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3 years ago

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A bowling ball traveling with constant speed hits the pins at the end of a bowling lane 16.5 m long. The bowler hears the sound

of the ball hitting the pins 2.79 s after the ball is released from his hands.

1 answer:

pshichka [43]3 years ago

4 0

Let , speed of bowling ball is v .

Time taken by ball to hit the pins , $t=\dfrac{16.5}{v}$ .

Speed of sound , u = 343 m/s .

Time taken by sound to reach ear , $T=\dfrac{16.5}{343}=0.048\ s$ .

Now ,

Time taken by sound to hear after release :

$2.79=\dfrac{16.5}{V}+0.048\\\\2.79-0.048=\dfrac{16.5}{V}\\\\2.742=\dfrac{16.5}{V}\\\\V=\dfrac{16.5}{2.742}\\\\V=6.02\ m/s$

Hence , this is the required solution .

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The average coefficient of linear expansion of copper is 1.7 10-5 (°c)−1. the statue of liberty is 93 m tall on a summer morning

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Let the rise in temperature be $5^0C$

The expansion in length due to change in temperature is given by the expression lαΔt , where l is the length, α is the coefficient of linear expansion, Δt is the change in temperature.

Here l = 93 m, α = $1.7*10^{-5}$ $^0C^{-1}$ , and Δt = $5^0C$

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After falling for 1 second, its speed was 9.8 m/s straight down (gravity).

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A billiard ball of mass m moving with speed v1=1 m/s collides head-on with a second ball of mass 2m which is at rest. What is th

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Answer:

The velocity of mass 2m is $v_B = 0.67 m/s$

Explanation:

From the question w are told that

The mass of the billiard ball A is =m

The initial speed of the billiard ball A = $v_1$ =1 m/s

The mass of the billiard ball B is = 2 m

The initial speed of the billiard ball B = 0

Let the final speed of the billiard ball A = $v_A$

Let The finial speed of the billiard ball B = $v_B$

According to the law of conservation of Energy

$\frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2$

Substituting values

$\frac{1}{2} m (1)^2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2$

Multiplying through by $\frac{1}{2}m$

$1 =v_A^2 + 2 v_B ^2 ---(1)$

According to the law of conservation of Momentum

$mv_1 + 2m(0) = mv_A + 2m v_B$

Substituting values

$m(1) = mv_A + 2mv_B$

Multiplying through by $m$

$1 = v_A + 2v_B ---(2)$

making $v_A$ subject of the equation 2

$v_A = 1 - 2v_B$

Substituting this into equation 1

$(1 -2v_B)^2 + 2v_B^2 = 1$

$1 - 4v_B + 4v_B^2 + 2v_B^2 =1$

$6v_B^2 -4v_B +1 =1$

$6v_B^2 -4v_B =0$

Multiplying through by $\frac{1}{v_B}$

$6v_B -4 = 0$

$v_B = \frac{4}{6}$

$v_B = 0.67 m/s$

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3 years ago

Help me plz.<br>Show workings

ddd [48]

Change in momentum: finial momentum - initial momentum
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The answer is 0.8Ns

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3 years ago

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A cylindrical container with a cross-sectional area of 66.2 cm2 holds a fluid of density 856 kg/m3 . At the bottom of the contai

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Answer:

A. h = 2.15 m

B. Pb' = 122 KPa

Explanation:

The computation is shown below:

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After solving this,

h = 2.15 m

Therefore the depth of the fluid is 2.15 m

b)

Given that

height of the extra fluid is

$h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}$

h' = 0.355 m

Now let us assume the pressure at the bottom is Pb'

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Pb' = 122 KPa

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3 years ago