Answers
Answer:
4. unbalanced and Accelerating
5. balance and rest
Related Questions
uppose that the terminal speed of a particular sky diver is 150 km/h in the spread-eagle position and 320 km/h in the nosedive position. Assuming that the diver's drag coefficient C does not change from one position to the other, find the ratio of the effective cross-sectional area A in the slower position to that in the faster position (Aslower / Afaster).
Answers
Answer:
4.55
Explanation:
The terminal speed of a diver is given by:
[tex]v_t=\sqrt{\frac{2mg}{C\rho A} } \\\\Where\ m=mass\ of \ driver,d=acceleration\ due\ to\ gravity,C=drag\ \\coefficient,A=cross\ sectional\ Area.\\\\Therefore:\\\\A=\frac{2mg}{C \rho v_t^2} \\\\For\ area\ with\ terminal\ speed\ in\ spread\ angle\ position(v_s):\\\\A_s=\frac{2mg}{C \rho v_s^2} \\\\For\ area\ with\ terminal\ speed\ in\ nose\ dive\ position(v_n):\\\\A_n=\frac{2mg}{C \rho v_n^2}\\\\Therefore\ since\ g,m,C,\rho\ are\ constant:\\\\[/tex]
[tex]\frac{A_s}{A_n}= \frac{\frac{2mg}{C \rho v_s^2}}{\frac{2mg}{C \rho v_n^2}}\\\\\frac{A_s}{A_n}= \frac{v_n}{v_s} \\\\v_n=320\ km/h,v_s=150\ km/h\\\\\frac{A_s}{A_n}=\frac{320^2}{150^2} =4.55[/tex]
A woman accidentally drops a flowerpot from a windowsill at a height d above the street towards a man of height h standing below. The woman calls out to the man in just enough time for the man to move out of the way. If the man needs a time interval of Δt to respond to the warning, at what height above the street will the flowerpot be when the woman calls out the warning? (Use the following as necessary: d, h, Δt, v for the speed of sound, and g for gravitational acceleration.)
Answers
Answer:
h^2 - ( 2t_o v_s + 2v_s^2 /g) h + v_s^2 \ t_o^2 =0
The correct result is that of a positive height
Explanation:
For this exercise we use the kinematic relations, let's start by finding the time it takes for the sound to reach the man
v_s = y / t
t = [tex]\frac{y}{ v_s}[/tex]
this height is y = h
t = \frac{h}{ v_s}
the man has a response time of t = t₀, therefore
time to move is
t' = t - t₀
the initial height of flower pot is
y = y₀ + v₀ t' - ½ g t'²
when it reaches the floor the height is zero y = 0 and as the pot is dropped its initial velocity is zero v₀ = 0
0 = y₀ +0 - ½ g (t -t₀)²
if the initial height is i = h,
h = ½ g ([tex]\frac{h}{v_s}[/tex] - t₀)²2
[tex]\frac{2}{g} h[/tex] = [tex]\frac{h^2}{v_s^2}[/tex] - [tex]\frac{2t_o }{v_s} h[/tex] + t₀²
[tex]\frac{h^2}{v_s^2} - ( \frac{2t_o}{v_s} + \frac{2}{g} ) h + t_o^2 = 0[/tex]h2 / vs2 - (2nd / vs + 2 / g) h + to2 - = 0
[tex]h^2 - ( 2t_o v_s + 2v_s^2 /g) h + v_s^2 \ t_o^2 =0[/tex]
To know the height, you must solve the second degree equation, it is much easier with numerical values.
The correct result is that of a positive height
Potential energy is energy due to the:
a. motion of an object.
b. height of an object.
c. temperature of an object.
d. speed of an object.
Answers
Answer:I will say d
Explanation: because Potential energy is the energy stored within an object, due to the object's position, arrangement or state. Potential energy is one of the two main forms of energy, along with kinetic energy.
A motorcyclist is making an electric vest that, when connected to the motorcycle's 12 V battery, will warm her on cold rides. She is using 0.25-mm-diameter copper wire, and she wants a current of 4.2 A in the wire. Part A What length wire must she use
Answers
Answer:
L = 8.35 m
Explanation:
The lenght of a wire L can be calculated using the following expression:
L = R A/ρ (1)
Where:
R: resistance of the wire
A: Cross section area of the wire
ρ: resistivity of the copper wire.
