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GaryK [48]
2 years ago
11

20.0 -kg cannonball is fired from a cannon with muzzle speed of 1000m/s at an angle of 37.0° with the horizontal. A second ball

is fired at an angle of 90.0°. Use the isolated system model to find.(b) the total mechanical energy of the ball Earth system at the maximum height for each ball. Let y=0 at the cannon.
Physics
1 answer:
Dovator [93]2 years ago
5 0

The mechanical energy for the first and the second ball is

10 ^{7}  \: joules.

Mass of the first ball = 20 kg

The initial speed at which a cannonball is fired from a cannon =1000 m/s

The angle made by the cannonball while being fired from the cannon = 37°

The maximum height reached by the first ball is,

=   \frac{ u {}^{2} _{1}sin {}^{2} θ}{2g}

=    \frac{ {1000}^{2} sin {}^{2}37°}{2 \times 9.8}

= 18478.69 \: m

The maximum height of the first cannonball is 17478.69 m.

The initial speed at which a cannonball is fired from a cannon =1000 m/s

The angle made by the cannonball while being fired from the cannon = 90 °

=   \frac{ u {}^{2} _{2}sin {}^{2} θ}{2g}

=   \frac{ 1000{}^{2}sin^{2} 90°}{2 \times 9.8}[tex] = 51020.41 \: m

For the first ball, total mechanical energy= Potential energy at maximum height + kinetic energy at the maximum height

So, the total mechanical energy is,

= mgh \: + \frac{1}{2}mv {}^{2} _{x}[/tex]

= 20 \times 9.8 \times 18478.64  \times  \frac{20}{2} (1000 \: cos37 °)

= 10 ^{7}  The potential energy at the maximum height, = m _{2}gh

= 20 \times 9.8  \times 51020.41

= 10 ^{7} \:J

Therefore, the total mechanical energy for the first and the

\:second \:  cannonball \:  is  \: 10 ^{7}  \:joules.

To know about energy, refer to the below link:

brainly.com/question/1932868

#SPJ4

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A rocket fires two engines simultaneously. One produces a thrust of 725Ndirectly forward while the other gives a 513N thrust at
faust18 [17]

The magnitude of the resultant force, F = 1,190.3 acting at a direction X = 13.35°.

<h3>What is the resultant force the two engines exert on the rocket?</h3>

The resultant force on the rocket is calculated thus:

The 513N thrust is resolved into vertical and horizontal components;

Horizontal component: 513N cos(32.4°) = 433.14 N

Vertical component: 513N sin(32.4°) = 274.88 N

Total forward force on the rocket = 725 N + 433.14 N = 1,158.14 N

Total force at right angles:

0 + 274.88 N = 274.88 N

The resultant force (F) is then given as follows:

F² = a² + b²

F² = (1158.14 N)² + (274.88 N)²

F = √1,416,847.27

F = 1,190.3

To find the direction:

tan X 274.88 N / 1,158.14 N

X = tan⁻¹ 0.237346089419241

X = 13.35°

Therefore, the magnitude of the resultant force, F = 1,190.3 acting at a direction X = 13.35°.

In conclusion, the resultant force is obtained by resolving the forces into vertical and horizontal components.

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