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horrorfan [7]
4 years ago
15

ANSWER = BRAINLIEST

Physics
2 answers:
FinnZ [79.3K]4 years ago
6 0
In a circular motion the centripetal force pulls the object toward the centre of the circle and keeps the object in it's path.

Although the speed is constant, VELOCITY is not (it change allow the path to be always tangencial to the circular path).

Hence, A is the right answer
Tamiku [17]4 years ago
4 0

Answer: C. The forces are unbalanced and the resultant force acts towards the centre of the circle.

Explanation: Even though the vehicle is moving at a constant speed, it still changes its velocity and, subsequently, has an acceleration.

The acceleration is directed towards the center in a circular path and, since Newton's law states that if the object has an acceleration, it is submitted to a net force, <u>net force</u> also points to the center of the central path. This force is the <u><em>centripetal force</em></u>.

According to the <em><u>law of inertia,</u></em> an object tends to stay in motion with the same speed and direction unless an outside force act upon it. In case of centripetal force, the unbalanced forces are acting upon the car in order for it to move in circles,

So, the unbalanced forces are acting on the car for it to turn and the resultant force (net force) is acting towards the centre of the circle as the acceleration.

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On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
3 years ago
A circuit diagram with a power source labeled 12 V is connected to 3 resistors in series. The resistors are labeled 1.0 Ohms, 2.
nevsk [136]

Answer:

there it is fella u were right with ur answer

7 0
3 years ago
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