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ziro4ka [17]
3 years ago
11

A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 5.70 x 10^5 m/s^2 for 9.

60 x 10^−4 s. What is its muzzle velocity (in m/s) (that is, its final velocity)? (Enter the magnitude.)
Physics
1 answer:
s2008m [1.1K]3 years ago
3 0

Answer:

Final speed, v = 547.2 m/s

Explanation:

Initial velocity of the bullet, u = 0

Acceleration of the bullet, a=5.7\times 10^5\ m/s^2

Time taken, t=9.6\times 10^{-4}\ s

Let v is the final velocity of the muzzle. It can be calculated using the first equation of motion as :

v=u+at

v=at      

v=5.7\times 10^5\ m/s^2\times 9.6\times 10^{-4}\ s    

v = 547.2 m/s

So, the final speed of the muzzle is 547.2 m/s. Hence, this is the required solution.                                

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Answer:

Explanation:

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8 0
1 year ago
physics A river flows at a speed vr = 5.37 km/hr with respect to the shoreline. A boat needs to go perpendicular to the shorelin
vova2212 [387]

Answer: Vb is the vector  (-5.37m/s,  8.59 m/s), with a module 10.13m/s

then the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9

Explanation:

We can use the notation (x, y) where the river flows in the x-axis and the pier is on the y-axis.

We have Vr = (5.37m/s, 0m/s)

Now, if the boat wants to move only along the y-axis (perpendicularly to the shore).

The velocity of the boat Vb will be:

Vb = (-c*sin(32). c*cos(32))

Then we should have that:

5.37 m/s - c*sin(32) = 0

c = (5.37/sin(32))m/s = 10.13 m/s

the velocity in the y-axis is:

10.13m/s*cos(32) = 8.59 m/s

So Vb = (-5.37m/s,  8.59 m/s)

the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9 where i used Vb as the module of the boat's velocity.

7 0
3 years ago
According to Bode’s Law a planet is missing between Jupiter and Saturn. True False
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6 0
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k) A stream of warm water is produced in a steady-flow mixing process by combining 1.0 kg s-1of cool water at 25°C with 0.8 kg s
Irina18 [472]

Answer:

T_ww = 43,23°C

Explanation:

To solve this question, we use energy balance and we state that the energy that enters the systems equals the energy that leaves the system plus losses. Mathematically, we will have that:

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The energy associated to a current of fluid can be defined as:

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Replacing the values given on the statement, we have:

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Solving for the temperature Tww, we have:

(1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8 kg/s*4,18 kJ/(kg°C)*75°C-30 kJ/s)/(1.8 kg/s*4,18 kJ/(kg°C))=T_WW

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Have a nice day! :D

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