]
[tex]\lambda = 0.800 \ m[/tex]
Now, to estimate the respective amplitude of the string, we need to approach it by using the concept of wave equation which is:
y = A sin kx
here;
A = amplitude of the standing wave
k = wave number
x = maximum displacement
y = distance from center
recall that:
[tex]k = \dfrac{2 \pi}{\lambda}[/tex]
∴
[tex]y = A sin \dfrac{2 \pi}{\lambda }x[/tex]
Now;
for A = 0.400 cm ; [tex]\lambda[/tex] = 0.800 m ; k = 40 cm
Then;
[tex]y =(0.400 \ cm ) \ sin \dfrac{2 \pi}{0.800 m }\times (40 \ cm \times \dfrac{10^{-2} \ m}{1 \ cm } )[/tex]
y = 0.400 sin π
y = 0 cm
At distance 40 cm; the amplitude = 0 cm
Thus, it is a node.
For k = 20cm
Then:
[tex]y =(0.400 \ cm ) \ sin \dfrac{2 \pi}{0.800 m }\times (20 \ cm \times \dfrac{10^{-2} \ m}{1 \ cm } )[/tex]
y = 0.400 sin π/2
y = 0.400 cm
At distance 20 cm; the amplitude = 0.400 cm
Thus, it is antinode.
For k = 10cm
Then:
[tex]y =(0.400 \ cm ) \ sin \dfrac{2 \pi}{0.800 m }\times (10 \ cm \times \dfrac{10^{-2} \ m}{1 \ cm } )[/tex]
4y = 0.400 sin π/2
y = 0.283 cm
At distance 10 cm; the amplitude = 0.283 cm
b)
The required time taken to go through the displacement( i.e. from largest upward to downward) is the time required to cover half of the wavelength.
This is expressed as:
[tex]T = \dfrac{1}{2} \times \dfrac{1}{f}[/tex]
[tex]T= \dfrac{1}{2} \times \dfrac{1}{240 \ Hz}[/tex]
T = 0.00208
T = 2.08 × 10⁻³ s
A 35 kg box initially sliding at 10 m/s on a rough surface is brought to rest by 25 N
of friction. What distance does the box slide?
Answer:
the distance moved by the box is 70.03 m.
Explanation:
Given;
mass of the box, m = 35 kg
initial velocity of the box, u = 10 m/s
frictional force, F = 25 N
Apply Newton's second law of motion to determine the deceleration of the box;
-F = ma
a = -F / m
a = (-25 ) / 35
a = -0.714 m/s²
The distance moved by the box is calculated as follows;
v² = u² + 2ad
where;
v is the final velocity of the box when it comes to rest = 0
0 = 10² + (2 x - 0.714)d
0 = 100 - 1.428d
1.428d = 100
d = 100 / 1.428
d = 70.03 m
Therefore, the distance moved by the box is 70.03 m.
A 5-kg object is moving with a speed of 4 m/s at a height of 2 m. The potential energy of the object is approximately
J.
Answer:
P.E = 98 Joules
Explanation:
Given the following data;
Mass = 5kg
Speed = 4m/s
Height = 2m
We know that acceleration due to gravity is equal to 9.8m/s²
To find the potential energy;
Potential energy can be defined as an energy possessed by an object or body due to its position.
Mathematically, potential energy is given by the formula;
[tex] P.E = mgh[/tex]
Where, P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
Substituting into the equation, we have;
[tex] P.E = 5*9.8*2[/tex]
P.E = 98 Joules
How does speed and mass effect kinetic energy ?
Answer:
in fact, kinetic energy is directly proportional to mass: if you double the mass, then you double the kinetic energy. Second, the faster something is moving, the greater the force it is capable of exerting and the greater energy it possesses. ... Thus a modest increase in speed can cause a large increase in kinetic energy.
Explanation:
Answer: The more mass of an object has, the more Kinetic energy it has.
Explanation:
Kinetic energy is comparable to mass. If you double the mass then you double the kinetic energy. The faster the object is moving the greater the energy possesses. A large increase in speed can have a large increase in kinetic energy.
