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2 years ago

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Which energy source is formed when organic matter is trapped underground without exposure to air or moisture?. . A. natural gas.

B. coal. C. oil. D. biomass fuels

1 answer:

nikklg [1K]2 years ago

8 0

Answer:

The correct answer is option B. coal

Explanation:

Coal is made of remains of organic material including trees and other vegetation which got trapped beneath the earth’s surface or at the bottom of the swamps. After burial below the ground the organic material was acted upon by the high temperature and pressure in the absence of air to form peat. Peat after further processing for a longer period of time converted into coal

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An 80.0-kg man jumps from a height of 2.50 m onto a platform mounted on springs. As the springs compress, he pushes the platform

frosja888 [35]

Answer:

6.16 m/s

0.0105 m

Explanation:

Let the ground 0 for potential reference be at where the spring is compress 0.24 m. The the man would jump from a height h = 2.5 + 0.24 = 2.74 m from it. We can apply the law of energy conservation knowing that as the man jumps, his potential energy converts to kinetic energy, then finally to elastic energy:

$E_p = E_e$

$mgh = kx^2/2$

where m = 80 kg is the man mass, g = 9.81 m/s2 is the gravitational acceleration, h = 2.74 m is the potential distance he travels, k N/m is the spring constant and x = 0.24 is the distance it compresses

$80*9.81*2.74 = k0.24^2/2$

$2150.352 = 0.0288k$

$k = 74665 N/m$

Similarly at the position where it compresses by 0.12 m, it's 0.24 - 0.12 = 0.12 m far from ground 0.

$E_p = E_{e2} + E_k + E_{p2}$

$mgh = kx_2^2 + mv^2/2 + mgh_2$

$2150.352 = 74665*0.12^2/2 + 80v^2/2 + 80*9.81*0.12$

$2150.352 = 537.588 + 40v^2 + 94.176$

$40v^2 = 1518.588$

$v^2 = 37.9647$

$v = \sqrt{37.9647} = 6.16 m/s$

When he steps gently, then his gravity force would equal to his spring force

$mg = kx_3$

$x_3 = mg/k = 80*9.81/74665 = 0.0105m$

7 0

2 years ago

Vinny is on a motorcycle at rest, 200 m away from a ramp that jumps over a gully. Calculate the minimum constant acceleration Vi

soldier1979 [14.2K]

Answer: 6.25 m/s^2

Explanation:

The distance between Vinny and the ramp is 200m

And he has 8 seconds (At max) to reach that distance.

The initial velocity is 0m/s

The initial position is 0m

Now, we want to find the constant acceleration in order to do this, so suppose that we have a constant acceleration A.

a(t) = A.

To have the velocity, we must integrate over time, and remember that the constant of integration is equal to zero because the initial velocity is zero.

v(t) = A*t

For the position, we integrate again over time.

p(t) = 0.5*A*t^2

And we want to travel 200m in 8 seconds, then:

p(8s) = 200m

0.5*A*(8s)^2 = 200m

A*32s^2 = 200m

A = 200m/32s^2 = 6.25 m/s^2

This is the minimum acceleration in order to do this, if Vinny has a larger acceleration he will travel the 200m in a smaller time.

7 0

2 years ago

A construction worker pushes a crate horizontally on a frictionless floor with a net force of 10\, \text 10N, start text, N, end

Strike441 [17]

Answer:

$k_f$ = 40J

Explanation:

The work and energy theorem says that:

$W_f =k_f-k_i$

where $W_f$ is the work of the force, $k_f$ the final kinetic energy and $k_i$ the initial kinetic energy.

Addittionally, the work of the force is calculate as force multiply by distance and if the crate inittialy is at rest, the initial kinetic energy is zero, so:

$Fd = k_f$

where F is the force and d the distance. Then, replacing values, we get:

$(10N)(4m) = k_f$

$40J = k_f$

it means that the system gain 40J of kinetic energy.

