an object of 5kg is attached to a rope of length 4m is Rotating horizontally at 8m/s horizontally 20m above the ground if the rope is suddenly cut what is the horizontal distance travelled by the object? Please guys help

Answers

Answer 1

Answer:

16 meters

Explanation:

When the rope is suddenly cut the object moving tangent at the circle. In that moment the gravity act in the object making it falls.

First we need to find how much time de object take to reach at the ground.

VERTICALLY EQUATION:[tex]h(t)=h-v*t-\frac{g}{2} t^{2} \\[/tex]

g=acceleration of gravity=10m/s²

v= vertical velocity =0m/s

h=vertical altitude =20m

We will find t such that h(t)=0

[tex]0=20-5t^{2} \\\\5t^{2} =20\\\\t^{2} =4\\\\t=2s[/tex]

HORIZONTALLY EQUATION:*horizontally we do not have acceleration[tex]D(t)=v*t[/tex]

v=horizontal velocity

D(t=2) is the horizontal distancetravelled by the object:

[tex]D(2)=8*2\\\\D(2)=16m[/tex]

An Object Of 5kg Is Attached To A Rope Of Length 4m Is Rotating Horizontally At 8m/s Horizontally 20m

Related Questions

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

If a total 50 J of work are done on an object, it's energy...

Answers

Answer:

0.0119502868 kilocalorie

Explanation:

Answer:

increases by 50

Explanation:

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).

Answers

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.

125 cm of gas are collected at 15 °C and
755 mmHg pressure. Calculate the volume of
the gas at s.t.p.​

Answers

The new volume will be ≈26mL
rounded to two significant figures.
Explanation:
This question involves the combined gas law. The equation to use is:
When working with gas laws, the temperature is always in Kelvins. To get Kelvins from a Celsius temperature, add 273.15 to the Celsius temperature.

the ____ is a particle with one unit of positive change

a. proton
b. positron
c. electron
d. nucleus

Answers

Answer:

a proton because it has a positive charge

Answer:

The answer is

B)  

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.

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.

The force of gravity on a person or object on the surface of a planet is called
A. gravity
ОВ.
B. free fall
OC
c. terminal velocity
D. weight

Answers

Answer:

D. Weight

Explanation:

Hope that helps:)

D. Weight is the correct answer.
HOPE IT HELPS U!!!!!!

Please help I will fail

Answers

Answer:

6. A

7. C

8. B

9. The average speed is the distance (a scalar quantity) per time ratio. Speed is ignorant of direction. The average velocity is the displacement or position change (a vector quantity) per time ratio.

Which one of Newton’s Laws best explains a bottle flip?

Answers

the first one gravity

An ideal gas in a 50.0 L tank has a
pressure of 2.45 atm at 22.5°C.
How many moles of gas are in
the tank?

Answers

Answer:

5.05225 moles

Explanation:

The computation of the number of moles of gas in the tank is shown below:

Given that

Volume = V = 50 L = 50.0 × 10^-3m^3

Pressure = P = 2.45 atm = 2.45 × 101325

Temperature = T = 22.5°C  = (22.5 + 273)k = 295.5 K

As we know thta the value of gas constant R is 8.314 J/mol.K

Now

PV = nRT

n = PV ÷ RT

= ((2.45 × 101325) (50.0 × 10^-3)) ÷ ((8.314) (295.5))

= 5.05225 moles

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

A car hits a tree with a force of 45 N, the mass of the tree is 65g. What is the resulting acceleration?
a. 0.69 m/s2
b. 692 m/s2
c. 2,925 m/s2
d. 2.93 m/s2

Answers

Answer:

i think 692m/s2 is the correct answer

PLEASE HELP! WILL GIVE BRANILIEST TO FIRST REAL ANSWER If one marble is rolling three times as fast as a second marble of the same mass, the kinetic energy of the first marble is how many times larger when compared to the kinetic energy of the second marble?
a) 4
b) 9
c) 6
d) 3
(i already know its not 3)

Answers

Answer:

9

Explanation:

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

Answers

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.

What is the correct definition of amplitude

Answers

Answer:

In my textbook's words-

Amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. It is equal to one-half the length of the vibration path.

Explanation:

The correct definition of amplitude is that it is a maximum displacement

that occurs on a vibrating body from one point to the other.

The initial point of the wave is regarded as its equilibrium position which is

equal to one-half the length of the vibration path.

Amplitude helps to calculate the peak value of different types of waves

such as water waves and in electrical appliances so as to know the peak

current suitable for it.

Read more on https://brainly.com/question/21632362

Potential energy is energy due to motion.
True or False?

Answers

Answer:

true

Explanation:

Answer:

true

Explanation:

please give brainlest need 1

An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.70 g/cm3. Assume each aluminum atom supplies one conduction electron per atom. Find the drift speed of the electrons in the wire.

Answers

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

[tex] v = \frac{I}{nqA} [/tex]

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

[tex] n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3} [/tex]                  

Now, we can find the drift speed:

[tex]v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s[/tex]              

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!      

