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garri49 [273]
3 years 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
Physics
1 answer:
olga nikolaevna [1]3 years ago
3 0

Answer:

horizontal distance on moon > horizontal distance on earth

Explanation:

Let the time of fall is t and the height from which they fall is h.

The time of fall is

t=\sqrt{\frac{2h}{g}}

Time of fall on earth

t=\sqrt{\frac{2h}{g_{e}}

Time of fall on moon

t'=\sqrt{\frac{2h}{g_{m}}

As the gravity on moon is less than the gravity on earth

So, t' > t

Horizontal distance = horizontal velocity x time

So, the horizontal distance on moon is more than the horizontal distance on earth.

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Which of the following best states the difference between a description and an explanation? A. A description is a statement that
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Block B (of mass m) is initially at rest. Block A (of mass 3m) travels toward B with an initial speed v0 (vee nought) and collid
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Answer:

The maximum height reached by the two blocks is approximately 0.1147959 × v₀²

Explanation:

The mass of block B = m

The mass of block A = 3·m

The initial velocity of block B, v₂ = 0 m/s

The initial velocity of block A, v₁ = v₀

The amount of friction between the blocks and the surface = Negligible friction

By the Law of conservation of linear momentum, we have;

Total initial momentum = Total final momentum

3·m·v₁ + m·v₂ = (3·m + m)·v₃ = 4·m·v₃

Plugging in the values for the velocities gives;

3·m × v₀ + m × 0 = (3·m + m)·v₃ = 4·m·v₃

∴ 3·m × v₀ =  4·m·v₃

\therefore v_3 = \dfrac{3}{4} \times v_0 = 0.75 \times v_0

The kinetic energy, K.E. of the combined blocks after the collision is given as follows;

K.E. = 1/2 × mass × v²

\therefore K.E. = \dfrac{1}{2} \times 4\cdot m \times \left (\dfrac{3}{4} \cdot v_0 \right )^2 = \dfrac{9}{8} \cdot m\cdot v_0^2

The potential energy, P.E., gained by the two blocks at maximum height = The kinetic energy, K.E., of the two blocks before moving vertically upwards

The potential energy, P.E. = m·g·h

Where;

m = The mass of the object at the given height

g = The acceleration due to gravity

h = The height at which the object of mass, 'm', is located

Therefore, for h = The maximum height reached by the two blocks, we have;

P.E. = K.E.

m \cdot g \cdot h =  \dfrac{9}{8} \cdot m\cdot v_0^2

h = \dfrac{\dfrac{9}{8} \cdot m\cdot v_0^2}{m \cdot g }  = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ g }  =  \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ 9.8} = \dfrac{42}{392} \cdot  v_0^2 \approx 0.1147959   \cdot  v_0^2

The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².

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he type of vegetation growing in the buffer affects its usefulness to wildlife through the availability of food, foraging and nesting sites, and other habitat needs (Johnson and Beck 1988). The more diverse the habitat, the greater its utility to many species of animals.

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