Answer:
Technician A
Explanation:
camshafts rotate at one-half the crankshaft speed
the object distance of both lenses are positive.
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:
P = 4000 [Pa]
Explanation:
Pressure is defined as the relationship between Force and the area where the body rests.
The support area is equal to:
![A=50*20=1000[cm^{2} ]](https://tex.z-dn.net/?f=A%3D50%2A20%3D1000%5Bcm%5E%7B2%7D%20%5D)
But we must convert from square centimeters to square meters.
![1000[cm^{2}]*\frac{1^{2}m^{2} }{100^{2}m^{2} }=0.1[m^{2} ]](https://tex.z-dn.net/?f=1000%5Bcm%5E%7B2%7D%5D%2A%5Cfrac%7B1%5E%7B2%7Dm%5E%7B2%7D%20%20%7D%7B100%5E%7B2%7Dm%5E%7B2%7D%20%20%7D%3D0.1%5Bm%5E%7B2%7D%20%5D)
And the pressure is:
![P=\frac{F}{A} \\P=400/0.1\\P=4000[N/m^{2} ]or 4000[Pa]](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BF%7D%7BA%7D%20%5C%5CP%3D400%2F0.1%5C%5CP%3D4000%5BN%2Fm%5E%7B2%7D%20%5Dor%204000%5BPa%5D)
Answer:
The car would travel after applying brakes is, d = 14.53 m
Explanation:
Given that,
The time taken to apply brakes fully is, t = 0.5 s
The velocity of the car, v = 29.06 m/s
The distance traveled by the car in 0.5 s, d = ?
The relation between the velocity, displacement, and time is given by the formula
d = v x t m
Substituting the values in the above equation,
d = 29.06 m/s x 0.5 s
= 14.53 m
Therefore, the car would travel after applying brakes is, d = 14.53 m