Answer:
10 miles per minute.
How many elements in oxygen gas? PLEASE ANSWER!
Answer:
8
Explanation:
Answer:
Chemical Properties of Oxygen
At standard temperature and pressure (STP), two atoms of the element bind to form dioxygen, a colorless, odorless, tasteless diatomic gas with the formula O2. Oxygen is a member of the chalcogen group on the periodic table and is a highly reactive nonmetallic element.
Explanation:
Hitting a ball off a tee would be considered which of the following?
A.
A closed serial skill
B.
An open continuous skill
C.
An open discrete skill
D.
A closed discrete skill
The overall length of a piccolo is 30.0 cm. The resonating air column is open at both ends. (a) Find the frequency of the lowest note a piccolo can sound. (Assume that the speed of sound in air is 343 m/s.) Hz (b) Opening holes in the side of a piccolo effectively shortens the length of the resonant column. Assume the highest note a piccolo can sound is 3 000 Hz. Find the distance between adjacent antinodes for this mode of vibration.
Answer:
(a) the frequency of the lowest note the piccolo can sound is 571.7 Hz
(b) the distance between adjacent antinodes is 5.72 cm
Explanation:
(a)
Given;
length of piccolo, L = 30 cm = 0.3 m
the speed of sound in air is 343 m/s
The wavelength of a pipe open at both ends, for the first harmonic is given;
L = A → N + N → A
L = λ / 4 + λ / 4
L = λ / 2
λ = 2L
λ = 2 x 0.3 = 0.6 m
The fundamental frequency (lowest frequency) is given by;
f₀ = v / λ
f₀ = (343 / 0.6)
f₀ = 571.7 Hz
(b)
Given;
highest note, f = 3000 Hz
the distance between adjacent antinodes is given by;
[tex]d = \frac{v}{2f}\\\\ d = \frac{343}{2*3000}\\\\ d = 0.0572 \ m\\\\d = 5.72 \ cm[/tex]
Fossil fuels like gasoline or coal are examples of
Answer:
Nonrenewable Resources
Answer:
fossil fuels
Explanation:
Physics question URGENT PLEASE!!!
[tex]\left(95\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{231}{1.00}\dfrac{\mathrm{in}^3}{\rm gal}\right)\left(\dfrac{1.0}{12}\dfrac{\rm ft}{\rm in}\right)^3\left(\dfrac{1.0}{60}\dfrac{\rm min}{\rm s}\right)\approx0.211661\dfrac{\mathrm{ft}^3}{\rm s}[/tex]
Multiplying 95 gal by 231 converts it to in^3. Dividing by 12^3 converts in^3 to ft^3. Dividing by 60 converts min to s.
A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.394 m down the incline in 2.00 s. The 0.200 kg block is replaced by a 0.400 kg block. If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s
Answer:
The distance traveled by the second block is 0.197 m
Explanation:
Given;
mass of the small block, m₁ = 0.2 kg
distance traveled by the block, d₁ =0.394 m
time of travel, t = 2 s
mass of the second block, m₂ = 0.4 kg
distance traveled by the second block, d₂ = ?
The work done per unit time on the inclined plane is given by;
[tex]\frac{f_1d_1}{t} = \frac{f_1d_1}{t}\\\\f_1d_1 = f_2d_2\\\\m_1gd_1 = m_2gd_2\\\\m_1d_1 = m_2d_2\\\\d_2 = \frac{m_1d_1}{m_2} \\\\d_2 = \frac{0.2*0.394}{0.4}\\\\d_2 = 0.197 \ m[/tex]
Therefore, the distance traveled by the second block is 0.197 m
When a gas undergoes an isothermal process, there is
(A) no heat added to the gas.
(B) no work done by (or on) the gas.
(C) no change in the temperature of the gas.
(D) no change in the volume of the gas.
(E) no change in the pressure of the gas.
Answer:
(C) no change in the temperature of the gas.
