Temperature is a measure of the average kinetic energy of the particles of a substance.
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Answer:
a) y₂ = 49.1 m
, t = 1.02 s
, b) y = 49.1 m
, t= 1.02 s
Explanation:
a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero
² =
² - 2 g (y –yo)
The origin of the coordinate system is on the floor and the ball is thrown from a height
y-yo =
=
- g t
t =
/ g
t = 10 / 9.8
t = 1.02 s
b) the maximum height
y- 44.0 =
² / 2 g
y - 44.0 = 5.1
y = 5.1 +44.0
y = 49.1 m
The time is the same because it does not depend on the initial height
t = 1.02 s
Answer:
a) 5.851× 10¹⁰m/s²
b) 2.411×10⁻¹¹s
c) 1.70×10⁻¹¹m
d) 1.661×10⁻²⁷KJ
Explanation:
A proton in the field experience a downward force of magnitude,
F = eE. The force of gravity on the proton will be negligible compared to the electric force
F = eE
a= eE/m
= 1.602×10⁻¹⁹ × 610/1.67×10⁻²⁷
= 5.851× 10¹⁰m/s²
b)
V = u + at
u= 0
v= 1.4106m/s
v= (0)t + at
t= v/a
= 1.4106m/s/5.851 ×10¹⁰
= 2.411×10⁻¹¹s
c)
S = ut + at²
= (o)t + 5.851×10¹⁰×(2.411×10⁻¹¹)²
= 1.70×10⁻¹¹m
d)
Ke = 1/2mv²
= (1.67×10⁻²⁷×)(1.4106)²/2
= 1.661×10⁻²⁷KJ
Answer:Expression given below
Explanation:
Given mass of spring
Compression in the spring
Let the spring constant be K
Using Energy conservation
potential energy stored in spring =Kinetic energy of Block


now conserving momentum


where
is the final velocity
Explanation:
a. Net force is mass times acceleration (Newton's second law).
∑F = ma
∑F = (5.0 kg) (2.0 m/s²)
∑F = 10 N
b. The net force is the sum of the individual forces.
10 N = F − 5 N
F = 15 N
c. Friction force here is mgμ.
mgμ = 5 N
(5.0 kg) (10 m/s) μ = 5 N
μ = 0.1