An unbalanced force is required to accelerate an object according to Newton's Second Law of Motion.
<h3>
What does Newton's Second Law of Motion state?</h3>
It states that the force applied to the object is equal to the product of mass and acceleration.
- An object will accelerate when the net force applied on the object is more than zero or unbalanced.
- The acceleration is the change in the direction or speed of the object. To achieve acceleration the force must be greater in a direction.
- When force is greater in one the object move in that direction which is known as acceleration.
Therefore, an unbalanced force is required to accelerate an object according to Newton's Second Law of Motion.
Learn more about Newton's Second Law of Motion.:
brainly.com/question/25810165
Answer:
White hole is an impossible object in universe. ... This means that in a hypothetical universe where there is a black and a white hole, in a short time after their first interaction the white hole will become another black hole so that the system will end up with two black holes.
Answer:
0.61°
Explanation:
Since the box move at constant velocity, it means there is no acceleration then we can say it has a balanced force system.
Pulling force= resistance force
From the formula for pulling force,
F(x)= Fcos(θ)
= 425×cos(35.2)
=347N
The force exerted downward at an angle of 35.2° below the horizontal= Fsin(θ)= 425sin(35.2)
=425×0.567=245N
Resistance force= (325N+ 245N) (α)= 570N(α)
We can now equates the pulling force to resistance force
570 (α)= 347N
(α)= 347/570
= 0.61
Explanation:
4a)the displacement is the distance moved in a direction but since no direction is given, the displacement is equal to the distance
b) the distance moved is 400m because that's the length of the track
Answer:
a. 2v₀/a b. 2v₀/a
Explanation:
a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.
Since the dragster starts from rest with an acceleration, a, using
s' = ut + 1/2at² where u = 0 and s' = distance moved by dragster
s' = 0t + 1/2at²
s' = 1/2at²
Since the distance moved by me and the dragster must be the same,
s = s'
v₀t. = 1/2at²
v₀t. - 1/2at² = 0
t(v₀ - 1/2at) = 0
t= 0 or v₀ - 1/2at = 0
t= 0 or v₀ = 1/2at
t= 0 or t = 2v₀/a
So the maximum time tmax = 2v₀/a
b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s = v₀(2v₀/a)
= 2v₀/a