Explanation:
isothermal process of a gas, can result to compression of the gas or expansion of the gas at a constant temperature (ΔT = 0).
For isothermal gas expansion, work is done in reducing the pressure of the gas by increasing its volume at a constant temperature. The change in the internal energy of this process is zero (ΔU = 0).
For isothermal gas compression, work is done in decreasing the volume of the gas by increasing its pressure at a constant temperature. The change in the internal energy of this process is also zero (ΔU = 0).
Therefore, when a gas undergoes an isothermal process, there is no change in the temperature of the gas.
A car travels 20 meters east
in 1.0 second and then travels
10 meters west in 20 seconds. The
displacement of the car at the end
of this 3.0-second interval is-
Answer:
Alright the other day.
g uppose that 3 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 45 cm. (a) How much work is needed to stretch the spring from 35 cm to 37 cm? (Round your answer to two decimal places.) .02 Incorrect: Your answer is incorrect. J (b) How far beyond its natural length will a force of 10 N keep the spring stretched? (Round your answer one decimal place.)
Answer:
W= 0.319992 J and distance is 3.75 cm
Explanation:
The energy needed to stretch the spring from 30 cm to 45 cm = 3 J
Now we required to find the requirement of energy when the spring is stretched from 35 cm to 37 cm.
So first find the work done to stretch the spring from 35 cm to 45 cm.
Work done, w = (1/2) kx^2
3 = (1/2)k(0.45 – 0.30)^2
k = 266.66 N/m
now, x1 = 0.35 – 0.30 = 0.05m
x2 = 0.37 – 0.30 = 0.07m
Now the amount of work done to stretch from 35cm to 37.
w = (1/2) k (x2^2 – x1^2)
w = (1/2) (266.66) (0.07^2 – 0.05^2 )
w= 0.319992 J
(b). Given F = 10 N
F = kx
x = F / k
x = 10 / 266.66
x = 0.0375m
x = 3.75 cm
Thus, distance is 3.75cm
What does a diagonal upward sloping line on a position time graph mean?
diagonal line means the velocity is constant.
A cyclist accelerates from rest to 8 m/s in 3 seconds. How far did the cycles travel in 3 seconds?
Answer:
[tex] \boxed{\sf Distance \ travelled = 12 \ m} [/tex]
Given:
Initial speed (u) = 0 m/s (Accelerates from rest)
Final speed (v) = 8 m/s
Time taken (t) = 3 seconds
To Find:
Distance travelled by cyclist (s)
Explanation:
From equation of motion of object moving with uniform acceleration in straight line we have:
[tex] \boxed{ \bold{s = (\frac{v + u}{2} )t}}[/tex]
By substituting value of v, u & t in the equation we get:
[tex] \sf \implies s = ( \frac{8 + 0}{2} ) \times 3 \\ \\ \sf \implies s = \frac{8}{2} \times 3 \\ \\ \sf \implies s = 4 \times 3 \\ \\ \sf \implies s = 12 \: m[/tex]
[tex] \therefore[/tex]
Distance travelled by cyclist (s) = 12 m
Answer:
s(distance) =36m
Explanation:
u(initial velocity) =0 m/s
a =8 m/s^2
t=3s
s=ut+1/2at^2
s=1/2(8)(3)^2
s=1/2(8)(3)(3)
s=4(9)
s=36m
For any object in projectile motion, the vertical velocity is independent of gravity?
Answer:
the y axis, every moment is subjected to a vertical acceleration directed towards the center of the Earth, which is called the acceleration of gravity
Explanation:
In projectile launching, the movement is separated into two movements, one on the x-axis and the other on the y-axis, related through time.
In horizontal movement, the speed is constant, because there is no acceleration in this axis, the effect of air friction is almost always eliminated.
In the other movement on the y axis, every moment is subjected to a vertical acceleration directed towards the center of the Earth, which is called the acceleration of gravity.
