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bixtya [17]
3 years ago
9

A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It s

trikes a target 1.7 seconds later. What is the vertical distance from where the projectile was launched to where it hit the target.
Physics
1 answer:
Zarrin [17]3 years ago
7 0

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± \frac{1}{2} at^2               ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± \frac{1}{2} at^2               ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

<em>From the question;</em>

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s   (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - \frac{1}{2} at^2

h = 26.54(1.7) - \frac{1}{2} (10)(1.7)^2

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m        

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An object of mass 5 m is lifted to a height of 5 h above the floor. What is the potential energy the object gained in this proce
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Answer:

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Explanation:

Given that,

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E=m\times g\times h

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The noble gases neon (atomic mass 20.1797 u) and krypton (atomic mass 83.798 u) are accidentally mixed in a vessel that has a te
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Answer:

(a) Kav Ne = Kav Kr = 7.29x10⁻²¹J

(b) v(rms) Ne= 659.6m/s and v(rms) Kr= 323.7m/s

Explanation:

(a) According to the kinetic theory of gases the average kinetic energy of the gases can be calculated by:

K_{av} = \frac{3}{2}kT (1)        

<em>where K_{av}: is the kinetic energy, k: Boltzmann constant = 1.38x10⁻²³J/K, and T: is the temperature </em>

<u>From equation (1), we can calculate the</u><u> average kinetic energies for the krypton and the neon:</u>

K_{av} = \frac{3}{2} (1.38\cdot 10^{-23} \frac{J}{K})(352.2K) = 7.29\cdot 10^{-21}J  

(b) The rms speeds of the gases can be calculated by:

K_{av} = \frac{1}{2}mv_{rms}^{2} \rightarrow v_{rms} = \sqrt \frac{2K_{av}}{m}  

<em>where m: is the mass of the gases and v_{rms}: is the root mean square speed of the gases</em>

For the neon:

v_{rms} = \sqrt \frac{2(7.29 \cdot 10^{-21}J)}{20.1797 \frac{g}{mol} \cdot \frac {1mol}{6.022\cdot 10^{23}molecules} \cdot \frac{1kg}{1000g}} = 659.6 \frac{m}{s}          

For the krypton:

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Have a nice day!

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