The correct answer is option B, representational
All the painters in Peale family were involved in paintings which represent the day today life activities or were portraits or mimic some natural forms.
Charles Willson Peale , the head of the Peale family was known for painting sixty portraits of the first American president, George Washington. He also painted portraits of portraits of notable people of the society such as Benjamin Franklin, Thomas Jefferson etc.
Most of the paintings of peale family were based on the theme of family, art and science. Six of Peale’s son were known for their renaissance paintings. His oldest son Raphelle was known for still life paintings.
Titian Ramsay Peale, Charles’ youngest son was a naturalist painter.
A jet fighter flies from the airbase A 300 km East to the point M. Then 350 km at 30° West of North.
It means : at 60° North of West. So the distance from the final point to the line AM is :
350 · cos 60° = 350 · 0.866 = 303.1 km
Let`s assume that there is a line N on AM.
AN = 125 km and NM = 175 km.
And finally jet fighter flies 150 km North to arrive at airbase B.
NB = 303.1 + 150 = 453.1 km
Then we can use the Pythagorean theorem.
d ( AB ) = √(453.1² + 125²) = √(205,299.61 + 15,625) = 470 km
Also foe a direction: cos α = 125 / 470 = 0.266
α = cos^(-1) 0.266 = 74.6°
90° - 74.6° = 15.4°
Answer: The distance between the airbase A and B is 470 km.
Direction is : 15.4° East from the North.
Answer:
44.4 m/s
Explanation:
d = 80 m, t = 1.8 s, find v
d = v*t
v = d/t = 80/1.8 = 44.4 m/s
Answer:
453 gm
Explanation:
<u>Immersed </u>objects are buoyed up by force equal to mass of displaced liquid
400 + 53 = 453 gm in air
Part a
Answer: NO
We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.
Using the second equation of motion:
where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.
It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, 
Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.
Part b
Answer: 29.6 m/s
The maximum distance that car can travel is 
The acceleration is same,
The final velocity, v=0
Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:
Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.
