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son4ous [18]
2 years ago
13

NEED TODAY

Physics
2 answers:
never [62]2 years ago
5 0

Answer:

well,they may or they may not. It all depends on the vertical components of the speed with which the projectile is thrown.

Explanation:

to derive the time of flight :

Considering the vertical motion (along y axis) , the journey from the ground to highest point

v=u+at ,where v=final velocity , u=initial velocity , acceleration due to gravity i.e g and t is the time taken by the particle to reach the highest point

o=u(sin theta) + (- g)t

t=u(sin theta) - g

Hence total time of flight is twice of that i.e

T=2(vertical components of speed) / g

So, if two projectile have some vertical components of velocity , they will have equal time of flight.

lisabon 2012 [21]2 years ago
3 0

Answer:

No, projectile launched at 20° landed first.

Explanation:

70° projectile had a vertical velocity of v*sin(70), which is significantly bigger than v*sin(20). Since v=vx+at, we can see that Vyi*sin(20)-gt < Vyi*sin(70)-gt

solve for t, you can check that by substituting velocity for any number.

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Starting at (0,0) an object travels 36 meters north and then it covers 20 meters east. What is
Svetradugi [14.3K]

Answer:

Explanation:

Using the pythagoras theorem, the displacement is expressed as;

d² = x²+y²

y = 36m (north)

x = 20m east

Substitute;

d² = 36²+20²

d² = 1296+400

d² = 1696

d = √1696

d = 41.18m

For the direction;

theta = tan^-1(y/x)

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theta = tan^-1(1.8)

theta = 60.95°

Hence the magnitude is 41.18m and the direction is 60.95°

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3 years ago
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mario62 [17]

Answer:

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If it wants types of small/average stars, then go with

Small star names:

OGLE-TR-122B

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The vapor pressure of ethanol at 293 K is 5.95 kPa and at 336.5 K it is 53.3 kPa. Calculate the enthalpy of vaporization of etha
denis-greek [22]

Answer:

H=41.3kJmol^{-1}

Explanation:

The equation relating the the enthalphy, pressure and temperature is expressed as

ln(\frac{P_{2}}{P_{1}} )=\frac{H}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}} ) \\

Where P is the pressure, H is the enthalphy, and T is the temperature.

since the given values are

T_{1}=293k, \\T_{2}=336.5k\\P_{1}=5.95kPA\\p_{2}=53.3kPA\\and R=8.314J.K^{-1}mol_{-1}

if we insert values, we arrive at

ln(\frac{53.3}{5.95} )=\frac{H}{8.314}(\frac{1}{293}-\frac{1}{336.5} )\\2.19=\frac{H}{8.314}(0.00044)\\H=(2.19*8.314)/0.00044\\H=41,268.8Jmol^{-1}\\H=41.3kJmol^{-1}

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