A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near side a distance d = 0.45 m from the origin. A charge Q2 = 10.4 μC is uniformly spread over the length of the rod.Part (a) Consider a thin slice of the rod, of thickness dx, located a distance x away from the origin. What is the direction of the force on the charge located at the origin due to the charge on this thin slice of the rod? Part (b) Write an expression for the magnitude of the force on the point charge, |dF|, due to the thin slice of the rod. Give your answer in terms of the variables Q1, Q2, L, x, dx, and the Coulomb constant, k. Part (c) Integrate the force from each slice over the length of the rod, and write an expression for the magnitude of the electric force on the charge at the origin. Part (d) Calculate the magnitude of the force |F|, in newtons, that the rod exerts on the point charge at the origin.

Answers

Answer:

a) attractiva, b) dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex], c)  F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex], d) F = -1.09 N

Explanation:

a) q1 is negative and the charge of the bar is positive therefore the force is attractive

b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x

           dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex]

where k is a constant, Q₁ the charge at the origin, x the distance

c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L

         ∫ dF = [tex]k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2[/tex]

as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density

          λ = dQ₂ / dx

          DQ₂ = λ dx

we substitute

         F = [tex]k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}[/tex]

         F = k Q1 λ ([tex]-\frac{1}{x}[/tex])  

we evaluate the integral

        F = k Q₁ λ [tex](- \frac{1}{d+L} + \frac{1}{d} )[/tex]

        F = k Q₁ λ  [tex]( \frac{L}{d \ (d+L)})[/tex]

we change the linear density by its value

      λ = Q2 / L

       F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex]

d) we calculate the magnitude of F

       F =9 10⁹ (-4.2 10⁻⁶)   [tex]\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}[/tex]

       F = -1.09 N

the sign indicates that the force is attractive

Answer:

a)Toward the rod

b)|dF| = k|Q1|Q2(dx/L)/x^2

c)|F| = k|Q1|Q2/(d(d+L))

d)Plug in for answer c and solve

Explanation:

A)

Q1 is negative and Q2 is positive so it is an attractive force to  where the rod is located.

B)

The formula for Force due to electric charges is F=kQ1Q2/r^2

In this case, Q2 is distrusted through the length of the rod as opposed to a single point charge. As such Q2 is actually Q2*dx/L as dx is a small portion of the full length, L.

The radius between Q1 and Q2 depends on the section of the rod taken so r will be the variable x distance from Q1.

The force is only from a small portion of the rod so more accurately, we are finding |dF| as opposed to the full force, F, caused by the whole rod.

The final formula is |dF| = k|Q1|Q2(dx/L)/x^2

C)

Integrating with respect to the only changing variable, x, which spans the length of the rod, from radius = d to d+L we get this:

F = integral from d to d+L of k|Q1|Q2(dx/L)/x^2

factor out constants

F = kQ1Q2/L * integral d to d+L(1/x^2)dx

F = kQ1Q2/L * (-1/x)| from d to d+L

F = kQ1Q2/L * (-1/d+L - -1/d)

F = kQ1Q2/L * (-d/(d(d+L)) + (d+L)/(d(d+L))

F = kQ1Q2/L * (L)/(d(d+L))

F = kQ1Q2/(d(d+L))

D)

Plug in the given values into c and you have your answer.

When rubbing a balloon against your head, you notice the balloon pulling your hair away from your head. What best explains why the balloon and your hair are attracted to each other?

They become oppositely charged, which causes them to be attracted to each other.

They become similarly charged, which causes them to be attracted to each other.

They stick together because of the friction between the two objects.

They are made of different materials, which is why they attract each other.

Answers

Answer:

They become oppositely charged, which causes them to be attracted to each other.

Answer: number A

Explanation:

If a positive charge and a negative charge interact, their forces act in the same direction, from the positive to the negative charge. As a result opposite charges attract each other: The electric field and resulting forces produced by two electrical charges of opposite polarity. Have a nice day <3


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

3 because
12=2’2x3
21=3x7
30=2x3x5

Two people each do 100 joules of work by pushing a crate to the right. During this process, 50 joules of heat is generated from the
friction between the floor and the crate. How much energy is gained by the crate during this process?

Answers

Because of friction, net work on the crate is less. ∆KE = Net work = net force x distance = (100 N – 70 N)(10 m) = 300 J.

what is permittivity​

Answers

Answer:

Permittivity, also called electric permittivity, is a constant of proportionality that exists between electric displacement and electric field intensity.

Need help on another homework question

Rick places the blue lens of one pair of 3D glasses over the red lens of another pair. He then looks through both lenses at the same time. What color will he see?

A. blue
B. black
C. red
D. white

Answers

Black
Explanation : I've tried

Answer:

a

Explanation:

Which hand is negatively charged?

