Answer:
a) a = - 0.0833 m / s², b) t = 4.4 s
Explanation:
a) this is a kinematics exercise where the acceleration is along the inclined plane
v = v₀ - a t
a = v₀ - v / t
a = 3 - 8/60
a = - 0.0833 m / s²
b) in this case the final velocity is zero
v = v₀ - a t
0 = v₀ - at
t = v₀ / a
t = 28 / 6.4
t = 4.375 s
t = 4.4 s
Answer:
Newton's first law
Explanation:
Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force. Therefore, when the ice is smooth, friction gets lesser, and the force acted on that Puck will be decreased.
Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.
Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W
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.
Answer:
the gauge pressure at the upper face of the block is 116 Pa
Explanation:
Given the data in the question;
A cubical block of wood, 10.0 cm on a side.
height h = 1.50 cm = ( 1.50 × ( 1 / 100 ) ) m = 0.0150 m
density ρ = 790 kg/m³
Using expression for the gauged pressure;
p-p₀ = ρgh
where, p₀ is atmospheric pressure, ρ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.
we know that, acceleration due to gravity g = 9.8 m/s²
so we substitute
p-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m
= 116.13 ≈ 116 Pa
Therefore, the gauge pressure at the upper face of the block is 116 Pa
