Answer:
b. 0.45 meters
Explanation:
Given the following data;
Spring constant, k = 330 N/m
Force = 150 N
To find the extension of the spring;
Mathematically, the force exerted on a spring is given by the formula;
Force = spring constant * extension
Substituting into the formula, we have;
150 = 330 * extension
Extension, e = 150/330
Extension, e = 0.45 meters
The length of the slope of a mountain is 2780 m, and it makes
its base?
angle of 14.1° with the horizontal. What is the height of the mountain, relative to
Additional Materials
Reading
Answer:
677 m
Explanation:
Using the definition of the sine of an angle, we can write
sin 14.1 = (height of mountain) / (slope length of mountain)
sin 14.1 = H / (2780 m) ---> H = (2780 m) x sin 14.1
= 677 m
The height of the mountain is 677.21 m
The given parameters;
length of the slope, L = 2780 m
angle of inclination, Ф = 14.1°
let the height of the mountain, = h
A simple sketch of the problem is given below;
↓P
↓
↓ h
↓ 14.1°
↓------------------------------------------------Q
A straight line joining PQ is the hypotenuse of the right triangle.
The height of the right triangle is calculated as follows;
[tex]sin(14.1) = \frac{h}{PQ} \\\\\h = PQ \times sin(14.1)\\\\h = 2780 \times sin(14.1)\\\\h = 677.21 \ m[/tex]
Thus, the height of the mountain is 677.21 m
Learn more here: https://brainly.com/question/4326804
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standard power cycle for which the working fluid is evaporated, passed through a turbine, and subsequently condensed. The system is to be used in very special locations for which the oceanic water temperature near the surface is approximately 300 K, while the temperature at reasonable depths is approximately 280 K. The warmer water is
Answer:
Explanation:
Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.
To objective is to find the:
(i) required heat exchanger area.
(ii) flow rate to be maintained in the evaporator.
Given that:
water temperature = 300 K
At a reasonable depth, the water is cold and its temperature = 280 K
The power output W = 2 MW
Efficiency [tex]\zeta[/tex] = 3%
where;
[tex]\zeta = \dfrac{W_{out}}{Q_{supplied }}[/tex]
[tex]Q_{supplied } = \dfrac{2}{0.03} \ MW[/tex]
[tex]Q_{supplied } = 66.66 \ MW[/tex]
However, from the evaporator, the heat transfer Q can be determined by using the formula:
Q = UA(L MTD)
where;
[tex]LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}[/tex]
Also;
[tex]\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K[/tex]
[tex]\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K[/tex]
[tex]LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}[/tex]
[tex]LMTD = \dfrac{8}{In (5)}[/tex]
LMTD = 4.97
Thus, the required heat exchanger area A is calculated by using the formula:
[tex]Q_H = UA (LMTD)[/tex]
where;
U = overall heat coefficient given as 1200 W/m².K
[tex]66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\ A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\ \mathbf{A = 11178.236 \ m^2}[/tex]
The mass flow rate:
[tex]Q_{H} = mC_p(T_{in} -T_{out} ) \\ \\ 66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{ 66.667 \times 10^6}{4.18 \times 8} \\ \\ \mathbf{m = 1993630.383 \ kg/s}[/tex]
The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically accessible quantum state of a system has equal probability of being populated, which in turn leads to the Boltzmann distribution for a system in thermal equilibrium.
a. True
b. False
Answer:
Hello! Your answer would be, A) True
Explanation:
Hope I helped! Ask me anything if you have any questions. Brainiest plz!♥ Hope you make a 100%. Have a nice morning! -Amelia♥
Se transmiten ondas transversales en una cuerda tensada orientada sobre el eje x. La función de ondas correspondientes es y= 5.00 sen(6.05x + 5.19t + 1.57). Donde y y x están en metros y t en segundos. Cual es la rapidez de las ondas que se transmiten en dicha cuerda?
Answer:
Explanation:
Whats the entire question?
A 10kg block is Pulled along a horizontal
Surface by a force
of 50N at an angles
of 37° with the horizontal If the
coefficient of sliding friction b/n the
block and the surface is o.2
(g=10m/s^2 Sin 37=O.6 and cos 37 = 0.8)
A, what frictional forces acting on the block?
