Answer:
Both objects travel the same distance.
(c) is correct option
Explanation:
Given that,
Mass of first object = 4.0 kg
Speed of first object = 2.0 m/s
Mass of second object = 1.0 kg
Speed of second object = 4.0 m/s
We need to calculate the stopping distance
For first particle
Using equation of motion
Where, v = final velocity
u = initial velocity
s = distance
Put the value in the equation
....(I)
Using newton law
Now, put the value of a in equation (I)
Now, For second object
Using equation of motion
Put the value in the equation
....(I)
Using newton law
Now, put the value of a in equation (I)
Hence, Both objects travel the same distance.
The distance covered in the last second of motion is 737.5 m
Explanation:
The motion of the car is a uniformly accelerated motion, so we can use the suvat equations.
First of all, we have to find the velocity of the car when the last second of motion starts, that is the velocity of the car after t = 29 s. We can use the equation:
v = u + at
where
u = 0 is the initial velocity
is the acceleration
Substituting t = 29 s,
Now we can find the distance covered in the last second of motion by using
where
u = 725 m/s is the velocity at the beginning of the last second
t = 1 s is the time interval considered
is the acceleration
Substituting,
Note that the acceleration of 
is not realistic for a car, but I still have used the data of the problem.
Learn more about acceleration:
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Answer:
=9.72 m/s
Explanation:
From the Newton's laws of motion;
x=2(v²cos∅sin∅)/g
Using geometry we see that 2 cos∅sin∅ = sin 2∅
Therefore, x= (v²sin 2∅)g, where v is the take off speed x the range and ∅ the launch angle.
Making v the subject of the formula we obtain the following equation.
v=√{xg /(sin 2∅)}
x=7.80
∅=27.0
v=√{7.8×9.8/sin(27×2)}
v=√94.485
v=9.72 m/s
Answer:
The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.
Explanation:
The answer is true. The table does show an object moving with changing speed.
