Answer:
86.33m/s^2
Explanation:
Acceleration = Force/Mass
= 25.9/0.3
= 86.33
Specify whether the boiling point, as determined in the miniscale boiling-point apparatus, is the temperature a.of the liquid at the timebubbles first emerge slowly from the liquid. b.at the vapor-liquid interface above the surface of the boiling liquid while a drop of liquid c.is suspended from the thermometer. d.of the liquid at the timebubbles emerge rapidly from the liquid. e.of the heating source at the timebubbles emerge rapidly from the liquid.
Answer:
a. of liquid at the time bubbles first emerge slowly from the liquid.
Explanation:
Boiling point of liquid happens due to heat energy. This is an exothermic reaction as heat is released in to the environment. The initial boiling vapors slowly move away from the liquid and as the temperature increases the vapors start moving quickly.
In a physics lab experiment for the determination of moment of inertia, a team weighs an object and finds a mass of 2.15 kg. They then hang the object on a pivot located 0.163 m from the object's center of mass and set it swinging at a small amplitude. As two of the team members carefully count 113 cycles of oscillation, the third member measures a duration of 241 s. What is the moment of inertia of the object with respect to its center of mass about an axis parallel to the pivot axis
Answer:
0.339 kgm²
Explanation:
We know the period of this pendulum, T = 2π√(I/mgh) where I = moment of inertia of the object about the pivot axis, m = mass of object = 2.15 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.163 m.
Since T = 2π√(I/mgh), making I subject of the formula, we have
I = mghT²/4π²
Now since it takes 241 s to complete 113 cycles, then it takes 241 s/113 cycles to complete one cycle.
So, T = 241 s/113 = 2.133 s
So, Substituting the values of the variables into I, we have
I = mghT²/4π²
I = 2.15 kg × 9.8 m/s² × 0.163 m × (2.133 s)²/4π²
I = 15.63/4π² kgm²
I = 0.396 kgm²
Now from the parallel axis theorem, I = I' + mh² where I' = moment of inertia of object with respect to its center of mass about an axis parallel to the pivot axis
I' = I - mh²
I' = 0.396 kgm² - 2.15 kg × (0.163 m)²
I' = 0.396 kgm² - 0.057 kgm²
I' = 0.339 kgm²
it is an organic compound and an essential micronutrient that the body needs in small amounts.
Answer:
Nutrients the body needs in relatively small amounts are called micronutrients. They include vitamins and minerals. Vitamins are organic compounds that are needed by the body to function properly. ... Vitamins and minerals do not provide energy, but they are still essential for good health
Explanation:
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?
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
What can you conclude about the electric potential and the field strength at the two noted points between the two electrodes? What can you conclude about the electric potential and the field strength at the two noted points between the two electrodes? The potential is greater at point B; the field strength is greater at point A. The potential is greater at point B; the field strength is greater at point B. The potential is greater at point A; the field strength is greater at point B. The potential is greater at point A; the field strength is greater at point A
Answer: The potential is greater at point B; the field strength is greater at point B.
Explanation:
The thing that can be concluded about the electric potential and the field strength at the two noted points between the two electrodes is that the potential is greater at point B; the field strength is greater at point B.
We should note that electrodes are used in the provision of current which typically takes place through a nonmetal objects.
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.
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.
Which hand is negatively charged?
Answer:
Left
Explanation:
Answer:
Your Left hand is negatively charged, and receives energy. It emits the energy that "allows things to happen''.
Explanation:
did some research
Suppose that you connect the terminals of two batteries of different emfs positive to positive and negative to negative (opposing each other) in a circuit. If you wanted to add in a capacitor to charge it from the batteries, would you be able to get more charge onto the capacitor or less charge, than if there was only one battery. (hint: start this problem by aligning the batteries positive to negative, and think of it from conservation of energy perspective).
Answer:
Answer is explained in the explanation section below.
Explanation:
This question is very basic and easy. The answer to this question is:
Answer: If both batteries are connected we would get less amount of charge as compared to connected a single battery.
Reasoning:
If both batteries are connected in a manner of positive terminal to positive terminal and negative terminal to negative terminal then a capacitor is added to charge it from the batteries then, total electromotive force (emf) would decrease.
As a result, the capacitor added would get less amount of charge stored. But capacitor added will get more amount of charge stored when a single battery is connected.
