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3 years ago

6

An airplane accelerat s down a runway at 3.20 m/s(squared) for 32.8s until is finally lifts off the ground. Presume to he plane

starts from the rest. Determine the distance traveled before takeoff.
The answer is 1721.3 but I want to know how to get it

1 answer:

rusak2 [61]3 years ago

5 0

SORRRY not this smart

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Please answer the questions... I will surely mark you as the brainliest according to me :)

frutty [35]

Answer:

(a) You can tell that have the same strength because they have attracted the same amount of paper clips.

(b) Iron is used in electromagnets because steel retained magnetic properties after the power was turned off, but in the iron, the paper clips dropped off right away.

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3 years ago

Question 8 of 10

-BARSIC- [3]

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2 years ago

A stack of 55 business cards is 1.85 cm tall. Use this information to determine the thickness of one business

Sindrei [870]

To find the answer, take 55 and divide it by 1.85 to get the thickness of one card. In this case the answer would be 29.72973 cm. each.

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3 years ago

Read 2 more answers

classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 128 feet per

ddd [48]

Answer:

$t_1 = 0.28 s$

$t_2 = 7.72 s$

Explanation:

Given that height of the projectile as a function of time is

$h = -16 t^2 + 128 t + 112$

here we know that

h = 147 ft

so from above equation

$147 = -16 t^2 + 128 t + 112$

$16 t^2 - 128 t + 35 = 0$

now by solving above quadratic equation we know that

$t_1 = 0.28 s$

$t_2 = 7.72 s$

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3 years ago

A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.

LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

$\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}$

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

$W = \frac{1}{2}(50)(5.61^2) = 786.8025 J$

Now, we can define work:

$\large\boxed{W = Fdcos\theta}}$

Now, plug in the values:

$786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta$

Solve for theta:

$cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}$

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3 years ago