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1 year ago

5

A fighter plane is descending at 30 m/s. The pilot ejects, and the ejector seat accelerates him upwards at 120 m/s2 for 2 second

s. What is his velocity at the end of the 2 seconds?

1 answer:

katrin [286]1 year ago

4 0

Answer:

210m/s upwards

Explanation:

u= -30m /s

a= 120m/s^2

t=2s

v=u+at

=(-30)+(120x2)

=240-30

=210m/s

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A 0.5 kg baseball is hit by a bat delivering an impulse of 180 kg x m/s. If the bat is in contact with the baseball for 0.4 seco

yan [13]

Answer:

450N

Explanation:

Impulse = mv ---> 180kgm/s = (0.5kg)v

v = 360 m/s

Acceleration = v/t ---> a = (360 m/s)/(0.4s) = 900 m/s^2

Force = ma ---> F = (0.5kg)(900 m/s^2) = 450N

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2 years ago

Which test was conducted on the skin found under the victim’s nails

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3 years ago

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A reconnaissance plane flies 404 km awayfrom its base at 730 m/s, then flies back to its base at 1095 m/s.What is it’s average s

elena-14-01-66 [18.8K]

The average speed of the plane is 875.999 m/s.

Average speed can be defined as the ratio of total distanced travelled by the object to that of total time taken to cover the distance.

Mathematically, Average speed = Av = $\frac{Total Distance}{Total Time}$

According to the question,

Speed of the plane away from its base V₁ = 730 m/s

Speed of the plane when it flies back V₂ = 1095 m/s

Plane flies the distance D = 404 km

Total Distance covered by the plane S = 404 * 2 km

(because the distance travelled by the plane when going away from the base and then flying back to the base is same)

Therefore S = 808 km = 808 ˣ 10³ m

Time taken by the plane while flying away from the base T₁ = $\frac{D}{V1}$

T₁ = $\frac{404000}{730}$ = 553.425 s

Time taken by the plane while flying back to the base T₂ = $\frac{D}{V2}$

T₂ = $\frac{404000}{1095}$ = 368.949s

Total Time T = T₁ + T₂ = 922.375 s

Therefore Av = $\frac{Total Distance}{Total Time}$

= $\frac{D}{T}$ m/s

= $\frac{808000}{922.375}$ m/s

= 875.999 m/s

The average speed of the plane will be 875.999 m/s.

To know more about Average speed,

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11 months ago

It took 1700 J of work to stretch a spring from its natural length of 2m to a length of 5m. Find the spring's force constant.

Alenkasestr [34]

The work to stretch a spring from its rest position is

   (1/2) (spring constant) (distance of the stretch)²

   E = 1/2 k x² .

You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write

   1700 joules = 1/2 k (3m)²

1 joule = 1 newton-meter

   1700 N-m = 1/2 k (3m)²

Multiply each side by 2: 3400 N-m = k · 9m²

Divide each side by 9m²    k = 3400 N-m / 9m²

   = (377 and 7/9)   newton per meter

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3 years ago

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To determine the height of a tall building such as Sears Tower in Chicago, Illinois a ball was dropped from the top of the build

Darya [45]

Answer:

The height of Sears Tower is 1448.5 feet.

Explanation:

<h3>We apply the free fall formula to the ball: </h3><h3> $y=v_{o} *t+\frac{1}{2} *g*t^{2}$ </h3><h3>y: The vertical distance the ball moves at time t </h3><h3> $v_{o}$ i: Initial speed </h3><h3>g=Gravity acceleration= $9.8*(\frac{\frac{1ft}{0.305m} }{s^{2} } )$ </h3>

Known information

We know that the vertical distance (y) that the ball moves in 9,5s is equal to height of Sears Tower (h).

Too we know that the ball is released from rest, then, $v_{0}$ =0

Height of Sears Tower calculation:

We replace in the equation 1 the data following;

$y=h$

$v_{o} =0$

$g=32,1\frac{ft}{s^{2} }$

$t= 9,5s$

$h=0*9.5+\frac{1}{2} *32.1*9.5^{2}$

$h=1448.5 ft$

Answer: The height of Sears Tower is 1448.5 ft

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3 years ago