For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:
a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
t is the time
The solution is as follows;
a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
a = 36,000 mph²
2(36,000 mph²)(x) = 60² - 30²
Solving for x,
x = 0.0375 miles
Answer:
Objective: It is raining. Subjective: I love the rain!
Explanation:
Anything objective sticks to the facts, but anything subjective has feelings. Objective and subjective are opposites.
(Hope this helps can I pls have brainlist (crown)☺️)
Answer:
a) v₁fin = 3.7059 m/s (→)
b) v₂fin = 1.0588 m/s (→)
Explanation:
a) Given
m₁ = 0.5 Kg
L = 70 cm = 0.7 m
v₁in = 0 m/s ⇒ Kin = 0 J
v₁fin = ?
h<em>in </em>= L = 0.7 m
h<em>fin </em>= 0 m ⇒ U<em>fin</em> = 0 J
The speed of the ball before the collision can be obtained as follows
Einitial = Efinal
⇒ Kin + Uin = Kfin + Ufin
⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0
⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))
⇒ v₁fin = 3.7059 m/s (→)
b) Given
m₁ = 0.5 Kg
m₂ = 3.0 Kg
v₁ = 3.7059 m/s (→)
v₂ = 0 m/s
v₂fin = ?
The speed of the block just after the collision can be obtained using the equation
v₂fin = 2*m₁*v₁ / (m₁ + m₂)
⇒ v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)
⇒ v₂fin = 1.0588 m/s (→)
Answer:
ya we can write the imaginary character's name .
So that we can identify these imaginary people, as we cannot simply write the conversation and leave it .
Or maybe sometimes the reader will get confused as there is no name for the two people .
So, i suggest that you should write the names
Explanation:
You can even ask to your class teacher for further clarification
Given:
u(initial velocity)=0
a=5.54m/s^2
v(final velocity)=2 m/s
v=u +at
Where v is the final velocity.
u is the initial velocity
a is the acceleration.
t is the time
2=0+5.54t
t=2/5.54
t=0.36 sec
