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3 years ago

When a bullet is fired, does gunpowder only travels back towards the shooter

1 answer:

3 0

It seems that you have missed the necessary options for this question, but anyway the correct answer for this would be FALSE. It is not true that when a bullet is fired, gunpowder only travels back towards the shooter. Though gunpowder residue can remain on the shooter's hand for a few hours. Hope this answer helps.

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Elanso [62]

The direction of work.........

4 0

3 years ago

Atmospheric pressure is due to the weight _______

Kryger [21]

Of the gravitational pull and other things like mass. Planet earth it,s self as you said sir.

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4 years ago

Active listening includes all of the following EXCEPT: A. paraphrasing B. clarifying C. ignoring D. empathizing Please select th

scoray [572]

Answer:

C: Ignoring

Explanation:

On edge 2021

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3 years ago

The mass of a ship before launch is 55,000 metric tons. The ship is launched down a ramp and drops a total of 10 vertical meters

skelet666 [1.2K]

Answer:

ΔT = 17.11 °C

Explanation:

In this case, we have a ship standing on a place with a given mass and it's about to be launched to a lock containing water.

At first, before launch, the ship has a potential energy, and when the ship hits the water after being launched, this potential energy is transformed into kinetic energy.

So, let's calculate first the potential energy of the ship:

E = mgh (1)

We have the mass, gravity and height, so we need to replace the given data here. Before we do that, let's remember to use the correct units. A ton is 1000 kg, so replacing and converting we have:

E = (55000 ton * 1000 kg/ton) * (9.8 m/s²) * 10 m

E = 5.39x10⁹ J

Now this energy will be the same when the ship hits the water, only that is kinetic energy that will result in the rise of temperature. To get this rise we use the following expression:

E = m * C * ΔT (2)

We have the energy, the mass of water (assuming density of water as 1 kg/m³) and the specific heat, so, replacing in (2) and solving for ΔT we have:

ΔT = E / m * C (3)

ΔT = 5.39x10⁹ / 4200 * 75000

Hope this helps

5 0

3 years ago

An accepted value for the acceleration due to gravity is 9.801 m/s2. In an experiment with pendulums, you calculate that the val

Fed [463]

g Generally the accepted value of acceleration due to gravity is 9.801 $m/s^2$

as per the question the acceleration due to gravity is found to be 9.42 $m/s^2$ in an experiment performed.

the difference between the ideal and observed value is 0.381.

hence the error is - $\frac{0.381}{9.801} *100$

=3.88735 percent

the error is not so high,so it can be accepted.

now we have to know why this occurs-the equation of time period of the simple pendulum is give as- $T=2\pi\sqrt[2]{l/g}$

$g=4\pi^2\frac{l}{T^2}$

As the experiment is done under air resistance,so it will affect to the time period.hence the time period will be more which in turn decreases the value of g.

if this experiment is done in a environment of zero air resistance,we will get the value of g which must be approximately equal to 9.801 $m/s^2$

5 0

4 years ago