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1 year ago

when you park alongside a curb, the front and back wheels must be parallel and no more than ....... inches from the curb

Physics

1 answer:

Rainbow [258]1 year ago

7 0

Between 12 to 18 inches.

The front and back wheels must be parallel and within at least 12 to 18 inches of the curb while parking next to a curb on a level street. If there is no curb, park parallel to the road.

➤ The following specific parking guidelines apply to painted colored curbs:

Only stop at a white light long enough to pick up or drop off mail or passengers.

Green-Park only for a short while. For time limits, look for a sign that is put adjacent to the green zone or the time limit that is painted on the curb.

Yellow-Stop for no longer than the indicated amount of time to load or unload cargo or passengers. Noncommercial vehicle drivers are typically expected to stay behind the wheel.

No stopping, standing, or parking is permitted (buses may stop at a red zone marked for buses).

Blue-parking is only allowed for a disabled person or the driver of a disabled person who has a placard or special license plate for disabled people or disabled veterans. A no-parking zone is defined as a crosshatched (diagonal lines) area adjacent to a designated disabled parking space.

Find more on parking related questions at : brainly.in/question/7229826

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What is the volume of an object that has a density of 65g/cm3 and a mass of 130g.

lora16 [44]

Density ρ is mass m per unit volume v, or

ρ = m / v

Solving for v gives

v = m / ρ

So the given object has a volume of

v = (130 g) / (65 g/cm³) = 2 cm³

5 0

3 years ago

An airplane starts from rest at the end of a runway and accelerates at a constant rate. In the first second, the airplane travel

Licemer1 [7]

Answer:

$v=4.44\frac{m}{s}$

Explanation:

Given that the airplane starts from the rest (this is initial velocity equals to zero) and accelerates at a constant rate, position can be described like this: $x=v_{0}t +\frac{1}{2} at^{2}$ where x is the position, t is the time a is the acceleration and $v_{0}$ is initial velocity. In this way acceleration can be found. $a=\frac{2(x-v_{0}t) }{t^{2} } =\frac{2(1.11m-0)}{1s^{2} } =2.22\frac{m}{s^{2} }$ .

Now we are able to found velocity at any time with the formula: $v=v_{0} +at = 0\frac{m}{s} +(2.22\frac{m}{s^{2}}.2s)=4.44\frac{m}{s}$

3 0

3 years ago

The isomerization of cyclopropane to form propene is a first-order reaction. At 760 K, 85% of a sample of cyclopropane changes t

Nataliya [291]

Answer:

so rate constant is 4.00 x 10^-4 $s^{-1}$

Explanation:

Given data

first-order reactions

85% of a sample

changes to propene t = 79.0 min

to find out

rate constant

solution

we know that

first order reaction are

ln [A]/[A]0 = -kt

here [A]0 = 1 and (85%) = 0.85 has change to propene

so that [A] = 1 - 0.85 = 0.15.

that why

[A] / [A]0= 0.15 / 1

[A] / [A]0 = 0.15

here t = (79) × (60s/min) = 4740 s

k = - {ln[A]/[A]0} / t

k = -ln 0.15 / 4740

k = 4.00 x 10^-4 $s^{-1}$

so rate constant is 4.00 x 10^-4 $s^{-1}$

3 0

3 years ago

Cynthia forgot to put the fabric softener in the wash. As her socks tumbled in the dryer, they became charged. If a small piece

polet [3.4K]

Answer:

E = 24000 N/C = 24 KN/C

Explanation:

The electric field experienced by a test charge is given by the following formula:

$E = \frac{F}{q}\\\\$

where,

E = Electric Field = ?

F = Force of attraction = 3 x 10⁻⁶ N

q = Charge on piece of lint = 1.25 x 10⁻¹⁰ C

Therefore, using these values in the equation, we get:

$E = \frac{3\ x\ 10^{-6}\ N}{1.25\ x\ ^{-10}\ C}\\\\$

E = 24000 N/C = 24 KN/C

6 0

3 years ago

Stronger acids have more _______

mario62 [17]

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7 0

3 years ago