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ohaa
16 days ago
7

Identical guns fire identical bullets horizontally at the same speed from the same height above level planes, one on the Earth a

nd one on the Moon. Which of the following three statements is/are true?
I. The horizontal distance traveled by the bullet is greater for the Moon.
II. The flight time is less for the bullet on the Earth.
III. The velocities of the bullets at impact are the same.
Physics
2 answers:
serg [2.5K]16 days ago
8 0

Answer:

I. The bullet covers a longer horizontal distance on the Moon.

II. The bullet's flight duration is shorter on Earth.

Explanation:

The behavior of bullets is characterized by projectile motion, showing uniform motion in the horizontal plane and free-fall motion vertically.

Bullets drop due to gravity, and since earth's gravity is stronger, bullets descend more swiftly, leading to reduced flight time on Earth. Consequently, with shorter flight time, there is less duration for horizontal travel prior to falling, thus making the horizontal distance achieved lesser on Earth.

Sav [2.2K]16 days ago
3 0

Answer:

I. For the Moon, the bullet travels a longer horizontal distance.

II. The bullet experiences a shorter flight time on Earth.

Explanation:

The distance covered horizontally depends on initial velocity, height, and gravitational force. Bullets possess the same initial speed and are fired from an identical height. Hence, in this scenario, the horizontal distance solely relies on gravity, which has an inverse relationship. This leads to the conclusion that less gravity results in greater horizontal distance. Gravity impedes bullet movement and influences its arrival at the ground. The increased gravity on Earth causes the bullet to hit the ground more quickly.

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ValentinkaMS [2425]

Answer:

The rotational angular speed is measured at 1.34 rad/s.

Explanation:

Considering the following parameters,

Length = 3.40 m

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Angle = 45.0°

We are tasked with finding the angular speed of rotation

Using the balance equation

Horizontal component

T\cos\theta=mg

T=\dfrac{mg}{\cos\theta}

Vertical component

T\sin\theta=m\omega^2 r

Substituting the tension value

mg\tan\theta=m\omega^2(d+L\sin\theta)

\omega=\sqrt{\dfrac{g\tan\theta}{(d+L\sin\theta)}}

Substituting the value into the equation

\omega=\sqrt{\dfrac{9.8\tan45.0}{5.90+3.40\sin45.0}}

\omega=1.34\ rad/s

Thus, the angular speed of rotation computes to 1.34 rad/s.

7 0
1 month ago
Alicia intends to swim to a point straight across a 100 m wide river with a current that flows at 1.2 m/s. She can swim 2.5 m/s
Sav [2226]

Answer:

θ = 61.3°

Alicia must swim at an angle of 61.3°

Explanation:

Parameters given include:

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The angle she needs to swim can be determined by combining the velocities, taking into account the current's influence.

Her swimming speed aimed against the current must offset the current's velocity;

2.5cosθ - 1.2 = 0

2.5cosθ = 1.2

cosθ = 1.2/2.5

θ = cosinverse(1.2/2.5)

θ = 61.3°

4 0
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Softa [2029]

The correct choice is D!

Clarification:

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2*3.5 = 7m/s

You need to multiply the acceleration by the time (which must both be in seconds; if not, convert them to the same units).

7 0
9 days ago
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The second question necessitates a figure to provide an answer. For the initial question
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