With this expression we realize that we do not have the area of the cross section, and the resistance of the wire either.
To calculate the area we can use the following expression:
A = πr² (2)
If the diameter is 0.25 mm, then the radius is half, 0.125 mm. Converting this in meter it will have to be:
0.125 /1000 = 0.000125 m
Replacing we have:
A = π(0.000125)²
A = 4.91x10⁻⁸ m²
The reported resistivity of a copper wire is 1.68x10⁻⁸ Ω.m, so we just need to determine the resistance, which can be found using Ohm's law:
R = V/I (3)
Replacing (3) into (1) we have:
L = (V * A) / (I * ρ) (4)
So finally, the length of the copper wire will be:
L = (12 * 4.91x10⁻⁸) / (4.2 * 1.68x10⁻⁸)
L = 8.35 mHope this helps
1. Clara stops for 10 minutes to catch up with a friend.
Answers
Answer:
Clara has speed of 80m/min
Explanation:
Clara was jogging at 600 m in 5 minutes. She stopped suddenly which reduced her velocity and then she waited for 10 minutes so that her friends comes near her. She stopped to catch her friend. During this 10 minutes the velocity of Clara is zero. She started to walk again at a slower speed of 80m/min.
The cylinder with piston locked in place is immersed in a mixture of ice and water and allowed to come to thermal equilibrium withthe mixture. The piston is then moved inward very slowly, that thegas is always in thermal equilibrium with the ice-water mixture,what happens to the following(increase, decrease, same)?
a. volume of gas
b. temperature of gas
c. internal energy of gas,
d. pressure of gas
Answers
Answer:
a. volume of gas: (decreases)
b. temperature of gas: (same)
c. internal energy of gas: (same)
d. pressure of gas: (increases)
Explanation:
We have a gas (let's suppose that is ideal) in a piston with a fixed volume V.
Then we put in a reservoir at 0°C (the mixture of water and ice)
remember that the state equation for an ideal gas is:
P*V = n*R*T
and:
U = c*n*R*T
where:
P = pressure
V = volume
n = number of mols
R = constant
c = constant
T = temperature.
Now, we have equilibrium at T = 0°C, then we can assume that T is also a constant.
Then in the equation:
P*V = n*R*T
all the terms in the left side are constants.
P*V = constant
And knowing that:
U = c*n*R*T
then:
n*R*T = U/c
We can replace it in the other equation to get:
P*V = U/c = constant.
Now, the piston is (slowly) moving inwards, then:
a) Volume of the gas: as the piston moves inwards, the volume where the gas can be is smaller, then the volume of the gas decreases.
b) temperature of the gas: we know that the gas is a thermal equilibrium with the mixture (this happens because we are in a slow process) then the temperature of the gas does not change.
c) Internal energy of the gas:
we have:
P*V = n*R*T = constant
and:
P*V = U/c = constant.
Then:
U = c*Constant
This means that the internal energy does not change.
d) Pressure of the gas:
Here we can use the relation:
P*V = constant
then:
P = (constant)/V
Now, if V decreases, the denominator in that equation will be smaller. We know that if we decrease the value of the denominator, the value of the quotient increases.
And the quotient is equal to P.
Then if the volume decreases, we will see that the pressure increases.
A vertical wire carries a current straight up in a region of the magnetic field directed north. What is the direction of the magnetic force on the current due to the magnetic field
Answers
Answer:
The direction of the force on the vertical wire is towards the East or right.
Explanation:
Using Fleming's right hand rule, the current is the middle finger pointing straight up, the magnetic field is the fore-finger pointing Northwards and then the thumb is the direction of the force on the vertical wire.
Following these conventions, the thumb points towards the East. So, the direction of the force on the vertical wire is towards the East or right.
1. Three centimeters of water evaporated from a 200-hectare vertical walled reservoir during 24 hours. Storm water was added to the reservoir at a constant rate of 3 m3/s during this period. Determine the volume in ha-cm of water released during the period (through the bottom of the reservoir) if the water level was the same at the beginning and the end of the day.
Answers
Answer:
25920 ha-cm
Explanation:
Since water evaporates from the reservoir at a rate of 3 cm in 24 hours, its height changes at a rate of 3 cm/24 × 3600 s = 3 cm/86400s = 3.472 10⁻⁵ cm/s.
