<u>Answer:</u>
Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.
<u>Explanation:</u>
Let the east point towards positive X-axis and north point towards positive Y-axis.
Speed of truck = 25 m/s north = 25 j m/s
Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s
= (1.43 i + 1.00 j) m/s
Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j
Magnitude of velocity = 26.04 m/s
Angle from positive horizontal axis = 86.85⁰
So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.
Answer:
Anything below 7.0 is acidic, so the range would be 0 to 7.
Neutral is simply 7, in the middle of the scale.
Lastly, anything above 7.0 is basic or alkaline, so that would be 7 to 14.
Good luck, I hope this helps
Answer:
Here's what I got:
Let's assume that N and E are + directions while S and W are - directions.
Wind is blowing from SW; thus, it is blowing towards NE (or at 45 deg N of E).
Dividing the wind's speed into components:y-component: +70.71 km/h; x-component: +70.71 km/h
Dividing the airplane's speed into components:y-component: -600 km/h; x-component: 0 km/h
Adding the components to get the resulting components:y-component: -529.29 km/h; x-component: +70.71
Using the Pythagorean Theorem to find the resulting speed:v^2 = y^2 + x^2 so v = 533.99 km/h
To find the angle of direction, use arctan (y/x):arctan (529.29/70.71) = 82.39 deg
ANSWER: velocity = 533.99 km/h at 82.39 deg S of E
Explanation:
