Answer: Lever
A wheelbarrow consists of a lever to be able to lift the material, and a wheel to be able to move it horizontally. So in a sense, a wheelbarrow is a complex machine consisting of two simple machines.
The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

Substitute numerical values:

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
Learn more about Gaussian sphere here:
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Density is the ratio of a substance's mass to its volume. On the other hand, according to Archimedes' principle, the volume of water displaced is equal to the volume of the object placed on the water. Thus, the density of the metal is equal to 8.39 mL. So, the density would be
Density = 32.5 g/8.39 mL = 3.87 g/mL
So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.
<h3>Introduction</h3>
Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

With the following condition :
= angular frequency (rad/s)
= change of angle value (rad)- t = interval of the time (s)
<h3>Problem Solving</h3>
We know that :
= change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.- t = interval of the time = 54.9 s.
What was asked :
= angular frequency = ... rad/s
Step by step :



<h3>Conclusion :</h3>
So, the angular frequency of the blades approximately 36.43π rad/s.
E=(mV^2)/2
m=1000kg, V=20m/s
then, E=(1000kg*(20m/s)^2)/2
E=(1000*400)/2 J = 200000J