Answer:
A first-class lever has the fulcrum between the load and the effort. A second-class lever has the load between the effort and the fulcrum. A third-class lever has the effort between the load and the fulcrum. A see-saw is an example of a first-class lever.
Explanation:
<span>FIRST SECTION
You should use the formula for uniformly accelerated linear movement.
Initial speed is 0 because it starts from rest.
d=(1/2)*a*t^2+vo*t =(1/2)*(4.0 m/s^2)*(3s)^2+0*3s=(1/2)*(4.0 m/s^2)*3^2*s^2+0=2.0 m*9=18m
You can calculate the final speed with the other formula:
v=a*t+vo=(4.0 m/s^2)*(3s)+0=(4.0 m/s)*(3)=12m/s
SECOND SECTION
You should use the formula for uniform linear movement.
Velocity is a constant: it remains in 12m/s.
d=v*t=12m/s*2s=12m*2=24m
THIRD SECTION
We should use the same formulas as the first section, but with different numbers.
Initial velocity will be 12m/s, and then velocity will start to decrease until it gets to 0.
We don’t know what the time is for this section.
Acceleration is negative, because it’s slowing down.
v=a*t+vo
0=-3.0 m/s^2*t+12m/s
3.0 m/s^2*t=12m/s
t=(12m/s)/(3.0 m/s^2)=4(1/s)/(1/s^2)=4s^2/s=4s
Now let’s use that time in the other formula:
d=(1/2)*a*t^2+vo*t =(1/2)*(-3.0 m/s^2)*(4s)^2+(12m/s)*3s=(-1.5 m/s^2)*4^2*s^2+12*3m*s/s=-1.5 m*4^2+36m=-1.5*16m+36m=-24m+36m=12m
Now let’s add the 3 stages:
d=18m+24m+12m=54m
</span>
(AB) & (CD)
Most machines are fueled by gasoline and electricity
Answer:
<h2>17.1 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question
3800 g = 3.8 kg
We have
force = 3.8 × 4.5
We have the final answer as
<h3>17.1 N</h3>
Hope this helps you
