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Sphinxa [80]
2 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]2 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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user100 [1]

Answer:

1.04\times 10^7\ J.

Explanation:

In the question given :

Pressure is constant

Therefore, Work done, W=P\times\Delta V

Pressure, P=1.01 × 105 Pa.

Final volume, V_f=8.50\ m^3.

Initial volume, V_i=\dfrac{Mass}{density}=\dfrac{5}{958}=5.22\times10^-3\ m^3.

Therefore, W=8.58\times 10^{5}\ J.

Also, Heat Given, Q=m\times L=5\times 2.26\times 10^{6}\ J=1.13\times 10^7\ J.

Also, according to First law of thermodynamics:

\Delta U=Q-W=(1.13\times 10^7)-(8.58\times 10^5)=1.04\times 10^7\ J.

Hence, this is the required solution.

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Answer:

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Firlakuza [10]

Answer:

Given,

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Solution,

We can simply solve this numerical problem by using the following process.

Now,

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Total time taken to display the given number of frames (ie. 25 frames) = 1 second

To calculate the time interval between one frame and next, we need to divide the time taken to display total number of frames by total number of frames.

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<em></em>

Answer:

1. The magnitude of the force from the spring on the object is zero on <em>Equilibrium.</em>

2. The magnitude of the force from the spring on the object is a maximum on <em>The top and bottom.</em>

3. The magnitude of the net force on the object is zero on <em>The Bottom.</em>

4. The magnitude of the force on the object is a maximum on <em>the Top.</em>

Explanation:

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<em>2. Because of delta X.</em>

<em>3. Beacuse, the force of gravity and the force of the spring oppose each other to keep the block at rest, away from the equilibrium position.</em>

<em>4. Because, the force of the spring from compressiom and the force of gravity both act on the mass.</em>

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