6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a driving speed of 50 mi/h. When the driver is sober, a stop can be made just in time to avoid hitting an object that is first visible 385 ft ahead. After a few drinks under exactly the same conditions, the driver fails to stop in time and strikes the object at a speed of 30 mi/h. Determine the driver's perception/reaction time before and after drinking. (Assume practical stopping distance.)
Answer:
a. 10.5 s b. 6.6 s
Explanation:
a. The driver's perception/reaction time before drinking.
To find the driver's perception time before drinking, we first find his deceleration from
v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m
So, a = v² - u²/2s
substituting the values of the variables into the equation, we have
a = v² - u²/2s
a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)
a = - 499.52 m²/s²/234.7 m
a = -2.13 m/s²
Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time
So, t = (v - u)/a
Substituting the values of the variables into the equation, we have
t = (0 m/s - 22.35 m/s)/-2.13 m/s²
t = - 22.35 m/s/-2.13 m/s²
t = 10.5 s
b. The driver's perception/reaction time after drinking.
To find the driver's perception time after drinking, we first find his deceleration from
v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m
So, a = v² - u²/2s
substituting the values of the variables into the equation, we have
a = v² - u²/2s
a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)
a = 179.83 m²/s² - 499.52 m²/s²/234.7 m
a = -319.69 m²/s² ÷ 234.7 m
a = -1.36 m/s²
Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time
So, t = (v - u)/a
Substituting the values of the variables into the equation, we have
t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²
t = - 8.94 m/s/-1.36 m/s²
t = 6.6 s
Anyone can help me out with this question ? Just number 2,
Answer:
- 21⁰C .
Explanation:
Speed of jet = 2.05 x 10³ km /h
= 2050 x 1000 / (60 x 60 ) m /s
= 569.44 m / s
Mach no represents times of speed of sound , the speed of jet
1.79 x speed of sound = 569.44
speed of sound = 318.12 m /s
speed of sound at 20⁰C = 343 m /s
Difference = 343 - 318.12 = 24.88⁰C
We know that 1 ⁰C change in temperature changes speed of sound
by .61 m /s
So a change in speed of 24.88 will be produced by a change in temperature of
24.88 / .61
= 41⁰C
temperature = 20 - 41 = - 21⁰C .
A flat circular mirror of radius 0.100 m is lying on the floor. Centered directly above the mirror, at a height of 0.920 m, is a small light source. Calculate the diameter of the bright circular spot formed on the 2.70 m high ceiling by the light reflected from the mirror.
Answer:
the diameter of the bright circular spot formed is 0.787 m
Explanation:
Given that;
Radius of the flat circular mirror = 0.100 m
height of small ight source = 0.920 m
high ceiling = 2.70 m
now;
Diameter(mirror) = 2×r = 2 × 0.100 = 0.2 m
D(spot) = [Diameter(mirror) × ( 2.70m + 0.920 m)] / 0.920 m
so
D(spot) = 0.2m × 3.62m / 0.920 m
D(spot) = 0.724 m / 0.920 m
D(spot) = 0.787 m
Therefore, the diameter of the bright circular spot formed is 0.787 m
Determine the magnitude of the electric field at the point P. Express your answer in terms of Q, x, a, and k. Express your answer in terms of the variables Q, x, a, k, and appropriate constants.
Complete Question
The question image is in the first uploaded image
Answer:
[tex]E=\frac{KQ*4xa}{(x^2-a^2)^2}[/tex]
Explanation:
From the question we are told that
Distance b/w Q mid point and P is given as x
Generally the equation for magnitude of the electric field at the point P is given as
[tex]E=\frac{kQ}{d^2}[/tex]
where
[tex]k=\frac{1}{4\pi e_0}[/tex]
[tex]d=x^2-a^2[/tex]
Therefore
[tex]E= \frac{1}{4\pi e_0} \frac{Q}{(x^2-a^2)^2}- \frac{1}{4\pi e_0} \frac{Q}{(x^2+a^2)^2}[/tex]
[tex]E= \frac{Q}{4\pi e_0} (\frac{1}{(x^2-a^2)^2}- \frac{1}{(x^2+a^2)^2})[/tex]
Therefore equation for magnitude of the electric field at the point P is
[tex]E=\frac{KQ*4xa}{(x^2-a^2)^2}[/tex]
A remote controlled airplane moves 7.2 m in 2.5seconds what is the plane’s velocity
Answer:
2.88m/s
Explanation:
Given parameters:
Displacement = 7.2m
Time taken = 2.5s
Unknown:
Velocity of the plane = ?
