Over the past 150 years, what has happened to the amount of forest cover in Minnesota?

Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w

Tatiana [17]

Answer:

The average velocity is

$266\frac{m}{s},274\frac{m}{s}$ and $117\frac{m}{s}$ respectively.

Explanation:

Let's start writing the vertical position equation :

$L(t)=2t^{3}+t^{2}-5t+1$

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

$V_{avg}=\frac{Displacement}{Time}$ = Δx / Δt = $\frac{x2-x1}{t2-t1}$

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

$t2-t1=8s-5s=3s$

For the position variation we use the vertical position equation :

$x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m$

$x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m$

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

$\frac{798m}{3s}=266\frac{m}{s}$

For the second time interval :

t1 = 4 s → t2 = 9 s

$x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m$

$x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m$

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

$\frac{1370m}{5s}=274\frac{m}{s}$

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

$x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m$

$x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m$

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

$\frac{702m}{6s}=117\frac{m}{s}$

5 0

3 years ago

A 50.0-kg girl stands on a 9.0-kg wagon holding two 13.5-kg weights. She throws the weights horizontally off the back of the wag

Kobotan [32]

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5 0

3 years ago

A thin hoop is hung on a wall, supported by a horizontal nail. The hoop's mass is M=2.0 kg and its radius is R=0.6 m. What is th

boyakko [2]

Answer:

Explanation:

Given that,

Mass of the thin hoop

M = 2kg

Radius of the hoop

R = 0.6m

Moment of inertial of a hoop is

I = MR²

I = 2 × 0.6²

I = 0.72 kgm²

Period of a physical pendulum of small amplitude is given by

T = 2π √(I / Mgd)

Where,

T is the period in seconds

I is the moment of inertia in kgm²

I = 0.72 kgm²

M is the mass of the hoop

M = 2kg

g is the acceleration due to gravity

g = 9.8m/s²

d is the distance from rotational axis to center of of gravity

Therefore, d = r = 0.6m

Then, applying the formula

T = 2π √ (I / MgR)

T = 2π √ (0.72 / (2 × 9.8× 0.6)

T = 2π √ ( 0.72 / 11.76)

T = 2π √0.06122

T = 2π × 0.2474

T = 1.5547 seconds

T ≈ 1.55 seconds to 2d•p

Then, the period of oscillation is 1.55seconds

6 0

3 years ago

Calculate the kinetic energy of a 40 kg rock if the rock rolls down the hill with a velocity of 8 m/s.

ella [17]

Answer:

1280 J

Explanation:

see pic

7 0

3 years ago

A spring balance used to weigh candy is built with a spring. The spring stretches 3.00 cm when a 15 N weight is placed in the pa

OleMash [197]

Answer:

5.9 cm

Explanation:

We'll begin by calculating the spring constant (K) of the spring. This can be obtained as follow:

Force (F) = 15 N

Extention (e) = 3 cm

Spring constant (K) =.?

F = Ke

15 = K × 3

Divide both side by 3

K = 15 / 3

K = 5 N/cm

Therefore, the spring constant (K) is 5 N/cm.

Now, we shall determine how far the spring will stretch when a 3 kg candy is placed on the spring balance.

This can be obtained as follow:

First, we shall determine the weight of the object.

Mass of object = 3 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Weight (W) =?

W = mg

W = 3 × 9.8

W = 29.4 N

Finally, we shall determine how far the spring will stretch as follow:

Weight (W) = Force (F) = 29.4 N

Spring constant (K) = 5 N/cm

Extention (e) =?

F = ke

29.4 = 5 × e

Divide both side by 5

e = 29.4 / 5

e = 5.88 ≈ 5.9 cm

Therefore, the spring will stretch 5.9 cm when a 3 kg candy is placed on the spring balance.

6 0

3 years ago