Answer:
D: It shows that Frida Kahlo used art to cope with her pain.
Explanation:
Within the text given it shows her emotions being lonely, immobile and in pain. But it all shows her asking her father for art which states that art is her sort of relief and happy place.
Your answer is 8. You add 2 + 1 + 5.3 to get 8.3. You round down to 8 because of the sig fig rules.
Answer:
Distance = 13.9 meters
Explanation:
Given the following data;
Maximum speed = 150 km/hr to meters per seconds = 150 * 1000/3600 = 41.67 m/s
Decelerating speed = 3m/s
To find the distance travelled with this speed;
Distance = maximum speed/decelerating speed
Distance = 41.67/3
Distance = 13.9 meters
Therefore, the bus would travel a distance of 13.9 meters before stopping.
Two things can happen to <u>old satellites</u>: For the closer satellites, engineers will use its last bit of fuel to slow it down so it will fall out of orbit and burn up in the atmosphere. Further satellites are instead sent even farther away from Earth.
Answer:
She can swing 1.0 m high.
Explanation:
Hi there!
The mechanical energy of Jane (ME) can be calculated by adding her gravitational potential (PE) plus her kinetic energy (KE).
The kinetic energy is calculated as follows:
KE = 1/2 · m · v²
And the potential energy:
PE = m · g · h
Where:
m = mass of Jane.
v = velocity.
g = acceleration due to gravity (9.8 m/s²).
h = height.
Then:
ME = KE + PE
Initially, Jane is running on the surface on which we assume that the gravitational potential energy of Jane is zero (the height is zero). Then:
ME = KE + PE (PE = 0)
ME = KE
ME = 1/2 · m · (4.5 m/s)²
ME = m · 10.125 m²/s²
When Jane reaches the maximum height, its velocity is zero (all the kinetic energy was converted into potential energy). Then, the mechanical energy will be:
ME = KE + PE (KE = 0)
ME = PE
ME = m · 9.8 m/s² · h
Then, equallizing both expressions of ME and solving for h:
m · 10.125 m²/s² = m · 9.8 m/s² · h
10.125 m²/s² / 9.8 m/s² = h
h = 1.0 m
She can swing 1.0 m high (if we neglect dissipative forces such as air resistance).