B,what is the acceleration of the block?
Answer:
hope u can understand the method
A solenoid that is 66.2 cm long has a cross-sectional area of 18.0 cm2. There are 1300 turns of wire carrying a current of 8.15 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy in joules stored in the magnetic field there (neglect end effects).
Answer:
(a) Energy Density = 160.94 J/m³
(b) Energy Stored = 0.192 J
Explanation:
(a)
The energy density of the magnetic field inside the solenoid is given by the following formula:
[tex]Energy\ Denisty = \frac{B^2}{2\mu_o}\\[/tex]
where,
B = magnetic field strength of solenoid = [tex]\frac{\mu_oNI}{l}[/tex]
Therefore,
[tex]Energy\ Density = \frac{\mu_oN^2I^2}{2l^2}[/tex]
where,
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
N = No. of turns = 1300
I = current = 8.15 A
L = length = 66.2 cm = 0.662 m
Therefore,
[tex]Energy\ Density = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(1300)^2(8.15\ A)^2}{2(0.662\ m)^2}[/tex]
Energy Density = 160.94 J/m³
(b)
Energy Stored = (Energy Density)(Volume)
Energy Stored = (Energy Density)(Area)(L)
Energy Stored = (160.94 J/m³)(0.0018 m²)(0.662 m)
Energy Stored = 0.192 J
Which of the following is a category of mechanical wave?
O A. Transverse
B. Frictional
C. Parallel
D. Perpendicular
Answer:
a
because the mechanical wave is when it goes over and over again
Answer:
The answer is a like i said 3hrs ago i dont know if this guy copied me tbh
Explanation:
True or false. When a girl walks the action of pushing and the equal amd opposite reaction is being projected forward
This is true I think
It applies to Newton's Laws
it's true because it's a part of newtons law
what is the maximum distance we can shoot a dart,from ground level provided our toy dart gun gives a maximum initial velocity of 2.7m/s and air resistance is negligible
Answer:
R = v^2 sin 2 theta / g
The range provides the distance a projectile can travel
R(max) = v^2 / g if theta = 45 deg
R = 2.7^2 / 9.8 = .74 m
On a distance-time graph, what is shown when the curve is flat going from left to the right?
A. a negative speed
B. no speed
C. a positive speed
D. It does not mean anything.
Please help me !!im on a test
A soccer ball is released from rest at the top of a grassy incline. After 6.2 seconds, the ball travels 47 meters. One second later, the ball reaches the bottom of the incline.
(a) What was the balls acceleration?(assume that the acceleration was constant).
(b) How long was the incline?
Answer:
(a) a = 2.44 m/s²
(b) s = 63.24 m
Explanation:
(a)
We will use the second equation of motion here:
[tex]s = v_it+\frac{1}{2}at^2[/tex]
where,
s = distance covered = 47 m
vi = initial speed = 0 m/s
t = time taken = 6.2 s
a = acceleration = ?
Therefore,
[tex]47\ m = (0\ m/s)(6.2\ s)+\frac{1}{2}a(6.2\ s)^2\\\\a = \frac{2(47\ m)}{(6.2\ s)^2}[/tex]
a = 2.44 m/s²
(b)
Now, we will again use the second equation of motion for the complete length of the inclined plane:
[tex]s = v_it+\frac{1}{2}at^2[/tex]
where,
s = distance covered = ?
vi = initial speed = 0 m/s
t = time taken = 7.2 s
a = acceleration = 2.44 m/s²
Therefore,
[tex]s = (0\ m/s)(6.2\ s)+\frac{1}{2}(2.44\ m/s^2)(7.2\ s)^2\\\\[/tex]
s = 63.24 m
In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it elastically strikes another block that is only one-half its mass. Find the velocity of block M at the bottom of the incline before the collision with the small block m.
