All categories

3 years ago

Which style of painting is the Peale family known for?

Physics

2 answers:

Misha Larkins [42]3 years ago

8 0

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.

lord [1]3 years ago

7 0

Answer:

Representational

Explanation:

Charles Willson Peale was the head of the Peale family and a famed painter, soldier, inventor, politician and naturalist. The style of painting used by him and later on by his family members is "representational" and mainly centers around family, science and art.

He is best known for painting portraits of American leaders and American revolution in 18th century. Peale family gave the world a lot of famous painters of 18th and 19th century.

You might be interested in

suppose two masses are connected by a spring. compute the formula for the trajectory of the center of mass of the two mass oscil

Ymorist [56]

Answer:

The center mass (Xcm) of the two mass = (M₁X₁ + M₂X₂)/(M₁ +M₂)

Explanation:

let the first mass = M

let the position of second = M

8 0

3 years ago

If two objects are moving at the same speed, and Object 1 has four times as much mass as Object 2, how much momentum does Object

Margaret [11]

The answer is number 1

7 0

3 years ago

Five different forces act on an object. Is it possible for the net force on the object to be zero?

bazaltina [42]

No because there must be an even # if their is an even amount one of the forces isn’t being cancelled

4 0

3 years ago

On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the

Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

$\frac{F_E}{F_G}=\frac{qE}{mg}$ (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

$\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764$

Then, the gravitational force is 0.764 times the electric force on the particle

The acceleration of the particle is obtained by using the second Newton law:

$F_E-F_G=ma\\\\a=\frac{qE-mg}{m}$

you replace the values of all variables:

$a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}$

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0

3 years ago

Freight car A with a gross weight of 200,000 lbs is moving along the horizontal track in a switching yard at 4 mi/hr. Freight ca

zhenek [66]

Answer: a) 4.7 mi/hr. b) 86,500 lbs. mi²/Hr²

Explanation:

As in any collision, under the assumption that no external forces exist during the very small collision time, momentum must be conserved.

If the collision is fully inelastic, both masses continue coupled each other as a single mass, with a single speed.

So, we can write the following:

p₁ = p₂ ⇒m₁.v₁ + m₂.v₂ = (m₁ + m₂). vf

Replacing by the values, and solving for vf, we get:

vf = (200,000 lbs. 4 mi/hr + 100,000 lbs. 6 mi/hr) / 300,000 lbs = 4.7 mi/hr

If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

K₁ = 1/2 (m₁.v₁² + m₂.v₂²) = 3,400,000 lbs. mi² / hr²

After the collision, total kinetic energy is as follows:

K₂ = 1/2 ((m₁ + m₂) vf²) = 3,313,500 lbs. mi²/hr²

So we have an Energy loss, equal to the difference between initial kinetic energy and final kinetic energy, as follows:

DE = K₁ - K₂ = 86,500 lbs. mi² / hr²

This loss is due to the impact, and is represented by the work done by friction forces (internal) during the impact.

8 0

3 years ago