Answer:243joules
Explanation:
Mass(m)=54kg
Velocity(v)=3m/s
Kinetic energy =(m x v^2)/2
Kinetic energy =(54 x 3^2)/2
Kinetic energy =(54 x 9)/2
Kinetic energy =486/2
Kinetic energy =243joules
Answer:
The Kinetic energy and mass are _directly_ proportional.
Explanation:
We know that Kinetic Energy is basically termed as the capacity of a body to do work.
Kinetic energy is often used to associate with moving objects, therefore, K.E is normally termed as the energy of motion.
The formula of K.E of an object of mass and velocity is defined
K.E = 1/2mv²
From the formula, it is clear that K.E is directly proportional to its mass and also directly proportional to the square of its velocity.
For example,
If A toy plane with a mass of 10 kg is flying at 20 m/s. Its K.E will be:
K.E = 1/2mv²
= 1/2(10)(20)²
= 1/2(10)(400)
= 5(400)
= 2000 J
Now, let suppose, if we double the mass of a toy plane i.e.
m = 20 kg
so
K.E = 1/2mv²
= 1/2(20)(20)²
= 1/2(20)(400)
= 10(400)
= 400 J
Therefore, the K.E is doubled when doubled the mass.
Therefore, the Kinetic energy and mass are _directly_ proportional.
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:
brainly.com/question/2004529
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