Solution :
Given :
M = 0.35 kg
[tex]$m=\frac{M}{2}=0.175 \ kg$[/tex]
Total mechanical energy = constant
or [tex]$K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$[/tex]
But [tex]$K.E._{top} = 0$[/tex] and [tex]$P.E._{bottom} = 0$[/tex]
Therefore, potential energy at the top = kinetic energy at the bottom
[tex]$\Rightarrow mgh = \frac{1}{2}mv^2$[/tex]
[tex]$\Rightarrow v = \sqrt{2gh}$[/tex]
[tex]$=\sqrt{2 \times 9.8 \times 0.35}$[/tex] (h = 35 cm = 0.35 m)
= 2.62 m/s
It is the velocity of M just before collision of 'm' at the bottom.
We know that in elastic collision velocity after collision is given by :
[tex]$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$[/tex]
here, [tex]$m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$[/tex]
∴ [tex]$v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}[/tex]
[tex]$=\frac{0.175}{0.525}+0$[/tex]
= 0.33 m/s
Therefore, velocity after the collision of mass M = 0.33 m/s
The density of table sugar is 1.59g/cm3 what is the volume of 7.85g of sugar?
Answer: 4.94cm³
Explanation:
Data;
ρ = 1.59g/cm³
mass = 7.85g
volume = ?
density = mass / volume
ρ = m / v
v = m / ρ
v = 7.85 / 1.59
v = 4.94cm³
Electroconvulsive therapy would be done under the
supervision of a counseling psychologist, where high level
of electric shock would be admistered.
Select one:
True
False
Answer:
the answer of this question is true
An 800 kg charging bull rams through a wooden fence. It was travelling at
5 m/s, now it's travelling at 3 m/s. How much impulse did the bull
experience by smashing the fence?
Answer:
J = 1600 kg-m/s
Explanation:
Given that,
The mass of charging bull rams, m = 800 kg
Initial speed, u = 5 m/s
Final speed, v = 3 m/s
We need to find the impulse the bull experience by smashing the fence. Let it is J. We know that, impulse is equal to the change in momentum such that,
J = m(v-u)
Put all the values,
J = 800(3-5)
= 800(-2)
= -1600 kg-m/s
Hence, the magnitude of impulse is equal to 1600 kg-m/s.
Man-made climate change is an example
of...
An object of mass 45 kg is observed to accelerate at the rate of 6 m/s2. Calculate the force required to produce this acceleration
Calculate the potential difference across the 8 ohm resistor
Explanation:
if the current is 1A
V=iR
V= 1 × 8
V = 8volts
An astronaut is a distance L from her spaceship, and is at rest with respect to the ship, when she discovers that her tether has broken. She tosses a wrench with a speed Vw in the opposite direction of the ship to propel herself back to the ship. The astronaut has mass MA, and the wrench has mass Mw.
Required:
a. Draw a sketch, showing the subsequent motion of the astronaut and the wrench.
b. What is the initial momentum (before toss) of the astronaut plus wrench system? What is the final momentum?
c. Use conservation of momentum to solve for the speed of the astronaut VA, relative to the ship, in terms of MA, Mw and Vw.
d. How long does it take her to reach the ship in terms of L, MA, Mw and Vw?
e. How far has the wrench traveled from its original position when the astronaut reaches the ship? Express your answer in terms of L, MA and Mw.
Answer:
B) I₀ = I_f= 0, C) vₐ = [tex]\frac{m_w}{m_a} \ v_w[/tex] , D) t = [tex]\frac{m_a}{m_w} \ \frac{L}{v_w}[/tex]
Explanation:
A) in the attachment you can see a diagram of the movement of the key and the astronaut that is in the opposite direction to each other.
B) Momentum equals the change in momentum in the system
I = ∫ F dt = Δp
since the astronaut has not thrown the key, the force is zero, so the initial impulse is zero
I₀ = 0
The final impulse of the two is still zero, since it is a vector quantity, subtracting the impulse of the two gives zero, since it is an isolated system
I_f = 0
C) We define the system formed by the astronaut and the key, for which the forces during the separation are internal and the moment is conserved
initial instant.
