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spayn [35]
3 years ago
9

A car traveling at 27 m/s slams on its brakes to come to a stop. It decelerates at a rate of 8 m/s2 . What is the stopping dista

nce of the car?
Physics
1 answer:
qaws [65]3 years ago
5 0

<em>v</em>² - <em>u</em>² = 2 <em>a</em> ∆<em>x</em>

where <em>u</em> = initial velocity (27 m/s), <em>v</em> = final velocity (0), <em>a</em> = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆<em>x</em> = stopping distance.

So

0² - (27 m/s)² = 2 (-8 m/s²) ∆<em>x</em>

∆<em>x</em> = (27 m/s)² / (16 m/s²)

∆<em>x</em> ≈ 45.6 m

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How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential d
Archy [21]

Answer:

The voltage is   V =   0.993V_b

Explanation:

From the question we are told that

   The time that has passed is  t = \frac{\tau}{2}

 Here \tau is know as the time constant

    The voltage of the  power source is   V_b

Generally the voltage equation for charging a capacitor is mathematically represented as

       V =  V_b  [1 - e^{- \frac{t}{\tau} }]

=>   V =  V_b  [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }]

=>   V =  V_b  [1 - e^{- \frac{\tau}{2\tau} }]

=>   V =  V_b  [1 - e^{- \frac{1}{2} }]

=>   V =   0.993V_b    

5 0
3 years ago
Why will a sheet of paper fall slower than one that is crumbled into a ball
Zarrin [17]
It’s coming in contact with more air molecules than I would if it was in a ball because there is less surface area
4 0
3 years ago
Read 2 more answers
g Which of the following is true about magnetic field lines? A. All magnetic field lines are always parallel to the Earth’s magn
Goshia [24]

Answer:B

Explanation:

Magnetic field lines form close loops and never intercept

3 0
3 years ago
Una carga de -10Mc está situada a 20cm delante de otra carga de 5 Mc. Calcular la fuerza electrostática en Newton ejercida por u
tino4ka555 [31]

Answer:

a)  force between them is attraction,   b)  F = 1.125 10⁻² N

Explanation:

In this case the electric force is given by Coulomb's law

          F =k \frac{q_1q_2}{r^2}

           

In the exercise they give us the values ​​of the loads

          q1 = - 10 mC = -10 10⁻³ C

          q2 = 5 mC = 5 10⁻³ C

           d = 20 cm = 0.20 m

let's calculate

          F = 9 10⁹ \frac{10 \ 10^{-3} \ 5 \ 10^{-3}}{0.20^2}

          F = 1.125 10⁻² N

To find the direction of the force we use that charges of the same sign repel each other, as in this case there is a positive and a negative charge, the force between them is attraction

7 0
3 years ago
44. A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down
Vladimir [108]

Answer:

18.4 m

Explanation:

(a)

The known variables in this problem are:

u = 1.40 m/s is the initial vertical velocity (we take downward direction as positive direction)

t = 1.8 s is the duration of the fall

a = g = 9.8 m/s^2 is the acceleration due to gravity

(b)

The vertical distance covered by the life preserver is given by

d=ut + \frac{1}{2}at^2

If we substitute all the values listed in part (a), we find

d=(1.40 m/s)(1.8 s)+\frac{1}{2}(9.8 m/s^2)(1.8 s)^2=18.4m

8 0
3 years ago
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