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A projectile is launched at an angle 60° to the horizontal

Katena32 [7]

Explanation:

We have,

A projectile is launched at an angle 60° to the horizontal with some initial speed 45 m/s.

The cliff is 265 m high.

(A) The final velocity of the projectile is given by :

$v_f^2-v_i^2=2gs\\\\v_f^2=2gs+v_i^2\\\\v_f^2=2\times 9.8\times 265+45^2\\\\v_f=84.96\ m/s$

(B) The maximum height of the projectile is given by :

$H=\dfrac{v_i^2\sin^2\theta}{2g}\\\\H=\dfrac{45^2\times \sin^2(60)}{2\times 9.8}\\\\H=77.48\ m$

(C) The horizontal range of the motion is given by :

$R=\dfrac{v_i^2\sin 2\theta}{g}\\\\R=\dfrac{(45)^2\times \sin 2(60)}{9.8}\\\\R=178.94\ m$

7 0

4 years ago

Calculate A, E, μ, cv and S for 1 mole of Kr at 298 K and 1 atm (assuming ideal behavior)

vaieri [72.5K]

Answer:

internal energy E 3716.35 j

cv = 12.47 J/K

S = 12.47 J/K

A = 0.29 J

$\mu =3716.35 J/mole$

Explanation:

given data:

Kr atomic number = 36

degree of freedom = 3

1) internal energy E = $\frac{f}{2} n R T$

= $\frac{3}{2} *1*8.314*298 = 3716.35 j$

2) $cv = \frac{E}{T}$

$= \frac{3716.35}{298} = 12.47 J/K$

3) $S = cv =\frac{E}{T} = 12.47 J/K$

4) A, Halmholtz free energy = E -TS = 37146.35 - 12.47*298 = 0.29 J

5) $chemcial potential \mu = \frac{energy}{mole} = 3716.35 J/mole$

6 0

3 years ago

True or False: Translations are rigid transformations.

GaryK [48]

Answer:

Explanation:

Yes, translations are rigid transformations. You can verify that the distances between the points are preserved.

8 0

3 years ago

Read 2 more answers

Please List 3 categories of properties used to classify solids.

icang [17]

Electrical conductivity
Magnetism
Texture

Hope this helps!

8 0

3 years ago

The term ________ commonly is used to refer to a string that is part of another string.

Soloha48 [4]

Commonly is used to refer to a string that is part of another string.its a substring

8 0

4 years ago