PLEASEEE HELPPPPP does anyone know these answers?

Hawks and gannets soar above the ground and, when they spot prey, they fold their wings and essentially drop like a stone. They

denis-greek [22]

Answer:

v = 54.2 m / s

Explanation:

Let's use energy conservation for this problem.

Starting point Higher

Em₀ = U = m g h

Final point. Lower

$Em_{f}$ = K = ½ m v²

Em₀ = Em_{f}

m g h = ½ m v²

v² = 2gh

v = √ 2gh

Let's calculate

v = √ (2 9.8 150)

v = 54.2 m / s

3 0

3 years ago

Imagine that there is a small rocky body caught by Earth’s gravity. Draw a comic-strip cartoon

vladimir1956 [14]

To create this comic strip you can use a narration describing each step and illustrate each one with one image or drawing.

Creating a comic strip involves using images and short texts to explain a specific idea or phenomenon. In the case of the process for a meteor to enter Earth you can use the following ideas.

A meteoroid approaches the Earth at high speed and draw a meteor traveling near to different planets and approaching Earth.
What is that? and draw the Earth wondering who or what is approaching.
The meteoroid enters the atmosphere of the Earth and becomes a meteor and draw the rocky body burning
The rocky body crashes with the surface becoming a meteorite and draw the zone where the meteorite crashed.

Learn more about comic in: brainly.com/question/1418309

6 0

2 years ago

1.20 Newton force is working on a 250 gram object. What is the acceleration?

Leokris [45]

Answer:

The answer is B

Explanation:

250g = 0.25kg

F = m × a

a = F/m

= 1.2/0.25

= 4.8m/s²

7 0

3 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$