Zaiden and lisa are testing music effects on plants

A student is skateboarding down a ramp that is 6.0 m long and inclined at 180 with respect to the horizontal. The initial speed

dlinn [17]

Answer:

The speed at the bottom of the ramp is 2.6m/s

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

$v_{f}^{2} = v_{i}^{2} + 2ad$

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$ (1)

Notice that it is necessary to found the acceleration that can be done by means of Newton's second law:

The acceleration can be found by means of Newton's second law:

$\sum F_{net} = ma$

Where $\sum F_{net}$ is the net force, m is the mass and a is the acceleration.

$Fx + Fy = ma$ (2)

<em>Force in the x axis:</em>

$F_{x} = W_{x}$

The component of the weight in the x axis can be gotten by means of trigonometric:

$\frac{OC}{H} = sen \theta$

$\frac{W_{x}}{W} = sen \theta$

$W_{x} = W sen \theta$

$W_{x} = mgsen \theta$

$F_{x} = mgsen \theta$ (3)

<em>Forces in the y axis: </em>

$F_{y} = N - W_{y}$ (4)

The component of the weight in the y axis can be gotten by means of trigonometric:

$\frac{AC}{H} = cos \theta$

$\frac{W_{y}}{W} = cos \theta$

$W_{y}= W cos \theta$

Remember that the weight is defined as:

$W = mg$

$W_{y}= mgcos \theta$

The normal force can be obtained from equation (4)

$N - W_{y} = 0$

$N = W_{y}$

$N = mgcos \theta$

Therefore, equation 4 can be rewritten as:

$F_{y} = mgcos \theta - mgcos \theta$

$F_{y} = 0$ (5)

Then, replacing equation 3 and equation 5 in equation 2 it is gotten:

$mgsen \theta + 0 = ma$

$mgsen \theta = ma$ (6)

However, a can be isolated from equation 6

$a = \frac{mgsen \theta}{m}$

$a = gsen \theta$ (7)

Finally, equation 7 can be replaced in equation 2:

$v_{f} = \sqrt{v_{i}^{2} + 2d(gsen \theta)}$

$v_{f} = \sqrt{(2.6m/s)^{2} + 2(6.0m)(9.8m/s^{2})sen 180^{\circ})}$

$v_{f} = 2.6m/s$

Hence, the speed at the bottom of the ramp is 2.6m/s

7 0

3 years ago

The muzzle velocity of an armor-piercing round fired from an M1A1 tank is 1770 m/s (nearly 4000 mph or mach 5.2). A tank is at t

loris [4]

<h2>Height of cliff is 66.43 meter.</h2>

Explanation:

Consider the horizontal motion of shell,

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 1770 m/s

Acceleration, a = 0 m/s²

Displacement, s = 6520 m

Substituting

s = ut + 0.5 at²

6520 = 1770 x t + 0.5 x 0 xt²

t = 3.68 s

Time of flight of shell = 3.68 s.

Now consider the vertical motion of shell,

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Time, t = 3.68 s

Substituting

s = ut + 0.5 at²

s = 0 x 3.68 + 0.5 x 9.81 x 3.68²

s = 66.43 m

Height of cliff is 66.43 meter.

8 0

3 years ago

Ian walks 2 km to his best friend’s house, then walks 0.5 km to the library. He then makes a 2.5 km walk home. The entire walk t

Sonbull [250]

Average speed = (total distance covered) / (time to cover the distance)

Ian's total distance = (2km + 0.5km + 2.5km) = 5 km

Ian's total time = 3 hours

Average speed = (5km) / (3 hours)

Average speed = 1.66 km/hr

4 0

3 years ago

The maximum amount of work that can be extracted is equal to the changein Gibbs free energy___________.

steposvetlana [31]

Answer:

The statement is true

Explanation:

The Gibbs free energy is defined as:

ΔG = ΔH - TΔS

Where ΔH : Change in the enthalpy

T : The temperature

ΔS : Change in the entropy

And the Gibbs free energy is a thermodynamic potential that can be used to calculate the maximum of reversible work that may be performed by a thermodynamic system at a constant temperature and pressure.

If the thermodynamic process is isothermal and isobaric, then the statement is true.

3 0

4 years ago

A student pushes a 12.0 kg box with a horizontal force of 20 N. The friction force on the box is 9.0 N. Which of the following i

antiseptic1488 [7]

Fnet=m*a
a= Fnet/m

a=(20N[E]+9.0N[E])/ 12.0kg

a=2.4167m/s^2

a ≈ 2.4m/s^2

8 0

4 years ago

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