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gayaneshka [121]
3 years ago
14

A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average forc

e (in N) exerted on a 0.0200 kg bullet to accelerate it to a speed of 700 m/s in a time of 1.50 ms (milliseconds)
Physics
1 answer:
Natali5045456 [20]3 years ago
7 0

Answer:

<h2>9.3kN</h2>

Explanation:

Step one:

given data

mass of bullet= 0.02kg

speed v=700m/s

time taken =1.5ms= 0.0015 seconds

Step two:

we know that from the first law

F=ma-----1  first law of motion

also, we know that

a=v/t----2

put a=v/t in equation 1 we have

F=mv/t

Step three:

substitute our given data to find force

F=0.02*700/0.0015

F=14/0.0015

F=9333.33N

F=9.3kN

<u>The average force exerted is 9.3kN</u>

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1. Due to the first law of Newton: there is a force that accelerate the body,

then the net force for 2 forces in opposite directions can calculated as

F_{net}=F_{1}-F_{2}=26-8=14N

2. For a force make an angle with the horizontal plane, we need to analysis the force into 2 perpendicular forces:

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F_{x}=Fsin ( θ )= 67sin(73)=64N

then, as Q(1),

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3. Because of the non change in the velocity, there is no acceleration

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F_{Tension}=420 N

4. A) As we solve the Q(2), we will analysis the un horizontal or vertical forces into 2 forces

F_{x}=Fsin ( θ )= 600sin(45)=424.264N

F_{y} = F sin ( θ )= 600cos45)=424.264N

Hence we have 2 opposite horizontal forces. also, 2 opposite vertical forces.

now, we calculate the net forces in each direction

F_{net in x}=F_{1}-F_{2}=424.264-250=174.264 N in east

F_{net in y}=F_{1}-F_{2}=-500-424.264=75.736 N in south

The total net force for perpendicular forces is calculated from the relation:  

F_{total}=\sqrt{F_{x} ^{2}+F_{y}^{2}}=\sqrt{174.264^{2}+75.736^{2}}=190N

B) To find the direction, we need to calculate the angle which the total force is made with the east direction of the x-axis (horizontal)

θ = tan^{-1}(\frac{F_{y}}{F_{x} })= tan^{-1}(\frac{-75.736}{174.264})=-23.48^{o}

That meas that the total force is made with the east direction of the x-axis angle equal to 23.48 for the south (the 4th quarter)

5. According to the third law of Newton, to make the balance the net force  equal to zero (for each action there is a reaction with the same amplitude and in opposite direction)

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Star a has an absolute magnitude of 8. what is its apparent magnitude at a distance of 100 pc
tresset_1 [31]
From the definition of apparent magnitude, we know that:
m_{1} - m_{2} = -2.5Log( \frac{F_{1}}{F_{2}} )
where:
m = apparent magnitude
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We also know that the flux is given by the formula:
F =  \frac{L}{4 \pi  d^{2} }
where:
L = luminosity
d = distance

Therefore:
\frac{F_{1} }{F_{2}} = (\frac{L_{1} }{4 \pi d_{1}^{2} })(\frac{4 \pi d_{2}^{2} }{L_{2} }) \\ =  \frac{L_{1}d_{2}^{2}}{L_{2}d_{1}^{2}}

Now, let's apply these formulae to the same star (therefore, same luminosity), using apparent magnitude and absolute magnitude (which is defined as the apparent magnitude the star would have if it were at a distance of 10pc):
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Now, let's solve for m:
m = M + 2.5 Log ( \frac{d}{10})^{2}
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= 13

Hence, the apparent magnitude of the star would be m = +13
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