Now, the volume loss is dV/dt = dV/dh × -dh/dt
= dV/dt × -3.472 × 10⁻⁵ cm/s
= -3.472 × 10⁻⁵ cm/sdV/dh
The reservoir increases in volume at a rate of 3 m³/s = 3 × 10⁶ cm³/s in 24 hours.
So, the net rate of volume change per unit time of the reservoir is
3 × 10⁶ cm³/s - 3.472 × 10⁻⁵ cm/sdV/dh = Adh/dt where A = area of vertical walled reservoir and dh/dt = change in height of the reservoir with respect to time
So, 3 × 10⁶ cm³/s - 3.472 × 10⁻⁵ cm/sdV/dh = Adh/dt
Since dh/dt = 0 in 24 hours(since the water level remains the same after 24 hours, that is dh = 0)
3 × 10⁶ cm³/s - 3.472 × 10⁻⁵ cm/sdV/dh = Adh/dt
3 × 10⁶ cm³/s - 3.472 × 10⁻⁵ cm/sdV/dh = A × 0
3 × 10⁶ cm³/s - 3.472 × 10⁻⁵ cm/sdV/dh = 0
3.472 × 10⁻⁵ cm/sdV/dh = 3 × 10⁶ cm³/s
dV/dh = 3 × 10⁶ cm³/s ÷ 3.472 × 10⁻⁵ cm/s
dV/dh = 8.64 × 10¹¹ cm²
dV = (8.64 × 10¹¹ cm²)dh
Integrating both sides with V from 0 to V and h from h = 0 to h = 3 cm, we have
∫dV = ∫(8.64 × 10¹¹ cm²)dh
∫dV = (8.64 × 10¹¹ cm²)∫dh
V = (8.64 × 10¹¹ cm²)[h]₀³
V = (8.64 × 10¹¹ cm²)[3 cm - 0 cm]
V = (8.64 × 10¹¹ cm²)(3 cm)
V = 25.92 × 10¹¹ cm³
V = 2.592 × 10¹² cm³
V = 2.592 × 10¹² cm² × 1 cm
Since 1 ha = 10⁸ cm²,
V = 2.592 × 10¹² cm² × 1 ha/10⁸ cm² × 1 cm
V = 2.592 × 10⁴ ha-cm
V = 25920 ha-cm
How do you think that changing the mass of the pendulum bob will affect the period of the pendulum swing?
Answers
Four electrons and one proton are at rest, all at an approximate infiitne distance away from each other. This original arrangment of the four particles is defined as having zero electrical potential energy No work is required to bring one electron from infitinty to a location defined as the origin, while the other three particles remain at infiniuty. This is because no voltage exists near the origin until the first electron arrives. (a) Now, with the first electron remaining fixed at the origin, how much work is required to bring one of the remaining electrons from infinity to the coordinate (0 m, 2.00 m)? The other three particles remain at infinity. If this second electron was subsequently released, how fast would it be traveling once it returned to infinity? (b) Nļw, considering the two electrons fixed 2.00 m apart, how much work is required to bring the third electron from infinity to the coordinate (3.00 m, 0 m)? The other two particles remain at infinity. If this third electron was subsequently released, how fast would it be traveling once it returned to infinity? (c) Now considering the three fixed electrons at the coordinates described above. How much work is required to bring the last electron from infinity to the coordinate (3.00 m, 4.00 m)? If this forth electron was subsequently released, how fast would it be traveling once it returned to infinity? (d) Now considering the three fixed electrons at the coordinates described above. Finally, how much work is required to bring the proton from infinity to a coordinate of (1.00 m, 1.00 m)? If the proton is subsequently released and we assume that minimum separation distance between a proton and an electron is 1.00 pm, then how fast will the proton be traveling once it crashes into an electron?
