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4 years ago

If your accelerator is stuck while driving

Physics

1 answer:

sertanlavr [38]4 years ago

5 0

Answer: put in neutral, apply the breaks, turn on hazards, pull off the road, last turn off the ignition.

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Water is made of two hydrogen atoms and one oxygen atom bonded together. Julia is describing how water undergoes a physical chan

FrozenT [24]

Everything that Julia said is correct EXCEPT the words
"the Oxygen and Hydrogen separate". They do not.

The "particles" that she mentioned are water molecules,
each one composed of two Hydrogen atoms and one Oxygen
atom bonded together. THOSE are the particles that shift to be
farther apart, but the Hydrogen and Oxygen in each one remain
bonded together.

Julia gets 30 ounces of merit badges, out of a possible quart.

6 0

3 years ago

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

A particle of ink in a ink-jet printer carries a charge of -8x 10^-13 C and is deflected onto paper force of 3.2x10^-4. Find the

vitfil [10]

Beef and cheddar is your answer

3 0

3 years ago

QUESTION 1

frez [133]

Go90 day would work better if they could just pay ate a little bit more than we

4 0

3 years ago

Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards th

yuradex [85]

Answer:

Explanation:

Formula E=F/C also E=V/d

In this case use the second formula; E=V/d

Data given; E=4N/C d=8m

So v=E X d

V=4x8=32V

k.e=eV= 2X32=64eV

3 0

4 years ago