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Darina [25.2K]
1 year ago
9

It has been suggested that rotating cylinders about 14.5 mi long and 4.78 mi in diameter be placed in space and used as colonies

. What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth?
Physics
1 answer:
taurus [48]1 year ago
7 0

Angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth is 0.0466 rad/s.

Length of the cylinder, L = 9 mi

= (14.5mi)(1609.344 m / 1 mi)

= 23,335.48 m

Diameter of the cylinder, D = 4.78 mi

= (4.78 mi)(1609.344 m / 1 mi)

= 7692.645 m

Radius of the cylinder, r = D / 2

= ( 7692.645 m) / 2

=  3846.32 m

Centripetal acceleration is given by, ac = v^2 / r

We have the relation between linear velocity (v) and the angular velocity(ω) as

 v = r ω

Then, ac = v2 / r

= (rω)2 / r

ac = ω^2 r

If the centripetal acceleration is equals the free fall acceleration of earth, then

ω^2 r = g

ω=0.0466 rad/s

  • Angular acceleration is the term used to describe the rate of change in angular velocity. If the angular velocity is constant, the angular acceleration is constant.

To know more about  angular velocity visit : brainly.com/question/12446100

#SPJ1

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An electric heater draws a steady 15.0A on a 120-V
morpeh [17]

Answer:

1. 1800 W

2. $ 17.3

Explanation:

From the question given above, the following data were obtained:

Current (I) = 15 A

Voltage (V) = 120 V

Time (t) = 20 h per day

Duration = 31 days

Cost = 15.5 cents per kWh

1. Determination of the power.

Current (I) = 15 A

Voltage (V) = 120 V

Power (P) =?

P = IV

P = 15 × 120

P = 1800 W

Thus, 1800 W of power is required.

2. Determination of the cost per month (31 days).

We'll begin by converting 1800 W to KW.

1000 W = 1 KW

Therefore,

1800 W = 1800 W × 1 KW / 1000 W

1800 W = 1.8 KW

Next, we shall determine the energy consumption for 31 days. This can be obtained as follow:

Power (P) = 1.8 KW

Time (t) = 2 h per day

Time (t) for 31 days = 2 × 31 = 62 h

Energy (E) =?

E = Pt

E = 1.8 × 62

E = 111.6 KWh

Finally, we shall determine the cost of consumption. This can be obtained as follow:

1 KWh = 15.5 cents

Therefore,

111.6 KWh = 111.6 KWh × 15.5 cents / 1 KWh

111.6 KWh = 1729.8 cents

Converting 1729.8 cents to dollar, we have:

100 cents = $ 1

Therefore,

1729.8 cents = 1729.8 cents × $ 1 / 100 cents

1729.8 cents = $ 17.3

Thus, it will cost $ 17.3 per month to run the electric heater.

4 0
3 years ago
A charge of -8.00 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.05 cm.
gavmur [86]

Answer:

(a) E = -1.02 \times 10^5~N/C

(b) E = -9.7 \times 10^4~N/C

Explanation:

(a) The electric field for a point charge is given by the following formula:

\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\^r

Since this formula is valid for point charges, we have to choose an infinitesimal area, da, from the disk. Then we will calculate the E-field (dE) created by this small area using the above formula, then we will integrate over the entire disk to find the E-field created by the disk.

dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{(\sqrt{z^2 + r^2})^2}

Here, z = 0.025 m. And r is the distance of the infinitesimal area from the axis. dQ is the charge of the small area, and should be written in terms of the given variables.

In cylindrical coordinates, da = r dr dθ. So,

\frac{Q}{\pi R^2} = \frac{dQ}{da}\\\frac{Q}{\pi R^2} = \frac{dQ}{rdrd\theta}\\dQ = \frac{Qrdrd\theta}{\pi R^2}

Hence, dE is now:

dE = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}

The surface integral over the disk can now be taken, but there is one more thing to be considered. This dE is a vector quantity, and it needs to be separated its components.

It has two components, one in the vertical direction and another in the horizontal direction. By symmetry, the horizontal components cancel out each other in the end (since it is a disk, each horizontal vector has an equal but opposite counterpart), so only the vertical component should be considered.

Let us denote the angle between dE and the horizontal axis as α. This angle can be found by the geometry of the triangle formed by dE, vertical axis of the disk, and horizontal plane. So,

\sin(\alpha) = \frac{z}{\sqrt{z^2 + r^2}}

Therefore, vertical component of dE now becomes

dE_z = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}\frac{z}{\sqrt{z^2+r^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}\\E_z =  \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 \int\limits^R_0 {\frac{rdrd\theta}{(z^2+r^2)^{3/2}}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2} 2\pi(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}})

Substituting the parameters, z = 0.025 m, Q = - 8 x 10^(-9) C, and R = 0.0105 m, yields the final result:

E_z = \frac{1}{2\epsilon_0}\frac{Qz}{\pi R^2}(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}}) = -1.02 \times 10^5~N/C

(b) We will have a similar approach, but a simpler integral.

dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{z^2 + R^2}\\\frac{Q}{2\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{2\pi}\\dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\frac{z}{\sqrt{z^2+R^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qzd\theta}{2\pi(z^2 + R^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta  = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}2\pi

E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{(z^2 + R^2)^{3/2}} = -9.07\times 10^4~N/C

Note that, in this case the source object is a one dimensional hoop rather than a two dimensional disk.

3 0
3 years ago
How many times can a three-dimensional object that has a radius of 1,000 units fit something with a radius of 10 units inside of
Mekhanik [1.2K]

Answer:

# _units = 1000

Explanation:

This exercise we can use a direct proportion rule.

If a volume of radius r = 1 is one unit, how many units can fit in a volume of radius 10?

    # _units = V₁₀ / V₁

The volume of a body of radius 1 is

       V₁ = 4/3 π r₁³

        V₁ = 4/3π

the volume of a body of radius r = 10

        V₁₀ = 4/3 π r₂³

        V10 = 4/3 π 10³

     

the number of times this content is

         #_units = 4/3 π 1000 / (4/3 π 1)

        # _units = 1000

8 0
3 years ago
A 9.0 kg test rocket is fired vertically from Cape Canaveral. Its fuel gives it a kinetic energy of 1905 J by the time the rocke
mrs_skeptik [129]

Answer:

21.6m

Explanation:

Since the rocket engine burns all the fuel hence the kinetic energy will be converted to potential energy

Potential Energy = mass × acceleration due to gravity × height

Given

PE = 1905J

Mass = 9.0kg

Acceleration due to gravity =9.8m/s²

Required

Height h

Substitute into the formula

1905 = 9(9.8)h

1905 = 88.2h

h =1905/88.3

h = 21.6m

Hence the required height is 21.6m

5 0
3 years ago
Which type of energy is thermal energy a form of?
KATRIN_1 [288]
It's thermal Kinetic energy.
3 0
4 years ago
Read 2 more answers
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