Answers
Answer:
a) W = 1.63 10⁻²⁸ J, b) W = 1.407 10⁻²⁷ J, c) W = 1.68 10⁻²⁸ J,
d) W = - 4.93 10⁻²⁸ J
Explanation:
a) In this problem we have an electron at the origin, work is requested to carry another electron from infinity to the point x₂ = 0, y₂ = 2.00m
If we use the law of conservation of energy, work is the change in energy of the system
W = ΔU = U_∞ -U
the potential energy for point charges is
U =k [tex]\sum \frac{q_i q_j}{r_{ij} }[/tex]
in this case we only have two particles
U = k [tex]\frac{q_1q_2}{r_{12} }[/tex]
the distance is
r₁₂ = [tex]\sqrt{(x_2-x_1)^2 + ( y_2-y_1)^2 }[/tex]
r₁₂ =[tex]\sqrt{ 0 + ( 2-0)^2}[/tex]Ra 0 + (2-0)
r₁₂ = √2= 1.4142 m
we substitute
W = k \sum \frac{q_i q_j}{r_{ij} }
let's calculate
W = [tex]\frac{ 9 \ 10^9 (1.6 \ 10^{-19})^2 }{1.4142}[/tex] 9 109 1.6 10-19 1.6 10-19 / 1.4142
W = 1.63 10⁻²⁸ J
b) the two electrons are fixed, what is the work to bring another electron to x₃ = 3.00 m y₃ = 0
in this case we have two fixed electrons
U = k [tex]( \frac{q_1q_3}{r_{13} } + \frac{q_2q_3}{r_{23} } )[/tex]
in this case all charges are electrons
q₁ = q₂ = q₃ = q
W = U = k q² [tex]( \frac{1}{r_{13} } + \frac{1}{r_{23} } )[/tex]
the distances are
r₁₃ = [tex]\sqrt{(3-0)^2 + 0}[/tex]RA (3.00 -0) 2 + 0
r₁₃ = 3
r₂₃ = [tex]\sqrt{ 3^2 + 2^2}[/tex]Ra (3 0) 2 + (2 0) 2
r₂₃ = √13
r₂₃ = 3.606 m
let's look for the job
W = U
let's calculate
W =[tex]{9 \ 10^3 ( 1.6 10^{-19})^2 }({\frac{1}{3} + \frac{1}{3.606} } )[/tex]
W = 1.407 10⁻²⁷ J
c) the three electrons are fixed, we bring the four electron to x₄ = 3.00m,
y₄ = 4.00 m
W = U = k [tex]( \frac{q_1q_4}{r_{14 }} + \frac{q_2q_4}{r_{24} } + \frac{q_3q_4}{r_{34} } )[/tex]
all charges are equal q₁ = q₂ = q₃ = q₄ = q
W = k q² [tex](\frac{1}{r_{14} } + \frac{1}{r_{24} } + \frac{1}{r_{34} } )[/tex]
let's look for the distances
r₁₄ = [tex]\sqrt{3^2 +4^2}[/tex]
r₁₄ = 5 m
r₂₄ = [tex]\sqrt{3^2 + ( 4-2)^2}[/tex]
r₂₄ = √13 = 3.606 m
r₃₄ = [tex]\sqrt{(3-3)^2 + (4-0)^2}[/tex]
r₃₄ = 4 m
we calculate
W = 9 10⁹ (1.6 10⁻¹⁹)² [tex]( \frac{1}{5} + \frac{1}{3.606} + \frac{1}{4} )[/tex]
W = 1.68 10⁻²⁸ J
d) we take the proton to the location x5 = 1m y5 = 1m
W = U = k [tex]( \frac{q_1q_5}{r_{15} } + \frac{q_2q_5}{r_{25} } + \frac{q_3q_5}{r_{35} } + \frac{q_4q_5}{r_{45} } )[/tex]
in this case the charges have the same values but charge 5 is positive and the others negative, so the products of the charges give a negative value
W = - k q² [tex]( \frac{1}{r_{15} } + \frac{1}{r_{25} } + \frac{1}{r_{35} } + \frac{1}{r_{45} } )[/tex]
we look for distances
r₁₅ = [tex]\sqrt{ 1^2 +1^2}[/tex]Ra (1-0) 2 + (1-0) 2
r₁₅ = √ 2 = 1.4142 m
r₂₅ = [tex]\sqrt{ (2-1)^2 +1^2}[/tex]
r₂₅ = √2 = 1.4142 m
r₃₅ = [tex]\sqrt{ ( 3-1)^2 +1^2}[/tex]
r₃₅ = √5 = 2.236 m
r₄₅ = [tex]\sqrt{ (3-1)^2 + (4-1)^2}[/tex]
r₄₅ = √13 = 3.606 m
we calculate
W = - 9 10⁹ (1.6 10⁻¹⁹)² [tex]( \frac{1}{1.4142} +\frac{1}{1.4142} + \frac{1}{2.236} + \frac{1}{3.606} )[/tex]
W = - 4.93 10⁻²⁸ J
Dereck is looking at how electrically charged objects can attract other objects without touching. What control would he need to use?