If 0.5 C charge passes through a wire in 10 seconds, what will be the value of the current flowing through the wire? *
20 mA
30 mA
50 mA
60 mA
Answer:
electric current passing through it will be 50mA
Explanation:
electric current = charge / time
I = Q / TI = 0.5 / 10 I = 0.05 amperecurrent = 0.05 A = 50mA
If 0.5C charge passes through a wire in 10 seconds, then 50mA current is flowing through the wire. Thus, the correct option is C.
What is Electric current?
Electric current is the flow of electricity in an electronic circuit. It is the amount of electricity flowing through a electronic circuit. It is generally measured in amperes (A). The larger the value in amperes, the more electricity is flowing in that circuit.
The formula for calculation of Electric current is:
I = Q/T
where, I = electric current,
Q = amount of charge,
T = time required
Therefore, the current flowing in the wire is:
I = 0.5C/ 10 seconds
I = 0.05 A or 50mA (1mA = 10⁻³A)
Therefore, the correct option is C.
Learn more about Electric current here:
https://brainly.com/question/2264542
#SPJ6
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
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
2. One tin for weight control is to:
Eat alone
Eat slowly
Answer:
Eat slowly
Explanation:
If you eat slower, you'll chew your food better, which leads to better digestion. Digestion actually starts in the mouth, so the more work you do up there, the less you'll have to do in your stomach. This can help lead to fewer digestive problems. Less stress.
g Suppose that you seal an ordinary 60W lightbulb and a suitable battery inside a transparent enclosure and suspend the system from a very sensitive balance. (a) Compute the change in the mass of the system if the lamp is on continuously for one year at full power. (b) What difference, if any, would it make if the inner surface of the container were a perfect reflector
Answer:
kekemeeimdeiddnekem
Explanation:
mdjdjdiddmjd jjeneeiej
Please help! Will mark brainliest.
Answer:
1122.8
Explanation:
12.73 kg x 9.8 m/s^2 x 9m
=1122.786
Rounded=1122.8
3.
What part of your eye is responsible for regulating the amount of light that enters your eye?
Answer:
Iris
Explanation:
The iris seems to be the illuminated portion of the eyes which really covers the pupil. It controls the amount of light reaching the eye. The lens is indeed a translucent layer of the retina that serves to concentrate light and objects on the lens.
Answer:
I hope this helps.
Explanation:
Which two statements below are central ideas in the article, "How Gross Is Your Bathroom"?
a. What you can't see might hurt you.
b. Different numbers of bacteria are hiding on various surfaces around your bathroom.
c. Most bacteria are harmless, and some are even good for you.
d. Your bathroom is filled with germs that you might not know anything about, including
viruses and bacteria.
What are some technological limitations that currently prevent humans from traveling to distant planets?
Answer:
Propulsion system, antigravitational tech
Explanation:
Fuel is extremely inefficient and expensive not to mention it weighs a lot. You really only need to reach escape velocity to leave earth. The rest is just a little amount of boosting to alter course and slow down for landing. I couldn't really think of much. Once we have an antigravitational system then you could say the whole rocket is holding you back because the design would be different. Nobody really knows how to defy gravity but that would be a technolgical limitation for sure.
The next four questions refer to the situation below.
A person is swimming in a river with a current that has speed vR with respect to the shore. The swimmer first swims downstream (i.e. in the direction of the current) at a constant speed, vS , with respect to the water. The swimmer travels a distance D in a time tOut . The swimmer then changes direction to swim upstream (i.e. against the direction of the current) at a constant speed, vS , with respect to the water and returns to her original starting point (located a distance D from her turn-around point) in a time tIn .
What is tOut in terms of vR, vS, and D, as needed?
Answer:
t_{out} = [tex]\frac{v_s - v_r}{v_s+v_r}[/tex] t_{in}, t_{out} = [tex]\frac{D}{v_s +v_r}[/tex]
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is
[tex]v_{sg 1} = v_{sr} + v_{rg}[/tex]
The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
[tex]v_{sg1}[/tex] = D / [tex]t_{out}[/tex]
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point
[tex]v_{sg 2} = v_{sr} - v_{rg}[/tex]
[tex]v_{sg 2}[/tex] = D / [tex]t_{in}[/tex]
D = v_{sg 2} t_{in}
with the distance is the same we can equalize
[tex]v_{sg1} t_{out} = v_{sg2} t_{in}[/tex]
t_{out} = t_{in}
t_{out} = [tex]\frac{v_s - v_r}{v_s+v_r}[/tex] t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D / [tex]v_{sg2}[/tex]
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} = [tex]\frac{D}{v_s +v_r}[/tex]
20. For each improvement in glider design, engineers follow
O A. the written instructions that are provided in the hang glider build kit.
