To determine the energy equivalent of an object, we use the famous equation of Einstein which is E=mc^2 where m is the mass of the object and c is the speed of light (3x10^8 m/s). We calculate as follows:
E = mc^2
E = 1.83 kg (3x10^8 m/s)^2
E = 1.647x10^17 J
Answer:
Symptoms of excessive stress include all of the following EXCEPT: increased energy level.
Answer:
The magnitude of F1 is

The magnitude of F2 is

And the direction of F2 is

Explanation:
<u>Net Force
</u>
Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.
The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.
Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

The sum of the vertical components of F1 and F2 must equal the total force Ft

Solving for
in the first equation






The magnitude of F1 is

The magnitude of F2 is

And the direction of F2 is

Answer:
3.536*10^-6 C
Explanation:
The magnitude of the charge is expresses as Q = CV
C is the capacitance of the capacitor
V is the voltage across the capacitor
Get the capacitance
C = ε0A/d
ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m
A is the area = 0.2m²
d is the plate separation = 0.1mm = 0.0001m
Substitute
C = 8.84 x 10-12 * 0.2/0.0001
C = 1.768 x 10-8 F
Get the potential difference V
Using the formula for Electric field intensity
E = V/d
2.0 × 10^6 = V/0.0001
V = 2.0 × 10^6 * 0.0001
V = 2.0 × 10^2V
Get the charge on each plate.
Q = CV
Q = 1.768 x 10-8 * 2.0 × 10^2
Q = 3.536*10^-6 C
Hence the magnitude of the charge on each plate should be 3.536*10^-6 C