I believe the answer is b) slowly heating the surface
Answer:
The object will travel 675 m during that time.
Explanation:
A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.
In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.
In this case, the position is calculated using the expression:
x = xo + vo*t + ½*a*t²
where:
- x0 is the initial position.
- v0 is the initial velocity.
- a is the acceleration.
- t is the time interval in which the motion is studied.
In this case:
- x0= 0
- v0= 0 because the object is initially stationary
- a= 6
- t= 15 s
Replacing:
x= 0 + 0*15 s + ½*6 *(15s)²
Solving:
x=½*6 *(15s)²
x=½*6 *225 s²
x= 675 m
<u><em>
The object will travel 675 m during that time.</em></u>
The force needed to accelerate a 2500 kg car at a rate of 4m/s² will be 10,000 N.Option B is correct.
<h3>What is force?</h3>
Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.
Given data;
Force,F = ?
Mass,m = 2500 kg
Acceleration,a = 4 m/s²
The force is found as;
F=ma
F=2500 kg × 4 m/s²
F= 10,000 N
The force needed to accelerate a 2500 kg car at a rate of 4m/s² will be 10,000 N.
Hence option B is correct.
To learn more about the force refer to the link;
brainly.com/question/26115859
#SPJ1
When the oscillator is at maximum extension, we know all of its energy is in Potential Energy, so if the total oscillation energy is 4.1 J, we know that at maximum displacement of 0.2 m, that
<span>energy = 1/2 kA^2 where A= 0.2 m </span>
<span>k= 2E / A^2 = 2*4.1 J /0.2^2=200 N/m </span>
<span>the frequency of oscillation is (1/2pi) sqrt[k/m] </span>
<span>knowing k and m, we can substitute values and find frequency</span>
