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2 years ago

Which formula can be used to calculate the horizontal displacement not of a horizontally launched projectile

1 answer:

8 0

If you are looking to get an object up the highest, shoot it straight up. If you want to go for a specific horizontal displacement, use the range equation. R = v2sin(twice the launch angle)/ g. g is the gravitaional constant, 9.8 meters per second. Use degrees for the angle. v is the launch velocity. R is the horizontal displacement. This formula only works if your start altitude and end altitude are the same, i.e. you must shoot over a level field. it depends on the gravitational force of attraction of earth and air resistance. if we are neglecting air resistance, the max.horizontal distance is according to this formulae, V0/2 * sin (2theta) where V0 is the initial velocity theta is the angle with x axis and the projection. There are a number of ways that you could find a horizontally displaced object. You could for example just look.

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A starship passes Earth at 80% of the speed of light and sends a drone ship forward at half the speed of light rela- tive to its

alisha [4.7K]

Answer:

Check Explanation

Explanation:

Let the speed of the drone relative to earth be u = ?

Let the speed of the drone relative to the starship be u' = 0.5c

Let the speed of the starship relative to earth be v = 0.8c

In the theory of relativity,

The 3 speeds are related thus

u = (u' + v)/[1 + (u'v/c²)]

u = (0.5c + 0.8c)[1 + ((0.8c × 0.5c)/c²)]

u = 1.3c/[1 + (0.4c²/c²)]

u = 1.3c/1.4 = 0.9286 c = 0.93 c = 93% c

8 0

2 years ago

(a) A 15.0 kg block is released from rest at point A in the figure below. The track is frictionless except for the portion betwe

castortr0y [4]

Answer:

(a) coefficient of friction = 0.451

This was calculated by the application of energy conservation principle (the total sum of energy in a closed system is conserved)

(b) No, it comes to a stop 5.35m short of point B. This is so because the spring on expanding only does a work of 43 J on the block which is not enough to meet up the workdone of 398 J against friction.

Explanation:

The detailed step by step solution to this problems can be found in the attachment below. The solution for part (a) was divided into two: the motion of the body from point A to point B and from point B to point C. The total energy in the system is gotten from the initial gravitational potential energy. This energy becomes transformed into the work done against friction and the work done in compression the spring. A work of 398J was done in overcoming friction over a distance of 6.00m. The energy used in doing so is lost as friction is not a conservative force. This leaves only 43J of energy which compresses the spring. On expansion the spring does a work of 43J back on the block is only enough to push it over a distance of 0.65m stopping short of 5.35m from point B.

Thank you for reading and I hope this is helpful to you.

4 0

2 years ago

One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force fis applied to the other end of the

sergeinik [125]

Answer:The weight of the door creates a CCW torque given by Tccw = 145 N*3.13 m / 2 You need a CW torque that's equal to that Tcw = F*2.5 m*sin20

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2 years ago

Read 2 more answers

If you peel two strips of transparent tape off the same roll and immediately let them hang near each other, they will repel each

g100num [7]

Answer:

Firstly removing off one strip and it leaves electrons behind, so the strip becomes positively charged.

2. The roll however is not negatively charged because it is "earthed " by the hand holding it, thus excess negatives repel each other away through the hand.

3.Tearing off the next strip and once more it leaves electrons behind, the new strip is also positively charged and will repel the first strip.

4. Then, tear two strips apart and one will leave electrons behind on the other. Meaning that one strip is positive and the other is negative and they will attract each other.

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2 years ago

A submerged submarine alters its buoyancy so that it initially accelerates upward at 0.325 m/s^2. What is the submarine's averag

scoundrel [369]

You are given a submerged submarine accelerating upward at 0.325 m/s² and the density of sea water is 1.025x10³ kg/m³. The submarine's average density at this time is 22 kg/m³.

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2 years ago