This value of the eceleration of gravity is constant for small distances
compared to the radius of the Earth, for higher altitudes an expansion in beings of the distance is used, giving a linear dependence.
how do accelerating objects act
Explanation:
They speed up, slow down, or change direction.
Answer:In order for object to be accelerating, the net force (sum of all forces) acting on the object has to be non-zero. According to Newton's Second Law of Motion, Force = Mass x Acceleration. ... If the object is sitting motionless on the ground, then the net force (sum of all forces) acting on it is zero.
Explanation:
An unstrained horizontal spring has a length of 0.34 m and a spring constant of 180 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.024 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges. g
Answer:
a
[tex]q_1 = -1.389 *10^{-5} \ C [/tex] , [tex]q_2 = -1.389 *10^{-5} \ C [/tex]
OR
[tex]q_1 = 1.389 *10^{-5} \ C [/tex] , [tex]q_2 = 1.389 *10^{-5} \ C [/tex]
b
[tex]q_1 = 1.389 *10^{-5} \ C [/tex] and [tex]q_2 = 1.389 *10^{-5} \ C [/tex]
Explanation:
Generally the force exerted on the string is mathematically represented as
[tex]F = k * e[/tex]
substituting values 180 N/m for k and 0.024 m for e
[tex]F = 180 * 0.024[/tex]
[tex]F = 4.32 \ N[/tex]
This force can also equivalent to the electrostatic force between the charges i.e
[tex]F = k * \frac{q^2}{ r^2}[/tex]
substituting [tex]9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex] for k and ( 0.34 + 0.024 = 0.364 m) for r we have
[tex] 4.32= 9*10^{9} * \frac{q^2}{ (0.364)^2}[/tex]
[tex]q = \sqrt{1.929 *10^{-10}}[/tex]
[tex]q = 1.389 *10^{-5} \ C [/tex]
Given the spring was stretched it means that the force between the charges is a repulsive for which tell us that both charge are of the same sign thus the possible algebraic signs of the charges are
[tex]q_1 = -1.389 *10^{-5} \ C [/tex] , [tex]q_2 = -1.389 *10^{-5} \ C [/tex]
OR
[tex]q_1 = 1.389 *10^{-5} \ C [/tex] , [tex]q_2 = 1.389 *10^{-5} \ C [/tex]
You are jogging with a velocity of 2 meters/second to the west, when you feel thirsty and stops to drink water after 5 seconds. First find your velocity the moment you stopped running. Then, find your acceleration during that motion. What are the units for acceleration? What is the direction of acceleration in this motion? Show your work.
Answer:
The velocity when the runner stopped is v =0 m/s
The acceleration is [tex]a = 0.4 \ m/s [/tex]
Explanation:
From the question we are told that
The initial velocity is [tex]u = 2 \ m/s[/tex]
The time taken is [tex]t = 5 \ s[/tex]
Generally from the question we are told that the person stopped running so the person velocity at that moment is v =0 m/s
Now the acceleration is mathematically represented as
[tex]a = \frac{v - u}{t}[/tex]
=> [tex]a = \frac{0 - 2}{5}[/tex]
=> [tex]a = 0.4 \ m/s [/tex]
A woman walks 150 min 45 seconds. What was her average speed?
Answer:
0.3
Explanation: when you average out the number sthat is the answer you get
4. What is the velocity of an object that doesn't move?
It depends on the object b. it depends on the speed c. it depends on the height
O mis
help
Answer:
Acceleration /Speed
Explanation:
An objects Velocity can be determined by acceleration,
Please pay attention in your middle school class, speed and velocity quiz.
What was the current annual rate of inflation in 2009?
pls help !
Answer:
The 2009 inflation rate was -0.36%. The current year-over-year inflation rate (2019 to 2020) is now 0.99%
Explanation:
If a 500kg elephant is sliding across a frictionless patch of ice, how much force is needed to keep the elephant from slowing down?
