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3 years ago

What is common to all fossil fuels?

2 answers:

4 0

All fossil fuels are formed deep underground. They are all nonrenewable energy sources. They are all made of hydrocarbons.They all release carbon dioxide pollution when burned.

Hope this helps xox :)

natta225 [31]3 years ago

3 0

All fossil fuels are formed deep underground. they are all nonrenewable energy

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After fixing a flat tire on a bicycle you give the wheel a spin. Its initial angular speed was 5.45 rad/s and it rotated 14.4 re

Komok [63]

Answer:

(a) α = -0.16 rad/s²

(b) t = 33.2 s

Explanation:

(a)

Applying 3rd equation of motion on the circular motion of the tire:

2αθ = ωf² - ωi²

where,

α = angular acceleration = ?

ωf = final angular velocity = 0 rad/s (tire finally stops)

ωi = initial angular velocity = 5.45 rad/s

θ = Angular Displacement = (14.4 rev)(2π rad/1 rev) = 28.8π rad

Therefore,

2(α)(28.8π rad) = (0 rad/s)² - (5.45 rad/s)²

α = -(29.7 rad²/s²)/(57.6π rad)

α = -0.16 rad/s²

Negative sign shows deceleration

(b)

Now, we apply 1st equation of motion:

ωf = ωi + αt

0 rad/s = 5.45 rad/s + (-0.16 rad/s²)t

t = (5.45 rad/s)/(0.16 rad/s²)

t = 33.2 s

6 0

3 years ago

What Is a Sound Wave? Learning Goal: To understand the nature of a sound wave, including its properties: frequency wavelength, l

NISA [10]

Answer:

A) Propagation of pressure fluctuations in a medium

B) air is the medium in which the wave is transported,

Explanation:

Part A.

A sound wave is a longitudinal oscillation of the molecules that forms in a material medium, they can be solid, liquid or gases, therefore the wave propagates in the same direction as the oscillation of the particles.

The most correct answer is:

* Propagation of pressure fluctuations in a medium

Part b

air is the medium in which the wave is transported, otherwise it cannot propagate

8 0

3 years ago

a seismic wave has an amplitude of 0.012 Meters.If the amplitude of this wave reduces to 0.006 meters, what happens to the energ

irina1246 [14]

Answer:The energy of the wave by a factor of 4

Explanation:

5 0

3 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0

3 years ago

An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl

svet-max [94.6K]

Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Displacement, s = 192 m

Acceleration, a = 9.81 m/s²

Substituting

s = ut + 0.5 at²

192 = 0 x t + 0.5 x 9.81 x t²

t = 6.26 seconds

Now we need to find horizontal distance traveled in 6.26 seconds by the package.

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 41 m/s

Time, t = 6.26 s

Acceleration, a = 0 m/s²

Substituting

s = ut + 0.5 at²

s = 41 x 6.26 + 0.5 x 0 x 6.26²

s = 256.52 m

The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

Initial velocity, u = 0 m/s

Time, t = 6.26 s

Acceleration, a = 9.81 m/s²

Substituting

v = u+ at

v = 0 + 9.81 x 6.26 = 61.41 m/s

Vertical component of velocity = 61.41 m/s

4 0

3 years ago