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Karo-lina-s [1.5K]
3 years ago
9

A car is traveling at 70 km/h. It then uniformly decelerates to a complete stop in 12 s. Find its acceleration ( in m/s2 ).

Physics
2 answers:
Lera25 [3.4K]3 years ago
8 0

Answer:

- 1.62 m/s²

Explanation:

initial velocity, u = 70 km/h = 19.44 m/s

time, t = 12 s

final velocity, v  = 0

Acceleration is defined by the rate of change of velocity.

By use of first equation of motion

v = u + a t

where, a be the acceleration

0 = 19.44 + a x 12

a = - 1.62 m/s²

Thus, the acceleration is - 1.62 m/s².

Yakvenalex [24]3 years ago
5 0

Answer:

Acceleration will be -1.620m/sec^2

Explanation:

We have given initial speed of the car is 70 km/hr

We know that 1 km = 1000 m

And 1 hour = 3600 sec

So 70km/hr=70\times \frac{1000m}{3600sec}=19.444m/sec

It is given that car stops in 12 sec

So final speed of the car v = 0 m/sec

Time t = 12 sec

From first equation of motion v = u+at

So 0=19.444+a\times 12

a=\frac{-19.444}{12}=-1.620m/sec^2 ( negative sign indicates that speed of the car will constantly decrease )

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Answer:

all colours are absorbed except for the colour of the filter.

Explanation:

When white light passes through a coloured filter, all colours are absorbed except for the colour of the filter. For example, an orange filter transmits orange light but absorbs all the other colours. If white light is shone on an orange filter, only the orange wavelengths will be observed by the human eye.

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3 years ago
10 points
grandymaker [24]
The answer is B tell me if I am wrong.
6 0
2 years ago
a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&
Katen [24]

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

8 0
3 years ago
What is Newton's Second Law?
Rudiy27

Answer:

An object's acceleration depends on its mass and on the net force acting on it.

Explanation:

Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

8 0
3 years ago
Read 2 more answers
A,b, e are complete. Help on the others would be so appreciated!!
bixtya [17]

Answer:

serie Ceq=0.678 10⁻⁶ F  and the charge Q = 9.49 10⁻⁶ C

Explanation:

Let's calculate all capacity values

a) The equivalent capacitance of series capacitors

    1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3 + 1 / C4 + 1 / C5

    1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 1 / 6.2 + 1 / 6.2

    1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 2 / 6.2

    1 / Ceq = 0.666 + 0.3030 +0.1818 +0.3225

    1 / Ceq = 1,147

    Ceq = 0.678 10⁻⁶ F

b) Let's calculate the total system load

   Dv = Q / Ceq

   Q = DV Ceq

   Q = 14 0.678 10⁻⁶

   Q = 9.49 10⁻⁶ C

In a series system the load is constant in all capacitors, therefore, the load in capacitor 5.5 is Q = 9.49 10⁻⁶ C

c) The potential difference

   ΔV = Q / C5

   ΔV = 9.49 10⁻⁶ / 5.5 10⁻⁶

   ΔV = 1,725 ​​V

d) The energy stores is

    U = ½ C V²

    U = ½ 0.678 10-6 14²

    U = 66.4 10⁻⁶ J

e) Parallel system

   Ceq = C1 + C2 + C3 + C4 + C5

   Ceq = (1.5 +3.3 +5.5 +6.2 +6.2) 10⁻⁶

   Ceq = 22.7 10⁻⁶ F

f) In the parallel system the voltage is maintained

   Q5 = C5 V

   Q5 = 5.5 10⁻⁶ 14

   Q5 = 77 10⁻⁶ C

g) The voltage is constant V5 = 14 V

h) Energy stores

   U = ½ C V²

   U = ½ 22.7 10-6 14²

   U = 2.2 10⁻³ J

8 0
3 years ago
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