Protons are found togather in a

Natural gas, a nonrenewable fossil fuel, is often seen as an alternative fuel in the worldwide transition to a renewable energy

liubo4ka [24]

Answer:

a) Natural gas burns more efficiently and cleanly than other sources of carbon-based fossil fuels, thus contributing to a reduction in atmospheric carbon dioxide.

Explanation:

Although fossil fuels such as coal and oil have carbon content, natural gas also has it in smaller proportions, which is why it is a good idea to use it as a transitional fuel. Although natural gas does not have the same specific heat as gasoline, its production is simpler since it does not require the expensive refining processes such as petroleum derivatives, which is why its production is more efficient.

Another facility is the amount of natural gas that is obtained from oil wells, this can be extracted and stored efficiently to prevent its spread into the atmosphere.

3 0

3 years ago

Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad

love history [14]

Answer:

a) The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ , b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

$\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where:

$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

$\alpha$ - Angular acceleration, measured in radians per square second.

$r$ - Radius of rotation (Radius of a tire), measured in meters.

Given that $\alpha = 15\,\frac{rad}{s^{2}}$ and $r = 0.31\,m$ . The linear acceleration experimented by the car is:

$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

$\theta = 46.875\,rad$

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0

3 years ago

In a particular case of an object in front of a spherical mirror with a focal length of +12.0 cm, the magnification is +4.00.(a)

salantis [7]

Answer:

9 cm

-36 cm

Explanation:

u = Object distance

v = Image distance

f = Focal length = 12

m = Magnification = 4

$m=-\frac{v}{u}\\\Rightarrow 4=-\frac{v}{u}\\\Rightarrow v=-4u$

Lens equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{12}=\frac{1}{u}+\frac{1}{-4u}\\\Rightarrow \frac{1}{12}=\frac{3}{4u}\\\Rightarrow u=9\ cm$

Object distance is 9 cm

$v=-4\times 9=-36\ cm$

Image distance is -36 cm (other side of object)

7 0

3 years ago

A bicyclist in the Tour de France crests a mountain pass as he moves at 18 km/h. At the bottom, 4.0 km farther, his speed is at

Allisa [31]

We are given:

v0 = initial velocity = 18 km/h

d = distance = 4 km

v = final velocity = 75 km/h

a =?

We can solve this problem by using the formula:

v^2 = v0^2 + 2 a d

75^2 = 18^2 + 2 (a) * 4

5625 = 324 + 8a

a = 662.625 km/h^2

6 0

3 years ago

suppose that 273 g of one of the substances listed above displaces 26 mL of water. What is the substance?

guajiro [1.7K]

The unknown substance is silver. I don't see a list of available substances, but let's see if there's something reasonable available that will match. First, let's calculate the density of the unknown substance. Density is mass per volume, so 273 g / 26 mL = 10.5 g/mL Looking up a list of elements sorted by density, I see the following: 10.07 Actinium 10.22 Molybdenum 10.5 Silver 11.35 Lead And silver at 10.5 g/ml is a very nice match for the unknown substances' density of 10.5 g/ml.

6 0

3 years ago

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