What happens when a mixture is separated

Calculate the energy absorbed when 13 kg of liquid water raises from 18°C to 100°C and then boils at 100°C.

Cloud [144]

Answer:

the energy absorbed is 4.477 x 10⁶ J

Explanation:

mass of the liquid, m = 13 kg

initial temperature of the liquid, t₁ = 18 ⁰C

final temperature of the liquid, t₂ = 100 ⁰C

specific heat capacity of water, c = 4,200 J/kg⁰C

The energy absorbed is calculated as;

H = mcΔt

H = mc(t₂ - t₁)

H = 13 x 4,200(100 - 18)

H = 4.477 x 10⁶ J

Therefore, the energy absorbed is 4.477 x 10⁶ J

5 0

3 years ago

A bodybuilder loads a bar with 550 Newtons (~125 pounds) of weight and pushes the bar over her head 10 times. Each time she lift

mafiozo [28]

Work = (force) x (distance)

Each time she lifts the weight, she does

(550 N) x (0.5 m) = 275 joules of work against gravity.

Each time she lets the bar down gently, gravity does

(550 N) x (0.5 m) = 275 joules of work against her muscles.

If the human physical muscular system were 100% efficient, and capable
of absorbing work as well as spending it, then the bodybuilder would do
exactly zero work in the process of 1-up followed by 1-down.

3 0

3 years ago

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A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f

amid [387]

You need to first measure the angle of descent, i.e. the angle the hill makes with the ground. Then identify the forces acting on the sled, split them up into horizontal and vertical components, or into components that are parallel and perpendicular to the hill, and use Newton's second law to determine the components of the sled's acceleration vector.

There are at least 2 forces acting on the sled:

• its weight, pointing downward with magnitude W = m g

• the normal force, pointing perpendicular to the hill and away from the ground with mag. N

The question doesn't specify, but there might also be friction to consider, indicated in the attachment by the vector F pointing parallel to the slope of the hill and opposing the direction of the sled's motion with mag. F.

Splitting up the forces into parallel/perpendicular components is less work. By Newton's second law, the net force (denoted with ∑ or "sigma" here) in a particular direction is equal to the mass of the sled times its acceleration in that direction:

∑ (//) = W (//) = m a (//)

∑ (⟂) = W (⟂) + N = m a (⟂)

where, for instance, W (//) denotes the component of the sled's weight in the direction parallel to the hill, while a (⟂) denotes the component of the sled's acceleration perpendicular to the hill. If there is friction, you need to add -F to the first equation.

If the hill makes an angle of θ with flat ground, then W makes the same angle with the hill so that

W (//) = -m g sin(θ)

W (⟂) = -m g cos(θ)

So we have

-m g sin(θ) = m a (//) → a (//) = -g sin(θ)

-m g cos(θ) + N = m a (⟂) → a (⟂) = 0

where the last equality follows from the fact that the normal force exactly opposes the perpendicular component of the weight. This is because the sled is moving along the slope of the hill, and not into the air or into the ground.

Then the acceleration vector is

a = a (//)

with magnitude

||a|| = a = g sin(θ).