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Romashka-Z-Leto [24]
3 years ago
14

Gases are more prone to expansion and contraction than liquid . The biggest change in the volume in your thermometer was probabl

y in the water or air
Physics
1 answer:
KiRa [710]3 years ago
8 0

Answer:

In the air

Explanation:

There are three states of matter:

- Solids: in solids, the particles are tightly bond together by strong intermolecular forces, so they cannot move freely - they can only vibrate around their fixed position

- Liquids: in liquids, particles are more free to move, however there are still some intermolecular forces keeping them close to each other

- Gases: in gases, particles are completely free to move, as the intermolecular forces between them are negligible

For this reason, it is generally easier to compress/expand the volume of a gas with respect to the volume of a liquid.

In this problem, we are comparing water (which is a liquid) with air (which is a gas). From what we said above, this means that the change in volume is larger in the air rather than in the water.

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Jason launches a model rocket with a mass of 2.0 kg from his spring-powered rocket launcher with a spring constant of 800 N/m. H
Arada [10]

Answer:

121 Joules

6.16717 m

Explanation:

m = Mass of the rocket = 2 kg

k = Spring constant = 800 N/m

x = Compression of spring = 0.55 m

Here, the kinetic energy of the spring and rocket will balance each other

\frac{1}{2}mu^2=\frac{1}{2}kx^2\\\Rightarrow u=\sqrt{\frac{kx^2}{m}}\\\Rightarrow u=\sqrt{\frac{800\times 0.55^2}{2}}\\\Rightarrow u=11\ m/s

The initial velocity of the rocket is 11 m/s = u.

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² = g

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11^2}{2\times -9.81}\\\Rightarrow s=6.16717\ m

The maximum height of the rocket will be 6.16717 m

Potential energy is given by

P=mgh\\\Rightarrow P=2\times 9.81\times \frac{0^2-11^2}{2\times -9.81}\\\Rightarrow P=121\ J

The potential energy of the rocket at the maximum height will be 121 Joules

5 0
3 years ago
A 2.0-kg object is attached to a spring (k = 55.6 N/m) that hangs vertically from the ceiling. The object is displaced 0.045 m v
rusak2 [61]

Answer:

Maximum acceleration will be 1.251m/sec^2

Explanation:

We have given mass of the object m = 2 kg

Spring constant k = 55.6 N/m

Amplitude is given as A = 0.045 m

We know that maximum acceleration in SHM is given by

Maximum acceleration =A\omega ^2

We know that \omega ^2=\frac{k}{m}=\frac{55.6}{2}=27.8

So maximum acceleration = 27.8\times 0.045=1.251m/sec^2

8 0
3 years ago
Read 2 more answers
Select the correct answer. An object has one force acting on it. It is a 33-newton force pointing downward. To create a net forc
kiruha [24]

Answer:

33 Newton upwards to get a net force of zero.

4 0
3 years ago
Read 2 more answers
If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the in
wel

The question is incomplete! The complete question along with answer and explanation is provided below.

Question:

A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.

What is the change in the potential energy (in Joules) of the mass as it goes up the incline?  

If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?

Given Information:  

Mass = m = 0.5 kg

Horizontal distance = d = 40 cm = 0.4 m

Vertical distance = h = 7 cm = 0.07 m

Normal force = Fn = 1 N

Required Information:  

Potential energy = PE = ?

Work done = W = ?

Answer:

Potential energy = 0.343 Joules

Work done = 0.39 N.m

Explanation:

The potential energy is given by

PE = mgh

where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.

PE = 0.5*9.8*0.07

PE = 0.343 Joules

As you can see in the attached image

sinθ = opposite/hypotenuse

sinθ = 0.07/0.4

θ = sin⁻¹(0.07/0.4)

θ = 10.078°

The horizontal component of the normal force is given by

Fx = Fncos(θ)

Fx = 1*cos(10.078)

Fx = 0.984 N

Work done is given by

W = Fxd

where d is the horizontal distance

W = 0.984*0.4

W = 0.39 N.m

3 0
3 years ago
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son4ous [18]
Any charged object can<span> exert the force upon other objects ... i think tell me if im right</span>
8 0
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