Answer:
63.750KeV
Explanation:
We are given that
Initial velocity of second electron,
Radius,

1 m=100 cm
Magnetic field,B=0.0370 T
We have to determine the energy of the incident electron.
Mass of electron,
Charge on an electron,
Velocity,
Using the formula
Speed of electron,
Speed of second electron,

Kinetic energy of incident electron=
Kinetic energy of incident electron=
Kinetic energy of incident electron=
1KeV=1000eV
Answer:
This values shows a right angle triangle
Explanation:
Given;
a vector 4.0 km due East
a 3.0 km due north
the resultant vector is 5.0 km
The resultant vector can be obtained by Pythagoras theorem if the vectors form a right angle triangle.
R² = 4² + 3²
R² = 16 + 9
R² = 25
R = √25
R = 5 km (right angle triangle proved)
Therefore, this values shows a right angle triangle
Answer: A) mass on earth surface = 5.91kg
B) mass on surface of jupiter = 5.91kg
C) weight on surface of jupiter = 10.697N
Explanation:
The relationship between weight (W), mass (m) and acceleration due gravity (g) is given below
W=mg
From the question, g= 9.8m/s² and weight on the surface on the earth is 58N
A) The mass of watermelon on earth is
m = 58/ 9.8 = 5.91kg
B) the mass of the watermelon on jupiter is 5.91kg.
You will notice this is the same as the mass of watermelon on earth and that is so because mass is a scalar quantity that does not depends on the distance away from the center of the earth (unlike weight which is a vector) thus making it constant all through any location.
C) mass of watermelon is 5.91kg, g=9.8m/s² weight of watermelon on jupiter is given below as
W = mg
W = 5.91 x 9.8
= 10.697N.
A :-) it was given the name Newton (N). from this, the derived unit of energy (or work) is defined ,as the work produced when the unit of force causes a displacement equal to the unit of length of its point of application along its direction . It was given the name Joule (J).
Answer:
The y-component of the electric force on this charge is 
Explanation:
<u>Given:</u>
- Electric field in the region,

- Charge placed into the region,

where,
are the unit vectors along the positive x and y axes respectively.
The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

Thus, the y-component of the electric force on this charge is 