First, we have a change in the velocity from 85 to 164 m/s in 10 sec.
Then, we calculate the <u>acceleration </u>as:

Hence we need to calculate the velocity of the space vehicle at t = 2 sec using the first equation of motion:

Then, using the second equation of motion to calculate the distance:


Answer:
The horizontal distance covered by the firework will be 
Explanation:
Let acceleration due to gravity on the planet be g, initial velocity of the firework be u and angle made with the horizontal be ∅.
writing equation of motion in vertical direction:


and 
therefore 
writing equation of motion in horizontal direction:


therefore the equation becomes 
therefore horizontal distance traveled =
Answer:


Explanation:
<u>Net Force And Acceleration
</u>
The Newton's second law relates the net force applied on an object of mass m and the acceleration it aquires by

The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N
The components of the first force are


The components of the second force are


The net force is


The magnitude of the net force is


The acceleration has a magnitude of



The direction of the acceleration is the same as the net force:


Answer:
Explanation:
In wheel and axle. …with the system is the velocity ratio, or the ratio of the velocity (VF) with which the operator pulls the rope at F to the velocity at which the weight W is raised (VW). This ratio is equal to twice the radius of the large drum divided by the difference…