What is the magnitude of an electric field in which the electric force it exerts on a proton is equal in magnitude to the proton
1 answer:
Answer:
5.58 × 10⁻¹¹ N/C
Explanation:
Electric force on a charge, F = qE
where,
F is the electric force in Newton (N)
q is the unit charge in Coulomb (C)
E is the electric field intensity in N/C
Also, the weight of an object, W = mg
Where,
w is weight of an object in N
m is mass in kg
g is acceleration due to gravity in m/s²
Now from the question given, the electric force exerted on the proton is equal in magnitude to the proton's weight.
⇒ F = w
qE = mg
making electric field, E the subject of the equation,
E = (9.11 × 10⁻³¹ kg × 9.8 m/s) ÷ 1.6 × 10⁻¹⁹ C
E = 55.8 × 10⁻¹² N/C
E = 5.58 × 10⁻¹¹ N/C
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Answer:
The second projectile was 1.41 times faster than the first.
Explanation:
In the ballistic pendulum experiment, the speed (v) of the projectile is given by:
<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum. </em>
To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:
(1)
<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively. </em>
Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):
Therefore, the second projectile was 1.41 times faster than the first.
I hope it helps you!