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Scilla [17]
4 years ago
14

On dry concrete, a car can decelerate at a rate of 7.00 m/s2 , whereas on wet concrete it can decelerate at only 5.00 m/s2 . Fin

d the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h) (a) on dry concrete and (b) on wet concrete. (c) Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the brake.
Physics
1 answer:
MariettaO [177]4 years ago
3 0

Answer:

Explanation:

a ) Let the distance required in former case be d₁ .

initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 7.00 m /s²

v² = u² - 2 a s

0 = 30² - 2 x 7 x d₁

d₁ = 64.28 m

b) initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 5.00 m /s²

v² = u² - 2 a s

0 = 30² - 2 x 5 x d₂

d₂ = 90  m

c)

t = .5 s

s₁  = ut - .5 at²

= 30 x .5 - .5 x 7 x .5²

= 15 - .875

= 14.125 m

t = .5 s

s₂  = ut - .5 at²

= 30 x .5 - .5 x 5 x .5²

= 15 - .625

= 14.375  m

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grigory [225]

Answer:

The box displacement after 6 seconds is 66 meters.

Explanation:

Let suppose that velocity given in statement represents the initial velocity of the box and, likewise, the box accelerates at constant rate. Then, the displacement of the object (\Delta s), in meters, can be determined by the following expression:

\Delta s = v_{o}\cdot t+\frac{1}{2}\cdot a\cdot t^{2} (1)

Where:

v_{o} - Initial velocity, in meters per second.

t - Time, in seconds.

a - Acceleration, in meters per square second.

If we know that v_{o} = 5\,\frac{m}{s}, t = 6\,s and a = 2\,\frac{m}{s^{2}}, then the box displacement after 6 seconds is:

\Delta s = 66\,m

The box displacement after 6 seconds is 66 meters.

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3 years ago
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What is the period of a wave traveling 5 m/s if its wavelength is 20 m/s
Ilia_Sergeevich [38]

Speed of wave is given as

v = 5 m/s

Wavelength of the wave is given as

\lambda = 20 m

now from the formula of wave time period we can say

speed = \frac{wavelength}{time period}

5 = \frac{20}{T}

T = \frac{20}{5}

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How to understand that an inductor behaves like short circuit and capacitor like an open circuit in steady state in a network?
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3 years ago
PE=30J, m=?, g=10m/s2, h=10m
OleMash [197]
Based on the given, this is probably a gravitational potential energy problem (PEgrav). The formula for PEgrav is:

PEgrav = mgh

Where:
m = mass (kg)
g = acceleration due to gravity
h = height (m)

With this formula you can derive the formula for your unknown, which is mass. First put in what you know and then solve for what you do not know.

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30J=m(10)(10[tex] \frac{30}{100} =m)[/tex]

Do operations that you can with what is given first.

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Transpose the 100 to the other side of the equation. Do not forget that when you transpose, you do the opposite operation.

\frac{30}{100} =m

m = 0.30kg

5 0
3 years ago
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