Answers

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

It has been suggested that rotating cylinders several miles in length and several miles in diameter be placed in space and used as colonies. Inhabitants of the space colonies would live on the inside surface of the cylinder. Inertial effects would resemble gravity's influence and keep them 'plastered to the surface.' Suppose that you are an inhabitant of a space colony which is 1070 miles in length and 4.86 miles in diameter. How many revolutions per hour must the cylinder have in order for the occupants to experience a centripetal acceleration equal to the acceleration of gravity

Answers

Answer:

the required revolution per hour is 28.6849

Explanation:

Given the data in the question;

we know that the expression for the linear acceleration in terms of angular velocity is;

[tex]a_{c}[/tex] = rω²

ω² = [tex]a_{c}[/tex] / r

ω = √( [tex]a_{c}[/tex] / r )

where r is the radius of the cylinder

ω is the angular velocity

given that; the centripetal acceleration equal to the acceleration of gravity a [tex]a_{c}[/tex]  = g = 9.8 m/s²

so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

so we substitute

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²  

ω = 0.0500647 ≈ 0.05 rad/s  

we know that; 1 rad/s = 9.5493 revolution per minute

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM  

1 rpm = 60 rph  

so  

ω = 0.478082 × 60

ω = 28.6849  revolutions per hour  

Therefore, the required revolution per hour is 28.6849

The required revolution per hour is 28.6849

Calculation of revolution per hour:

The expression for the linear acceleration with respect to the angular velocity is

= rω²

So,

ω² =  / r

ω = √(  / r )

Here r represent the radius of the cylinder

ω represent the angular velocity

Now

diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

And,

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

So,

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²  

ω = 0.0500647 ≈ 0.05 rad/s  

Now

1 rad/s = 9.5493 revolution per minute

So,

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM  

Now

1 rpm = 60 rph  

so  

ω = 0.478082 × 60

ω = 28.6849  revolutions per hour  

Learn more about the cylinder here: https://brainly.com/question/17262276

Which statement describes what will most likely occur when warm air cools and the temperature drops to the
point?
A.air will contain more water vapor. B. Dew will form on leaves C.clouds will disappear. D. Water vapor in the air will evaporate

Answers

Answer:

The person who did the comment is correct, it B - Dew will form on leaves

Explanation:

Answer:

B

Explanation:

Valeriie.07 tap in g

Which example is correctly matched with its type of friction?
A. Pushing a car that isn't moving is an example of slkiding friction.
B. A plane flying through the air is an example of static friction.
OC.
A skateboard wheels on cement is an example of rolling friction.
OD.
A sled sliding down a grassy hill is an example of fluid friction.

Answers

Answer:

A. pushing a car that isn't moving

A block of mass m = 150 kg rests against a spring with a spring constant of k = 880 N/m on an inclined plane which makes an angle of θ degrees with the horizontal. Assume the spring has been compressed a distance d from its neutral position.

Required:
a. Set your coordinates to have the x-axis along the surface of the plane, with up the plane as positive, and the y-axis normal to the plane, with out of the plane as positive.
b. Denoting the coefficient of static friction by μs, write an expression for the sum of the forces in the x-direction just before the block begins to slide up the inclined plane. Use defined quantities and g in your expression ΣFx = 25%
c. Assuming the plane is frictionless, what will the angle of the plane be, in degrees, if the spring is compressed by gravity a distance 0.1 m?
d. Assuming θ = 45 degrees and the surface is frictionless, how far will the spring be compressed, d in meters?

Answers

Answer:

b)  k Δx - W cos θ - μ mg cos θ = m a ,  c)  θ = 86.6º, d)  Δx = 1.18 m

Explanation:

a) In the attachment we can see a diagram of the forces in this problem and the coordinate axes for its decomposition.

F is the force applied by the spring, while it is compressed, this force disappears when the block leaves the spring

b) Let's apply Newton's second law for when the spring is compressed

let's use trigonometry to break down the weight

            sin θ = Wₓ / W

            cos θ = W_y / W

             Wₓ = W sin θ

             W_y = W cos θ

Y axis  

               N - W_y = 0

               N = W_y

               N = W cos θ

X axis

           F -Wₓ -fr = ma

the force applied by the spring is given by hooke's law

           F = k Δx

friction force has the expression

           fr = μ N

           fr = μ W cos θ

we substitute

            k Δx - W cos θ - μ mg cos θ = m a           ( 1)

c) If the plane has no friction, what is the angle so that Δx = 0.1m

             

We write the equation 1, with fr = 0 and since the system is still a = 0

            k Δx - W cos θ -0 = 0

            cos θ = [tex]\frac{k \Delta x}{ m g}[/tex]

            cos θ = [tex]\frac{880 \ 0.1}{ 150 \ 9.8}[/tex]

            cos θ = 0.0598

            θ = cos⁻¹ 0.0598

            θ = 86.6º

d) In this part they give the angle θ = 45º and there is no friction, they ask the compression

the acceleration is zero, we substitute in 1

            k Δx - W cos θ - 0 = 0

            Δx = [tex]\frac{mg \ cos \ \theta}{k}[/tex]

            Δx = [tex]\frac{ 150 \ 9.8 \ cos45}{880}[/tex]

            Δx = 1.18 m

help me pls it’s a usa test prep pretty easy

Answers

Answer:

Im 99.99999% sure its c

Explanation:

i cant see the pictures too well

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