Answer:
-22.2 m/s²
Explanation:
The equation for position x for a constant acceleration a, time t and initial velocity v₀, initial position x₀:
(1)
For rocket A the initial and final position: x = x₀= 0. Using these values in equation 1 gives:
(2)
Solving for time t:
(3)
The times for both rockets must be equal, since they start and end at the same location. Using equation 3 for rocket A and B gives:
(4)
Solving equation 4 for acceleration of rocket B:
(5)
Option C.
Ceres is considered the smallest dwarf planet of the solar system.
Answer:
Option B
3
Explanation:
The two dimensions have the following significant numbers
4.56- 3 significant numbers
5.668- 4 significant figures
When responding, we give area as per the most precise number given hence the smallest (most precise) significant figures in these two numbers is 3. Therefore, the area will be given to 3 significant figures ie
4.56*5.668=25.84608 square meters
Area=25.8 square meters
Answer:
Explanation:
Use Archimedes' principle, that states something of this nature: "The buoyant force acting on an object immersed in a fluid is equal to the weight of the fluid displaced"
Say, you take a cube of wood(for example) and place it in a bucket of water. Your cube is bound to sink, until, the upthrust force equals its weight.
Where does this upthrust come from?
As the cube sinks, it displaced some volume of water(if the bucket were full you would see water pouring out)
Archimedes simply stated that: Upthrust,
U=mg
- Example: One way to measure the volume of any irregular object (in your case, a stone) is to submerge it completely under water and measure the change in the height of the water level. This change in the water level (let's say it goes from 50 mL to 65 mL) indicates that the stone has a volume of 15 mL.
- Example:Subtract the first volume from the second volume to calculate the volume of the stone. For example, if you recorded 40 fluid ounces the first time, and 50 fluid ounces the second time, the stone volume is 10 fluid ounces.
1. Height of the object is 5 cm
The object distance (u) is 18 cm
The focal length is 10 cm
To calculate the image distance (v) we use;
1/f=1/v+1/u
1/v = 1/10-1/18
= 8/180
Therefore v = 180/8
= 22.5 cm
2. Height of the image
Magnification = v/u = Height of image/height of object
Therefore;
22.5/ 18= x/5
x = (22.5×5)/18
x = 6.25 cm
Therefore; the height of the image is 6.25 cm