7 0

2 years ago

A motorcyclist heading east through a small Iowa town accelerates after he passes a signpost at x=0 marking the city limits. His

adoni [48]

Answer:

1) $v_2=23\ m.s^{-1}$ & $x_2=43\ m$ east of sign post

2) $x'=55\ m$ east of sign post

3) $x_n=205\ m$ east of the signpost.

4) $v_z=35\ m.s^{-1}$

Explanation:

Given:

position of motorcyclist on entering the city at the signpost, $x_0=0\ m$

time of observation after being at x=5m east of the signpost, $t_m=0\ s$

constant acceleration of the on entering the city, $a=4\ m.s^{-2}$

distance of the motorcyclist moments later after entering, $s_m=5\ m$

velocity of the motorcyclist moments later after entering, $u_m=15\ m.s^{-1}$

<u>Now the initial velocity on at the sign board:</u>

$u_m^2=u^2+2.a.x_m$

where:

$u=$ initial velocity of entering the city at the signpost

Putting respective values:

$15^2=u^2+2\times 4\times 5$

$u=13.6015\ m.s^{-1}$

1)

Position at time $t_2=2\ s$ sec.:

Using equation of motion,

$x_2=u_m.t_2+\frac{1}{2} a.(t_2)^2+5$ because it has already covered 5m before that point

$x_2=15\times 2+0.5\times 4\times 2^2+5$

$x_2=43\ m$ east of sign post

Velocity at time $t_2=2\ s$ sec.:

$v_2=u_m+a.t_2$

$v_2=15+4\times 2$

$v_2=23\ m.s^{-1}$

2)

Position when the velocity is $v'=25\ m.s^{-1}$ :

using equation of motion,

$v'^2=u_m^2+2.a.x'+5$

$25^2=15^2+2\times 4\times x'+5$

$x'=55\ m$ east of sign post

3)

Given that:

acceleration be, $a_n=2\ m.s^{-2}$

time, $t_n=5\ s$

Position after the new acceleration and the new given time:

using equation of motion,

$x_n=u_m.t_n+\frac{1}{2} a_n.(t_n)^2+5$

$x_n=15\times 5+0.5\times 2\times 5^2+5$

$x_n=205\ m$ east of the signpost.

4)

now time of observation, $t_z=5\ s$

$v_z=u_m+a.t_z$

$v_z=15+4\times 5$

$v_z=35\ m.s^{-1}$

8 0

2 years ago

A bird is flying in a room with a velocity field of . Calculate the temperature change that the bird feels after 9 seconds of fl

Korvikt [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The temperature change is $\frac{dT}{dt} = 1.016 ^oC/m$

Explanation:

From the question we are told that

The velocity field with which the bird is flying is $\vec V = (u, v, w)= 0.6x + 0.2t - 1.4 \ m/s$

The temperature of the room is $T(x, y, u) = 400 -0.4y -0.6z-0.2(5 - x)^2 \ ^o C$

The time considered is t = 10 \ seconds

The distance that the bird flew is x = 1 m

Given that the bird is inside the room then the temperature of the room is equal to the temperature of the bird

Generally the change in the bird temperature with time is mathematically represented as

$\frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 * (5-x)] [-\frac{dx}{dt} ]$

Here the negative sign in $\frac{dx}{dt}$ is because of the negative sign that is attached to x in the equation

So

$\frac{dT}{dt} = -0.4v_y -0.6v_z -0.2[2 * (5-x)][ -v_x]$

From the given equation of velocity field

$v_x = 0.6x$

$v_y = 0.2t$

$v_z = -1.4$

So

$\frac{dT}{dt} = -0.4[0.2t] -0.6[-1.4] -0.2[2 * (5-x)][ -[0.6x]]$

substituting the given values of x and t

$\frac{dT}{dt} = -0.4[0.2(10)] -0.6[-1.4] -0.2[2 * (5-1)][ -[0.61]]$

$\frac{dT}{dt} = -0.8 +0.84 + 0.976$

$\frac{dT}{dt} = 1.016 ^oC/m$

5 0

2 years ago