Answer:
Option A. 40 mi/h
Explanation:
To obtain the average speed of the vehicle, we'll begin by calculating the distance travelled by the vehicle in each case. This is illustrated below:
Case 1:
Speed = 30 mi/h
Time = 2 h
Distance =...?
Speed = Distance /Time
30 = Distance /2
Cross multiply
Distance = 30 × 2
Distance = 60 mi
Case 2:
Speed = 60 mi/h
Time = 1 h
Distance =...?
Speed = Distance /Time
60 = Distance /1
Cross multiply
Distance = 60 × 1
Distance = 60 mi
Finally, we shall determine the average speed of the vehicle as follow:
Total distance travelled = 60 + 60
Total distance travelled = 120 mi
Total time = 2 + 1
Total time = 3 h
Average speed =..?
Average speed = Total Distance travelled /Total time
Average speed = 120/3
Average speed = 40 mi/h
Therefore, the average speed of the vehicle is 40 mi/h
Answer:
u/2 √(1 + 3 cos² θ)
Explanation:
The object is thrown at an angle θ, so the velocity has two components, vertical and horizontal.
Initially, the vertical component is u sin θ and the horizontal component is u cos θ.
At the maximum height, the vertical component is 0 and the horizontal component is u cos θ.
The mean vertical velocity is:
(u sin θ + 0) / 2 = u/2 sin θ
The mean horizontal velocity is:
(u cos θ + u cos θ) / 2 = u cos θ
The net mean velocity can be found with Pythagorean theorem:
v² = (u/2 sin θ)² + (u cos θ)²
v² = u²/4 sin² θ + u² cos² θ
v² = u²/4 (1 − cos² θ) + u² cos² θ
v² = u²/4 (1 − cos² θ) + u²/4 (4 cos² θ)
v² = u²/4 (1 − cos² θ + 4 cos² θ)
v² = u²/4 (1 + 3 cos² θ)
v = u/2 √(1 + 3 cos² θ)
Hey I hope this helps in any way.
As the front<span> moves through, cool, fair </span>weather<span> is likely to follow. Warm </span>front<span> Forms when a moist, warm air mass slides up and over a cold air mass. As the warm air mass rises, it condenses into a broad area of clouds. A warm </span>front<span> brings gentle rain or light snow, followed by warmer, milder </span>weather<span>.</span>
Please, you have to apply the formula below:
<span>Q=c∗m∗Δt
</span><span>
where Q is the energy lost,
c is the specific heat of water,
m is the mass of water involved,
so m=3.75 *10^-1 Kg
c=4,184 J/(Kg*°C)
<span>delta t=37.5 °C
I hope my answer has come to your help. God bless and have a nice day ahead!
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