Answer:
The Earth is traveling around the sun with an orbital velocity of 30 kilometers per second. This is exactly the speed it needs to be going to counteract the force of gravity from the sun pulling it inward. If the sun were to suddenly disappear, Earth would travel in a perfectly straight line at 30 km/s
Explanation:
Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?
Answer:
Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?
Average velocity of Mike Phelps swimming 100 m race in the 50 m long pool (2 laps) is approximately equal to *
A. 0 m/s
B. 1 m/s
C. 2 m/s
D. 4 m/s
Answer:
2
Explanation:
Calculate the force a 75 kg high jumper must exert in order to produce an acceleration that is 3.2 times the acceleration due to gravity.
Answer:
Explanation
According to Newton's second law of motion,
F = ma
m is the mass
a is the acceleration
If the acceleration is 3.2 times the acceleration due to gravity, then a = 3.2g
The formula becomes;
F = m(3.2g)
F = 3.2mg
m= 75kg
g = 9.81m/s²
F = 3.2(75)(9.81)
F = 2,354.4N
Hence the force exerted by the jumper is 2,354.4N
Suppose that your favorite AM radio station broadcasts at a frequency of 1600 kHz. What is the wavelength in meters of the radiation from the station
Answer:
187.5
Explanation:
Which statement best describes an atom? (2 points)
оа
Protons and neutrons grouped in a specific pattern
Ob
Protons and electrons spread around randomly
ос
A group of protons and neutrons that are surrounded by electrons
Od
A ball of electrons and neutrons surrounded by protons
Answer:
A group of protons and neutrons that are surrounded by electrons I think that's the answer...
Explanation:
) A 1000-gallon tank currently contains 100.0 gallons of liquid toluene and a gas saturated with toluene vapor at 85°F and 1 atm. (a) What quantity of toluene (lbm) will enter the atmosphere when the tank is filled and the gas displaced? (b) Suppose that 90% of the displaced toluene is to be recovered by compressing the displaced gas to a total pressure of 5 atm and then cooling it isobarically to a temperature T(°F). Calculate T.
Answer:
A) m[tex]_{T}[/tex] = 0.3025 * 0.0476 * 92.13 = 1.327 Ibm
B) T= 63.32°F
Explanation:
Given data:
1000 gallon tank currently contains 100.0 gallons of liquid toluene
and A gas saturated with toluene vapor at 85°F and 1 atm
A) Calculate quantity of toluene ( Ibm ) that will enter the atmosphere when the tank is filled
m[tex]_{T}[/tex] = [tex]n_{gas} * Y_{T} * M_{T}[/tex]
[tex]n_{gas}[/tex] (total mole of gas) = 0.3025 Ib-mole ( calculated using : [tex]\frac{PV}{RT}[/tex] )
[tex]Y_{T}[/tex] (mole fraction of toluene) = 0.0476 ( calculated using [tex]\frac{P_{T} }{P}[/tex] )
M[tex]_{T}[/tex] = 92.13 Ibm/Ib-mole
therefore: m[tex]_{T}[/tex] = 0.3025 * 0.0476 * 92.13 = 1.327 Ibm
B) using Antoine equation to solve for T
Antoine equation : [tex]log_{10} (P_{T} ) = A - \frac{B}{T+C}[/tex]
PT( partial pressure ) = 18.95 ( calculated using : [tex]y_{tb} * P[/tex] )
A = 6.95805
B = 1346.773
T = ?
C = 219.693
to calculate T make T the subject the subject of the equation
T + 219.693 = 1346.773 / 5.68044
∴ T = 17.40°C
convert T to Fahrenheit
T = 1.8 * 17.40 +32
= 63.32°F
The scientific method is the only way of learning about Nature used by scientist today *
A. true
B. false
Answer:
false
Explanation:
On top of a cliff of height h, a spring is compressed 5m and launches a projectile perfectly horizontally with a speed of 75 m s . It hits the ground with speed 90 m s . How high above the ground was the cliff? (Hint: use energy conservation to make the problem easier!)