Solution:
Velocity is the displacement divided by the time taken.
Velocity = [tex]\frac{displacement}{time taken}[/tex]
So;
Velocity = [tex]\frac{7.2}{2.5}[/tex] = 2.88m/s
If the speed of an object does NOT change, the object is traveling at a
constant speed
increasing speed
decreasing speed
Answer:
If the speed does not change at all, the object would be moving at a constant speed.
An object of height 2.7 cm is placed 29 cm in front of a diverging lens of focal length 16 cm. Behind the diverging lens, and 12 cm from it, there is a converging lens of the same focal length.
A. Find the location of the final image, in centimeters beyond the converging lens.
B. What is the magnification of the final image?
Answer:
A) q = -8.488 cm , B) m = 0.29
Explanation:
A) For this exercise in geometric optics, we will use the equation of the constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where p and q are the distance to the object and image, respectively and f is the focal length
in our case the distance the object is p = 29 cm the focal length of a diverging lens is negative and indicates that it is f = - 12 cm
[tex]\frac{1}{q} = \frac{1}{f} - \frac{1}{p}[/tex]
we calculate
[tex]\frac{1}{q} = - \frac{1}{12} - \frac{1}{29}[/tex]
[tex]\frac{1}{q}[/tex] = - 0.1178
q = -8.488 cm
the negative sign indicates that the image is virtual
B) the magnification is given
[tex]m = \frac{h'}{h} = - \frac{q}{p}[/tex]
we substitute
m = [tex]- \frac{-8.488}{29}[/tex]
m = 0.29
the positive sign indicates that the image is right
On the periodic table , the vertical columns that extend down the periodic table are called ?
Answer:
groups
Explanation:
Answer: Groups
Explanation: They are in the same group! Like the Alkaline Metals are all the group. They all lose an electron. :)
According to Newton's law of universal gravitation, which statements are true?
As we move to higher altitudes, the force of gravity on us decreases.
O As we move to higher altitudes, the force of gravity on us increases,
O As we gain mass, the force of gravity on us decreases.
O Aswe gain mass, the force of gravity on us increases.
DAs we move faster, the force of gravity on us increases.
A 30 kg male emperor penguin under a clear sky in the Antarctic winter loses very little heat to the environment by convection; its feathers provide very good insulation. It does lose some heat through its feet to the ice, and some heat due to evaporation as it breathes; the combined power is about 12 W. The outside of the penguin's body is a chilly −22∘C, but its surroundings are an even chillier −38∘C. The penguin's surface area is 0.56 m2, and its emissivity is 0.97. What is the rate of energy loss by radiation
Answer:
Rate of energy loss by radiation is 28.31 Watt
Explanation:
Given that;
m = 30 kg
power p = 12 W
emissivity e = 0.97
Surface Area A = 0.56 m²
outside of the penguin's body T = −22°C
surroundings Temperature Ts = -38°C
the rate of energy loss by radiation = ?
Now, using Stefan-Boltzmann law;
P = σeA [ T⁴ - Ts⁴ ]
Stefan's constant σ = 5.67 × 10⁻⁸
so we substitute
P = 5.67 × 10⁻⁸ × 0.97 × 0.56 [ (-22 + 273 k)⁴ - (-38 + 273 k )⁴]
= 3.079944 × 10⁻⁸ [ 919325376]
= 28.31 Watt
the rate of energy loss by radiation is 28.31 Watt
Energy from the Sun is transferred from the Earth’s surface to the atmosphere, resulting in
atmospheric convection currents that produce winds. How do physical properties of the air
contribute to convection currents?
a -The warmer air sinks because it is more dense than cooler air.
b -The warmer air rises because it is more dense than cooler air.
c- The warmer air sinks because it is less dense than cooler air.
d -The warmer air rises because it is less dense than cooler air.
For a flourish at the end of her act, a juggler tosses a single ball high in the air. She catches the ball 3.2 s later at the same height from which it was thrown. What was the initial upward speed of the ball?
