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3 years ago

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A bullet is fired horizontally at a height of 2 meters at a velocity of 930 m/s. Assume no air resistance. How far did the bulle

t travel horizontally when it hit the ground?
595.2 m
478.4 m
364.0 m
247.2 m

2 answers:

Vesnalui [34]3 years ago

5 0

It’s most likely A , i’m not so sure though

USPshnik [31]3 years ago

4 0

Answer:

im pretty sure its a

Explanation:

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Air enters the compressor of a gas-turbine engine at 300 K and 100 kPa, where it is compressed to 700 kPa and 580 K.Heat is tran

Natasha2012 [34]

Answer:

- the fraction of the turbine work output used to drive the compressor is 50%

- the thermal efficiency is 29.6%

Explanation:

given information:

$T_{1}$ = 300 k

$P_{1}$ = 700 kPa

$T_{2}$ = 580 K

Qin = 950 kj/kg

efficiency, η = 86% = 0.86

Assume, Ideal gas specific heat capacities of air, $C_{p}$ = 1.005

k = 1.4

(a) the fraction of the turbine work output used to drive the compressor

Qin = $C_{p}$ ( $T_{3}$ - $T_{2}$ )

$T_{3}$ - $T_{2}$ = Qin / $C_{p}$

$T_{3}$ = Qin / $C_{p}$ + $T_{2}$

thus,

$T_{3}$ = $\frac{950 kJ/kg}{1.005 kj/kgK} + 580 K
[tex]T_{3}$ = 1525 K

$\frac{T_{4} }{T_{3} }$ = $(\frac{P_{4} }{P_{3} }) ^{k-1/k}$ \

$\frac{P_{4} }{P_{3} }$ = $\frac{P_{1} }{P_{2} }$

so,

$\frac{T_{4} }{T_{3} }$ = $(\frac{P_{1} }{P_{2} }) ^{k-1/k}$ \

$T_{4}$ = $T_{3}$ $(\frac{P_{1} }{P_{2} }) ^{k-1/k}$ \

= 1525 K $(\frac{100}{700} )^{1.4-1/1.4}$

= 875 K

$W_{in}$ = $C_{p}$ ( $T_{2}$ - $T_{1}$ )

= 1.005 kj/kgK (580 K - 300 K)

= 281.4 kJ/kg

$W_{out}$ = η $C_{p}$ ( $T_{3}$ - $T_{4}$ )

= 0.86 (1.005 kj/kgK) (1525 k - 875 K)

= 562.2 kJ/kg

therefore,

the fraction of turbine = $\frac{W_{in} }{W_{out} }$

= $\frac{281.4}{562.2}$

= 50%

(b) the thermal efficiency

η = ΔW/ $Q_{in}$

= ( $W_{out}$ - $W_{in}$ )/ $Q_{in}$

= $\frac{562.2}{281.4}$ /950

= 29.6%

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3 years ago

The strength of an object's gravitational force is affected by both____<br> and____

DochEvi [55]

Answer:

the answers are "mass" and "distance"

Explanation:

hope this helps, plz mark brainiest :)

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3 years ago

a bullet of mass 4g when fired with a velocity of 50m/s can enter a wall upto a depth of 10cm how much will be the average resis

jok3333 [9.3K]

Mass, m = 4g = 0.004 kg
Velocity, = 50cm/s = 0.5m/s
Distance, 10cm = 0.1m
The wall would have to resist the energy acquired by the bullet.

Kenetic Energy of bullet = Resistance offered by the wall.

1/2 mv² = Resistance Force * Distance

(1/2) * 0.04 * 0.5 * 0.5 = F * 0.1

0.5 * 0.04 * 0.5 * 0.5 = F * 0.1

0.5 * 0.04 * 0.5 * 0.5/0.1 = F

0.05 = F

Therefore, Resistance offered by the wall = 0.05 N

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5 What is the maximum speed at which a car round a curve of 25m radius on a level road if the coefficient of static friction bet

pshichka [43]

Hi there!

On a level road:

∑F = Ff (Force due to friction)

The net force is the centripetal force, so:

mv²/r = Ff

Rewrite the force due to friction:

mv²/r = μmg

Cancel out the mass:

v²/r = μg

Solve for v:

v = √rμg

v = √(25)(9.81)(0.8) = 14.01 m/s

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Please help!! The graph in the figure shows the position of a particle as it travels along the x-axis. What is the magnitude of

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Answer:

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Explanation:

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