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3 years ago

Why is foam a good insulator?

1 answer:

7 0

Air or any sort of gas is not that good in conducting heat in relation to other materials.
<span>since foam consists of many bubbles with trapped gas it both stops convection by simply being in position, and prevents conduction of heat through it, due to the gas inside</span>

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A block with mass M = 3 kg is moving on a flat surface with constant speed v1 = 12 m/s. A bullet with mass m = 0,1 kg is shot at

adelina 88 [10]

The speed does the block move after it is hit by the bullet that remains stuck inside the block will be 23.7 m/sec and it takes 12.07 seconds to stop.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

Apply the law of conservation of momentum principle;

m₁v₁+m₂v₂cosΘ =(m₁+m₂)V

3 kg × 12 m/s + 0,1 kg × 400 m/s cos 20° = (3+0.1)V

36 + 40 cos 20° = 3.1 V

V=23.7 m/sec

The time it takes to stop when the friction coefficient between the block and the surface is 0.2 is found as;

V = u +at

V = 0+ μgt

23..7=0.2× 9.81 ×t

t=12.07 sec

Hence, it takes 12.07 seconds to stop.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

4 0

3 years ago

A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water

elena-14-01-66 [18.8K]

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ( $\dot Q$ ), measured in watts, that is, joules per second, by the following formula:

$\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}$ (1)

Where:

$m$ - Mass of the sphere, measured in kilograms.

$c$ - Specific heat of the material, measured in joules per kilogram-degree Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures of the sphere, measured in degrees Celsius.

$\Delta t$ - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

$\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}$ (2)

Where:

$m_{I}$ , $m_{X}$ - Masses of the iron and unknown spheres, measured in kilograms.

$\Delta t_{I}$ , $\Delta t_{X}$ - Times of the iron and unknown spheres, measured in seconds.

$c_{I}$ , $c_{X}$ - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

$c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}$

If we know that $\Delta t_{I} = 6.35\,s$ , $\Delta t_{X} = 4.59\,s$ , $m_{I} = 0.515\,kg$ , $m_{X} = 1.263\,kg$ and $c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}$ , then the specific heat of the unknown material is:

$c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)$

$c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}$

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

3 0

3 years ago

The equation shows neutralization of an acid and a base to produce a salt and water.

Masja [62]

The answer is; 2.

To balance a chemical equation, the moles on one side of the equation has to be the same as that on the other side. This ensures that the law of conservation is observed because matter or energy can't be created or destroyed but can only be transformed from one form to another.

In this equation, putting 2 in front of NaCl ensures that there are 2 moles of Na and CL just as there are 2 moles of Na and CL in the reactants side.

6 0

3 years ago

Read 2 more answers

Some instruments differentiate individual quanta of electromagnetic radiation based on their energies. assume such an instrument

lesya [120]

There is a relationship between the energy of a photon and its wavelength. This can be expressed as a mathematical equation shown below:

E = hc/λ
where
h is the Planck's constant equal to 6.62607004 × 10⁻³⁴ m²<span> kg / s
c is the speed of light equal to 3</span>× 10⁸ m/s
λ is the wavelength

3.5×10⁻¹⁶ J = (6.62607004 × 10⁻³⁴ m² kg / s)(3× 10⁸ m/s)/λ
Solving for λ,
λ = 56.8×10⁻⁹ m or<em> 56.8 nm</em>

5 0

4 years ago

The coefficient of linear expansion of iron is 1.0 × 10−5 per C◦ . The surface area of an iron cube, with an edge length of 5.0

Allushta [10]

The linear expansion of iron is a = 1× 10∧ -5 /°C
Length edge is 5cm
Initial surface area of cube A = 6L ∧²
= 150cm∧²
The edge length of cube after being heated from
10°C to 60°C is
L = L[I +a dt] where the dt is = 60-10 = 50°C
5.0025cm
A surface area of the cube after heated is
A' = 6L' ∧²
= 150.1500375 cm∧²
The increased area is = A' -A = 0.15cm∧²

7 0

3 years ago