Where danger is and how to avoid it as well as to locate other whales
Answer:
Magnetic force is equal to 
Explanation:
We have given electron is accelerated with a potential difference of 81700 volt.
Magnetic field B = 0.508 T
Angle between magnetic field and velocity 
Mass of electron 
Charge on electron 
By energy conservation.



Magnetic force on electron



Answer:
q = 400 nC
the correct answer is b
Explanation:
The expression for the electric potential of a point charge is
V = k q / r
they ask us for the electrical charge
q = V r / k
let's calculate
Q = 600 6.0 / 9 10⁹
Q = 4 10⁻⁷ C
let's reduce to nC
Q = 4 10⁻⁷ C (10⁹ nC / 1C)
q = 4 10² nC = 400 nC
the correct answer is b
Traslate
La expresión para el potencial eléctrico de una carga puntual es
V = k q/r
nos piden la carga eléctrica
q= V r /k
calculemos
Q= 600 6,0 / 9 10⁹
Q= 4 10⁻⁷ C
reduzcamos a nC
Q = 4 10⁻⁷ C(10⁹ nC/1C )
q = 4 10² nC = 400 nC
la respuesta correcta es b
Newton's second law allows calculating the response for the person's acceleration while leaving the trampoline is:
-4.8 m / s²
Newton's second law says that the net force is proportional to the product of the mass and the acceleration of the body
F = m a
Where the bold letters indicate vectors, F is the force, m the masses and the acceleration
The free body diagram is a diagram of the forces without the details of the body, in the attached we can see the free body diagram for this system
-W = m a
Whera
is the trampoline force
Body weight is
W = mg
We substitute
- mg = ma
a =
Let's calculate
a = 
a = -4.8 m / s²
The negative sign indicates that the acceleration is directed downward.
In conclusion using Newton's second law we can calculate the acceleration of the person while leaving the trampoline is
-4.8 m / s²
Learn more here: brainly.com/question/19860811
Answer:
Directly Proportional
Explanation:
Gravitational force can be calculated with the equation F = g(m1 * m2)/ r^2
So if we increase mass, force will also increase because mass is in the numerator.