An electrically charged object
An uncharged object
A positively charged object
A negatively charged object
Answers
Answer:
its An uncharged object.
if its not charged the electrically wont go on it
Answer:
uncharged object
Explanation:
PLEASE HELP QUICK which statement describes a primary difference between an electromagnetic wave and mechanical wave?
A. electromagnetic waves can travel through empty space
B. electromagnetic waves can be transverse longitudinal or surface waves
C. electromagnetic waves can only travel through solids liquids or gases
D. electromagnetic waves need a medium to transfer energy
Answers
Answer:
A.
Explanation:
An electromagnetic wave is produced by the interaction between a variable electric field, and a magnetic electric field, which propagates in space, even in vaccuum, at a fixed speed, whilst the mechanical waves require a medium in order to transfer energy.Answer: A
Explanation:
How much force is needed to accelerate a Kia Soul with a
mass of 1200 kg to 5 m/s2?
Answers
Answer:
[tex]\boxed {\boxed {\sf 6,000 \ Newtons}}[/tex]
Explanation:
Force is the product of mass and acceleration.
[tex]F=ma[/tex]
The mass of the Kia Soul is 1200 kilograms and its acceleration is 5 meters per square second.
[tex]m= 1200 \ kg \\a= 5 \ m/s^2[/tex]
Substitute the values into the formula.
[tex]F= 1200 \ kg * 5 \ m/s^2[/tex]
Multiply.
[tex]F= 6000 \ kg*m/s^2[/tex]
1 kilgram meter per square second is equal to 1 Newton. Our answer of 6000 kg*m/s² equals 6000 N[tex]F= 6000 \ N[/tex]
Answer:
Given :-Mass = 1200 kgAcceleration = 5 m/s²To Find :-Force
Solution :-We know that
F = ma
F = Force
m = mass
a = acceleration
F = 1200 × 5
F = 6000 N
[tex] \\ [/tex]
What are regular and irregular reflection of light? plz help its
urgent..
Answers
Explanation:
Regular reflection: It is the reflection from a smooth surface such that the light rays are evenly parallel to each other and an image is formed. ... Irregular reflection: It is the diffused reflection from uneven surface such that the light rays are not parallel to each other and do not form an image.
PLEASE HELP ASAP! WILL GIVE BRAINLIEST TO CORRECT ANSWER! HELP!! HELP!!
The diagram shows the structure of an animal cell.
The image of an animal cell is shown with some organelles labeled numerically from 1 to 6. The outer double layer boundary of the cell is labeled 1. A stacked disc like structure is labeled 2. A broad rod shaped structure with an irregular shape inside it is labeled 3. The entire plain section that forms the background of the cell and is within the outer boundary is labeled 4. A small circular shape within the large circular shape is labeled 5. The large central circular shape is labeled 6.
Which number label represents the cell membrane?
1
2
4
6
(this is middle school science)
Answers
Answer:
1. cell membrane
2. golgi body
3. mitochondrion
4. cytoplasm
5. nucleolus
6. nucleus
Explanation:
The correct answer to this question is Option A; 6.
Why?
In a plant cell, the nucleus surrounds the nucleolous, which would be number 5. Therefore, number 6 would be your correct answer.
~Thank you~
An iron block of 12 kg undergoes a process during which there is a heat gain from the block at 2 kJ/kg, an elevation increase of 32 m, and a decrease in velocity from 40 m/s to 7 m/s. During the process, which also involves work transfer, the internal energy of the block increases by 70 kJ. Suppose the total energy of the system remains constant. Determine the work transfer during the process in kJ and indicate whether the work is done on/by the system.
Answers
Answer:
Explanation:
Total heat gain by the block ΔQ = 2 x 12 kJ = 24 kJ .
Gain of potential energy = mgh = 12 x 9.8 x 32 = 3.763 kJ
Decrease in kinetic energy KE = 1/2 x 12 ( 40² - 7² )
= 9.306 kJ
increase in internal energy ΔE = 70 kJ
ΔQ = ΔE + PE - KE + W , W is work done by the gas
Putting the values
24 = 70 + 3.763 - 9.306 + W
W = - 40.457 kJ .
Since W is negative that means work is done on the system .
A toy car can go 5 mph. How long would it take to go 12 miles?
