All categories

3 years ago

When fossil fuels are burnt, what 2 gases are given off?

1 answer:

6 0

Salutations!

When fossil fuels are burnt, what 2 gases are given off?

When fossil fuels are burnt, the following are one of the two gases when they are given off:

→ Carbon dioxide: Carbon dioxide is a green house gas, and plays a major role in the global warming.

→ Carbon Monoxide: Carbon monoxide is a gas that is dangerous to human beings when released.

Hope I helped (:

Have a great day!

You might be interested in

A block of mass

Gennadij [26K]

(a) The work done by the applied force is 26.65 J.

(b) The work done by the normal force exerted by the table is 0.

(d) The work done by the net force on the block is 26.65 J.

<h3>Work done by the applied force</h3>

W = Fdcosθ

W = 14 x 2.1 x cos25

W = 26.65 J

<h3>Work done by the normal force</h3>

W = Fₙd

W = mg cosθ x d

W = (2.5 x 9.8) x cos(90) x 2.1

W = 0 J

<h3>Work done force of gravity</h3>

The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.

<h3> Work done by the net force on the block</h3>

∑W = 0 + 26.65 J = 26.65 J

Thus, the work done by the applied force is 26.65 J.

The work done by the normal force exerted by the table is 0.

The work done by the force of gravity is 0.

The work done by the net force on the block is 26.65 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

6 0

2 years ago

A mass attached to a spring oscillates in simple harmonic motion with an amplitude of 10 cm. When the mass is 5.0 cm from its eq

timama [110]

When the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.

<h3>Total energy of the mass</h3>

The total energy possessed by the mass under the simple harmonic motion is calculated as follows;

U = ¹/₂kA²

where;

k is the spring constant
A is the amplitude of the oscillation

<h3>Potential energy of the mass at 5 cm from equilibrium point</h3>

P.E = ¹/₂k(Δx)²

<h3>Kinetic energy of mass</h3>

K.E = U - P.E

K.E = ¹/₂kA² - ¹/₂k(Δx)²

<h3>Percentage of its energy that is kinetic</h3>

$K.E (\%) = \frac{U - P.E}{U} \times 100\%\\\\K.E (\%) =\frac{\frac{1}{2}kA^2 - \frac{1}{2}k(\Delta x)^2 }{\frac{1}{2}kA^2} \times 100\%\\\\K.E (\%) = \frac{A^2 - (\Delta x)^2}{A^2} \times 100\%\\\\K.E (\%) = \frac{10^2 - (10-5)^2}{10^2} \times 100\%\\\\K.E (\%) = \frac{10^2 - 5^2}{10^2} \times 100\%\\\\K.E (\%) = 75\%$

Thus, when the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.

Learn more about kinetic energy here: brainly.com/question/25959744

3 0

2 years ago

an ant weighing 200 N and wearing platform heels with a total area of 7 square centimeters would exert how much pressure?

iVinArrow [24]

The pressure exerted is $2.86\cdot 10^5 Pa$

Explanation:

The pressure exerted by a force on a surface is given by

$p=\frac{F}{A}$

where

p is the pressure

F is the magnitude of the force

A is the area of the surface

In this problem, we have:

F = 200 N is the force exerted (the weight of the ant)

$A=7 cm^2 = 7\cdot 10^{-4} m^2$ is the area

Therefore, the pressure exerted is:

$p=\frac{200}{7\cdot 10^{-4}}=2.86\cdot 10^5 Pa$

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

7 0

4 years ago

The figure above represents a stick of uniform density that is attached to a pivot at the right end and has equally spaced marks

zavuch27 [327]

Answer:

After 2.0s the angular momentum is $L= 2(4A+3B+2C+D)x$

Explanation:

Let us call forces acting on the rod, A, B, C, and D, and the separation between them $x$ .

Then, the torque due to force A is

$\tau_a = 4Ax$ ,

due to the force B

$\tau_b = 3Bx$ ,

due to force C

$\tau_c = 2Cx$ ,

and the torque due to force D is

$\tau_d = Dx$ .

Therefore, the total torque on the the stick is

$\tau_{tot} =\tau_a+\tau_b+\tau_c+\tau_d$

$\tau_{tot} =4Ax+3Bx+2Cx+Dx$

$\tau_{tot} =x(4A+3B+2C+D)$

Now, this torque causes angular acceleration $\alpha$ according to the equation

$I \alpha = \tau_{tot}$

where $I$ is moment of inertia of the stick and it has the value

$I = \dfrac{1}{3} m(4x)^2$

Therefore the angular acceleration is

$\alpha = \dfrac{\tau_{tot} }{I}$

$\alpha =\dfrac{x(4A+3B+2C+D)}{\dfrac{1}{3}m(4x)^2 }$

$\boxed{\alpha =\dfrac{3(4A+3B+2C+D)}{16mx } .}$

Now, the angular momentum $L$ of the stick is

$L = I\omega$ ,

where $\omega$ is the angular velocity.

Since $\omega = \alpha t$ , we have

$L = \dfrac{1}{3}m (4x)^2 *\dfrac{3(4A+3B+2C+D)}{16mx }* t$

$L= (4A+3B+2C+D)x t$

Therefore, $t = 2.0s$ , the angular momentum is

$\boxed{ L= 2(4A+3B+2C+D)x. }$

5 0

3 years ago

An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9000 m/s an

Sergio [31]

Answer:

The distance between earthquake center and the measuring station is 1350 kilometers.

Explanation:

Let the earthquake center be at a distance of 'S' meters from the recording station.

Now from the basic relation of distance, speed and time we know that

$Distance=Speed\times Time$

For a Primary wave (P wave) let us assume that it appraoches the measuring station after $t_{1}$ minutes

Thus making use of the above relation we have

$Distance=V_{p}\times Time\\\\\therefore D=9000\times t_{1}.......(i)$

Now since it is given that the secondary wave (S wave) reaches the measuring spot after 2 minutes or 120 seconds thus the time taken by secondary waves to reach recorder equals $t_{1}+120$ making use of the same relation we get

$Distance=V_{s}\times Time\\\\\therefore D=5000\times (t_{1}+120).......(ii)$

Solving equation 'i' and 'ii' we get

$D=5000\times (\frac{D}{9000}+120)\\\\\therefore \frac{D}{5000}=\frac{D}{9000}+120\\\\\frac{D}{5000}-\frac{D}{9000}=120\\\\\therefore D=\frac{120}{\frac{1}{5000}-\frac{1}{9000}}=1350000meters=1350kilomerers$

8 0

3 years ago