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3 years ago

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A flying saucer lifts the Physical Science building 10,000 ft into the air before discovering it is useless and discards the rem

ains. If the building weighs 1,000,000 pounds and the ascent takes 20 seconds, what is one saucer power?

1 answer:

scZoUnD [109]3 years ago

7 0

Answer:

500000000 lbft/s

Explanation:

F = Force or weight = 1000000 lbf

s = Displacement = 10000 ft

t = Time taken = 20 seconds

Work done is given by

$W=Fs\\\Rightarrow W=1000000\times 10000\\\Rightarrow W=10000000000\ lb-ft$

Power is given by

$P=\dfrac{W}{t}\\\Rightarrow P=\dfrac{10000000000}{20}\\\Rightarrow P=500000000\ lbft/s$

One Saucer power is 500000000 lbft/s

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A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

8 0

3 years ago

You discover a binary star system in which one member is a 15 solar mass main-sequence star and the other star is a 10 solar mas

Sever21 [200]

Answer:

A star with 15 solar masses is too big to be a main-sequence star.

4 0

3 years ago

A ball is dropped from an aircraft flying at an altitude of 8,848 meters assuming gravity is 9.8m/s what is the total amount of

Dafna11 [192]

In this question, you're determining the time (t) taken for an object to fall from a distance (d).

The equation to represent this is:
Time equals the square root of 2 times the distance divided by the gravitational force of earth.
In equation from it looks like this (there isn't an icon to represent square root so just pretend like there's a square root there):
t = 2d/g (square-rooted)

d = 8,848m and g = 9.8m/s

Now plug in the information we have:
t = 2 x 8,848m/9.8m/s (square-rooted)

The first step is to multiply 2 times 8,848m:
t = 17,696m/9.8m/s (square-rooted)

Now divide 9.8m/s by 17,696m (note that the two m's (meters) cancels out leaving you with only s (seconds):
t = 1805.72s (square-rooted)

Now for the last step, find the square root of the remaining number:
t = 42.5s

So the time it takes the ball to drop from the height (distance) of 8,848 meters, and falling with the gravitational pull of 9.8 meters per second is 42.5 seconds.

I hope this helps :)

7 0

2 years ago

A weight lifter does 700J of work on a weight that he lifts in 3.1s. What is the power with

IgorLugansk [536]

Power = Work done/Time taken

So, keeping this in mind,we can solve it as follows:

= 700/3.1

= 7000/31

= 225.80 W

8 0

2 years ago

A stone is thrown with an initial speed of 12 m/s at an angle of 30o above the horizontal from the top edge of a cliff. If it ta

kherson [118]

Answer:

$d=58m$

Explanation:

From the question we are told that:

Initial Speed $U=12m/s$

Time $T=5.6s$

Angle $\theta=30$

Generally the Newton's equation for motion is mathematically given by

$d=d'+ut+\frac{at^2}{2}$

$d=12cos30*5.6$

$d=58m$

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2 years ago