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bekas [8.4K]
3 years ago
5

a car starts from rest and increases its speed to 15 m/s in 20 seconds. what is the cars acceleration

Physics
2 answers:
GalinKa [24]3 years ago
7 0

Answer:

In the picture above.

Explanation:

Good luck.

Bess [88]3 years ago
5 0

The acceleration of the car is 0.75m/s²

<u>Explanation:</u>

Given:

Speed, s = 15m/s

Time, t = 20 seconds

Acceleration, a = ?

We know,

Acceleration = \frac{speed}{time}

On substituting the value, we get:

a = \frac{15}{20} \\\\a = 0.75 m/s^2

Thus, the acceleration of the car is 0.75m/s²

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2 years ago
In which direction does a bag at rest move when a force of 20 newton's is applied from the right?
Olegator [25]

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8 0
3 years ago
A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at
yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)

Therefore, the work done by the worker in lifting the bucket is given as:

W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)

Now, plug in the values given and solve for 'W'. This gives,

W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

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