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3 years ago

Calculate the missing variable....

Physics

1 answer:

OlgaM077 [116]3 years ago

6 0

Answer:

option b

Explanation:

from the given formula, s=d/t

make t the subject of the formula we have

t=d/s

5/100

0.5

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A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 10 degrees to the right of her orig

r-ruslan [8.4K]

Answer:

2406 miles

Explanation:

Let A be the starting position, B the junction position and C the final position after flying the 3.5 hrs. Also, let b be the distance from the starting point:

$\angle ABC =(180-10) \textdegree=170\textdegree$

#Distance traveled in 1.5hrs is;

$c=690x1.5\\=1035mi$

#Distance traveled in next two hrs:

$a=690\times 2\\=1380mi$

#Now using the Cosine Rule:

$b^2=a^2+c^2-2ab\cos B\\\\=1380^2+1035^2-2(1380)(1035)cos170\textdegree\\\\b^2=5788.83\\\\b\approx 2406.00 \ mi$

Hence, the pilot is 2406 miles from her starting position.

4 0

3 years ago

PLS ANSWER ASAP

TiliK225 [7]

Answer: Answer is D

I took the test little while back.

3 0

3 years ago

Which of the following has the most potential energy?

Alex

A Car at the top of a hill.

It is because in that case, produce of mass and height is highest which is directly proportional to potential energy

In short, Your Answer would be Option A

Hope this helps!

3 0

3 years ago

Read 2 more answers

2. Kevin works as a janitor, and he is pushing a fully-

dybincka [34]

The time taken for him to move the bin 6.5 m is 2.30 s.

The given parameters;

weight of the load, w = 557 N
force applied , F = 410 N
angle of force, = 15°
coefficient of kinetic friction = 0.46
distance moved, d = 6.5 m

The net horizontal force on the recycling bin is calculated as follows;

$Fcos\theta - F_k = ma$

where;

m is the mass of the recycling bin
$F_k$ is the frictional force

W = mg

$557 = 9.8m\\\\m = \frac{557}{9.8} \\\\m = 56.84 \ kg$

The net horizontal force on the recycling bin is calculated as;

$Fcos \theta - F_k = ma\\\\Fcos\theta - \mu_kF_n = ma\\\\410\times cos(15) \ - \ 0.46(557) = 56.84 a\\\\139.8 = 56.84a\\\\a = \frac{139.8}{56.84} \\\\a = 2.46 \ m/s^2$

The time taken for him to move the bin 6.5 m is calculated as follows;

$s = v_0t + \frac{1}{2} at^2\\\\6.5 = 0 + \frac{1}{2} \times 2.46\times t^2\\\\6.5 = 1.23 t^2\\\\t^2 = \frac{6.5 }{1.23} \\\\t^2 = 5.285\\\\t = \sqrt{5.285} \\\\t = 2.30 \ s$

Thus, the time taken for him to move the bin 6.5 m is 2.30 s.

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7 0

3 years ago

A falling back in mid air has what

Umnica [9.8K]

Acceleration I think if I’m not mistaken

4 0

4 years ago