Hunter works to fix wires and paneling. Hunter is a(n)

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

3 0

2 years ago

A small plane flies 37.0 km in a direction 45° north of east and then flies 28.0 km in a direction 25° north of east.

Karo-lina-s [1.5K]

Answer:

d= 64.1 km θ = 36.4º

Explanation:

a) In order to find the plane's straight-line distance from the starting point, we need to know the coordinates of the final and initial position of the plane, so we can find the total displacement, as the difference between the final and initial position.

If we choose to put our origin at the initial point of trajectory, we have that (x₀, y₀) = (0, 0)

In order to find the position of the plane after finishing the flight, we need to find its final coordinates (x₁, y₁).

In order to get x₁, we need to add the x-coordinate after flying 45º north of east, and the Δx after completing the flight in a direction 25º of east, that we can find applying trigonometry, as follows:

x₁ = 37.0 km * cos 45º + 28.0 km* cos 25º = 51.6 Km

Appying the same considerations for the y-coordinate, we have:

y₁ = 37.0 km * sin 45º + 28.0 km* sin 25º = 38.0 km

Now, as the initial position coincides with the origin, the distance in a straight line from this point to the origin, is just the hypotenuse of the triangle determined by the coordinates (x₁, y₁) and (0,0), as follows:

$d = \sqrt{x1^{2}+y1^{2}} =\sqrt{(51.6km)^{2}+(38km)^{2}} =64.1 km$

The geographic direction of the displacement vector (which coincides in magnitude with the distance we have just found), is just the angle that this distance forms with the east axis, that we can find getting the tangent of this angle as follows:

tg θ = $\frac{y1}{x1} = \frac{38km}{56.1km} =0.736$

⇒ θ = 36.4º North of East (counterclockwise from the east axis).

7 0

2 years ago

A 0.150-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa

Hitman42 [59]

Answer:

The magnitude of the change in momentum of the stone is 5.51kg*m/s.

Explanation:

the final kinetic energy = 1/2(0.15)v^2

1/2(0.15)v^2 = 70%*1/2(0.15)(20)^2

v^2 = 21/0.075

v^2 = 280

v = 16.73 m.s

if u is the initial speed and v is the final speed, then:

u = 20 m/s and v = - 16.73m/s

change in momentum = m(v-u)

= 0.15(- 16.73-20)

= -5.51 kg*m.s

Therefore, The magnitude of the change in momentum of the stone is 5.51kg*m/s.

3 0

3 years ago

Determine the kinetic energy of a 1000-kg roller coaster car that is moving with a speed of 20.0 m/s.

ankoles [38]

Answer:

The correct answer is - 200000 J

Explanation:

We use the formula of kinetic energy:

The formula to calculate kinetic energy is,

Here,

Ec =1/2 x m x v^2

The mass of the roller coaster is, m = 1000 k g

The speed of the roller coaster is, v = 20.0 m/s

Therefore,

Ec=1/2 x 1000kg x (20m/s)^2 = 200.000Joule

200,000J

7 0

3 years ago

An astronaut on another planet drops a 1-kg rock from rest and finds that it falls a vertical distance of 2.5 meters in one seco

kow [346]

X=.5(a)t^2 can be used: 2.5m=.5(g)(1), g=5m/s^2.

4 0

3 years ago

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