Answer:
I'm not sure..but please refer to your teacher later.
Answer: Based on Newton's First law of motion (where inertia is involved), smooth ice increases the forceused to accelerate the hockey puck.
Explanation;
- smooth ice reduces the resistances between the surface of the figure skates and the ice itself.
- based on inertia theory ; the heavier the weight, the larger the inertia.. which explains it takes alot of force to move a heavier object than the lighter ones.. it also hard to *stop* the motion of heavier objects than the lighter ones.
- now let's look at the design of the player shoe itself, they have a sharp blade at the bottom of the figure stakes.. which takes us to the law of the force.. the smaller the surface area, the more forces acting on it. So, players force (weight, F= mg) acts on the tip of the blade and on the ice
- high inertia (run fast) and high force (attack opponent and pass puck) enables them to perform well in playing hockey
- Thus if there's no resistance and the inertia of the player is high then they could run and pass the puck quickly
Answer:
47.8 °C
Explanation:
Use the heat equation:
q = mCΔT
where q is the heat absorbed/lost,
m is the mass of water,
C is the specific heat capacity,
and ΔT is the change in temperature.
Here, q = 100 kJ, m = 0.5 kg, and C = 4.184 kJ/kg/°C.
100 kJ = (0.5 kg) (4.184 kJ/kg/°C) ΔT
ΔT = 47.8 °C
Hello There!
It takes the planet Mars around 24 hours, 37 minutes, 23 seconds to rotate on its axis. This is around the same amount of time that it takes our planet to rotate once on its axis.
Since this is a projectile motion problem, break down each of the five kinematic quantities into x and y components. To find the range, we need to identify the x component of the displacement of the ball.
Let's break them down into components.
X Y
v₁ 32 cos50 m/s 32 sin50 m/s
v₂ 32 cos50 m/s ?
Δd ? 0
Δt ? ?
a 0 -9.8 m/s²
Let's use the following equation of uniform motion for the Y components to solve for time, which we can then use for the X components to find the range.
Δdy = v₁yΔt + 0.5ay(Δt)²
0 = v₁yΔt + 0.5ay(Δt)²
0 = Δt(v₁ + 0.5ayΔt), Δt ≠ 0
0 = v₁ + 0.5ayΔt
0 = 32sin50m/s + 0.5(-9.8m/s²)Δt
0 = 2<u>4</u>.513 m/s - 4.9m/s²Δt
-2<u>4</u>.513m/s = -4.9m/s²Δt
-2<u>4</u>.513m/s ÷ 4.9m/s² = Δt
<u>5</u>.00s = Δt
Now lets put our known values into the same kinematic equation, but this time for the x components to solve for range.
Δdₓ = v₁ₓΔt + 0.5(a)(Δt)²
Δdₓ = 32cos50m/s(<u>5</u>.00s) + 0.5(0)(<u>5</u>.00)²
Δdₓ = 32cos50m/s(<u>5</u>.00s)
Δdₓ = 10<u>2</u>.846
Therefore, the answer is A, 102.9m. According to significant digit rules, neither would be correct, but 103m is the closest to 102.9m so I guess that is what it is.