Answers
The electric field 30cm from a van de Graaff generator is measured to be 28,300N/C. What is the charge of the van de Graaf?
Answers
Answer:
14
Explanation:
EWAN KO LANG DIN BASTA YAN ALAM KO
Two very small +3.00-μC charges are at the ends of a meter stick. Find the electric potential (relative to infinity) at the center of the meter stick.
Answers
Answer:
The electric potential at the center of the meter stick is 54 KV.
Explanation:
Electric potential (V) is given as:
i.e V = [tex]\frac{kq}{r}[/tex]
Where: k is the Coulomb constant, q is the charge and r is the distance.
Given: q = 3.0 μC = 3.0 x [tex]10^{-6}[/tex] C, r = 0.5 m
So that,
V = [tex]\frac{9*10^{9}*3.0*10^{-6} }{0.5}[/tex]
= [tex]\frac{2.7*10^{4} }{0.5}[/tex]
V = 54000
= 54 000 volts
The electric potential at the center of the meter stick is 54 KV.
Assuming 84.0% efficiency for the conversion of electrical power by the motor, what current must the 13.0-V batteries of a 716 kg electric car be able to supply to climb a 3.00 x 102 m high hill in 2.00 min at a constant 22.0 m/s speed while exerting 7.00 x 102 N of force to overcome air resistance and friction
Answers
Answer:
[tex]\mathbf{ current(I) =1766.67 \ A}[/tex]
Explanation:
Given that:
The air resistance and friction = 700 N
The gravity caused force = 716 × 9.8 = 7016.8
Total force = (7016.8 + 700) N
Total force = 7716.8 N
∴
[tex]13 \times current(I) \times 0.84 = \dfrac{7716.8 \times 300}{2 \times 60}[/tex]
[tex]current(I) \times 10.92= 19292[/tex]
[tex]current(I) = \dfrac{19292}{10.92}[/tex]
[tex]\mathbf{ current(I) =1766.67 \ A}[/tex]
Velocity time graph and how to draw it
Answers
Answer:
Velocity time graph
Explanation:
Draw on graph paper two straight lines originating at the same point and perpendicular to each other. This is the x-y axis. The x-axis is the horizontal line and the y-axis is the vertical line.
Mark appropriate equally-spaced time intervals on the x-axis so that you can easily graph the time values from the table.
Mark appropriate velocity increments on the y-axis so that you can easily graph the velocity values from the table. If you have negative velocity values, extend the y-axis downward.
Find the first time value from the table and locate it on the x-axis. Look at the corresponding velocity value and find it on the y-axis.
Put a dot where a straight line vertically drawn up through the x-axis value and a straight line horizontally drawn through the y-axis value intersect.
Plot in similar fashion for all other velocity-time pairs in your table.
Draw a straight line with a pencil, connecting each dot you have put down on the graph paper, going from left to right
Find the GCF of each set of numbers.
12, 21, 30
Math
Answers
Answer:
3 is the GCF for all these numbers if thats what you're asking
12=2’2x3
21=3x7
30=2x3x5
Tyler and Jim race each other up a mountain on their bicycles. Tyler rides a road bike on the switchbacks of the twisting and turning mountain road. Jim rides a mountain bike and follows a direct, but steeper, straight-line path up the mountain. They start at the same time and place at the bottom of the mountain and finish at the same time and place at the top of the mountain. From start to finish a. whose distance traveled was longer? b. whose displacement was longer? c. which rider had the faster average speed? d. which rider had the faster average velocity? e. who won the race?
Answers
Answer:
Explanation:
Displacement is minimum distance between initial and final point .
Distance is total length of path covered in a journey .
a )
Tyler covered a longer distance in the journey because total length of path covered by him is longer due to curved path .
b )
Both have same displacement , because minimum distance between initial and final point in both the case is same .
c )
average speed = distance / time
as time is same for both the case ,
average speed ∝ distance
As distance covered by Tyler is more , his average speed is more .
d )
average velocity = displacement / time
As both displacement and time are same in both the case , average velocity in both the case is same .
e )
They start at the same time and place at the bottom of the mountain and finish at the same time , both have tie and nobody won the race , in spite of speed of Tyler being greater .
