Answer:
Colour of any object is by the colour of light it reflects i.e if white light is incident on the object,it will reflect blue color.so it will appear blue.But if red light is incident on it,it will not reflect that and absorb it.so as it will not reflect any light it will appear black.
Explanation:
Answer:
A) 17.7 m/s
B) 15.98 m
C) Zero
E) 9.8 m/s²
Explanation:
given information
distance, h = - 34 m
time, t = 5 s
A) What is the initial speed of the egg?
h - h₀ = v₀t - 
t², h₀ = 0
- 34 = v₀ 5 - \frac{1}{2} 9.8 5²
- 34 = 5 v₀ - 122.5
v₀ = 122.5 - 34/5
= 17.7 m/s
B) How high does it rise above its starting point?
v² = v₀² - 2gh
v = 0 (highest point)
2gh = v₀²
h = v₀²/2g
= 17.7²/2 (9.8)
= 15.98 m
C) What is the magnitude of its velocity at the highest point?
v = 0 (at highest point)
E) What are the magnitude and direction of its acceleration at the highest point?
g= 9.8 m/s², since the egg is moved vertically, the acceleration is the same as the gravitational acceleration.
Answer:
See the answer below
Explanation:
<em>The best thing one can do in this case would be to return the microscope's objective to low power and then </em><em>re-center the specimen </em><em>before switching back to high-dry power.</em>
Most of the time, <u>what makes the specimen under the microscope to be out of focus at higher objective powers after being in focus at low power is because they are not properly centered on the stage</u>. Hence, before calling on the instructor, it would be wise to first return to low power, re-center the specimen and bring it into focus after which the high power objective can be returned to and the fine focus adjusted to bring the image back to focus.
After doing the above and the specimen still does not come into focus, then the instructor can be called upon.
Based on your problem, what you are looking for is the quantity of heat. To solve for it, you will need this formula:
Q = mc(T2-T1)
Where: Q = Quantity of heat
m = mass of the substance
c = Specific heat
T2 = Final temperature
T1 = Initial temperature
Now the specific heat of water is 4.184 J/(g°C), meaning that is how much energy is required to raise the temperature of 1g of liquid water by 1 degree Celsius.
Since your mass is in kilograms, let us convert that into grams, which will be equal to 50,000 grams. Now we can put our given into the equation:
Q = mc(T2-T1)
= 50,000g x 4.184 J/(g°C) x (80°C - 45°C)
= 50,000 g x 4.184 J/(g°C) x 35°C
= 7,322,000 J or 7,322 kJ or 7.322 MJ
