What subatomic particle has a nuetral charge?

Goryan [66]

Answer:

The braking distance would be about nine times as long (assuming that acceleration during braking stays the same.)

Explanation:

Let $u$ denote the initial velocity of the vehicle ( $20\; \text{mph}$ or $60\; \text{mph}$ ) and let $v$ denote the velocity of the vehicle after braking ( $0\; \text{mph}$ ). Let $x$ denote the braking distance.

Assume that the acceleration during braking are both constantly $a$ in both scenarios. The SUVAT equations would apply. In particular:

$\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a}\end{aligned}$ .

Since $v = 0$ <em> </em>(the vehicle has completely stopped), the equation becomes $x = (-u^{2}) / (2\, a)$ .

Assuming that $a$ (braking acceleration) stays the same, the braking distance $x$ would be proportional to $u^{2}$ , the square of the initial velocity.

Hence, increasing the initial speed from $20\; \text{mph}$ to $60\; \text{mph}$ would increase the braking distance by a factor of $3^{2} = 9$ .

7 0

1 year ago

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What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from

ElenaW [278]

When is at the end of the runway the velocity of the plane is given by the equation $vf^{2}=0+2*a*s$ where s=1800 m is the runway length. Thus
$vf^{2}=2*5*1800=18000 (m/s)^{2}$
$vf =134.164 (m/s)$

At half runway the velocity of the plane is
$v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}
$
$v= \sqrt{9000}=94.87 ( \frac{m}{s})$

Therefore at midpoint of runway the percentage of takeoff velocity is
‰ $P= \frac{v}{vf}= \frac{94.87}{134.164}=0.707$

6 0

3 years ago

Visible light travels at a speed 3.0 × 108 of m/s. If red light has a wavelength of 6.5 × 10–7 m, what the frequency of this lig

gayaneshka [121]

i think its d on edge

4 0

3 years ago

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Standard solar panels transform 3,600,000 J of input light energy into 720,000 J of useful electrical energy. The new solar pane

dangina [55]

Answer:

5

Explanation:

we divide 3600000J by 720,000J

and after that the answer will be 5

3600,000J

---------------

720,000J

=5answer

7 0

3 years ago

A car is traveling at 104 km/h when the driver sees an accident 50 m ahead and slams on the brakes. What minimum constant decele

Solnce55 [7]

Answer:

–8.35 m/s²

Explanation:

We'll begin by converting 104 km/h to m/s. This can be obtained as follow:

3.6 Km/h = 1 m/s

Therefore,

104 km/h = 104 km/h × 1 m/s / 3.6 Km/h

104 km/h = 28.89 m/s

Thus, 104 km/h is equivalent to 28.89 m/s.

Finally, we shall determine the deceleration of the car. This can be obtained as follow:

Initial velocity (u) = 28.89 m/s

Final velocity (v) = 0 m/s

Distance (s) = 50 m

Deceleration (a) =?

v² = u² + 2as

0² = 28.89² + (2 × a × 50)

0 = 834.6321 + 100a

Collect like terms

0 – 834.6321 = 100a

–834.6321 = 100a

Divide both side by 100

a = –834.6321 / 100

a = –8.35 m/s²

Thus, the deceleration of the car is –8.35 m/s².

5 0

3 years ago

What subatomic particle has a nuetral charge?​

What subatomic particle has a nuetral charge?