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2 years ago

A student heats a liquid on a burner. What happens to the portion of liquid that first begins to warm? What all applys

2 answers:

5 0

For water, as the temperature goes up the density of the liquid water decreases therefore it becomes less dense so it tends to rise upward. And since the temperature here is higher, heat flows from the higher temperature to the lower so it releases energy to the environment. Therefore, the correct statements are 2, 3 and 5.

sineoko [7]2 years ago

5 0

Answer:

It becomes less dense.

It absorbs energy from the environment.

It rises upward.

Explanation:

Took the test!

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For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the fir

abruzzese [7]

Case A :

A .75 kg 65 N/m 1.2 m

m = mass of car = 0.75 kg

k = spring constant of the spring = 65 N/m

h = height of the hill = 1.2 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s

Case B :

B .60 kg 35 N/m .9 m

m = mass of car = 0.60 kg

k = spring constant of the spring = 35 N/m

h = height of the hill = 0.9 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s

Case C :

C .55 kg 40 N/m 1.1 m

m = mass of car = 0.55 kg

k = spring constant of the spring = 40 N/m

h = height of the hill = 1.1 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

v = 5.1 m/s

Case D :

D .84 kg 32 N/m .95 m

m = mass of car = 0.84 kg

k = spring constant of the spring = 32 N/m

h = height of the hill = 0.95 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s

hence closest is in case C at 5.1 m/s

7 0

2 years ago

Read 2 more answers

A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2. the tangential acceleration of a point on its ri

vodka [1.7K]

We know that tangential acceleration is related with radius and angular acceleration according the following equation:
at = r * aa
where at is tangential acceleration (in m/s2), r is radius (in m) aa is angular acceleration (in rad/s2)
So the radius is r = d/2 = 1.2/2 = 0.6 m
Then at = 0.6 * 5 = 3 m/s2
Tangential acceleration of a point on the flywheel rim is 3 m/s2

5 0

2 years ago

For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser poin

nydimaria [60]

Answer:

The width of slide is 0.092 mm.

Explanation:

Given that,

Wave length = 680 nm

Distance between slit and screen D= 5.4 m

Distance of bright band 2y = 7.9 cm

$y =\dfrac{7.9}{2}$

$y=3.95\ cm$

We need to calculate the width of slide

Using formula of width

$d=\dfrac{\lambda D}{y}$

Where, d = width

D =Distance between slit and screen

y = Distance of bright band

Put the value into the formula

$d=\dfrac{680\times10^{-9}\times5.4}{3.95\times10^{-2}}$

$d=0.092\times10^{-3}\ m$

$d=0.092\ mm$

Hence, The width of slide is 0.092 mm.

4 0

2 years ago

A boy stands at a certain distance from a large building and blows a whistle. After 2.3s he hears the echo of the sound. He move

ddd [48]

Answer:

I) 57.5 m

Il) 50 m/s

Explanation:

Given that a boy stands at a certain distance from a large building and blows a whistle.

After 2.3s he hears the echo of the sound. The speed V of the sound will be:

V = 2X / T

Where X is the distance covered

V = 2X / 2.3

V = 0.87X ...... (1)

He moves 50m towards the building and blows his whistle again; this time the echo reaches him after 2.0s.

V = (2×50) / 2

V = 50 m/s

Substitutes the V into equation 1

50 = 0.87X

X = 50 / 0.87

57.5 m

Therefore, the Boys original distance from the building is 57.5m and the

Speed of sound in air is 50m/s

6 0

2 years ago

You are bungee jumping from a bridge. Initially, while you are falling the slack bungee cord isn’t exerting any forces or torque

harina [27]

Answer:

he fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

Explanation:

Let's pose the solution of this problem, to be able to analyze the firm affirmations.

When the person is falling, the weight acts on them all the time, initially the rope has no force, but at the moment it begins to lash it exerts a force towards the top that is proportional to the lengthening of the rope.

The equation for this part is

Fe - W = m a

k x - mg = m a

As the axis of rotation is located at the top where they jump, there is a torque.

What is it

Fe y - W y = I α

angular and linear acceleration are related

a = α r

Fe y - W y = I a / r

In the fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

4 0

2 years ago