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2 years ago

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A person shooting at a target from a distance of 450 metres finds that the sound of the bullet hitting the target comes 1 / 2 se

conds after he fired. A person equidistant from the target and shooting point hears the bullet hit 3 seconds after he heard the gun. The speed of sound is ___.

1 answer:

White raven [17]2 years ago

3 0

Answer:

1350 m/s

Explanation:

Speed of bullet

Distance traveled by the bullet = 450 m

Distance traveled by sound = 450 m

Using S= V × t

==> t= S/V

So, time for bullet t1=450/vb

time for sound t2=450/vs

Because it takes 1/2 sec from firing to hearing of sound by the shooter

==> t1+t2= 1/2 sec

==> 450/vb+450/vs=0.5sec ---- (1)

Now a person at distance 'x' each from gun and target takes 3 sec to hear sound from firing to hitting the target.

Fire sound duration

Distance =x m

Speed of sound = V

time =T1

==> T1 = x/vs

Bullet hitting + target sound duration

During this duration bullet travels a distance of 450 m and sound of hitting travels a distance 'x'

time taken by sound = 450/vb

time taken by sound to travel distance 'x'= x/vs

so let T2= 450/vb + x/vs

But all this happens in 3 sec i-e T = 3 sec

and because firing occurs before hitting the target so he hears strike sound in time T = T2-T1= 450/vb + x/vs -x/vs

But T= 3sec

==> 3= 450/vb

==> vb= 1350 m/s

From eq 1 and 2

again using same relation as above, i-e within 3 seconds he watches bullet traveling a distance of 450 m and hearing sound

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Answer:

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0.7 kg is 109.4 ms on the x axis

b) Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

sin 40 = voy / v₀

$v_{oy}$ = v₀ sin 40

$v_{oy}$ = 50.0 sin40

$v_{oy}$ = 32.14 m / s

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Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and $v_{y}$ = -10.0 m / s

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Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

p₀ₓ = m v₀ₓ

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After the break

X axis

$p_{fx}$ = m₂ v₂

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$p_{fy}$ = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis

0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

v₃ = - m₁ v₁ / m₃

v₃ = - (-10) 1 / 0.3

v₃ = 33.3 m / s

X axis

M v₀ₓ = m₂ v₂

v₂ = v₀ₓ M / m₂

v₂ = 38.3 2 / 0.7

v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

$v_{fy}$ ² = $v_{oy}$ ² - 2 gy

0 = $v_{oy}$ ²-2gy

y = $v_{oy}$ ² / 2g

y = 32.14² / 2 9.8

y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

$v_{f}$ ² = $v_{y}$ ² - 2 g y₂

0 = $v_{y}$ ² - 2 g y₂

y₂ = $v_{y}$ ² / 2g

y₂ = 33.3²/2 9.8

y₂ = 56.58 m

Total body height is

Y = y + y₂

Y = 52.7 + 56.58

Y = 109.3 m

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A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m

ArbitrLikvidat [17]

Answer:

<h2>206.67N</h2>

Explanation:

The sum of force along both components x and y is expressed as;

$\sum Fx = ma_x \ and \ \sum Fy = ma_y$

The magnitude of the net force which is also known as the resultant will be expressed as $R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}$

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

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Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

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3 years ago