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ohaa
2 months 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 [3.5K]2 months 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 [3.1K]2 months 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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Hawks and gannets soar above the ground and, when they spot prey, they fold their wings and essentially drop like a stone. They
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Answer:

  v = 54.2 m/s

Explanation:

We can utilize conservation of energy to solve this issue.

Initial condition Higher

         Em₀ = U = m g h

Final condition. Lower

        Em_{f} = K = ½ m v²

        Em₀ = Em_{f}

        m g h = ½ m v²

         v² = 2gh

         v = √ (2gh)

Now let's perform the calculation

         v = √ (2 * 9.8 * 150)

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2 months ago
An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.
Softa [3030]

Answer:

The plane's speed in relation to the ground is 300.79 km/h.

Explanation:

Provided details include:

Wind speed = 75.0 km/hr

Plane's airspeed = 310 km/hr

Next, we must find the ground speed of the plane

Calculating the angle

Using the angle formula

\sin\theta=\dfrac{v'}{v}

Where v' represents the wind speed

v represents the plane's speed

We will substitute the values into the formula

\sin\theta=\dfrac{75}{310}

\theta=\sin^{-1}(\dfrac{75}{310})

\theta=14.0^{\circ}

Now, we must find the resultant speed

Using the resultant speed formula

\cos\theta=\dfrac{v''}{v}

Insert the values into the formula

\cos14=\dfrac{v''}{310}

v''=\cos14\times310

v''=300.79\ km/h

Consequently, the plane's speed in relation to the ground equals 300.79 km/h.

6 0
2 months ago
 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer
serg [3582]

Response:

a) The mug makes contact with the ground 0.7425m from the bar's end. b) |V|=5.08m/s θ= -72.82°

Clarification:

To address this issue, we begin with a diagram depicting the situation. (refer to the attached illustration).

a)

The illustration shows that the problem involves motion in two dimensions. To determine how far from the bar the mug lands, we need to find the time the mug remains airborne by examining its vertical motion.

To compute the time, we utilize the following formula with the known values:

y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}

We have y_{f}=0 and v_{y0}=0, allowing us to simplify the equation to:

0=y_{0}+\frac{1}{2}at^{2}

Now, we can calculate for t:

-y_{0}=\frac{1}{2}at^{2}

-2y_{0}=at^{2}

\frac{-2y_{0}}{a}=t^{2}

t=\sqrt{\frac{-2y_{0}}{a}}

We know y_{0}=1.20m and a=g=-9.8m/s^{2}

The negative gravity indicates the downward motion of the mug. Hence, we substitute these values into the provided formula:

t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}

Which results in:

t=0.495s

This time helps us evaluate the horizontal distance the mug traverses. Since:

V_{x}=\frac{x}{t}

Solving for x, we have:

x=V_{x}t

Substituting the known values yields:

x=(1.5m/s)(0.495s)

This calculates to:

x=0.7425m

b) With the time determining when the mug strikes the ground established, we can find the final velocity in the vertical direction using the formula:

a=\frac{v_{f}-v_{0}}{t}

The initial vertical velocity being zero simplifies our calculations:

a=\frac{v_{f}}{t}

Thus, we can determine the final velocity:

V_{yf}=at

Given that the acceleration equates to gravity (showing a downward effect), we substitute that alongside the previously found time:

V_{yf}=(-9.8m/s^{2})(0.495s)

This leads to:

V_{yf}=-4.851m/s

Now, we ascertain the velocity components:

V_{xf}=1.5m/s and V_{yf]=-4.851m/s

Next, we find the speed by calculating the vector's magnitude:

|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}

<pThus:

|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}

Yielding:

|V|=5.08m/s

Lastly, to ascertain the impact direction, we apply the equation:

\theta = tan^{-1} (\frac{V_{y}}{V_{x}})

<pFulfilling this provides:

\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})

<pLeading to:

\theta = -72.82^{o}

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2 months ago
The difference between the two molar specific heats of a gas is 8000J/kgK. If the ratio of the two specific heats is 1.65, calcu
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Answer:

I apologize

Explanation:

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