Can you use scalars and vectors to describe a home run?

Calculate the mass of wood that has the same energy as 1 g of oil

san4es73 [151]

2.3 grams of wood has the same energy as 1g of oil.

The majority of wood and wood waste used as fuel in the United States is used by industry. Manufacturers of paper and wood products are the biggest industrial users. They produce steam and electricity from waste from paper and lumber mills, which saves money by lowering the number of other fuels and electricity they need to buy to run their facilities.

There are numerous power plants that primarily burn wood to produce electricity in the electric power sector, and some coal-burning power plants burn wood chips along with coal to cut down on sulfur dioxide emissions. The primary purpose of wood consumption in the business sector is heating.

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3 0

1 year ago

ANSWER ASAP

shtirl [24]

5.6 g/ml. That is the density.

5 0

3 years ago

Read 2 more answers

What is the impulse needed to stop a 45-kg boy who is running at 6 m/s in 3 seconds?

Tju [1.3M]

Impulse = change of momentum
Impulse = 45 x 6 = 270 Ns

6 0

3 years ago

65. The weight of a body when totally immersed in a liquid is 4.2N if he weight of the liquid displaced is 2.5N. Find the weight

Anna35 [415]

Answer:

Given, Apparent weight(W₂)=4.2N

Weight of liquid displaced (u)=2.5N

Let weight of body in air = W₁

Solution,

U=W₁-W₂

W₁=4.2=2.5=6.7N

∴Weight of body in air is 6.7N

5 0

2 years ago

A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an

3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

$a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2$

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

$F_s = am = 1.31*16 = 20.92 N$

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

$F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N$

So the maximum acceleration on the block is

$a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2$

4)As the box slides, it is now subjected to kinetic friction, which is

$F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N$

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0

3 years ago