Answer:
Exercise 1;
The centripetal acceleration is approximately 94.52 m/s²
Explanation:
1) The given parameters are;
The diameter of the circle = 8 cm = 0.08 m
The radius of the circle = Diameter/2 = 0.08/2 = 0.04 m
The speed of motion = 7 km/h = 1.944444 m/s
The centripetal acceleration = v²/r = 1.944444²/0.04 ≈ 94.52 m/s²
The centripetal acceleration ≈ 94.52 m/s²
Answer:
3. all of the above
4. d=m/v
5. gas has well separated particles with no particular arrangement
liquid has particles close together but still with no particular arrangement
and solid has particles that are tightly packed, usually in a regular pattern
6. (idk what number it is but)
the density of the rock is 2.4 g/ml
d=m/v so d=12/v according to the question the mass is 12
to find volume subtract 20-15 because before the rock was 15 and after the rock was 20, so then you get 5 for volume
d=12/5 and when you calculate that, you would get 2.4g/ml ; hope this helped :-)
Answer:
7.1934 x 10^12 V/m.s
Explanation:
In order to do this exercise, you need to use the correct formula. Besides that, we need to identify our data.
First we have the radius of the plates which are circular, and it's 0.1 m. The current of the loop (I) is 2.0 A, and the radius of the loop is 0.2 m.
Now with this data, we use the next formula:
I = dE/dt Eo A
Where:
dE/dt = rate of electric field
Eo = constant of permittivity of free space
A = Area of circle
Solving for dE/dT:
dE/dt = I / Eo*A
Now, the area of the circle is A = πr²
A = 3.1416 * (0.1)² = 0.031416 m²
Now solving the electric field:
dE/dt = 2 / (8.85x10^-12 * 0.031416)
dE/dt = 7.1934 x 10^12 V/m.s
C. It was mainly concerned about wildlife.
Birds and other wild animals can be scared away from the area. This would affect the ecosystem and the local webchains.
Hey!
Given ,
Acceleration, a = 5 m/s^2
time , t = 10 seconds
Initial velocity,u = 0
Final velocity , v = ?
We have ,
v = u + at
=> v = (0)+(5)(10)
=> v = 50 m/s