O B. an iterative process of testing, modifying, retesting, and modifying again.
O C. a complicated process of checks and balances while obtaining financing.
O D. a mathematical process, rejecting designs that don't follow blueprint dimensions.
Turn In
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).
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]
Animals conduct_______.
A. cellular respiration
B. photosynthesis
C. both cellular respiration and photosynthesis
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
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:
The force between two charges when they are 2 cm apart is
0.036 N. If the sum of two charges is 10uC, what are the
charges? (1/4ttɛo=9x109 Nm-C-2).
Answer:
[tex]q_1=9.9998\mu C[/tex] and [tex]q_2=0.0002\mu C[/tex]
Or
[tex]q_1=0.00016\mu C[/tex] and [tex]q_2=9.99984\mu C[/tex]
Explanation:
We are given that
Force between two charges=0.036 N=[tex]36\times 10^{-3}N[/tex]
Distance between two charges, r=2cm=[tex]2\times 10^{-2}[/tex]m
1m=100cm
Sum of two charges=[tex]10\mu C[/tex]
Let one charge=[tex]q_1=q\mu C=q\times 10^{-6}C[/tex]
[tex]q_2=(10-q)\times 10^{-6} C[/tex]
We know that
Electric force between two charges
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Where [tex]k=\frac{1}{4\pi \epsilon_0}=9\times 10^{9}[/tex]
Using the formula
[tex]36\times 10^{-3}=9\times 10^{9}\times \frac{q\times 10^{-6}\times(10-q)\times 10^{-6}}{(2\times 10^{-2})^2}[/tex]
[tex]\frac{144\times 10^{-7}}{9\times 10^{9}\times 10^{-12}}=q(10-q)[/tex]
[tex]0.0016=10q-q^2[/tex]
[tex]q^2-10q+0.0016=0[/tex]
[tex]10000q^2-100000q+16=0[/tex]
[tex]q=\frac{100000\pm\sqrt{(100000)^2-4\times 10000\times 16}}{2\times 10000}[/tex]
Using the formula
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]q=9.999[/tex] and [tex]q=0.00016[/tex]
[tex]q_2=10-9.9998=0.0002[/tex]
[tex]q_2=10-0.00016=9.99984[/tex]
Hence, two charges are
[tex]q_1=9.9998\mu C[/tex] and [tex]q_2=0.0002\mu C[/tex]
Or
[tex]q_1=0.00016\mu C[/tex] and [tex]q_2=9.99984\mu C[/tex]
Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown to the right. The mass of the left block m1 = 1.4 kg and the mass of the right block m2 = 4.9 kg. The angle between the applied force and the horizontal is θ = 54°. The coefficient of kinetic friction between the blocks and the surface is μ = 0.38. Each block has an acceleration of a = 3.6 m/s2 to the right.
Answer:
Explanation:The Mass Of The Left Block M1 = 1.3 Kg And The Mass Of The Right Block M2 = 3.1 Kg. The Angle Between The String And The Horizontal Is ... (10%) Problem 8: Two blocks connected by a string are pulled across a horizontal surface by a ... m m, 50% Part (a) Write an equation for the magnitude of the force exerted by the ...
For the questions below, include units if applicable. If necessary, use a separate sheet of paper for 1, 6c and 7c. Tire pressure is in part a function of the temperature of the tire.
1. Based on everyday experience, state (in words) the relationship between tire pressure and temperature. Look at the data below and see if the numbers support your statement.
2. Prepare a hand-drawn plot of the two variables on the reverse side of this worksheet. Include a title, axis labels (with units), and a trendline. Estimate the tire pressure when the temperature is 18.6°C: Estimate the temperature of the air in the tire when the pressure is 37.0 psi: 3.
a. Prepare a plot using graphing software. Include a title, axis labels (with units), the equation of the best-fit Line and the R? value on the graph.
b. Re-write the equation of the best-fit line substituting "Temperature" for x and "Pressure" for y directly on the graph.
c. Attach the fully labeled graph to this worksheet.
4. What is the value of the slope for the relationship between temperature and pressure?
5. Determine the percent error using the definition of percent error: Use 0.145 psi/" for the "Actual" value of the slope. %error = Actual-Experimental % Error Actual
6. Based on your computer-generated graph,
a. visually estimate the tire pressure when the temperature is 18.6°C:
b. calculate the tire pressure at this temperature using the equation of the best fit line: the graph to ensure that this value is reasonable.
c. compare the calculated pressure to the two visually interpolated values (Steps 2 and 6a). Comment on any discrepancies.
