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qaws [65]
3 years ago
5

How might heating the molecules inside the bottle change the motion of the molecules?

Physics
1 answer:
swat323 years ago
6 0
Heat causes the molecules to move faster, (heat energy is converted to kinetic energy ) which means that the volume of a gas increases more than the volume of a solid or liquid. However, gases that are contained in a fixed volume cannot expand - and so increases in temperature result in increases in pressure.
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Describe briefly how you would<br> determine the density an<br> irregular object (stone)
4vir4ik [10]

Answer:

by measuring calender we can find

Explanation:

6 0
2 years ago
A 15.0-kg block is dragged over a rough, horizontal surface by a 70.0-N force acting at 20.0° above the horizontal. The block is
Tatiana [17]

Answer:

Explanation:

from the question we are told that

Load L=15kg

ForceF=70N

Angle of inclination =20.0 degres

Displacement m=5 meters

coefficient of kinetic friction \alpha  =0.300

3 0
3 years ago
Arrow_forward
garri49 [273]

Explanation:

(a) Hooke's law:

F = kx

7.50 N = k (0.0300 m)

k = 250 N/m

(b) Angular frequency:

ω = √(k/m)

ω = √((250 N/m) / (0.500 kg))

ω = 22.4 rad/s

Frequency:

f = ω / (2π)

f = 3.56 cycles/s

Period:

T = 1/f

T = 0.281 s

(c) EE = ½ kx²

EE = ½ (250 N/m) (0.0500 m)²

EE = 0.313 J

(d) A = 0.0500 m

(e) vmax = Aω

vmax = (0.0500 m) (22.4 rad/s)

vmax = 1.12 m/s

amax = Aω²

amax = (0.0500 m) (22.4 rad/s)²

amax = 25.0 m/s²

(f) x = A cos(ωt)

x = (0.0500 m) cos(22.4 rad/s × 0.500 s)

x = 0.00919 m

(g) v = dx/dt = -Aω sin(ωt)

v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)

v = -1.10 m/s

a = dv/dt = -Aω² cos(ωt)

a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)

a = -4.59 m/s²

3 0
3 years ago
1. This heating system maintains room temperature at or near a particular value, known as the .
stiv31 [10]

Answer:

Explanation:

1. This heating system maintains room temperature at or near a particular value, known as the set point.

A temperature setpoint is the level at which the body attempts to maintain its temperature. When the setpoint is raised, the result is a fever.

2. You open the window, and a blast of icy air enters the room. The temperature drops to 17 degrees Celsius, which acts as a STIMULUS to the heating system.

3. The thermostat is a SENSOR that detects the stimulus and triggers a response.

Thermostat is use to turn off or on a switch, when the temperature is high or low

4. The heater turns on, and the temperature in the room INCREASE until it returns to the original setting.

This is the work of the thermostat above, when the temperature of the room is below a certain temperature the thermostat triggered the switch and keep increasing the temperature of the room until normal setting.

5. The response of the heating system reduces the stimulus. This is an example of NEGATIVE feedback.

Body temperature is regulated by negative feedback. The stimulus is when the body temperature exceeds 37 degrees Celsius, the sensors are the nerve cells with endings in the skin and brain, the control is the temperature regulatory center in the brain, and the effector is the sweat glands throughout the body.

6. The way this heating system maintains a stable room temperature is similar to the way an animal's body controls many aspects of its internal environment. The maintenance of a relatively constant internal environment is known as HOMEOSTASIS.

Humans rely on homeostasis to keep their core temperature hovering around 98.6 degrees Fahrenheit, so that their bodies can maintain proper function. Homeostasis is the ability to maintain a relatively stable internal state that persists despite changes in the world outside.

5 0
4 years ago
A sled of mass m is given a kick on a frozen pond. The kick imparts to the sled an initial speed of 2.00 m/s . The coefficient o
kobusy [5.1K]

2.04 meters distance is traveled by the sled before stopping.

Mass of the sled = m

The initial speed of the sled = 2 m/s

Coefficient of kinetic friction between sled and ice = 0.100

Let the distance the sled moves before it stops be d.

Gravity = 9.8 m/ s²

Let the initial kinetic energy sled be

= K _{i}

K_{i} =  \frac{1}{2} mv ^{2}

The work done by the frictional force is,

Work \: done \:  by \: frictional \: force =W_{f}

W _{f} = μ_{k}mgd

Work done by frictional force= Initial kinetic energy of the sled

W_{f} = K_{i}

μ_{k}mgd= \frac{1}{2} mv ^{2}

So, the distance traveled by the sled before stopping is

d= \frac{1mv ^{2} }{2 \:μ_{k}mg}

d= \frac{1v ^{2} }{2 \:μ_{k}g}

d= \frac{2^{2} }{2  \times \:0.100 \times 9.8}

d= 2.04 \: m

Therefore, the distance traveled by the sled before stopping is 2.04 meters.

To know more about work done, refer to the below link:

brainly.com/question/13662169

#SPJ4

5 0
2 years ago
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