Answer:
15.68 m/s
Explanation:
Given that,
She catches the ball 3.2 s later at the same height from which it was thrown.
When it reaches the maximum height, its height is equal to 0.
It will move under the action of gravity.
[tex]t=\dfrac{2u}{g}[/tex]
2 here comes for the time of ascent and descent.
So,
[tex]u=\dfrac{tg}{2}\\\\u=\dfrac{3.2\times 9.8}{2}\\\\u=15.68\ m/s[/tex]
So, the initial upward speed of the ball is 15.68 m/s.
A particle with charge Q and mass M has instantaneous speed uy when it is at a position where the electric potential is V. At a later time, the particle has moved a distance R away to a position where the electric potential is V2 ) Which of the following equations can be used to find the speed uz of the particle at the new position?
a. 1/2M(μ2^2-μ1^2)=Q (v1-v2)
b. 1/2M(μ2^2-μ1^2)^2=Q(v1-v2)
c. 1/2Mμ2^2=Qv1
d. 1/2Mμ2^2=1/4πx0 (Q^2/R)
Answer:
A
Explanation:
Ke = 1/2 MV^2
Statement A: 2.567 km, to two significant figures. Statement B: 2.567 km, to three significant figures. Determine the correct relationship between the statements. View Available Hint(s) Determine the correct relationship between the statements. Statement A is greater than Statement B. Statement A is less than Statement B. Statement A is equal to Statement B.
Answer:
Statement A is greater than Statement B.
Explanation:
Statement A: 2.567 km, to two significant figures..
To 2 sig figures means only 2 whole numbers should be left after approximation. Thus, 2.567 to 2 significant figures is 2.6 km
Statement B: 2.567 km, to three significant figures. To 3 sig figures means only 3 whole numbers should be left after approximation. Thus, 2.567 to 3 significant figures is 2.57 km
Comparing both values, statement A is obviously greater than Statement B
Which of the following is a mixture?
a air
biron
Chydrogen
d nickel
Answer:
it will option option A hope it helps
what is momentum of a train that is 60,000 kg that is moving at velocity of 17m/s?
explain your answer
Consider the low-speed flight of a Space Shuttle as it is nearing a landing. If the air pressure and temperature at the nose of the shuttle are 1.05 atm and 300 K, respectively, calculate the density and specific volume. (Round the final answer to two decimal places.) The density is kg/m3. The specific volume is m3/kg.
Answer:
d = 1.24 kg/m³
v = 0.81 m³/kg
Explanation:
To do this, we need to analyze the given data and know the expressions we need to use here to do calculations.
We have a pressure of 1.05 atm and 300 K of temperature. To determine the density, we need to use a similar expression of an ideal gas. In this case, instead of using moles, we will use density:
P = dRT
d = P/RT (1)
Where:
R: universal constant of gases
d: density.
From here we can determine the specific volume by using the following expression:
v = 1/d (2)
Now, as we are looking for density, we need to convert the units of pressure in atm to Pascal (or N/m) and the conversion is the following:
P = 1.05 atm * 1.013x10⁵ N/m atm = 106,365 N/m
Now, using R as 287 the density would be:
d = 106,365 / (287 * 300)
d = 1.24 kg/m³Finally the specific volume:
v = 1 / 1.41
v = 0.81 m³/kgHope this helps
What is the period of an objects motion?
A group of 25 particles have the following speeds: two have speed 11 m/s, seven have 16 m/s , four have 19 m/s, three have 26 m/s, six have 31 m/s, one has 37 m/s, and two have 45 m/s.
Requiredd:
a. Determine the average speed.
b. Determine the rms speed.
c. Determine the most probable speed.
Answer:
a) Average speed is 24.04 m/s
b) the rms speed is 25.84 m/s
c) the most probable speed is 16 m/s
Explanation:
Given the data in the question;
a) Determine the average speed.