7. Based on your computer-generated graph,
a. visually estimate the temperature of the air in the tire when the pressure is 37.0 psi:
b. calculate the temperature of the air in the tire at this pressure: Use the graph to ensure that this value is reasonable.
c. compare the calculated temperature to the two visually interpolated values (Steps 2 and 7a). Comment on any discrepancies.
Data:
Temperature (x) Tire Pressure, psi (y)
12.9°C 3.39 x 10
15.4C 34.25
-2.10 F 2.68 x 10
19.5 °C 3.50 x 10
29.6 'F 36.53
Answer:
All answer are explained below in the explanation section.
Explanation:
1. The pressure varies proportionally with the change in temperature. It can also be observed in our daily lives.
As for example, a pressure cooker uses the same principal to cook food faster. With the increasing temperature, the pressure inside the cooker increases.
Thus after a while, the excess pressure inside is released through the top nozzle. The data shown below supports that pressure and temperature varies linearly.
2. Hand drawn plot is attached in the attachment please refer to the attachment for the hand drawn plot.
Tire pressure at temperature 18.6 degree C is ~ 35 psi.
Temperature at air pressure of 37 psi is ~26.1 degree C
3. a.) Necessary values are included in the stat box. It is attached in the attachment please refer to the attachment.
3. b) The equation becomes: Pressure = 0.176 x temperature + 32.32
3. c) It is already done in part a of this question.
4. The value of the slope estimated from the linear fit is 0.176 +/- 0.094.
5. % Error = [tex]\frac{Actual - Experiment}{Actual} x 100[/tex]
Plugging in the values, we get:
Actual = 0.145, Experimental = 0.176. Thus, percentage error is given by:
% Error = 21.33%
6. a.) Visual estimation of tire pressure at t = 18.6 degree C is ~ 35 psi
6. b.) Estimation of pressure from the best fit line is given by 35.6 psi, which is consistent with the eye estimation value.
6. c.) The eye estimation and the estimation from the line fit are quite comparable. The discrepancy of +/-0.5 psi is within the percentage error calculated in 5.
7. a.) Visual estimation of temperature of the air for a tire-pressure of 37 psi is ~ 26 degree C.
7. b.) Estimation of temperature from best fit value of line is = 26.64 degree C
7 c) The values from eye estimation and evaluated from the fit are quite consistent within a random fluctuation of +/- 0.64 degree C.
Student pushes a 50 N block across the floor for a distance of 15 m how much work was done to move the block
Answer:
750 J
Explanation:
We have a student that pushes a 50N block across the floor for a distance of 15m. The question is asking how much work was done to move the block.
To solve this, we must know that we are looking for a certain thing called joules. And to get the answer, we must follow the formula of W = FS
F being the force and S being the distance.
W = FS
W = (50)(15)
W = 750
Therefore, 750 joules is our answer.
Two blocks collide on a frictionless surface, as shown above. They have a combined mass of 10 kg and a speed of 2.5 m/s. Before the collision, one of the blocks was at rest. This block had a mass of 8 kg. What was the speed of the second block?
Answer:
12.5 m/s
Explanation:
Excuse my scribbles!
I had to work backwards using the inelastic collision formula for this problem.
Formula: V=(M₁V₁+M₂V₂)/(M₁+M₂)V= Combined SpeedM₁= Block 1's MassV₁= Block 1's Velocity M₂= Block 2's MassV₂= Block 2's VelocityStep 1: Substitute in the values provided in the problem
Combined mass: 10kgCombined speed: 2.5m/sBlock 1's mass: 8kgBlock 1's speed: 02.5=(8*0)+(?*?)/(8+?)
Step 2: Subtract block 1's mass from the combined mass to determine block 2's mass
10-8=2 Block 2's mass is 2.
2.5=(8*0)+(2*x)/(8+2)
Now simplify.
2.5=(2*x)/(8+2)
2.5=2x/10
Step 3: Multiply both sides by the reciprocal
(5)2.5=2x/10(5)
12.5=x
Answer is checked in the attached images!