Answer:
The height of the cliff is 121.276 m
Explanation:
Given;
initial velocity of the projectile, v₁ = 75 m/s
final velocity of the projectile, v₂ = 90 m/s
spring compression = 5 m
Apply the law of conservation of energy;
mgh₀ + ¹/₂mv₁² = mgh₂ + ¹/₂mv₂²
gh₀ + ¹/₂v₁² = gh₂ + ¹/₂v²
gh₁ - gh₂ = ¹/₂v₂² - ¹/₂v₁²
g(h₀ - h₂) = ¹/₂ (v₂² - v₁²)
h₀ - h₂ = ¹/₂g (v₂² - v₁²)
h₀ = h(cliff) + 5m
when the projectile hits the ground, Final height, h₂ = 0
[tex]h_o - 0 = \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} + 5= \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} = \frac{1}{2g}(v_2^2-v_1^2) - 5\\\\h_{cliff} = \frac{1}{2*9.8}(90^2-75^2) - 5\\\\h_{cliff} = 121.276 \ m[/tex]
Therefore, the height of the cliff is 121.276 m
Suppose astronomers discover a radio message from a civilization whose planet orbits a star 35 light-years away. Their message encourages us to send a radio answer, which we decide to do. Suppose our governing bodies take 2 years to decide whether and how to answer. When our answer arrives there, their governing bodies also take two of our years to frame an answer to us. How long after we get their first message can we hope to get their reply to ours? Enter your answer in years.
Answer:
The duration is [tex]T =72 \ years /tex]
Explanation:
From the question we are told that
The distance is [tex]D = 35 \ light-years = 35 * 9.46 *10^{15} = 3.311 *10^{17} \ m [/tex]
Generally the time it would take for the message to get the the other civilization is mathematically represented as
[tex]t = \frac{D}{c}[/tex]
Here c is the speed of light with the value [tex]c = 3.08 *10^{8} \ m/s[/tex]
=> [tex]t = \frac{3.311 *10^{17} }{3.08 *10^{8}}[/tex]
=> [tex]t = 1.075 *10^9 \ s[/tex]
converting to years
[tex]t = 1.075 *10^9 * 3.17 *10^{-8} [/tex]
[tex]t = 1.075 *10^9 * 3.17 *10^{-8} [/tex]
[tex]t = 34 \ years [/tex]
Now the total time taken is mathematically represented as
[tex]T = 2* t + 2 + 2[/tex]
=> [tex]T = 2* 34 + 2 + 2[/tex]
=> [tex]T =72 \ years /tex]
Which property of a substance can be determined using a pH indicator.
Answer:
Acidity
Explanation:
A pH indicator measures how acidic or basic a substance is.
Hope this helped :)
In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 4.7 kg book is pushed from rest through a distance of 0.85 m by the horizontal 42 N force from the broom and then has a speed of 1.0 m/s, what is the coefficient of kinetic friction between the book and floor
Answer:
μ_k = 0.851
Explanation:
We are given;
Mass of book; m_book = 4.7 kg
Horizontal force; F_horiz = 42 N
Distance; d = 0.85 m
Speed; v = 1 m/s
First of all let's find the acceleration using Newton's equation of motion;
v² = u² + 2ad
u is initial velocity and it's 0 m/s in this case.
Thus;
1² = (2 × 0.85)a
1 = 1.7a
a = 1/1.7
a = 0.5882 m/s²
Now, resolving forces along the vertical direction, we have;
W - N = 0
Thus,W = N
Where W is weight = mg and N is normal force
Thus; N = mg = 4.7 × 9.81 = 46.107 N
Now, resolving forces along the horizontal direction, we have;
F_horiz - ((μ_k)N) = ma
Where μ_k is coefficient of kinetic friction.
Thus;
42 - 46.107(μ_k) = 4.7 × 0.5882
42 - 46.107(μ_k) = 2.76454
μ_k = (42 - 2.76454)/46.107
μ_k = 0.851
All biomes don’t have the same level of biodiversity. What seems to be the optimal conditions for high biodiversity?
