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sesenic [268]
3 years ago
13

A bullet is dropped into a river from a very high bridge. At the same time, another bullet is fired from a gun straight down tow

ards the water. If air resistance is negligible, the acceleration of the bullets just before they strike the water:
(A) is greater for the dropped bullet
(B) is greater for the fired bullet
(C) is the same for both bullets
(D) depends on how high they started
Physics
1 answer:
anastassius [24]3 years ago
8 0

Answer:

Option (c)

Explanation:

Both the bullets have same acceleration because they both falls under the influence of acceleration due to gravity.

The bullet which is fired from the gun has some initial velocity but the bullet which is dropped has zero initial velocity.

the acceleration is acting on both the bullets which is equal to the acceleration due to gravity and they both in motion in the influence of gravity.

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On a cold winter day, the flow of heat is from the outside in. A.True B.False
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How to calculate kinetic energy given mass and velocity
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2 years ago
By what factor must the amplitude of a sound wave be increased in order to increase the intensity by a factor of 9?a. 9 b. 2 c.
Marta_Voda [28]

Answer:

option D

Explanation:

given,

intensity\ \alpha \ (Amplitude)^2

increase the intensity by factor of 9

    I₁ = I₀

    I₂ = 9 I₀

now,

\dfrac{I_1}{I_2}=\dfrac{A_1^2}{A_2^2}

\dfrac{I_0}{9I_0}=(\dfrac{A_1}{A_2})^2

(\dfrac{A_1}{A_2})^2=\dfrac{1}{9}

\dfrac{A_1}{A_2}=\dfrac{1}{3}

      A₂ = 3 A₁

hence, amplitude increase with the factor of 3

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4 0
3 years ago
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The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc
Mnenie [13.5K]

Answer:

The image distance is 20.0 cm.

Explanation:

Given that,

Power = 1.55 dp

Distance between book to eye = 26.0+3.00=29.0 cm

We need to calculate the focal length

Using formula of focal length

f = \dfrac{1}{P}

Put the value into the formula

f=\dfrac{1}{1.55}

f=0.645\ m

f=64.5\ cm

We need to calculate the image distance

Using lens formula

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{-u}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{64.5}-\dfrac{1}{-29}

\dfrac{1}{v}=\dfrac{187}{3741}

v=20.0\ cm

Hence, The image distance is 20.0 cm.

5 0
3 years ago
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