Greatest to least order
Answers
Answer:
Explanation:
FBEDAC
Galileo _____.
did not believe friction existed
believed that friction stopped objects in motion
believed that friction kept objects in motion
assumed that in a frictionless environment objects would never move
Answers
Answer:
friction help to slow motion in other word it oppose motion, but in a frictionless environment object would move with difficult stopping point.
A diet is to contain at least 2400 mg vitamin C, 1800mg Calcium, and 1200 calories every day. Two foods, a dairy-based meal and a vegan option are to fulfill these requirements. Each ounce of the dairy-based meal provides 50 mg vitamin C, 30 mg Calcium, and 10 calories. Each ounce of the vegan option provides 20 mg vitamin C, 20 mg Calcium, and 40 calories. If the dairy-based meal costs $0.042 per ounce and the vegan option costs $0.208 per ounce, how many ounces of each food should be purchased to minimize costs? What is that minimum cost (per day)?
Answers
Answer:
The answer is below
Explanation:
Let x represent the number of ounce of dairy based meal and let y represent the number of vegan option in ounce.
Since the diet must contain at least 2400 mg vitamin C, therefore:
50x + 20y ≥ 2400
Since the diet must contain at least 1800 mg Calcium, therefore:
30x + 20y ≥ 1200
Since the diet must contain at least 1200 calories, therefore:
10x + 40y ≥ 1200
Therefore the constraints are:
50x + 20y ≥ 2400
30x + 20y ≥ 1200
10x + 40y ≥ 1200
x > 0, y > 0
The graph was drawn using geogebra online graphing tool, and the solution to the problem is at:
C(30, 45) and D(48, 18)
dairy-based meal costs $0.042 per ounce and the vegan option costs $0.208 per ounce. The cost equation is:
Cost = 0.042x + 0.208y
At C(30, 45); Cost = 0.042(30) + 0.208(45) = $10.62
At C(48, 18); Cost = 0.042(48) + 0.208(18) = $5.76
The minimum cost is at (48, 18). That is 48 dairy based meal and 18 vegan
who has brown hair and brown eyes but is a boy
Answers
Answer:
I have strawberry blonde/brown hair blue eyes and a girl lol
Explanation:
A football quarterback runs 15.0 m straight down the playing field in 2.30 s. He is then hit and pushed 3.00 m straight backward in 1.74 s. He breaks the tackle and runs straight forward another 28.0 m in 5.20 s.
Calculate his average velocity (in m/s) for each of the three intervals. (Assume the quarterback's initial direction is positive. Indicate the direction with the sign of your answer.)
Answers
Answer:
6.52 m/s
1.72 m/s
5.38 m/s
Explanation:
this question requires us to find the average velocity.
1. velocity in straight down direction:
velocity = distance/time
= 15.0/2.30
= 6.52 m/s
2. velocity in straight backward direction:
velocity = distance/time
= 3.00 /1.74
= 1.72m/s
3. velocity in straight forward direction
velocity = distance/time
= 28.0/5.20
= 5.38 m/s
these are the his velocities for each if the intervals.
thank you!
he nucleus of 8Be, which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of 2 protons and 2 neutrons). (a) What is the force between the two alpha particles when they are 6.60 ✕ 10−15 m apart? N (b) What is the initial magnitude of the acceleration of the alpha particles due to this force? Note that the mass of an alpha particle is 4.0026 u. m/s2
Answers
Answer:
A) F = 21.134 N
B) a = 3180.76 × 10^(24) m/s²
Explanation:
A) We are given;
Mass of alpha particle; m = 4.0026 u
Now, 1u = 1.66 × 10^(-27) kg
Thus; m = 4.0026 × 1.66 × 10^(-27)
Distance apart; r = 6.60 × 10^(−15) m
Charge on the alpha particle is;
q = 2e = 2 × 1.6 × 10^(-19) C
Formula for the force between the two alpha particles is;
F = kq1.q2/r²
k = 8.99 × 10^(9) N.m²/C²
q1 = q2 = 2 × 1.6 × 10^(-19) C
F = 8.99 × 10^(9) × (2 × 1.6 × 10^(-19))²/(6.60 × 10^(−15))²
F = 21.134 N
B) acceleration is given by;
a = F/m
Thus; a = 21.134/(4.0026 × 1.66 × 10^(-27))
a = 3180.76 × 10^(24) m/s²
Name the state of matter that diffusion happens the fastest in.
Answers
Answer:
Liquids
Explanation:
Diffusion occurs fastest in liquids.