Answer:
See the answer below
Explanation:
The optimal conditions for high biodiversity seem to be a warm temperature and wet climates.
The tropical areas of the world have the highest biodiversity and are characterized by an average annual temperature of above 18 [tex]^oC[/tex] and annual precipitation of 262 cm. The areas are referred to as the world's biodiversity hotspots.
Consequently, it follows logically that the optimal conditions for high biodiversity would be a warm temperature of above 18 [tex]^oC[/tex] and wet environment with annual precipitation of not less than 262 cm.
The variation in temperature and precipitation across biomes can thus be said to be responsible for the variation in the level of biodiversity in them.
A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 98 m and acquired a velocity of The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The upward acceleration of the rocket during the burn phase is closest to:
29 m/s2
31 m/s2
33 m/s2
30 m/s2
32 m/s2
Explanation:
The question is incomplete. Here is the complete question.
A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 98 m and acquired a velocity of 30m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The upward acceleration of the rocket during the burn phase is closest to...
Given
initial velocity of rocket u = 0m/s
final velocity of rocket = 30m/s
Height reached by the rocket = 98m
Required
upward acceleration of the rocket
Using the equation of motion below to get the acceleration a:
[tex]v^2 = u^2+2as\\30^2 = 0^2 + 2(a)(98)\\900 = 196a\\a = \frac{900}{196}\\a = 4.59m/s^2[/tex]
Hence upward acceleration of the rocket during the burn phase is closest to 5m/s²
Note that the velocity used in calculation was assumed.
PLEASE SOLVE FAST!!! If the average American watches hours of TV every day , how many minutes will be spent in front of the TV by a person's 65th birthday? Solve using Dimensional Analysis.
Answer:
5694000 min
Explanation:
Let's suppose the average American watches 4 hours of TV every day. First, we will calculate how many minutes they watch per day. We will use the conversion factor 1 h = 60 min.
(4 h/day) × (60 min/1 h) = 240 min/day
They watch 240 minutes of TV per day. Now, let's calculate how many minutes they watch per year. We will use the conversion factor 1 year = 365 day.
240 min/day × (365 day/year) = 87600 min/year
They watch 87600 min/year. Finally, let's calculate how many minutes they spend watching TV in 65 years.
87600 min/year × 65 year = 5694000 min
A beam of protons is directed in a straight line along the positive zz ‑direction through a region of space in which there are crossed electric and magnetic fields. If the electric field magnitude is E=450E=450 V/m in the negative yy ‑direction and the protons move at a constant speed of v=7.9×105v=7.9×105 m/s, what must the direction and magnitude of the magnetic field be in order for the beam of protons to continue undeflected along its straight-line trajectory? Select the direction of the magnetic field BB .
Answer:
The magnitude is [tex]B = \frac{450}{7.9* 10^5}[/tex]
The direction is the positive x axis
Explanation:
From the question we are told that
The electric field is E = 450 V/m in the negative y ‑direction
The speed of the proton is [tex]v= 7.9* 10^5\ m/s[/tex] in the positive z direction
Generally the overall force acting on the proton is mathematical represented as
[tex]F_E = q(\vec E + \vec v * \vec B)[/tex]
Now for the beam of protons to continue un-deflected along its straight-line trajectory then [tex]F_E =0[/tex]
So
[tex] 0 = q( E (-y) + v(z) * \vec B)[/tex]
=> [tex]E\^y = v \^ z * \vec B[/tex]
Generally from unit vector cross product vector multiplication
[tex]\^ z \ * \ \^ x = \^ y[/tex]
So the direction of B (magnetic field must be in the positive x -axis )
