Check the attached image for the question~

Which source of power can reduce emission of carbon dioxide?

ki77a [65]

Answer:

wind

Explanation:

this is because the others when burnt releases large amounts of carbon of carbon dioxide as they are carbon based and oxidises when burnt. However wind is renewable and converts the kinetic energy in the wind to electrical energy by the use of a turbine

6 0

3 years ago

How many coffee beans selling for $1.20 per pound should be mixed with 3 pounds of coffee beans selling for $2.40 per pound to o

Ulleksa [173]

This is not really a physics question but I can still help. It’s easier if you make a table. I hope this helped.

8 0

3 years ago

What is the heat needed to raise the temperature of 24.7 kg silver from 14.0 degrees celsius to 30.0 degrees celsius? specific h

Citrus2011 [14]

The amount of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q=m C_s \Delta T$
where
m is the mass of the substance
$C_s$ is its specific heat capacity
$\Delta T$ is the increase of temperature

The sample of silver of our problem has a mass of $m=24.7 kg$ . Its specific heat capacity is $C_s = 236 J/g^{\circ}C$ and the increase in temperature is
$\Delta T=30.0^{\circ}-14.0^{\circ}C=16.0^{\circ}C$

Therefore, the amount of heat needed is
$Q=mC_s \Delta T=(24.7 kg)(236 J/g^{\circ}C)(16.0^{\circ}C)=9.32 \cdot 10^4 J$

8 0

4 years ago

Can someone who knows slot about engineering explain to me by using an example on how to properly read a triangular cone shaped

Vlad1618 [11]

srry cant help u there

4 0

3 years ago

Problem 1: Three beads are placed along a thin rod. The first bead, of mass m1 = 24 g, is placed a distance d1 = 1.1 cm from the

Svet_ta [14]

Answer:

b) $x_{cm}$ = 4.88 cm , c) $x_{cm}$ ’= 1/M (m₁ d₁ + m₃ d₃) and d)

$x_{cm}$ ’= 1.88 cm

Explanation:

The definition of mass center is

$x_{cm}$ = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

Body 1

They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

Body 2

They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)

D₂ = d₁ + d₂

D₂ = 1.1 + 1.9

D₂ = 3.0 cm

Body 3

Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin

D₃ = D₂ + d₂

D₃ = 3.0 + 3.9

D₃ = 6.9 cm

With this data we substitute and calculate the center of mass

M = m₁ + m₂ + m₃

M = 24 + 12 + 56

M = 92 g

$x_{cm}$ = 1/92 (1.1 24 + 3.0 12 + 6.9 56)

$x_{cm}$ = 1/92 (448.8)

$x_{cm}$ = 4,878 cm

$x_{cm}$ = 4.88 cm

This distance is from the left end of the bar

Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

It is at d1 = -1.9 cm

It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

d2 = 0 cm

The reference system for her

Body 3

d3 = 3.9 cm

Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

$x_{cm}$ ’= 1/M (m₁ d₁ + m₂ d₂ + m₃ d₃)

$x_{cm}$ ’= 1/M (m₁ d₁ + m₃ d₃)

Part d) Let's calculate the value of the center of mass

$x_{cm}$ ’= 1/92 ((24 (-1.9) +56 3.9)

$x_{cm}$ ’= 1/92 (172.8)

$x_{cm}$ ’= 1.88 cm

This distance is to the right of the central marble

3 0

3 years ago