To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.
In turn, we will resort to the application of Newton's second law.
PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

Where,
X = Desplazamiento
V = Velocity
t = Time
In this case there is no initial displacement or initial velocity, therefore

Clearing for time,



PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:
F = ma
Where,
m=mass
a = acceleration
Acceleration can also be written as,

Then





Negative symbol is because the force is opposite of the direction of moton.
PART C) Acceleration through kinematics equation is defined as




The gravity is equal to 0.8, then the acceleration is


Answer: 15.8
Explanation:
You are given that the
Object distance U = 32 cm
Focal length F = 30.1 cm
First calculate the image distance V by using the formula
1/F = 1/U + 1/V
Substitute F and V into the formula
1/30.1 = 1/32 + 1/V
1/V = 1/30.1 - 1/32
1/V = 0.00197259
Reciprocate both sides
V = 506.94 cm
Magnification M is the ratio of image distance to object distance.
M = V/U
substitute the values of V and U into the formula
M = 506.94/32
M = 15.8
Therefore, the magnification of the image is 15.8 or approximately 16.
By afforestation.
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Answer:
Force on front axle = 6392.85 N
Force on rear axle = 8616.45 N
Explanation:
As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels
Now we know that



now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle
so we can write torque balance about its center of mass


now from above equation

now we have

now the other force is given as

Answer:
The total mechanical energy does not change if the value of the mass is changed. That is, remain the same
Explanation:
The total mechanical energy of a spring-mass system is equal to the elastic potential energy where the object is at the amplitude of the motion. That is:
(1)
k: spring constant
A: amplitude of the motion = 2.0cm
As you can notice in the equation (1), the total mechanical energy of the system does not depend of the mass of the object. It only depends of the amplitude A and the spring constant.
Hence, if you use a mass of 0.40kg the total mechanical energy is the same as the obtained with a mas 0.20kg
Remain the same