So
[tex]E\^y = v \^ z * B\^ x [/tex]
=> [tex]E\^y = vB ( \^ z * \^ x) [/tex]
=> [tex]E\^y = vB ( \^y) [/tex]
=> [tex]E = vB [/tex]
=> [tex]B = \frac{450}{7.9* 10^5}[/tex]
=> [tex]B = 0.0005696 \ T [/tex]
The current is suddenly turned off. How long does it take for the potential difference between points a and b to reach one-half of its initial value
Complete Question
The complete question is shown on the first uploaded image
Answer:
Explanation:
From the question we are told that
The original voltage is [tex]V_o[/tex]
The new voltage is [tex]V =\frac{V_o}{2}[/tex]
The capacitance is [tex]C = 150\ nF = 150 *10^{-9} \ F[/tex]
The first resistance is [tex]R_i = 26 \Omega[/tex]
The second resistance is [tex]R_E = 200 \Omega[/tex]
Generally the equivalent resistance is
[tex]R_e = R_1 + R_E[/tex]
=> [tex]R_e = 26 +200 [/tex]
=> [tex]R_e = 226 \ \Omega [/tex]
Generally the time constant is mathematically represented as
[tex]\tau = RC[/tex]
=> [tex]\tau = 226 * 150 *10^{-9}[/tex]
=> [tex]\tau = 3.39 *10^{-5} \ s [/tex]
Generally the voltage is mathematically represented as
[tex]V = V_o e^{-\frac{t}{\tau} }[/tex]
=> [tex]\frac{V_o}{2} = V_o e^{-\frac{t}{\tau} }[/tex]
=> [tex]0.5 = e^{-\frac{t}{\tau} }[/tex]
=> [tex]ln(0.5) = {-\frac{t}{ 3.39 *10^{-5} } }[/tex]
=> [tex]ln(0.5) * 3.39 *10^{-5} = -t [/tex]
=> [tex]t = 2.35*10^{-5} \ s [/tex]
I NEED HELP PLEASEE ITS AN ECONOMICS QUESTION ABOVE
Answer:
I believe the answer is Property taxes
Explanation:
Answer: I'm pretty sure property taxes
Explanation:
A shell is fired with a horizontal velocity in the positive x direction from the top of an 80 m high cliff. The shell strikes the ground 1330 m from the base of the cliff. What is the speed of the shell as it hits the ground
Answer:
V = 331.59m/s
Explanation:
First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.
S = ut + 1/2at²
Given height of the cliff S = 80m
initial velocity u = 0m/s²
a = g = 9.81m/s²
Substitute
80 = 0+1/2(9.81)t²
80 = 4.905t²
t² = 80/4.905
t² = 16.31
t = √16.31
t = 4.04s
Next is to get the vertical velocity
Vy = u + gt
Vy = 0+(9.81)(4.04)
Vy = 39.6324
Also calculate the horizontal velocity
Vx = 1330/4.04
Vx = 329.21m/s
Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.
V² = Vx²+Vy²
V² = 329.21²+39.63²
V² = 329.21²+39.63²
V² = 108,379.2241+1,570.5369
V² = 109,949.761
V = √ 109,949.761
V = 331.59m/s
Hence the speed of the shell as it hits the ground is 331.59m/s
Part D
Next, we'll examine magnetic force. Bring the ends of your two magnets together. Explore the three
possible combinations. In two of the combinations, the two ends are the same. In one combination, the
two ends are different. Describe the force you feel in each combination
Answer:
i. The magnetic force of repulsion.
ii. The magnetic force of attraction.
Explanation:
A magnet is a material that has the attraction and repulsion capability. Magnets has two poles, north and south, thus would attract or repel another magnet in its neighborhood. It can either be a permanent or temporal magnet, and attracts ferrous metals.
i. In the case of two combinations where two ends are the same, it could be observed that the two ends (poles) repels each other. Thus since like poles repels, magnetic force of repulsion is felt.
ii. In the case of one combination in which the two ends are different, the two ends (poles) attract. Since unlike poles attracts, magnetic force of attraction is observed.
in a controlled experiment do none of the variables change?
Answer:
Yes
Explanation:
The variables change in and experiment.
Answer:
If you are carefully enough to control everything, then everything that could change the result of your experiment won't happen.
Explanation:
dimensional formula of stress is same as
Dimensional formula of stress is same as pressure.
