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Anestetic [448]
4 years ago
10

Convert 150,000mm to km

Physics
1 answer:
Gwar [14]4 years ago
5 0

.15 km.  It goes in this way from largest to least:  Kilometers, Hectometers, decameters, meters, decimeters, centimeters, and finally millimeters.  When you are going up in measures you move the decimal point one space to the left for each measure you pass.  It is vice-versa for going down these measures plz mark brainliest.

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In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope
zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude g.

If we have the initial velocity v_o and its angle \theta, we can obtain the vertical component of the velocity v_{oy} using trigonometry:

v_{oy}=v_osin\theta

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}

Finally, from the equation of horizontal motion with constant speed, we have that:

x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s}  }{9.8\frac{m}{s^{2} } } =7.49m

In words, the projectile travels 7.49m horizontally before it lands.

8 0
4 years ago
The 9 kg block is then released and accelerates
katrin [286]

Answer: 1458

explanation

7 0
3 years ago
Based on the concept of the wave-like nature of light, Huygens' theory of light __________ postulates that the more light was "b
DochEvi [55]
I believe it was Christaan huygens.

have a nice day!
7 0
3 years ago
Read 2 more answers
A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T
Nady [450]
The first collision because a greater amount of momentum must be taken and used in order to push the cart back, giving it a greater mass and impulse
6 0
4 years ago
An orienteer runs north at 5 m/s for 120 seconds, and then west at 4 m/s for 180 seconds. Provide a graphic solution to show the
Ivan

Answer:937.22 m

Explanation:

Given

First orienteer is running towards North with 5 m/s for 120 s

so he covers a distance of 5\times 120=600 m

After that He runs towards west at 4 m/s for 180 s

So he covers a distance of 4\times 180=720 m

Mathematically its net Displacement is

displacement=\sqrt{600^2+720^2}

displacement=937.22 m      

3 0
3 years ago
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