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3 years ago

A millimeter is _______ meter(s). A. 1,000 B. 100 C. 1/100th D. 1/1,000th

Physics

1 answer:

Zielflug [23.3K]3 years ago

7 0

1 millimetre is 0.001 metre.
1 metre is 1,000 millimetres.

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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

taurus [48]

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

7 0

10 months ago

A cannon of mass 6.43 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 7

Nata [24]

Answer:

The velocity of the shell when the cannon is unbolted is 500.14 m/s

Explanation:

Given;

mass of cannon, m₁ = 6430 kg

mass of shell, m₂ = 73.8-kg

initial velocity of the shell, u₂ = 503 m/s

Initial kinetic energy of the shell; when the cannon is rigidly bolted to the earth.

K.E = ¹/₂mv²

K.E = ¹/₂ (73.8)(503)²

K.E = 9336032.1 J

When the cannon is unbolted from the earth, we apply the principle of conservation of linear momentum and kinetic energy

change in initial momentum = change in momentum after

0 = m₁u₁ - m₂u₂

m₁v₁ = m₂v₂

where;

v₁ is the final velocity of cannon

v₂ is the final velocity of shell

$v_1 = \frac{m_2v_2}{m_1}$

Apply the principle of conservation kinetic energy

$K = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_1(\frac{m_2v_2}{m_1})^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_2v_2^2(\frac{m_2}{m_1}) + \frac{1}{2}m_2v_2^2 \\\\K = \frac{1}{2}m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\2K = m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\v_2^2 = \frac{2K}{M_2(\frac{m_2}{m_1} + 1)} \\\\v_2^2 = \frac{2*9336032.1}{73.8(\frac{73.8}{6430} + 1)}\\\\$

$v_2^2 = 250138.173\\\\v_2 = \sqrt{250138.173} \\\\v_2 = 500.14 \ m/s$

Therefore, the velocity of the shell when the cannon is unbolted is 500.14 m/s

3 0

3 years ago

A robotic arm lifts a stack of cafeteria trays weighing 62 newtons (N). It moves the trays 2 meters (m). How much work has the r

tigry1 [53]

work done = force * distance

work = 62 * 2 = 124 j

7 0

3 years ago

A fatigue test was conducted in which the mean stress was 50 MPa (7250 psi) and the stress amplitude was 225 MPa (32,625 psi).

Tems11 [23]

Answer:

275 MPa, -175 MPa

-0.63636

450 MPa

Explanation:

$\sigma_{max}$ = Maximum stress

$\sigma_{min}$ = Minimum stress

$\sigma_m$ = Mean stress = 50 MPa

$\sigma_a$ = Stress amplitude = 225 MPa

Mean stress is given by

$\sigma_m=\frac{\sigma_{max}+\sigma_{min}}{2}\\\Rightarrow \sigma_{max}+\sigma_{min}=2\sigma_m\\\Rightarrow \sigma_{max}+\sigma_{min}=2\times 50\\\Rightarrow \sigma_{max}+\sigma_{min}=100\ MPa\\\Rightarrow \sigma_{max}=100-\sigma_{min}$

Stress amplitude is given by

$\sigma_a=\frac{\sigma_{max}-\sigma_{min}}{2}\\\Rightarrow \sigma_{max}-\sigma_{min}=2\sigma_a\\\Rightarrow \sigma_{max}-\sigma_{min}=2\times 225\\\Rightarrow \sigma_{max}-\sigma_{min}=450\ MPa\\\Rightarrow 100-\sigma_{min}-\sigma_{min}=450\\\Rightarrow -2\sigma_{min}=350\\\Rightarrow \sigma_{min}=-175\ MPa$

$\sigma_{max}=100-\sigma_{min}\\\Rightarrow \sigma_{max}=100-(-175)\\\Rightarrow \sigma_{max}=275\ MPa$

Maximum stress level is 275 MPa

Minimum stress level is -175 MPa

Stress ratio is given by

$R=\frac{\sigma_{min}}{\sigma_{max}}\\\Rightarrow R=\frac{-175}{275}\\\Rightarrow R=-0.63636$

The stress ratio is -0.63636

Stress range is given by

$\sigma_{max}-\sigma_{min}=450\ MPa$

Magnitude of the stress range is 450 MPa

8 0

3 years ago

Juliette is driving her car when she sees a cat run across the road. If she is able to stop the car over a distance of 0.025km i

jek_recluse [69]

$velocity=\frac{distance}{time}=\frac{0,025km}{2,5s}=\frac{25m}{2,5s}=100\frac{m}{s}\\\\
acceleration=\frac{velocity}{time}=\frac{-10\frac{m}{s}}{2,5s}=-4\frac{m}{s^2}\\\\Solution\ is\ D.$

6 0

3 years ago