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3 years ago

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Engineering

1 answer:

creativ13 [48]3 years ago

4 0

Answer:

its d

Explanation:

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The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".

If we analyze the staritning point we see that the initial velocity can be founded like this:

$v_o =\omega r_{OIC}=\omega (0.15m)$

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

$H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af$

$0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)$

And if we integrate the left part and we simplify the right part we have

$1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega$

And if we solve for $\omega$ we got:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

8 0

3 years ago

5. A biscuit joint does not require glue.<br> True or False

OLga [1]

Answer:

false

Explanation:

mark me brainliest

5 0

3 years ago

Read 2 more answers

Show that y = '(t - s)f(s)ds is a solution to my" + ky = f(t). Use g' (0) = 1/m and mg" + kg = 0. 6.1) Derive y' 6.2) Using g(0)

Gre4nikov [31]

Answer:

Explanation:

Given that:

$y = \int^t_og'(t-s) f(s) ds \ \text{is solution to } \ my"ky= f(t)$

where;

$g'(0) = \dfrac{1}{m}$ and $mg"+kg = 0$

$\text{Using Leibniz Formula to prove the above equation:}$

$\dfrac{d}{dt} \int ^{b(t)}_{a(t)} \ f (t,s) \ ds = f(t,b(t) ) * \dfrac{d}{dt}b(t) - f(t,a(t)) *\dfrac{d}{dt}a(t) + \int ^{b(t)}_{a(t)}\dfrac{\partial}{\partial t} f(t,s) \ dt$

So, $y = \int ^t_0 g' (t-s) f(s) \ ds$

$\text{By differentiation with respect to t;}$

$y' = g'(o) f(t) \dfrac{d}{dt}t- 0 + \int^{t}_{0}g'' (t-s) f(s) ds \\ \\ y' = \dfrac{1}{m}f(t) + \int ^t_0 g'' (ts) f(s) \ ds$

$y'' = \dfrac{1}{m} f'(t) + g"(0) f(t) + \int^t_o g"'(t-s) f(s)ds --- (1)$

$Since \ \ mg" (t) +kg (t) = 0 \\ \\ \implies g" (t) = -\dfrac{k}{m} g(t) --- (111) \\ \\ put \ t \ =0 \ we \ get;\\g" (0) = - \dfrac{k}{m } g(0) \\ \\ g"(0) = 0 \ \ \ \ ( because \ g(0) =0) \\ \\$

$Now \ differentiating \ equation (111) \ with \ respect \ to \ t \\ \\ g"'(t) = -\dfrac{k}{m}g(t) \\ \\ replacing \ it \ into \ equation \ (1) \\ \\ y" = \dfrac{1}{m}f' (t) + 0 + \int ^t_o \dfrac{-k}{m}g' (t-s) f(s) \ ds \\ \\ y" = \dfrac{1}{m}f' (t) - \dfrac{k}{m} \int ^t_o g' (t-s) \ f(s) \ ds \\ \\ y" = \dfrac{1}{m}f'(t) - \dfrac{k}{m}y \\ \\ my" = f'(t)-ky \\ \\ \implies \mathbf{ my" +ky = f'(t)}$

7 0

3 years ago

Comparison of density values determines whether an item will float or sink in water. For each of the values below, determine the

geniusboy [140]

Answer:

a) the object floats

b) the object floats

c) the object sinks

Explanation:

when an object is less dense than in the fluid in which it is immersed, it will float due to its weight and volume characteristics, so to solve this problem we must find the mass and volume of each object in order to calculate the density and compare it with that of water

volumen for a cube

V=L^3

L=1.53in=0.0388m

V=0.0388 ^3=5.8691x10^-5m^3=58.69ml

density=m/v

density=13.5g/58.69ml=0.23 g/ml

The wooden block floats because it is less dense than water

m=111mg=0.111g

density=m/v

density=0.111g/0.296ml=0.375g/ml

the metal paperclip floats because it is less dense than water

V=0.93cups=220.0271ml

m=0.88lb=399.1613g

Density=m/v

density=399.1613/220.027ml=1.8141g/ml

the apple sinks because it is denser than water

4 0

4 years ago

Hidiesjcfkesudfmjhtredcvbnm,lkjhgfdsuytrkduefkvke

Grace [21]

Answer:

THANKS FOR THE POINTS!!!!!!!

3 0

3 years ago

Read 2 more answers