Due tomarrow! 1.car has a mass of 1.72×103kg.If the force acting on the car is 6.44×103N to the east,what is the car’s acceleration?Answer units of m/s2
3.What is the acceleration of a 56 kg block of
cement when pulled sideways with a net force
of 526 N?Answer in units of m/s2
4.Sprinting near the end of a race,a runner
with a mass 62 kg accelerates from a speed of
12m/s to a speed of 13m/s in 31s.To gain
speed the runner produces a backward force
on the ground,so that the ground pushes the
runner forward,providing the force necessary
for the acceleration.Calcullate average force.

Question

Due tomarrow! 1.car has a mass of 1.72×103kg.If the force acting on the car is 6.44×103N to the east,what is the car’s acceleration?Answer units of m/s2
3.What is the acceleration of a 56 kg block of
cement when pulled sideways with a net force
of 526 N?Answer in units of m/s2
4.Sprinting near the end of a race,a runner
with a mass 62 kg accelerates from a speed of
12m/s to a speed of 13m/s in 31s.To gain
speed the runner produces a backward force
on the ground,so that the ground pushes the
runner forward,providing the force necessary
for the acceleration.Calcullate average force.

deepolisnoob · Answer

1.) $$F=ma$$
Your $$F=6.44*10^{3}N$$
and your 
$$m=1.72*10^{3}kg$$
You plug those values into you equation,
$$ 6,440N=1,720kg*a$$
I simplified the scientific notation.
Get your a by itself, so divide both sides by 1,720kg.
$$\frac{ 6,440N }{ 1,720kg }=a$$
$$3.74m/s^{2}=a$$

3.) $$F=ma$$
$$526N=56kg*a$$
$$\frac{526N}{56kg}=a$$
$$a=5.39m/s^{2}$$

4.) $$F=ma$$ $$a=\frac{\Delta v}{\Delta t}$$
$$a=\frac{13m/s-12m/s}{31s}$$
$$a=\frac{1m/s}{31s}$$
$$a=0.032m/s^{2}$$
$$F=62kg*0.032m/s^{2}$$
$$F=1.984N$$

anonymous · Answer

it say # 1 was wrong and # 3 was too and the answer to 3 is figured out the same was but the answer is 9.39285