What are the standard international (si) units of distance
Answer:
meter
Explanation:
Answer: The International System of Units is a system of measurement based on 7 base units
Explanation: the metre, kilogram, second, ampere, Kelvin, mole, and candela. These base units can be used in combination with each other.
Astronomers have proposed the existence of a ninth planet in the distant solar system. Its semi-major axis is suggested to be approximately 600 AU. If this prediction is correct, what is its orbital period in years
Answer:
T = 1.4696 10⁴ years
Explanation:
For this exercise we must use Kepler's laws, specifically the third law which is the application of the universal law of gravitation to Newton's second law
F = ma
G m M / r² = m a_c = m v² / r
G M / r = v²
the speed of the circular orbit is
v = 2π r / T
we substitute
G M / r = 4π² r² / T²
T² = (4π² / G M) r³
Kepler proved that this expression is the same if the radius is changed by the semi-major axis of an ellipse
T² = (4π² /GM) a³
the constant is worth
(4π² / GM) = 2.97 10⁻¹⁹ s² / m³
let's reduce the distance to SI units
AU is the distance from the Earth to the Sun
a = 600 AU = 600 AU (1.496 10¹¹ m / 1 AU)
a = 8.976 10¹³ m
T² = 2.97 10⁻¹⁹ (8.976 10¹³)³
T² = 21.4786 10²²
T = 4.63 10¹¹ s
Let's reduce to years
T = 4.63 10¹¹s (1 h / 3600s) (1 day / 24 h) (1 year / 365 days)
T = 1.4696 10⁴ years
cameron drives his car 15 km north. He stops for lunch and then drives 12 km south. What is his displacement?
Answer:
Displacement is 3 km North
Explanation:
An LED lamp powered by a USB-based portable battery has an effective resistance of 500 Ohm. If the battery is rated for 10,000 mAh, then how long can the lamp be powered and what is the total power consumed by the lamp? The operating voltage of a USB powered device is 5V. Assume that the battery is also rated at 5V.
Answer:
Time = 1000 h
Power = 0.05 W = 50 mW
Explanation:
First we will find the current consumed by the lamp. For this purpose we can use the Ohm's Law. The equation of Ohm's Law is written as follows:
V = IR
I = V/R
where,
I = Current = ?
V = Voltage = 5 V
R = Resistance = 500 Ω
Therefore,
I = 5 V/500 Ω
I = 0.01 A = 10 mA
Now the time duration of the operation f lamp can be found by:
Time = Rating/Current
Time = 10000 mAh/ 10 mA
Time = 1000 h
The power consumption of lamp is given as follows:
Power = IV
Power = (0.01 A)(5 V)
Power = 0.05 W = 50 mW
A toy helicopter takes off and moves 6 m up and then 1 m back down. What
is the displacement of the helicopter?
A. 7 m
5 m up
C. 6 m
D. 1 m down
the answer is 5 m up AKA "B"
At a distance of 10.0 m from a loudspeaker, the sound intensity level is measured to be 70 dB. At what distance from the source will the intensity be 40 dB?
Answer:
At a distance of 100 m from the source the intensity will be 40 dB.
Explanation:
Sound intensity is the acoustic power transferred by a sound wave per unit area normal to the direction of propagation.
The sound intensity depends on the power of the sound source, where the higher the power the greater the intensity, the distance to the sound source, the greater the distance being the lower the intensity, and the nature of the transmission medium.
The conversion between intensity and decibels corresponds to:
[tex]L=10*log\frac{I}{I0}[/tex]
where I0 = 10⁻¹² W/m² and corresponds to a level of 0 decibels therefore.