Answer:
4905NExplanation:
The force needed to keep the elephant from slowing down is expressed as shown according to Newtons second law of motion.
Force = mass * acceleration due to gravity
Given
Mass of elephant = 500kg
acceleration due to gravity = 9.81m/s²
Force = 500*9.81
Force = 4905N
Hence the force needed to keep the elephant from slowing down is 4905N
lol
idk what this is
Answer:
I know
Explanation:
Physics= hard
Help me please! You will get the brainliest!
Answer:
0.5 m.
Explanation:
From the question given above, the following data were obtained:
Force (F) applied = 5 N
Work done (W) = 2.5 J
Distance (s) travalled by the book =?
We can obtain the distance travelled by the book by applying the formula of work done. This is illustrated below:
Work done (W) = Force (F) × Distance (s)
W = F × s
Force (F) applied = 5 N
Work done (W) = 2.5 J
Distance (s) travalled by the book =?
2.5 = 5 × s
Divide both side by 5
s = 2.5/5
s = 0.5 m
From the above calculation, the book will travel 0.5 m when a force of 5 N is applied.
An excited squirrel starts from rest and accelerates towards a nut at 1.0 m/s2 for 2.0 seconds. What is the squirrel’s displacement?
Answer:
[tex]x=8m[/tex]
Explanation:
Hello,
In this case, the most suitable kinematic equation for the calculation of the squirrel's displacement is:
[tex]x=v_ot+\frac{1}{2}at^2[/tex]
Whereas the initial velocity is zero as it starts from rest, thus, we obtain:
[tex]x=\frac{1}{2}*1.0\frac{m}{s^2}*(2.0s)^2\\ \\x=8m[/tex]
Best regards.
What is an example of intellectual development?
Answer:
learning new activities like riding a bike and playing sports. and also having the ability to collect information and remember them (memory). <3
Which value represents accuracy: the mean, median, mode, range from center, or distance between farthest beanbags?
Answer:
the mean
Explanation:
The mean (informally, the “average“) is found by adding all of the numbers together and dividing by the number of items in the set: 10 + 10 + 20 + 40 + 70 / 5 = 30. The median is found by ordering the set from lowest to highest and finding the exact middle. The median is just the middle number: 20.
Using the equation for for Newton's Second law, F=ma, solve the following problem. You have been given an object with a mass of 6g and an acceleration of 2 m/s2, what is the force?
Group of answer choices
A. 12N
B. 3N
C. 8N
D. 120N
Answer:
F = 0.012 N
Explanation:
Given that,
Mass of the object, m = 6 g
Acceleration, a = 2 m/s²
1 kg = 1000 grams
6 g = 0.006 kg
Force, F = ma
So,
[tex]F=0.006\ kg\times 2\ m/s^2\\F=0.012\ N[/tex]
So, the force is 0.012 N.
Where do you feel that you are traveling at the fastest speed when on the swing?
Answer:
C
Explanation:
I think it's C, because at that point, you are going fastest. Sorry if im wrong, hope this helps.
Answer:
In between and the middle one
Explanation:
Earth's atmosphere is constantly bombarded by cosmic ray protons that originate somewhere in space. If the protons all passed through the atmosphere, each square meter of Earth's surface would intercept protons at the average rate of 1800 protons per second. What would be the electric current in amperes intercepted by a 41 × 107 km2 area on the planet
Answer:
Electric current in amperes = 1.1808 A
Explanation:
Given:
Intercept protons rate = 1800 protons per second
Area = 41 × 10⁷ km²
Find:
Electric current in amperes
Computation:
Current density = Intercept protons rate × 1.6 × 10⁻¹⁹
Current density = 1800 × 1.6 × 10⁻¹⁹
Current density = 2.88 × 10⁻¹⁶
1 km² = 10⁶m²
So,
Electric current in amperes = 2.88 × 10⁻¹⁶ × 41 × 10⁷ × 10⁶
Electric current in amperes = 1.1808 A
5. Why does a properly adjusted head restraint help prevent head and neck injuries to occupants in
rear-end collisions? Explain your answer in terms of the law of conservation of momentum.