To determine the average speed, we simply divide total some of speed by number of particles;
Average speed = [(2×11 m/s)+(7×16 m/s)+(4×19 m/s)+(3×26 m/s)+(6×31 m/s)+(1×37 m/s)+(2×45 m/s)] / 25
= 601 / 25
= 24.04 m/s
Therefore, Average speed is 24.04 m/s
b) Determine the rms speed
we know that (rms speed)² = sum of square speed / total number of particles
so
(rms speed)² = [(2×11²)+(7×16²)+(4×19²)+(3×26²)+(6×31²)+(1×37²)+(2×45²)] / 25
(rms speed)² = 16691 / 25
(rms speed)² = 667.64
(rms speed) = √ 667.64
(rms speed) = 25.84 m/s
Therefore, the rms speed is 25.84 m/s
c) Determine the most probable speed.
Most particles (7) have velocity 16 m/s
i.e 7 is the maximum number of particle for a particular speed ,
Therefore, the most probable speed is 16 m/s
g Incandescent bulbs generate visible light by heating up a thin metal filament to a very high temperature so that the thermal radiation from the filament becomes visible. One bulb filament has a surface area of 30 mm2 and emits 60 W when operating. If the bulb filament has an emissivity of 0.8, what is the operating temperature of the filament
Answer:
2577 K
Explanation:
Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.
So, T = ⁴√(P/σεA)
Since P = 60 W, we substitute the vales of the variables into T. So,
T = ⁴√(P/σεA)
= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)
= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)
= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)
= ⁴√(0.00441 × 10¹⁶K⁴)
= 0.2577 × 10⁴ K
= 2577 K
which of the following is used to answer scientific questions?
A. Experiments
B. Intuition
C. Opinion polls
D. Imagination
The impulse given to a body of mass 1.5 kg, is 6.0 kg
• m•s-?
If the body was initially at rest, what
will its resulting kinetic energy be? Give your answer in J without units.
Show work
Answer:
12J
Explanation:
Given parameters:
Mass = 1.5kg
Impulse = 6kgm/s
let us start first by find the velocity with which this body moves;
Impulse = mass x velocity
Velocity = Impulse / mass = 6/ 1.5 = 4m/s
Initial velocity = 0m/s
Unknown:
Resulting kinetic energy = ?
Solution:
To solve this problem use the formula below:
K.E = [tex]\frac{1}{2}[/tex] m (v - u)²
m is the mass
v is the final velocity
u is the initial velocity
So;
K.E = [tex]\frac{1}{2}[/tex] x 1.5 x (4 - 0)²
K.E = 1.5 x 8 = 12J
A 2028 kg Oldsmobile traveling south on Abbott Road at 14.5 m/s is unable to stop on the ice covered intersection for a red light at Saginaw Street. The car collides with a 4146 kg truck hauling animal feed east on Saginaw at 9.7 m/s. The two vehicles remain locked together after the impact. Calculate the velocity of the wreckage immediately after the impact. Give the speed for your first answer and the compass heading for your second answer. (remember, the CAPA abbreviation for degrees is deg) -1.75
Answer:
v = 8.1 m/s
θ = -36.4º (36.4º South of East).
Explanation:
Assuming no external forces acting during the collision (due to the infinitesimal collision time) total momentum must be conserved.Since momentum is a vector, if we project it along two axes perpendicular each other, like the N-S axis (y-axis, positive aiming to the north) and W-E axis (x-axis, positive aiming to the east), momentum must be conserved for these components also.Since the collision is inelastic, we can write these two equations for the momentum conservation, for the x- and the y-axes:We can go with the x-axis first:[tex]p_{ox} = p_{fx} (1)[/tex]
⇒ [tex]m_{tr} * v_{tr}= (m_{olds} + m_{tr}) * v_{fx} (2)[/tex]
Replacing by the givens, we can find vfx as follows:[tex]v_{fx} = \frac{m_{tr}*v_{tr} }{(m_{tr} + m_{olds)} } = \frac{4146kg*9.7m/s}{2028kg+4146 kg} = 6.5 m/s (3)[/tex]
We can repeat the process for the y-axis:[tex]p_{oy} = p_{fy} (4)[/tex]
⇒[tex]m_{olds} * v_{olds}= (m_{olds} + m_{tr}) * v_{fy} (5)[/tex]
Replacing by the givens, we can find vfy as follows:[tex]v_{fy} = \frac{m_{olds}*v_{olds} }{(m_{tr} + m_{olds)} } = \frac{2028kg*(-14.5)m/s}{2028kg+4146 kg} = -4.8 m/s (6)[/tex]
The magnitude of the velocity vector of the wreckage immediately after the impact, can be found applying the Pythagorean Theorem to vfx and vfy, as follows:[tex]v_{f} = \sqrt{v_{fx} ^{2} +v_{fy} ^{2} }} = \sqrt{(6.5m/s)^{2} +(-4.8m/s)^{2}} = 8.1 m/s (7)[/tex]
In order to get the compass heading, we can apply the definition of tangent, as follows:[tex]\frac{v_{fy} }{v_{fx} } = tg \theta (8)[/tex]
⇒ tg θ = vfy/vfx = (-4.8m/s) / (6.5m/s) = -0.738 (9)
⇒ θ = tg⁻¹ (-0.738) = -36.4º
Since it's negative, it's counted clockwise from the positive x-axis, so this means that it's 36.4º South of East.The radius of the Sun is 6.96 x 108 m and the distance between the Sun and the Earth is roughtly 1.50 x 1011 m. You may assume that the Sun is a perfect sphere and that the irradiance arriving on the Earth is the value for AMO, 1,350 W/m2. Calculate the temperature at the surface of the Sun.
