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xenn [34]
2 years ago
12

You are asked to calculate an object's velocity, in order to do so you must know the object's ?

Physics
1 answer:
elixir [45]2 years ago
5 0
There are so many answers. It depends on which field of physics you are working on.
You could calculate velocity from momentum or kinetic energy knowing also the object's mass for example.
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This energy source uses hydrogen and oxygen to produce water. It produces almost no pollution, and it works similarly to a batte
Karolina [17]
The answer is fuel cell....

8 0
3 years ago
Help please? I dont understand what this means.
Alex Ar [27]
The answer is c for real
5 0
3 years ago
Which of the following is not an example of approximate simple harmonic motion? A. a ball bouncing on the floor
Reptile [31]
<h2>Answer:</h2>

Answer to this question is (A)

<h2>Explanation</h2>

A ball bouncing on the floor is not the example of simple harmonic motion. SHM is the special kind of to and fro motion in which a particle oscillate about its mean position in a straight line. The acceleration of the particle is always directed towards its mean position and is directly proportional to its displacement from its mean position.

In case of a ball bouncing on the ground, the motion of the ball is not SHM, as neither it’s a to and fro motion nor the acceleration is proportional to its displacement from its mean position.


3 0
3 years ago
What are some things that doesn't move in a circular motion or are non examples of circular motion?
photoshop1234 [79]

Answer:In non-uniform circular motion, an object's motion is along a circle, but the object's speed is not constant. In particular, the following will be true. The object's velocity vector is always tangent to the circle. The speed and angular speed of the object are not constant

Explanation:In non-uniform circular motion an object is moving in a circular path with a ... Since the speed is changing, there is tangential acceleration in ... Here is an example with an object traveling in a straight path

7 0
3 years ago
Read 2 more answers
A projectile is launched from ground level with an initial speed of 53.7 m/s. Find the launch angle (the angle the initial veloc
Simora [160]

Answer:\theta =75.52^{\circ}

Explanation:

Given

initial speed of Launch(u)=53.7 m/s

Range of Projectile =Maximum height of Projectile

Range is given by R=\frac{u^2\sin 2\theta }{g}

Maximum height is given by H_{max}=\frac{u^2\sin ^2\theta }{2g}

R=H_{max}

\frac{u^2\sin 2\theta }{g}=\frac{u^2\sin ^2\theta }{2g}

\frac{u^2\sin 2\theta }{g}-\frac{u^2\sin ^2\theta }{2g}=0

\frac{u^2\sin \theta }{g}\cdot \left [ 2\cos \theta -\frac{1}{2}\right ]=0

as u cannot be zero therefore

2\cos \theta -\frac{1}{2}=0

\cos \theta =\frac{1}{4}

\theta =75.52^{\circ}

6 0
3 years ago
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