p₀ = 0
final instant
p_f = [tex]m_a v_a - m_w v_w[/tex]
We used the subscript “a” for the astronaut and the subscript “w” for the key
the moment is preserved
po = p_f
0 = mₐ vₐ - m_w v_w
vₐ = [tex]\frac{m_w}{m_a} \ v_w[/tex]
D) as the astronaut goes at constant speed we can use the uniform motion relationships
vₐ = x / t
t = x / vₐ
t = [tex]\frac{m_a}{m_w} \ \frac{L}{v_w}[/tex]
Light of intensity I0 passes through 4 ideal polarizing sheets. Unpolarized light enters the 1st sheet that has a horizontal transmission axis. Light continues to the 2nd sheet that has its transmission axis at 25 degrees with respect to the 1st sheet, then to the 3rd sheet that has its transmission axis at 47 degrees with respect to the 1st sheet then to the 4th sheet that has its transmission axis at 10 degrees with respect to the 3d sheet. The intensity of the emerging light as percentage of I0 is close to:
Answer:
34.24 %
Explanation:
Since I₀ is the intensity of the un-polarized light, the intensity I₁ of the light polarized by the 1st sheet is (by the one-half rule) I₁ = I₀/2.
The intensity of polarized light I from a polarized source I' is I = I'cos²Ф where Ф is the angle between the direction of I' and I. Since the second sheet has its transmission axis at 25° with respect °o the 1st sheet, the intensity of light I₂ from the second sheet is I₂ = I₁cos²25°.
Also, the 3rd sheet has its transmission axis 47° with respect to the 1st sheet. So, the angle between the transmission axis of the 2nd sheet and 3rd sheet is 47° - 25° = 22°. So, the intensity I₃ from the 3rd sheet is I₃ = I₂cos²22°
Finally, the 4th sheet has its transmission axis 10° with respect to the 3rd sheet. So, the intensity I₄ from the 4th sheet is I₄ = I₃cos²10°.
So, I₄ = I₃cos²10°
I₄ = I₂cos²22°cos²10°
I₄ = I₁cos²25°cos²22°cos²10°
I₄ = (I₀/2)cos²25°cos²22°cos²10°
I₄/I₀ = cos²25°cos²22°cos²10°/2
I₄/I₀ = (cos25°cos22°cos10°)²/2
I₄/I₀ = (0.9063 × 0.9272 × 0.9848)²/2
I₄/I₀ = 0.8275²/2
I₄/I₀ = 0.6848/2
I₄/I₀ = 0.3424
So, as a percentage,
I₄/I₀ × 100% = 0.3424 × 100% = 34.24 %
You are playing in a volley ball game Your team has 12 and the other team has 18.
How many points does your team needs to win?
How many points does the other team needs to win?
Answer:
you need 7 points and the other team just needs to stop you from scoring
Explanation:
A 4.00-kg block hangs by a light string that passes over a massless, frictionless pulley and is connected to a 6.00-kg block that rests on a frictionless shelf. The 6.00-kg block is pushed against a spring to which it is not attached. The spring has a spring constant of 180 N/m, and it is compressed by 30.0 cm. Find the speed of the block after the spring is released and the 4.00-kg block has fallen a distance of 40.0 cm.
Answer:
v = 2.82 m/s
Explanation:
For this exercise we can use the conservation of energy relations.
We place our reference system at the point where block 1 of m₁ = 4 kg
starting point. With the spring compressed
Em₀ = K_e + U₂ = ½ k x² + m₂ g y₂
final point. When block 1 has descended y = - 0.400 m
Em_f = K₂ + U₂ + U₁ = ½ m₂ v² + m₂ g y₂ + m₁ g y
as there is no friction, the energy is conserved
Em₀ = Em_f
½ k x² + m₂ g y₂ = ½ m₂ v² + m₂ g y₂ + m₁ g y
½ k x² - m₁ g y = ½ m₂ v²
v² = [tex]\frac{k}{m_2} x^2 - 2 \frac{m_1}{m_2} \ g y[/tex]
let's calculate
v² = [tex]\frac{180}{6.00} \ 0.300^2 - 2 \ \frac{4.00}{6.00} \ 9.8 \ (- 0.400)[/tex]
v² = 2.7 + 5.23
v = √7.927
v = 2,815 m / s
using of significant figures
v = 2.82 m/s
The period of a sound wave coming from an instrument is 2 seconds. What 1 point
is the frequency of the sound? (f = 1/T) *
5 Hz
50 Hz
ОО
0.5 Hz
Answer:
b
Explanation:
How much heat is required to raise the temperature of 50 grams of water from 30 °C to 90 °C? C of water 4186 J / kg C.