Answer:
5766.7 K
Explanation:
We are given that
Radius of Sun , R=[tex]6.96\times 10^{8} m[/tex]
Distance between the Sun and the Earth, D=[tex]1.50\times 10^{11}m[/tex]
Irradiance arriving on the Earth is the value for AMO=[tex]1350W/m^2[/tex]
We have to find the temperature at the surface of the Sun.
We know that
Temperature ,T=[tex](\frac{K_{sc}D^2}{\sigma R^2})^{\frac{1}{4}}[/tex]
Where [tex]K_{sc}=1350 W/m^2[/tex]
[tex]\sigma=5.67\times 10^{-8}watt/m^2k^4[/tex]
Using the formula
[tex]T=(\frac{1350\times (1.5\times 10^{11})^2}{5.67\times 10^{-8}\times (6.96\times 10^{8})^2})^{\frac{1}{4}}[/tex]
T=5766.7 K
Hence, the temperature at the surface of the sun=5766.7 K
A uniform electric field is present in the region between infinite parallel plane plates A and B and a uniform electric field is present in the region between infinite parallel plane plates B and C. When the plates are vertical, is directed to the right and to the left. The signs of the charges on plates A, B and C may be
Answer:
the signs on the plates are A + B- C +
Explanation:
The electric field for an infinite plate is perpendicular to it, this field is outgoing if the charge on the plate is positive and incoming if the charge on the plate is negative.
Let's analyze the situation presented, we have two infinite plates A and B with the elective field directed to the right, therefore the charge of the plate must be
plate A positive charge
plate B Negative charge
We also have a third plate C and they indicate that the field between B and C is directed to the left, therefore
plate B negative charged
plate C positive charged
in short the signs on the plates are
A + B- C +
Surface currents are on the
of the Earth's oceans
Water flows through a first pipe of diameter 3 inches. If it is desired to use another pipe for the same flow rate such that the velocity head in the second pipe is twice the velocity head in the first pipe, determine the diameter of the second pipe.
Answer:
the diameter of the second pipe is 2.52 in
Explanation:
Given the data in the question;
We know that; the rate of flow is the same;
so
Av1 = Av2
v ∝ √h
[tex]\frac{A1}{A2}[/tex] = [tex]\frac{V2}{V1}[/tex]
[tex]\frac{A1}{A2}[/tex] = √( [tex]\frac{h2}{h1}[/tex] )
( π/4.D1² / π/4.D2² ) = √( [tex]\frac{h2}{h1}[/tex] )
( D1² / D2² ) = √( [tex]\frac{2h1}{h1}[/tex] ) since second is double of first
so
( D1² / D2² ) = √( [tex]\frac{2}{1}[/tex] )
3² / D2² = √2
D2²√2 = 9
D2² = 9/√2
D2² = 9 / 1.4142
D2² = 6.364
D2 = √ 6.364
D2 = 2.52 in
Therefore, the diameter of the second pipe is 2.52 in