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3 years ago

what is the mass of a cannon ball if a force 2500 N gives the cannon ball an acceleration of 200 m/s squared?

Physics

1 answer:

nignag [31]3 years ago

8 0

Using Newton's second law of motion:

F=ma ; [ F = force (N: kgm/s^2);m= mass (kg); a = acceleration (m/s^2)

Given: Find: Formula: Solve for m:

F: 2500N mass:? F=ma Eq.1 m=F/a Eq. 2

a= 200m/s^2

Solution:

Using Eq.2

m= (2500 kgm/s^2)/ (200m/s^2) = 12.5 kg

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3 years ago

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3 years ago

Under ideal conditions (no atmospheric interference of any kind), if I hit a golf ball at an angle of 25 degrees at an initial s

g100num [7]

Answer:

The required angle is (90-25)° = 65°

Explanation:

The given motion is an example of projectile motion.

Let 'v' be the initial velocity and '∅' be the angle of projection.

Let 't' be the time taken for complete motion.

Let 'g' be the acceleration due to gravity

Taking components of velocity in horizontal(x) and vertical(y) direction.

$v_{x}$ = v cos(∅)

$v_{y}$ = v sin(∅)

We know that for a projectile motion,

t = $\frac{2vsin(∅)}{g}$

Since there is no force acting on the golf ball in horizonal direction.

Total distance(d) covered in horizontal direction is -

d = $v_{x}$ ×t = vcos(∅)× $\frac{2vsin(∅)}{g}$ = $\frac{v^{2}sin(2∅) }{g}$ .

If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -

α +β = 90° as-

d = $\frac{v^{2}sin(2α) }{g}$ = $\frac{v^{2}sin(2[90-β]) }{g}$ = $\frac{v^{2}sin(180-2β) }{g}$ = $\frac{v^{2}sin(2β) }{g}$ .

∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.

∴ If α = 25° , then

β = 90-25 = 65°

∴ The required angle is 65°.

5 0

3 years ago

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