What does the nervous system allow?

A 980 kg roller coaster cart is traveling along a track at 17 m/s before it rolls down a 30 m tall hill (Point A). What will be

MrRissso [65]

The kinetic energy halfway the hill is $2.86\cdot 10^5 J$

Explanation:

If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:

$U_A +K_A = U_B + K_B$

where

$U_A=mgh_A$ is the initial potential energy, at point A, with

m = 980 kg (mass of the cart)

$g=9.8 m/s^2$ (acceleration of gravity)

$h_A = 30 m$ (height at point A)

$K_A=\frac{1}{2}mv_A^2$ is the initial kinetic energy, at point A , with

$v_A=17 m/s$ (velocity at point A)

$U_B=mgh_B$ is the final potential energy, at point B, where

$h_B = 15 m$ (height at point B)

$K_B=\frac{1}{2}mv_B^2$ is the final kinetic energy, at point B, where

$v_B$ is the velocity at point B

Here we are interested in finding $K_B$ , so by re-arranging the equation and substituting we find:

$K_B = U_A+K_B-U_B = mg(h_A-h_B)+\frac{1}{2}mv_A^2=(980)(9.8)(30-15)+\frac{1}{2}(980)(17)^2=2.86\cdot 10^5 J$

Learn more about kinetic energy:

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8 0

3 years ago

A 2kg object initially going 4m/s to the right is later going 8m/s. whats the change in velocity?

Nezavi [6.7K]

The change in velocity is +4 m/s to the right (or -4 m/s to the left).

The object's mass is irrelevant.

6 0

3 years ago

3. The center of mass (or center of gravity) of a two-particle system is at the origin. One particle

nika2105 [10]

Answer:

B) (-2.0 m, 0.0 m)

Explanation:

Given:

Mass of particle 1 is, $m_1=2.0\ kg$

Mass of particle 2 is, $m_2=3.0\ kg$

Position of center of mass is, $(x_{cm},y_{cm})=(0,0)$

Position of particle 1 is, $(x_1,y_1)=(3.0\ m,0.0\ m)$

Position of particle 2 is, $(x_2,y_2)=(?\ m,?\ m)$

We know that, the x-coordinate of center of mass of two particles is given as:

$x_{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

Plug in the values given.

$0=\frac{(2.0\times 3)+(3.0\times x_2)}{2.0+3.0}\\\\0=6+3x_2\\\\3x_2=-6\\\\x_2=\frac{-6}{3}=-2.0\ m$

We know that, the y-coordinate of center of mass of two particles is given as:

$y_{cm}=\frac{m_1y_1+m_2y_2}{m_1+m_2}$

Plug in the values given.

$0=\frac{(2.0\times 0)+(3.0\times y_2)}{2.0+3.0}\\\\0=0+3y_2\\\\3y_2=0\\\\y_2=\frac{0}{3}=0.0\ m$

Therefore, the position of particle 2 of mass 3.0 kg is (-2.0 m, 0.0 m).

So, option (B) is correct.

8 0

3 years ago

A heavy-duty shock absorber is compressed 4 cm from its equilibrium position by a mass of 700nbspkg. How much work is required

tia_tia [17]

Answer:

Explanation:

A mass of 700 kg will exert a force of

700 x 9.8

= 6860 N.

Amount of compression x = 4 cm

= 4 x 10⁻² m

Force constant K = force of compression / compression

= 6860 / 4 x 10⁻²

= 1715 x 10² Nm⁻¹.

Let us take compression of r at any moment

Restoring force by spring

= k r

Force required to compress = kr

Let it is compressed by small length dr during which force will remain constant.

Work done

dW = Force x displacement

= -kr -dr

= kr dr

Work done to compress by length d

for it r ranges from 0 to -d

Integrating on both sides

W = $\int\limits^{-4}_0 {kr} \, dr$

= [ kr²/2]₀^-4

= 1/2 kX16X10⁻⁴

= .5 x 1715 x 10² x 16 x 10⁻⁴

= 137.20 J

3 0

3 years ago

A stable air mass is most likely to have which characteristic? 1. turbulent air 2. poor surface visibility 3. showery precipitat

Ne4ueva [31]

A stable air mass is most likely to have POOR SURFACE VISIBILITY.
Stable air mass refers to those air mass that have marked stability in their lower layers. The characteristics of stable air mass include the following: cloud cover, smooth air, uninterrupted precipitation and low visibility.<span />

3 0

4 years ago