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mash [69]
3 years ago
14

A particle of mass.25 kg is moving with a speed of 7m/s due north.find it's kinetic energy.​

Physics
1 answer:
Ivenika [448]3 years ago
3 0

Answer:

Explanation:

612J

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Can we write names while writing conversation in board exam​
lora16 [44]

Answer:

ya we can write the imaginary character's name .

So that we  can identify these imaginary people, as we cannot simply write the conversation and leave it .

Or maybe sometimes the reader will get confused as there is no name for the two people .

So, i suggest that you should write the names

Explanation:

You can even ask to your class teacher for further clarification

5 0
3 years ago
قوة الجذب المركزي تكون في اتجاه
pochemuha

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تكون دائمًا متعامدة مع سرعة الجسم وتكون دائمًا في اتجاه مركز انحناء المسار

Explanation:

6 0
3 years ago
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Cheetahs can accelerate to a speed of 20.0 m/s in 2.50 s and can continue to accelerate to reach a top speed of 29.5 m/s. Assume
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Answer:

boy if you dont get t f outa here

4 0
3 years ago
If the load distance of a lever is 30 cm and the effort distance is 60 cm, calculate the amount of effort required to lift a loa
vladimir2022 [97]

Here,

Load distance (Ld) = 30 cm

Effort distance (Ed) = 60 cm

Load (L) = 200N

Effort (E) = ?

Now, By using formula,

or, E * Ed = L * Ld

or, E * 60 = 200 * 30

or, E = 6000/60

◆ E = 100N

This is a Right answer...

I hope you understand...

7 0
3 years ago
A body of mass 2 kg is moving in the positive X-Direction with a speed of 4 m/s collides head on with an another body of mass 3
Inga [223]
m_1=2 \\ m_2=3 \\ v_1=4 \\ v_2=1 \\ v\text{ =speed after collision (to be determined)}.

The momentul of the system preserves:

m_1v_1-m_2v_2=(m_1+m_2)v \ \ \ \ \ \Rightarrow \ \ \ \ \ v=\dfrac{m_1v_1-m_2v_2}{m_1+m_2}.

Ok, we found the speed after the collision.
Now, because the impact is plastic, it produces heat, sound energy and who knows what other forms of energy. We denote all this wasted energy with E.

Now, we write the energy conservation law:

\dfrac{m_1v_1^2}{2}+\dfrac{m_2v^2_2}{2}=\dfrac{(m_1+m_2)v^2}{2}+E

From the above equation, you find E,  and then conclude that the sound energy can certainly not be greater than this.
8 0
3 years ago
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