Answer:
Have you tried google
Explanation:
Answer:
protect against whisplasg injuries
A highway curve with radius R = 274 m is to be banked so that a car traveling v = 25.0 m/s will not skid sideways even in the absence of friction. At what angle should the curve be banked? 2. what is the correct equation for the sum of the forces in the x direction? What is the correct equation for the sum of the forces in the y direction? A. ΣFx:−Nsinθ=mv2/R; ΣFy:Ncosθ−mg=0 B. ΣFx:−Ncosθ=−mv2/R; ΣFy:Nsinθ−mg=0 C. ΣFx:Nsinθ=mv2/R; ΣFy:Ncosθ+mg=0 D. ΣFx:−Nsinθ=mv2/R; ΣFy:Ncosθ=0 E. ΣFx:−Nsinθ=−mv2/R; ΣFy:Ncosθ−mg=0
Answer:
A) θ = 13.1º , B) E
Explanation:
A) For this exercise, let's use Newton's second law, let's set a reference frame where the axis ax is in the radial direction and is horizontal, the axis y is vertical.
In this reference system the only force that we must decompose is the Normal one, let's use trigonometry
sin θ = Nₓ / N
cos θ = [tex]N_{y}[/tex] / N
Nₓ = N sin θ
Ny = N cos θ
x-axis (radial)
Nₓ = m a
where the acceleration is centripetal
a = v² / R
we substitute
-N sin θ = -m v² / R (1)
the negative sign indicates that the force and acceleration towards the center of the circle
y-axis (Vertical)
Ny - W = 0
N cos θ = mg
N = mg / cos θ
we substitute in 1
mg / cos θ sin θ = m v² / R
g tan θ = v² / R
θ = tan⁻¹ (v² / gR)
we calculate
θ = tan⁻¹ (25² / 9.8 274)
θ = 13.1º
B) when comparing the equations the correct one is E
An oil pump is drawing 44kW while pumping oil with a density of 860 kg/m3 at a rate of 0.1 m3/s. The inlet and outlet diameters of the pipe are 8cm and 12cm, respectively. If the pressure increases by 500kPa going through the pump and the motoreffi ciency is 90%, determine the mechanical efficiency of the pump
Answer:
The mechanical efficiency of the pump is 91.8 %
Explanation:
Given;
input power, p = 44 kw
density of oil, ρ = 860 kg/m³
motor efficiency, η = 90 %
inlet diameter, d₁ = 8 cm
outlet diameter, d₂ = 12 cm
volume flow rate, V = 0.1 m³/s
pressure rise, P = 500kPa
output power = motor efficiency x input power
output power = 0.9 x 44 = 39.6 kW
Thus, the mechanical input power = 39.6 kW
The mechanical output power is given by change in mechanical energy;
[tex]E = mgh + \frac{m}{2} (v_2^2 - v_1^2) \\\\E = \rho V g h + \frac{\rho V}{2} [(\frac{V_2}{\pi r_2^2} )^2 - (\frac{V_1}{\pi r_1^2})^2]\\\\E = PV + \frac{\rho V^3}{2\pi^2} [\frac{1}{ r_2^4} - \frac{1}{ r_1^4}]\\\\E = (500 *10^3)(0.1) + \frac{(860)(0.1)^3}{2\pi^2} [\frac{1}{ 0.06^4} - \frac{1}{ 0.04^4}]\\\\E = 50000 -13653.51\\\\E = 36346.48 \ W\\\\E = 36.347 \ kW[/tex]
The mechanical efficiency is given by
η = mechanical output power / mechanical input power
η = 36.347 / 39.6
η = 0.918
η = 91.8 %
Therefore, the mechanical efficiency of the pump is 91.8 %