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Sphinxa [80]
3 years ago
5

A car advertisement states that a car can accelerate at 2 m/sec. The car starts from rest and accelerates to 70 km/hr. How much

time does it take to accelerate to 70 km/hr.​
Physics
1 answer:
Reptile [31]3 years ago
7 0

Answer:

9.72s

Explanation:

\pink{\frak{Given}}\begin{cases}\textsf{ A car starts from rest.}\\\textsf{It has an acceleration of 2m/s$^2$.}\\\textsf{ The final velocity of the car is 70km/hr.}\end{cases}

Here we need to find out the time that the car took to achieve the velocity of 70km/hr after starting from rest . Firstly convert the final velocity in m/s by multiplying it by 5/18 , as ,

\longrightarrow\sf 70km/hr = 70 \times \dfrac{5}{18} m/s =\bf 19.44\  m/s

  • As we know that the rate of change of velocity is known as acceleration , so ;

\longrightarrow\sf a =\dfrac{v - u}{t}

Plug in the respective values ,

\longrightarrow\sf 2m/s^2 = \dfrac{ 19.4 m/s -0m/s}{t} \\

Simplify ,

\longrightarrow\sf 2m/s^2 =\dfrac{19.4m/s}{t}

Cross multiply ,

\longrightarrow\sf t = \dfrac{19.44m/s}{2m/s^2}

Simplify

\longrightarrow\sf \boxed{\bf t = 9.72 s}

<h3>Hence the required answer is 9.72s.</h3>
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A snow ball is thrown upward at a speed of 27 m/s. How high is the snow ball 3.5 seconds later? Also, how fast is it moving at t
JulijaS [17]

Answer:

Dy = 111.66 [m]

t  = 3.5 [s]

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To solve this problem we must use the equations of kinematics.

v_{f} = v_{o} - (g*t)\\

where:

Vf = final velocity [m/s]

Vo = initial velocity = 27 [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time = 3.5 [s]

Note: The negative sign of the equation means that the gravity acceleration goes in opposite direction

Vf = 27 - (9,81*3,5)

Vf = - 7.33 [m/s] (this negative sign indicates that at this moment the snowball is going downwards)

To find how high the snowball was we must use the following equation:

Dy=v_{o} *t+\frac{1}{2}*g*t

Dy = (27*3.5) + (0.5*9.81*3.5)

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A 50 gram meterstick is placed on a fulcrum at its 50 cm mark. A 20 gram mass is attached at the 12 cm mark. Where should a 40 g
alekssr [168]

Answer:

The 40g mass will be attached at 69 cm

Explanation:

First, make a sketch of the meterstick with the masses placed on it;

--------------------------------------------------------------------------

               ↓                    Δ                      ↓

             20 g.................50 cm.................40g

                         38 cm                  y cm  

Apply principle of moment;

sum of clockwise moment = sum of anticlockwise moment

40y = 20 (38)

40y = 760

y = 760 / 40

y = 19 cm

Therefore, the 40g mass will be attached at 50cm + 19cm = 69 cm

              12cm             50 cm              69cm

--------------------------------------------------------------------------

               ↓                    Δ                      ↓

             20 g.................50 cm.................40g

                         38 cm                 19 cm                                              

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                   

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