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MrRissso [65]
3 years ago
14

Physics, calculating net force. Please work it out for me

Physics
1 answer:
Margarita [4]3 years ago
3 0

Answer:

F = 2985.125 N

Explanation:

Given that,

The radius of curvature of the roller coaster, r = 8 m

Speed of Micheal, v = 17 m/s

Mass of body, m = 65 kg  We need to find the net force acting on Micheal. Net force act the bottom of the circle is given by :

F=\dfrac{mv^2}{r}+mg\\\\F=m(\dfrac{v^2}{r}+g)\\\\=65(\dfrac{17^2}{8}+9.8)\\\\=2985.125\ N

So, the net force is 2985.125 N.

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Choose a substance you're familiar with. what are its physical and chemical properties? How would you measure its density? What
Gekata [30.6K]
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6 0
3 years ago
A car is moving in uniform circular motion. If the cars speed were to double to keep the car moving with the same radius the acc
Stells [14]

Answer:

<em>The centripetal acceleration would increase by a factor of 4</em>

<em>Correct choice: B.</em>

Explanation:

<u>Circular Motion</u>

The circular motion is described when an object rotates about a fixed point called center. The distance from the object to the center is the radius. There are other magnitudes in the circular motion like the angular speed, tangent speed, and centripetal acceleration. The formulas are:

v_t=w\ r

\displaystyle a_c=\frac{v_t^2}{r}

If the speed is doubled and the radius is the same, then

\displaystyle a_c=\frac{(2v_t)^2}{r}

\displaystyle a_c=4\frac{v_t^2}{r}

The centripetal acceleration would increase by a factor of 4

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5 0
3 years ago
A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba
Lyrx [107]

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

      The mass of the rubber ball is m_r   =  2 \ kg

      The  initial  speed of the rubber ball is  u =  3 \ m/s

      The final speed at which it bounces bank v  - 3 \ m/s

      The mass of the clay ball  is  m_c =  2  \ kg

       The  initial  speed of the clay  ball is u = 3 \ m/s

       The final speed of the clay ball is  v = 0 \  m/s

Generally Impulse is mathematically represented as

       I  =  \Delta p

where \Delta  p is the change in the linear momentum so  

       I  =  m(v-u)

For the rubber  is  

        I_r  =  2(-3 -3)

       I_r  = -12\ kg \cdot  m/s

=>     |I_r|  = 12\ kg \cdot  m/s

For the clay ball

       I_c  =  2(0-3)

        I_c =  -6 \ kg\cdot \ m/s

=>    | I_c| =  6 \ kg\cdot \ m/s

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

       

8 0
4 years ago
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