In this case, you can apply the following relationship between two intensities and distance, considering that the intensity of the sound level decreases with distance:
[tex]L1 - L2=10*log\frac{I1}{I0} - 10*log\frac{I2}{I0}[/tex]
[tex]L1 - L2=10*(log\frac{I1}{I0} - *log\frac{I2}{I0})[/tex]
[tex]L1 - L2=10*[log(\frac{I1}{I0}\frac{I0}{I2})][/tex]
[tex]L1 - L2=10*[log(\frac{I1}{I2})][/tex]
Being L1= 70 dB and L2= 40 dB
[tex]70 dB - 40 dB=10*[log(\frac{I1}{I2})][/tex]
[tex]30=10*[log(\frac{I1}{I2})][/tex]
[tex]\frac{30}{10} =log(\frac{I1}{I2})[/tex]
[tex]3=log(\frac{I1}{I2})[/tex]
[tex]10^{3} =\frac{I1}{I2}[/tex]
[tex]1,000=\frac{I1}{I2}[/tex]
The intensity is inversely proportional to the square of the distance to the source. The relationship between the intensities I1 and I2 at distances d1 and d2 respectively is:
[tex]\frac{I1}{I2} =\frac{d2^{2} }{d1^{2} }[/tex]
Then:
[tex]1,000=\frac{d2^{2} }{d1^{2} }[/tex]
Being d1= 10 m
[tex]1,000=\frac{d2^{2} }{10^{2} }[/tex]
[tex]1,000=\frac{d2^{2} }{100}[/tex]
1,000*100= d2²
10,000= d2²
√10,000= d2
100 m= d2
At a distance of 100 m from the source the intensity will be 40 dB.
An object from a certain height falls freely. which of the following happens PE and KE when the object is half on its way down
Answer:
A. Loses PE and gains KE
Explanation:
Statement is incomplete. Complete statement of problem is:
1. An object from a certain height falls freely. Which of the following happens to PE and KE when the object is half on its way down?
A. Loses PE and gains KE
B. Gains PE and loses KE
C. Loses both PE and KE
D. Gains both PE and KE
If we neglect the effects of any conservative force, the application of the Principle of Energy Conservation is reduced to a sum of gravitational potential ([tex]U_{g}[/tex]) and translational kinetic energies, measured in joules. That is:
[tex]U_{g,1}+K_{1} =U_{g,2}+K_{2}[/tex] (Eq. 1)
Let assume that an object falls from a height [tex]h[/tex] with a speed of zero. By definitions of gravitational potential and translational kinetic energies the previous is expanded. If final height is the half of initial value, then:
[tex]m\cdot g\cdot h = 0.5\cdot m\cdot g\cdot h +K_{2}[/tex] (Eq. 1b)
[tex]K_{2} = 0.5\cdot m\cdot g\cdot h[/tex]
[tex]K_{2} = 0.5\cdot U_{g,1}[/tex]
In a nutshell, translational kinetic energy is increased at the expense of diminishing gravitational potential energy. The correct answer is A.
Problem I Marcella (see warmup problem, above) gets her car moving steadily at 4m/s but suddenly someone stops ahead to assist her and parks their car 14 meters from the front of her car. Marcella grabs the car bumper and pulls very hard, with 200 N of force. The work she does transfers energy out, it reduces the K of the car, as it gradually approaches the other car. a) What is the initial kinetic energy before she tries to stop the car? b) What is the final kinetic energy, when her car hits the other car? What is the speed? c) Suppose the other person also slowed her car, pushing it from the front. How much force would be needed to stop her car 1 meter from the other car? [1 m allows the person not to be crushed!]
Answer:
Explanation:
a) KE = (1/2) * m * ([tex]v^{2}[/tex]) = F * d = 14m * 200N = 2800 m/N or 2.8 * [tex]10^{3}[/tex] m/N
b) 0J and 0m/s (if Marcella stopped after going 14 meters)
c) Known from part (a) that KE = 2800 J = F1 * d1,
2800J = F1 * (14m - 1m) => F1 = 2800J/13m = 215.384 N
what chemical diverged from trees
a resin and turpentine
b sodium
c lead
d marcotting
Answer:
a. resin and turpentine
Explanation:
The chemicals that diverged from trees are resin and turpentine.
Resin are produced by special cells in trees, most times we see them when a tree is damaged or cut. They are usually derived from pines and firs.
Turpentine is obtained by distilling resin.
Turpentine has an antiseptic property that has different uses. They are used as cleansing agents and for producing sanitary materials.