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1 year ago

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The muzzle velocity of a gun is the velocity of the bullet when it leaves the barrel. The muzzle velocity of one rifle with a sh

ort barrel is greater than the muzzle velocity of another rifle that has a longer barrel. In which rifle is the acceleration of the bullet larger

1 answer:

Anvisha [2.4K]1 year ago

7 0

Answer:small barrel gun

Explanation:

Given

Muzzle velocity of bullet is greater in short barrel gun as compared to larger barrel gun

acceleration is given by change in velocity with respect to time

$a=\dfrac{\Delta v}{\Delta t}$

In case of short barrel bullet time taken by bullet to reach its muzzle velocity is less therefore acceleration of small barrel bullet is more compared to long barrel bullet.

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calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame

kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

$\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2$

0 - p × Area for trachea =

$\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

⇒ $-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}$

$\frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s$

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

3 0

1 year ago

Read 2 more answers

Of waterfalls with a height of more than 50 m , Niagara Falls in Canada has the highest flow rate of any waterfall in the world.

Vinil7 [7]

Answer:

Power output: W=1426.9MW

Explanation:

The power output of the falls is given mainly by its change in potential energy:

$Q=-P_{tot}=-(P_{2}-P_{1})$

The potential energy for any point can be calculated as:

$P=m*g*h$

If we consider the base of the falls to be the reference height, at point 2 h=0, so P2=0, and height at point 1 equals 52m:

$Q=P_{1}=m*g*h$

If we replace m with the mass rate M we obtain the rate of change in potential energy over time, so the power generated:

$W=M*g*h=2.8*10^{3}m^{3}/s*1*10^{3}kg/m^{3}*9.8m/s^{2}*52m =1426.9MW$

5 0

1 year ago

A standard 1 kilogram weight is a cylinder 51.0 mm in height and 42.0 mm in diameter. Determine the density of the material

worty [1.4K]

Answer:

14160 kg/m^3

Explanation:

First of all, we need to find the volume of the cylinder.

The volume of the cylinder is given by:

$V=\pi r^2 h$

where:

$r=\frac{d}{2}=\frac{42.0 mm}{2}=21.0 mm=0.021 m$ is the radius

$h=51.0 mm=0.051 m$ is the height

Substituting, we find

$V=\pi (0.021 m)^2 (0.051 m)=7.1 \cdot 10^{-5} m^3$

And the density is given by

$d=\frac{m}{V}$

where m = 1 kg is the mass. Substituting, we find

$d=\frac{1 kg}{7.1\cdot 10^{-5} m^3}=14,160 kg/m^3$

3 0

2 years ago

A ledge on a building is 18 m above the ground. A taut rope attached to a 4.0-kg can of paint sitting on the ledge passes up ove

True [87]

Answer:t=5.07 s

Explanation:

Given

height of Building h=18 m

mass of Paint can $m_1=4 kg$

mass of second can $m_2=3 kg$

let T be the Tension in the rope

For 4 kg can

$4g-T=4a$

$T=4(g-a)$ ----1

For 3 kg can

$T-3g=3a$

$T=3(g+a)$ ----2

From 1 and 2

$4(g-a)=3(g+a)$

$g=7a$

$a=\frac{g}{7}$

So time taken to cover 18 m is

$h=ut+\frac{at^2}{2}$

$18=0+\frac{g\cdot t^2}{7\times 2}$

$t^2=\frac{18\times 2\times 7}{g}$

$t=5.07 s$

5 0

1 year ago

Squid use jet propulsion for rapid escapes. A squid pulls water into its body and then rapidly ejects the water backward to prop

Setler [38]

Answer:

a. FTh = 30 N

b. Fw = 30 N

c. a = 200 m/s2

Explanation:

See full explanation in the picture. Please rate as brainliest

5 0

1 year ago