12558 J
12558000 J
125580 J
1255800 J
Answer:
12558 J
Explanation:
Please do mark as brainliest. Hope this helps! :)
Two band members are marching toward each other beginning from opposite end zone goal lines (100 yards apart). One band member is playing a picolo and marching at 5 ft/s and the other is playing a sousaphone and marching at 1 ft/s. At what yard line do they meet?
Answer:
17
Explanation:
answer is 16.6666 which is equal to 17
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.370 mm wide. The diffraction pattern is observed on a screen 3.70 m away. Define the width of a bright fringe as the distance between the minima on either side. You may want to review (Page) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Single-slit diffraction. Part A What is the width of the central bright fringe
Answer:
Δx = 6.33 x 10⁻³ m = 6.33 mm
Explanation:
We can use the Young's Double Slit Experiment Formula here:
[tex]\Delta x = \frac{\lambda L}{d}\\\\[/tex]
where,
Δx = distance between consecutive dark fringes = width of central bright fringe = ?
λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m
L = distance between screen and slit = 3.7 m
d = slit width = 0.37 mm = 3.7 x 10⁻⁴ m
Therefore,
[tex]\Delta x = \frac{(6.33\ x\ 10^{-7}\ m)(3.7\ m)}{3.7\ x \ 10^{-4}\ m}[/tex]
Δx = 6.33 x 10⁻³ m = 6.33 mm
(121 m/s, or 43
4. The pilot of an airplane, which has been diving at a speed of
540 km/h, pulls out of the dive at constant speed.
(a) What is the minimum radius of the plane's circular path in
order that the acceleration of the pilot at the lowest point
will not exceed 7 g?
(b) What force is applied on an 80 kg pilot by the plane seat at
the lowest point of the pull-out? (328 m, 6.3 X 10N)
Answer:
(a) r = 328 m
(b) F = 6271.8 N
Explanation:
speed, v = 540 km/h = 150 m/s
Acceleration, a = 7 g
(a) Let the radius is r.
[tex]a =\frac{v^{2}}{r}\\7\times 9.8 = \frac{150\times 150}{r}\\r = 328 m[/tex]
(b) The force on 80 kg pilot at the lowest point is
[tex]F =m \times g + m\frac{v^{2}}{r}\\F = 80\times9.8 +80\times \frac{150\times 150}{328}\\\\F =784 + 5487.8 = 6271.8 N[/tex]
Which of the following best defines climate? PLEASE HELP!!
When a particular hanging mass is suspended from the string, a standing wave with two segments is formed. When the weight is reduced by 2.2 kg, a standing wave with five segments is formed. What is the linear density of the string
Solution :
Mass is varied keeping frequency constant.
Wavelength, λ [tex]$=\frac{2l}{n}$[/tex]
where length of spring = l
number of segments = n
Velocity, v = λ x f
= [tex]$\sqrt{\frac{T}{\mu}}$[/tex]
[tex]$\mu $[/tex] = mass density, T = tension in string
[tex]$T=\frac{4 \mu l^2f^2}{n^2}$[/tex]
[tex]$T=mg = \frac{4 \mu l^2f^2}{n^2}$[/tex] , n = 2
[tex]$T = (m-2.2)g = \frac{4 \mu l^2f^2}{n^2}, n = 5$[/tex]
[tex]$\Rightarrow \frac{m}{m-2.2}=\frac{25}{4}$[/tex]
[tex]$\Rightarrow m = 2.619\ kg$[/tex]
Therefore, μ = 0.002785 kg/ m
Frequency is varied keeping T constant
[tex]$T=\frac{4 \mu l^2f^2}{n^2}, f=60 , \ \ n = 2$[/tex]
[tex]$T=\frac{4 \mu l^2f^2}{n^2}, f=? , \ \ n = 7$[/tex]
[tex]$\Rightarrow \frac{60^2}{4}=\frac{f^2}{49}$[/tex]
f = 210 Hz
Which is the best way to become familiar with your company's policies and procedures?
O
A. ask the person who hired you
O
B. look in the employee handbook
C. tell your supervisor you need help
D. visit the company's website