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ch4aika [34]
3 years ago
8

In addition to the conditions required for any solar eclipse, what must also be true in order for you to observe a total solar e

clipse?
Physics
1 answer:
PtichkaEL [24]3 years ago
7 0

Answer: The umbra, which is the dark shadow wherein the Sun's light is completely blocked, must pass over where one is for that person to observe a total solar eclipse.

Explanation:

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Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E
Kay [80]

Answer:

The true course: 40.29^\circ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

  • V_w = velocity of wind = 30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h
  • V_p = velocity of plane in still air = 100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h

Assume:

  • V_r = resultant velocity of the plane
  • \theta = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h

Let us find the direction of this resultant velocity with respect to east direction:

\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ

This means the the true course of the plane is in the direction of 40.29^\circ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h

Hence, the ground speed of the plane is 96.68 km/h.

5 0
3 years ago
I need help with these. Please show workings<br>​
Sauron [17]

Answer:

Imp = 25 [kg*m/s]

v₂= 20 [m/s]

Explanation:

In order to solve these problems, we must use the principle of conservation of linear momentum or momentum.

1)

(m_{1}*v_{1})+(F*t)=(m_{1}*v_{2})

where:

m₁ = mass of the object = 5 [kg]

v₁ = initial velocity = 0 (initially at rest)

F = force = 5 [N]

t = time = 5 [s]

v₂ = velocity after the momentum [m/s]

(5*0) +(5*5) = (m_{1}*v_{2}) = Imp\\Imp = 25 [kg*m/s]

2)

(m_{1}*v_{1})+(F*t)=(m_{1}*v_{2})\\(0.075*0)+(30*0.05)=(0.075*v_{2})\\v_{2}=20 [m/s]

8 0
2 years ago
Read 2 more answers
Action reaction forces are in which one of newtons laws
Klio2033 [76]

Answer:

the third law

Explanation:

could u vote me brainliest plz? thx :)

4 0
3 years ago
What formula should I use?
Elodia [21]
These nuts on your chin
7 0
2 years ago
Imagine you’re driving along a road and you approach a bridge. You notice a sign that reads, “Bridge freezes before road.” Why d
nydimaria [60]

<u>Answer:</u>

<h3>During wet and freezing temperatures, ice is able to form at a faster pace on bridges because freezing winds blow from above and below and both sides of the bridge, causing heat to quickly escape. The road freezes slower because it is merely losing heat through its surface.</h3>

<u>Sources:</u>

-- https://intblog.onspot.com/en-us/why-do-bridges-become-icy-before-roads

and

-- https://www.accuweather.com/en/accuweather-ready/why-bridges-freeze-before-roads/687262

I hope this helps you! ^^

6 0
3 years ago
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