The electric force on the proton is:
F = Eq
F = electric force, E = electric field strength, q = proton charge
The gravitational force on the proton is:
F = mg
F = gravitational force, m = proton mass, g = gravitational acceleration
Since the electric force and gravitational force balance each other out, set their magnitudes equal to each other:
Eq = mg
Given values:
q = 1.60×10⁻¹⁹C, m = 1.67×10⁻²⁷kg, g = 9.81m/s²
Plug in and solve for E:
E(1.60×10⁻¹⁹) = 1.67×10⁻²⁷(9.81)
E = 1.02×10⁻⁷N/C
There are two possible answers that I can see, C or D
Answer:
F = 12.5N
Explanation:
Force (F) = Mass (m) x Acceleration (a)
F = ma
F = (2.5kg) x (5m/s^2)
F = 12.5N
Answer: 0.73 g/cm3 and YES the questionable ball in this range of acceptable density
Explanation:
Answer:
d) I and III only.
Explanation:
Let be 
and 
the masses of the two laboratory carts and let suppose that 
. The expressions for each kinetic energy are, respectively:
and 
.
After some algebraic manipulation, the following relation is constructed:
Since 
, then 
. That is to say, 
.
The expressions for each linear momentum are, respectively:
and 
Since , then 
. Which proves that statement I is true.
According to the Impulse Theorem, the impulse needed by cart I is greater than impulse needed by cart II, which proves that statement II is false.
According to the Work-Energy Theorem, both carts need the same amount of work to stop them. Which proves that statement III is true.
