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34kurt
3 years ago
7

A car initially rolling at 12 m/s comes to a stop in 20 seconds. Assuming the car has uniform acceleration, how far does it trav

el during this time interval?
Physics
1 answer:
Elis [28]3 years ago
4 0
As the car comes to rest it’s final velocity v will be 0
So our information g is
u = 12 m /s
t = 20s
v = 0

So acceleration is v-u/t
So 0-12/20
= -12/20
= -0.6 m / s^2

S= ut + 1/2 at^2
S = 12 x 20 + 1/2 x -12/20 x 400
S= 240 - 120
S = 120

It travels 120 m during the time interval.
Please mark brainliest
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3 years ago
A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19s lat
victus00 [196]

Answer / Explanation

It is worthy to note that the question is incomplete. There is a part of the question that gave us the vale of V₀.

So for proper understanding, the two parts of the question will be highlighted.

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19s later. You may ignore air resistance.

a) What must the height of the building be for both balls to reach the ground at the same time if (i) V₀ is 6.0 m/s and (ii) V₀ is 9.5 m/s?

b) If Vo is greater than some value Vmax, a value of h does not exist that allows both balls to hit the ground at the same time.  

Solve for Vmax

Step Process

a)  Where h = 1/2g [ (1/2g - V₀)² ] / [(g - V₀)²]

Where V₀ = 6m/s,

We have,

           h = 4.9 [ ( 4.9 - 6)²] / [( 9.8 - 6)²]

                 = 0.411 m

Where V₀ = 9.5m/s

We have,

     h = 4.9 [ ( 4.9 - 9.5)²] / [( 9.8 - 9.5)²]

                 = 1152 m

b)  From the expression above, we got to realise that h is a function of V₀, therefore, the denominator can not be zero.

Consequentially, as V₀ approaches 9.8m/s, h approaches infinity.

Therefore Vₙ = V₀max = 9.8 m/s

4 0
3 years ago
A series RC circuit contains a 1,000 ohm resistor and a 0.025 microfarad capacitor. What is the time constant of this circuit?
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\tau = RC

Here,

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C = Capacitance

Replacing we have that

\tau = (1000)(0.025*10^{6})

\tau = 25*10^{-6}

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☁️ Answer ☁️

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