Are defined as subatomic particles that have more mass than neutrinos but do not interact with light.

Given that coal used by electric power plants has a heating value of 27.5 million btus metric ton (25 million btus per ton), det

fredd [130]

Answer:

• 36.4 kg of coal.

• 80 pounds of coal.

Explanation:

Using proportionality constant,

Mass of coal = 1,000,000/27,500,000 btus/metric ton

= 0.0364 metric tons of coal

Mass of coal = 1,000,000/25,000,000 btus/ton

= 0.04 tons of coal.

Converting metric tons to kilogram,

1 metric ton = 1000kg,

0.0364 metric ton;

= 36.4 kg of coal.

Converting tons to pounds,

1 ton = 2000 pounds,

0.04 metric ton;

= 80 pounds of coal.

3 0

3 years ago

A car interior is heated to 48 C by the sun What type of energy transfer?

Aneli [31]

Answer:

Radiation heat energy transfer

Explanation:

The type of heat transfer from the Sun is radiation heat transfer, which is the transfer of heat through electromagnetic radiation

The distance of the Sun to the Earth is several million kilometers away, with the space between being composes of vacuum and the nuclear reaction in the Sun's core generates vast amount of electromagnetic radiation that is transferred all across the universe and reaches the Earth as visible light and radiant energy at the speed of light

The radiant energy transferred from the Sun heats up the Earth, including the car's interior.

7 0

3 years ago

The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

A)

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

B)

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$