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2 years ago

The second is a flexible _________.

Physics

1 answer:

babunello [35]2 years ago

7 0

Answer:

static stretching

Explanation:

hope this helped ^^

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The current in one wire is 5.00<br> a. what is the current in the other wire.

Gemiola [76]

The wire of the the other wire is to connect

5 0

3 years ago

The closest approach distance between Mars and Earth is about 56 million km. Assume you can travel in a spaceship at 58,000 km/h

vladimir1956 [14]

Answer:

40.22 days

Explanation:

Given data:

Closest approach distance between Mars and Earth = 56 million km = 56 × 10⁶ km

Speed of the spaceship = 58000 km/h

Now, the time (t) is calculated as:

time = Distance / speed

on substituting the values, we get

t = 56 × 10⁶ km / (58000 km/h)

t = 965.517 hours

t = 965.517 / 24 days = 40.22 days

6 0

3 years ago

The Andromeda Galaxy (our nearest spiral neighbor) has spectral lines that show a blue shift. From this we may conclude that:

saw5 [17]

The universe is the collection of galaxies, The Andromeda Galaxy has spectral lines with blue shift. The conclusion is that the Universe has stopped expanding.

<h3>What is Andromeda Galaxy?</h3>

The Andromeda Galaxy is the nearest spiral neighbor that has spectral lines showing a blue shift.

Therefore, this concludes that, the Universe has stopped expanding. This galaxy is slowly shifting towards us.

Learn more about Andromeda Galaxy.

brainly.com/question/1499364

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8 0

2 years ago

A mass weighting 16 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 20 lbs when the

const2013 [10]

Answer:

The equation for the object's displacement is $u(t)=0.583cos11.35t$

Explanation:

Given:

m = 16 lb

δ = 3 in

The stiffness is:

$k=\frac{m}{\delta } =\frac{16}{3} =5.33lb/in$

The angular speed is:

$w=\sqrt{\frac{k}{m} } =\sqrt{\frac{5.33*386.4}{16} } =11.35rad/s$

The damping force is:

$F_{D} =cu$

Where

FD = 20 lb

u = 4 ft/s = 48 in/s

Replacing:

$c=\frac{F_{D} }{u} =\frac{20}{48} =0.42lbs/in$

The critical damping is equal:

$c_{c} =\frac{2k}{w} =\frac{2*5.33}{11.35} =0.94lbs/in$

Like cc>c the system is undamped

The equilibrium expression is:

$u(t)=u(o)coswt+u'(o)sinwt\\u(o)=7=0.583\\u'(o)=0\\u(t)=0.583coswt\\u(t)=0.583cos11.35t$

3 0

3 years ago

Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

7 0

3 years ago