Help plz I’ll mark brainliest
Answer:
It's A
Explanation:
sound waves are longitudinal they need a medium to travel through
2.) The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his/her head, forcing them to move quickly away from the net. Suppose that you loft the ball with an initial speed of 15m/s at an angle of 50 degrees from the horizontal. At this moment your opponent is 10m from the ball. They begin to run away from you 0.3 seconds after the ball was launched hoping to reach the ball and hit it back to you at a height of 2.1m above where you hit it. What is the minimum average speed that your opponent must move so that he is in position to hit this ball
Answer:
The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s
Explanation:
The given parameters of the ball are;
The initial speed of the ball = 15 m/s
The direction in which the ball is launched = 50° above the horizontal
The location of the other tennis player when the ball is launched = 10 m from the ball
The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched
The height at which the ball is hit back = 2.1 m above the height from which the ball is launched
The vertical position, 'y', at time, 't', of a projectile motion is given as follows;
y = (u·sinθ)·t - 1/2·g·t²
When y = 2.1 m, we have;
2.1 = (15·sin(50°))·t - 1/2·9.8·t²
∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0
Solving with the aid of a graphing calculator function, we get;
t = 0.199776187257 s or t = 2.14525782198 s
Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds
The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;
x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m
The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m
Therefore, we have;
The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s
The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m
The minimum average speed the other player has to move with, [tex]v_s[/tex] = d/t₂
∴ [tex]v_s[/tex] = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s
The minimum average speed the opponent must move so that he is in position to hit the ball, [tex]v_s[/tex] ≈ 5.79 m/s.
Drag the tiles to the correct boxes to complete the pairs
Match the particles with their characteristics.
subatomic particles with a positive charge
subatomic particles with a negative charge
subatomic particles with no charge
made of atoms
neutrons
electrons
protons
malaria
Answer:
1. Protons.
2. Electrons.
3. Neutrons.
4. Molecules.
Explanation:
1. Protons: subatomic particles with a positive charge. They are bound together in the nucleus of an atom due to strong nuclear forces.
2. Electrons: subatomic particles with a negative charge. Electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.
3. Neutrons: subatomic particles with no charge. The negative charge of the electrons cancels the positive charge of the protons.
4. Molecules: they are made of atoms.
Generally, molecules attach on the inside of a mineral to give it shape. Therefore, the molecule of a mineral is a crystal three-dimensional regular structure (arrangement) of chemical particles that are bonded together and determines its shape.
Due to the fact that these molecules are structurally arranged or ordered and are repeated by different symmetrical and translational operations they determine the shape of minerals.
The engine in an imaginary sportThe engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hour (mph). At full power, the car can accelerate from zero to 30.0 mph in time 1.00 s .s car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hour (mph).
Required:
a. At full power, how long would it take for the car to accelerate from 0 to 58.0mph?
b. A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 29.0mph in time 1.50s , how long would it take to go from zero to 58.0mph ?
Answer:
a. 1.93 s b. 3 s
Explanation:
a. At full power, how long would it take for the car to accelerate from 0 to 58.0mph?
Since the car accelerates from 0 to 30 mph in 1.00 s, we find its acceleration, a from a = (v - u)/t where u = 0 m/s, v = 30 mph and t = 1.00s = 1/3600 h
So, substituting the values of the variables into the equation, we have
a = (v - u)/t
a = (30 mph - 0 mph)/ 1/3600 h
a = 30 mph × 3600 /h
a = 108000 mph²
So, we find the time it takes the car to accelerate to 58 mph from 0 mph from
t = (v' - u')/a where u = 0 mph, v = 58 mph and a = 108000 mph²
So, substituting the value of the variables into the equation, we have
t = (v' - u')/a
t = (58 mph - 0 mph)/108000 mph²
t = 58 mph/108000 mph²
t = 5.37 × 10⁻⁴ h
t = 5.37 × 10⁻⁴ × 3600 s
t = 1.93 s
b. A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 29.0mph in time 1.50s , how long would it take to go from zero to 58.0mph ?
Since the car accelerates from 0 to 29 mph in 1.50 s, we find its acceleration, a from a = (v - u)/t where u = 0 m/s, v = 29 mph and t = 1.05s = 1/3600 h
So, substituting the values of the variables into the equation, we have
a = (v - u)/t
a = (29 mph - 0 mph)/ 1.5/3600 h
a = 29 mph × 3600/1.5 /h
a = 104400/1.5 mph²
a = 69600 mph²
So, we find the time it takes the car to accelerate to 58 mph from 0 mph from
t = (v' - u')/a where u = 0 mph, v = 58 mph and a = 69600 mph²
So, substituting the value of the variables into the equation, we have
t = (v' - u')/a
t = (58 mph - 0 mph)/69600 mph²
t = 58 mph/69600 mph²
t = 8.33 × 10⁻⁴ h
t = 8.33 × 10⁻⁴ × 3600 s
t = 3 s