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Julli [10]
3 years ago
12

An agitated llama may spit to assert dominance, or to ward off threats. llamas can spit a considerable distance, and people hand

ling them need to keep this in mind. part a if the spittle from a llama is launched from an initial height of 1.8 m with a speed of 5.5 m/s, and at an angle of 13 ?? above horizontal, how far will it travel horizontally?
Physics
1 answer:
zmey [24]3 years ago
5 0
The answer is 4.7 m for the spàthan. this is a very question! I hope this helps my friend! :-)
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6 0
3 years ago
Please help me with this physics prooblem
zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

x=v_0\cos20.0^\circ t+\dfrac12a_xt^2

y=v_0\sin20.0^\circ t+\dfrac12a_yt^2

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ

y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ

So we have enough information to solve for the components of the acceleration vector, a_x and a_y:

x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}

y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}

The acceleration vector then has direction \theta where

\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ

5 0
3 years ago
While traveling along a highway a driver slows from 24 m/sec to 15 m/sec in 12 seconds. What is the
Vladimir [108]
24-15=9 m/s slower in 12 seconds. So 9/12 m/s² slower. Therefore the acceleration is -0,75 m/s²
4 0
3 years ago
About how many weeks does it take the Moon to go from a full Moon phase to a third quarter phase? A. 1 B. 2 C. 3 D. 4
Lostsunrise [7]
The answer is D) About 4.
~Deceptiøn
3 0
3 years ago
A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 11.8 m/s directly upward.
Fantom [35]

Answer:

7.09683 m

1.20285 s

2.4057 s

11.8 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)

From equation of motion we have

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11.8^2}{2\times -9.81}\\\Rightarrow s=7.09683\ m

The maximum height above the ground that the ball reaches is 7.09683 m

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-11.8}{-9.81}\\\Rightarrow t=1.20285\ s

Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s

v=u+at\\\Rightarrow v=0+9.81\times 1.20285\\\Rightarrow v=11.8\ m/s

The velocity just before it hits the ground is 11.8 m/s

6 0
3 years ago
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