Complete question
A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.
Answer:
The car must travel 68.94 meters.
Explanation:
First, we are going to find the acceleration of the car using Newton's second Law:
(1)
with m the mass , a the acceleration and 
the net force forces that is:
(2)
with F the force provided by traction and f the resistive force:
(2) on (1):
solving for a:
Now let's use the Galileo’s kinematic equation
(3)
With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and 
the time took to achieve that velocity, solving (3) for :
Answer:
5.01 J
Explanation:
Info given:
mass (m) = 0.0780kg
height (h) = 5.36m
velocity (v) = 4.84 m/s
gravity (g) = 9.81m/s^2
1. First, solve for Kinetic energy (KE)
KE = 1/2mv^2
1/2(0.0780kg)(4.84m/s)^2 = 0.91 J
so KE = 0.91 J
2. Next, solve for Potential energy (PE)
PE = mgh
(0.0780kg)(9.81m/s^2)(5.36m) = 4.10 J
so PE = 4.10 J
3. Mechanical Energy , E = KE + PE
Plug in values for KE and PE
KE + PE = 0.91J + 4.10 J = 5.01 J
Formulas change from F to degree C : C = ( F - 32 )/1.8
so we have (77-32)/1.8 = 25 oC
ok done. Thank to me :>
Plugging in for the Earth's mass and for G, we have 11.2 km/s for the escape velocity for an object launched from the Earth's surface. This is about 25,000 miles per hour
Answer:
A- Astronomical body
C- Galaxy
D- Comet
B- Moon
Hope this helps you! Have a great day!
