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Over [174]
2 years ago
6

A body weighs 1000 lbf when exposed to a standard earth gravity of g=32.174 ft/s^2. a. What is its mass in kg? b. What is its we

ight in N if on moon (gmoon = 1.62 m/s^2)? c. How fast will it accelerate if under a force of 400 lbf on moon and on the earth in m/s.
Physics
1 answer:
allsm [11]2 years ago
4 0

Answer

given,

weight = 1000 lbf

g = 32.174 ft/s²

mass =\dfrac{1000}{32.174}

m = 31 slug

1 slug = 14.593 kg

m = 31 x 14.593

m = 452.383 Kg

b) weight on moon

  W = m g

   W = 452.383 x 1.62

   W= 732.86 N

c) we know,

   F = m  a

  400 lbf = 31 slugs x a

  a = 12.90 ft/s²

1 ft = 0.304 m

  a = 3.92 m/s²

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Temperature and kinetic energy are ___________ proportional. adirectly directly indirectly 2. Heat is a measure of _____________
Usimov [2.4K]

Explanation:

It is known that relation between kinetic energy and temperature is as follows.

        K.E \propto \frac{3}{2}kT

Hence, kinetic energy is directly proportional to temperature.

Thermal energy is defined as the energy present within the molecules of an object due to the motion of particles. Basically, thermal energy is internal energy of an object.

Thus, we can conclude that:

  • Temperature and kinetic energy are directly proportional.
  • Heat is a measure of thermal energy.
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7 0
2 years ago
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Why is the entropy change negative for ring closures?
Rom4ik [11]

Answer:ring closure result in fewer molecules

Explanation:

entropy is a measure of the amount of disorderliness in a system.

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3 0
2 years ago
The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
2 years ago
A carbon-dioxide laser emits infrared light with a wavelength of 10.6 μm. What is the length of a tube that will oscillate in th
alex41 [277]

Answer:

The length of a tube and number of rounds are 0.848 m and 1.77\times10^{8}\ trip\ per\ second.

Explanation:

Given that,

Wavelength \lambda= 10.6\mu m

m = 160000

We need to calculate the length

Using formula of wavelength

Laser tube behave like closed pipe

m\dfrac{\lambda}{2}=L

L=160000\times\dfrac{10.6\times10^{-6}}{2}

L=0.848\ m

Distance traveled by pulse of light in one back and fourth trip

d=2L

d=2\times0.848

d=1.696\ m

We need to calculate the time

Using formula for time

t = \dfrac{d}{c}

t=\dfrac{1.696}{3\times10^{8}}

t=5.653\times10^{-9}\ s

We need to calculate the number of round

Using formula of number of round

N=\dfrac{1}{t}

N= \dfrac{1}{5.653\times10^{-9}}

N=1.77\times10^{8}\ trip\ per\ second

Hence, The length of a tube and number of rounds are 0.848 m and 1.77\times10^{8}\ trip\ per\ second.

7 0
2 years ago
During the 440, a runner changes his speed as he comes out of the curve onto the home stretch from 18 ft/sec to 38 ft/sec over a
Sloan [31]

Answer:

6.67ft/s^2

Explanation:

We are given that

Initial velocity=u=18ft/s

Final velocity,v=38ft/s

Time=t=3 s

We have to find the average acceleration over that 3 s period.

We know that

Average acceleration,a=\frac{v-u}{t}{t}

Using the formula

Average acceleration,a=\frac{38-18}{3}ft/s^2

Average acceleration,a=\frac{20}{3}ft/s^2

Average acceleration,a=6.67ft/s^2

Hence, the average acceleration=6.67ft/s^2

5 0
2 years ago
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