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2 years ago

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A 3.00-kg rifle fires a 0.00500-kg bullet at a speed of 300 m/s. Which force is greater in magnitude:(i) the force that the rifl

eexerts on the bullet; or (ii) the force that the bulletexerts on the rifle?A. the force that the rifle exerts on the bulletB. the force that the bullet exerts on the rifleC. both forces have the same magnitudeD. not enough information given to decide

1 answer:

alekssr [168]2 years ago

7 0

Answer:

C. both forces have the same magnitude

Explanation:

Here the action force is equal to the reaction force in accordance with the Newton's third law of motion.

Also when we apply the conservation of momentum so that the momentum bullet and the momentum of the gun are equal and according to the second law of motion by Newton, we have force equal to the rate of change in momentum.

We have the equation for momentum as:

$p=m.v$

Newton's second law is Mathematically given as:

$F=\frac{dp}{dt}$

Momentum is constant and the reaction time is equal, so the force exerted will also be equal.

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3 years ago

Read 2 more answers

A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold

LUCKY_DIMON [66]

Answer:

(a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

$Q=\dfrac{V}{t}$

Put the value into the formula

$Q=\dfrac{405}{90}$

$Q=4.5\ cm^3/s$

(a). We need to calculate the coefficient of permeability

Using formula of coefficient of permeability

$Q=kiA$

$k=\dfrac{Q}{iA}$

$k=\dfrac{Ql}{Ah}$

Where, Q=discharge

l = length

A = cross section area

h=constant head causing flow

Put the value into the formula

$k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}$

$k=8.6\times10^{-3}\ cm/s$

The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(c). We need to calculate the discharge velocity during the test

Using formula of discharge velocity

$v=ki$

$v=\dfrac{kh}{l}$

Put the value into the formula

$v=\dfrac{8.6\times10^{-3}\times38}{17.5}$

$v=0.0187\ cm/s$

The discharge velocity during the test is 0.0187 cm/s.

(b). We need to calculate the volume of solid in the ample

Using formula of volume

$V_{s}=\dfrac{M_{s}}{V_{s}}$

Put the value into the formula

$V_{s}=\dfrac{4950\times10^{-3}}{2710}$

$V_{s}=1826.56\ cm^3$

We need to calculate the volume of the soil specimen

Using formula of volume

$V=A\times L$

Put the value into the formula

$V=\dfrac{\pi(17.5)^2}{4}\times17.5$

$V=4209.24\ cm^3$

We need to calculate the volume of the voids

$V_{v}=V-V_{s}$

Put the value into the formula

$V_{v}=4209.24-1826.56$

$V_{v}=2382.68\ cm^3$

We need to calculate the seepage velocity

Using formula of velocity

$Av=A_{v}v_{s}$

$v_{s}=\dfrac{Av}{A_{v}}$

$v_{s}=\dfrac{V}{V_{v}}\times v$

Put the value into the formula

$v_{s}=\dfrac{4209.24}{2382.68}\times0.0187$

$v_{s}=0.0330\ cm/s$

The seepage velocity is 0.0330 cm/s.

Hence, (a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

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Answer:

$5.22m/s^2$

Explanation:

One of the first propulsion characteristics given in the example is that all engines are equal.

In this way if we have 4 engines running at the same time, it means that its capacity is 100%.

Under this premise, if 100% is found, the Jet is capable of reaching a speed of 8.7m / s ^ 2.

However, the question is, what would happen if 2.4 "Engines" now work.

To do this then we make a simple equivalence,

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$X = \frac{2.4 * 100\%} { 4 }= 60\%$

In this way it would mean that the body could be driven to 60% of its total.

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