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Answer:
a) a = 3.06 10¹⁵ m / s
, b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N
Explanation:
This exercise will average solve using the moment relationship.
a ) let's use the relationship between momentum and momentum
I = ∫ F dt = Δp
F t = m
- m v₀
F = m (v_{f} -v₀o) / t
in the exercise indicates that the speed module is the same, but in the opposite direction
F = m (-2v) / t
if we use Newton's second law
F = m a
we substitute
- 2 mv / t = m a
a = - 2 v / t
let's calculate
a = - 2 4.59 10²/3 10⁻¹³
a = 3.06 10¹⁵ m / s
b) F= m a
F= 4.68 10⁻²⁶ 3.06 10¹⁵
F= 1.43 10⁻¹⁰ N
c) if we hit the wall for 1015 each exerts a force F
F_total = n F
F_total = n m a
F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵
F_total = 14.32 10⁻²⁶ N
Answer:
TAJUK
Explanation:
Sebab saya suka makan ayam goreng, esok saya nak pesan daripada kedai pak abu, terima kasih bosku
Given gravitational potential energy when he's lifted is 2058 J.
Kinetic energy is transferred to the person.
Amount of kinetic energy the person has is -2058 J
velocity of person = 7.67 m/s².
<h3>
Explanation:</h3>
Given:
Weight of person = 70 kg
Lifted height = 3 m
1. Gravitational potential energy of a lifted person is equal to the work done.

Gravitational potential energy is equal to 2058 Joules.
2. The Gravitational potential energy is converted into kinetic energy. Kinetic energy is being transferred to the person.
3. Kinetic energy gained = Potential energy lost = 
Kinetic energy gained by the person = (-2058 kg.m/s²)
4. Velocity = ?
Kinetic energy magnitude= 
Solving for v, we get

The person will be going at a speed of 7.67 m/s².