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Elodia [21]
3 years ago
6

Assume you built a really big machine that could launch the projectile a “significant” distance; for instance, several hundred m

iles. Write a brief essay discussing the issues that would need to be accounted for with a projectile with that type of range. Be sure to include how those issues affect the range of the projectile.

Physics
2 answers:
sasho [114]3 years ago
4 0

Answer:

Trebuchet

Explanation:

I've imagined building a Trebuchet. After that idea, there were some things to consider carefully on my project.

The payload should be deadly and also win the air resistance, which brings up the fact the speed and the acceleration must be produced by a great torque by the Trebuchet's counterweight.

My medieval arm is rigid and launches payloads at a steady position so that the loss of forces could be minimal. A stationery trebuchet at the shooting is vital for a greater range, as the speed and impact of its payload would play its role as well.

No friction on its parts (pivots, joints, guide chute) so that we can improve the effectiveness of constant shots.

In addition to this, some parts of the Trebuchet must have no significant mass.

Finally, at each shot, the acceleration of the projectile could be calculated differentiating the speed of the counterweight over time.

All of these matters ought to be considered, for the prototype

Gwar [14]3 years ago
3 0
<span>First, a problem would be the sheer amount of wind resistance. If an object travels as far as even just one hundred miles it could encounter different wind patterns that could change the trajectory of the object. Second would be the size of the projectile. This creates a problem because the bigger it is, the more momentum it could potentially pick up, given that it is not too big to complete the distance. This is another problem with size, how far the projectile can actually travel. You would have to actually calculate the ideal size of said object to make sure it could actually make the distance you're looking for. hope it helps:)</span>
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Answer:

Given that,

  • Power = 2000 W
  • time = 60 seconds
  • distance= 10m

Power = work done ÷ time

Here, since the movement is vertical, w = mgh

So,

Power = mgh÷t

2000 = (m × 9.8 ×10) ÷ 60

m = (2000 ×60) ÷98

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3 years ago
The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a
grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

\sin\theta=\dfrac{3}{5}

The horizontal component is

T_{x}=T\cos\theta

The vertical component is

T_{y}=T\sin\theta

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'

Put the value into the formula

0=190\times2+300\times 4-T\sin\theta\times 4

T\sin\theta\times 4=380+1200

T=\dfrac{1580\times5}{3\times 4}

T=658.33\ N

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

F_{x}=T\cos\theta

Put the value into the formula

F_{x}=658.33\times\dfrac{4}{5}

F_{x}=526.66\ N

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

F_{y}=F_{b}+F_{w}-T\sin\theta

Put the value into the formula

F_{y}=190+300-658.33\times\dfrac{3}{5}

F_{y}=95.002\ N

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

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Answer: A projectile which is fired horizontally is being constantly acted upon by acceleration due to gravity, acting vertically downwards. Hence, it does not follow a straight line path. Also Why a projectile fixed along the horizontal not follow a straight line path? Because the projectile fired horizontally is constantly acts upon by acceleration due to gravity acting vertically downwards.

Explanation:

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