Answer:
98.13m
Explanation:
Complete question
Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building
CHECK THE ATTACHMENT
From the figure, using trigonometry
Tan(θ ) = opposite/adjacent
Where Angle (θ )= 63°
Opposite= X = height of the building
Adjacent= 50 m
Then substitute the values we have
Tan(63)= X/50
1.9626= X/50
X= 1.9626 × 50
X= 98.13m
Hence, the height of the building is 98.13m
Answer:
D: Increase the distance between the objects.
E: Decrease the mass of one of the objects.
The first blank: HEAT
The second blank: ELECTRICAL
Answer:
6m/s
Explanation:
the original momentum = mass x velocity = 8x (60+10) = 560
momentum after = mass x velocity of the school bag + mass x velocity of the boy = 10x20 + 60x A
200+60A = 560
A=6
I'm assuming the question is what is the robin's speed relative to to the ground...
Create an equation that describes its relative motion.
rVg = rVa + aVg
Substitute values.
rVg = 12 m/s [N] + 6.8 m/s [E]
Use vector addition.
| rVg | = √ | rVa |² + | aVg |²
| rVg | = √ 144 m²/s² + 46.24 m²/s²
| rVg | = √ 19<u>0</u>.24 m²/s²
| rVg | = 1<u>3</u>.78 m/s
Find direction.
tanФ = aVg / rVa
tanФ = 6.8 m/s / 12 m/s
Ф = 29°
Therefore, the velocity of the robin relative to the ground is 14 m/s [N29°E]
