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3 years ago

Which disease do you think is most easily spread? 5 Answers

Physics

1 answer:

Anika [276]3 years ago

7 0

Answer:

coronavirus

sida

tuberculosis

sifilis

epatitis

Explanation:

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A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$

5 0

2 years ago

Help me, please with this queastion

sergejj [24]

Answer:

Kinetic energy is the energy due to motion. Potential energy is energy stored in matter. The joule (J) is the SI unit of energy and equals (kg×m2s2) ( kg × m 2 s 2 ) .

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7 0

2 years ago

What is the process called when earth's surface is broken down into smaller pieces

fiasKO [112]

You are LOVED and a Child of JESUS come back he has open arms God bless

6 0

2 years ago

The earplug can reduce the sound level to about 18 decibels (dB). What percentage reduction is this intensity?

zlopas [31]

Answer:

1 x 10 -10 whisper at 1m distance.

Explanation:

Properly fitted ear plugs an reduce noise form 15-30db. Although they are better for low frequency

8 0

3 years ago

Read 2 more answers

A 6.00 g lead bullet traveling at420 m/s is stopped by a large

Westkost [7]

Answer:

The increase in temperature of the bullet is 351.1 kelvin

Explanation:

First, we should find the kinetic energy of the bullet is:

$K=\frac{mv^{2}}{2}=\frac{(6\times10^{-3}\,kg)(420\,\frac{m}{s})^{2}}{2}$

with m the mass and v the velocity.

$K=529.2 J$

Now we know that half of the kinetic energy of the bullet is transformed into internal energy, by second's law of thermodynamics that means heat (Q) to raise bullet temperature (T), so:

$Q=\frac{K}{2}= 264.6\,J$