Answer:
the change in thermal energy of the projectile is 43.8 kJ
Explanation:
Given;
mass of the object, m = 5kg
initial velocity of the projectile, v₁ = 200 m/s
final velocity of the projectile, v₂ = 150 m/s
To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.
Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²
KE = Δ¹/₂mv² = ¹/₂m(v₁²-v₂²)
KE = ¹/₂ × 5(200²-150²) = 2.5(17500) = 43750 J = 43.8 kJ
Therefore, the change in thermal energy of the projectile is 43.8 kJ
Hi!
1 decimeter = 100 millimeters.
Therefore 2 decimeters = 200 millimeters.
Explanation:
Given:
t = 20 seconds
x = 3000 m
y = 450 m
a) To find the vertical component of the initial velocity
, we can use the equation

Solving for
,



b) We can solve for the horizontal component of the velocity
as

or

Answer:
A = [kg]
B = [m/s²]
Explanation:
E = ½ Av² + Bmx
Substitute the units:
[J] = ½ A [m/s]² + B [kg] [m]
A Joule written in base units is:
1 J = 1 Nm = 1 kg m²/s²
Each term must have the same units.
[kg m²/s²] = A [m/s]²
[kg m²/s²] = A [m²/s²]
A = [kg]
[kg m²/s²] = B [kg] [m]
B = [m/s²]
Answer:
The correct answer is D. s₂=4s₁
Explanation:
The distance of a particle is given by:

where
s₀ is the initial position when t=0
v₀ is the initial speed when t=0
a is the constant acceleration
t is the time in seconds
Then, the position s₁ is given by:

As the particle starts from rest v₀=0 and we consider s₀=0, s₁ will be:

Furthermore, the position s₂ is:

In this case t₁=10 s and t₂=20 s (10 seconds later). In other words t₂=2t₁.
We replace the value of t₂ in the second equation (s₂):

Finally, we divide s₂ by s₁ to get the ratio:

