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1 year ago

Bullets spin when shot from a rifle or handgun. What causes this spinning?

Physics

1 answer:

tamaranim1 [39]1 year ago

7 0

Answer:

The spark from the primer ignites the gunpowder. Gas converted from the burning powder rapidly expands in the cartridge. The expanding gas forces the bullet out of the cartridge and down the barrel with great speed. The rifling in the barrel causes the bullet to spin as it travels out of the barrel.

Explanation:

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1. A Passenger car (m= 10.0 kg) and a flat car (m= 7.5 kg) collide and move off as

grin007 [14]

Answer:

use a calculator to solve that

6 0

1 year ago

The world record for a 100 m race is 10.44 m/s. How fast is this in mi/h?

ZanzabumX [31]

Answer: actually the fastest 100m is 9.58, because i run at a 9.86

Explanation: but 10.44 is like 15 mph, 9.58 is 28 mph

6 0

2 years ago

The volume flow rate of blood leaving the heart to circulate throughout the body is about 5 L/min for a person at rest. All this

skelet666 [1.2K]

Answer:

$n=2.9\times 10^9$

$A=1.88\times 10^{-8}\ m^2$

Explanation:

Given that

Q= 5 L/min

1 L = 10⁻³ m³/s

1 min = 60 s

Q=0.083 x 10⁻³ m³/s

d= 6 μm

v= 1 mm/s

So the discharge flow through one tube

q = A v

$A=\dfrac{\pi}{4}d^2$

$A=\dfrac{\pi}{4}\times (6\times 10^{-6})^2\ m^2$

A=2.8 x 10⁻¹¹ m²

v= 1 x 10⁻³ m/s

q= 2.8 x 10⁻¹⁴ m³/s

Lets take total number of tube is n

Q= n q

n=Q/q

$n=\dfrac{0.083\times 10^{-3} }{ 2.8\times 10^{-14}}$

$n=2.9\times 10^9$

Surface area A

A= π d L

$A=\pi \times 6\times 10^{-6}\times 10^{-3}\ m^2$

$A=1.88\times 10^{-8}\ m^2$

7 0

2 years ago

The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of

kipiarov [429]

Answer: 0.4911 kg

Explanation:

We have the following data:

$\rho_{0\°C}= 730 kg/m^{3}$ is the density of gasoline at $0\°C$

$\beta=9.60(10)^{-4} \°C^{-1}$ is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to $m^{3}$ :

$V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3}$ (1)

Knowing density is given by: $\rho=\frac{m}{V}$ , we can find the mass $m_{1}$ of 8.50 gallons:

$m_{1}=\rho_{0\°C}V$

$m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg$ (2)

Now, we have to calculate the factor $f$ by which the volume of gasoline is increased with the temperature, which is given by:

$f=(1+\beta(T_{f}-T_{o}))$ (3)

Where $T_{o}=0\°C$ is the initial temperature and $T_{f}=21.7\°C$ is the final temperature.

$f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C))$ (4)

$f=1.020832$ (5)

With this, we can calculate the density of gasoline at $21.7\°C$ :

$\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)$

$\rho_{21.7\°C}=745.207 kg/m^{3}$ (6)

Now we can calculate the mass of gasoline at this temperature:

$m_{2}=\rho_{21.7\°C}V$ (7)

$m_{2}=(745.207 kg/m^{3})(0.0323 m^{3})$ (8)

$m_{2}=24.070 kg$ (9)

And finally calculate the mass difference $\Delta m$ :

$\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg$ (10)

$\Delta m=0.4911 kg$ (11) This is the extra mass of gasoline

6 0

2 years ago

What happens when you change the number of electrons in an atom

aalyn [17]

<h2>Answer: It becomes an Ion </h2>

When an atom has gained or lost electrons (negative charge), it becomes an ion.

In this sense:

<h2>Ions are atoms that have <u>gained or lost</u> electrons in their electronic cortex. </h2><h2> </h2>

If a neutral atom <u>loses electrons</u>, it remains with an excess of positive charge and transforms into a positive ion or <u>cation</u>, whereas if a neutral atom <u>gains electrons</u>, it acquires an excess of negative charge and transforms into a negative ion or <u>anion</u>.

It is then how ions form bonds with other atoms differently depending on the number of electrons they have.

8 0

2 years ago