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Neporo4naja [7]
3 years ago
14

The net external force on a golf cart is 411 N north of tha cart has a total mass of 281 kg what is tha cart acceleration

Physics
1 answer:
Alisiya [41]3 years ago
8 0

1.46m/s²

Explanation:

Given parameters:

Mass of cart = 281kg

Net external force = 411N

Unknown:

Acceleration of the cart = ?

Solution:

According to newton's second law " the net force on a car is the product of its mass and acceleration";

            Force = mass x acceleration

Since acceleration of the cart is unknown;

     Acceleration = \frac{force}{mass} = \frac{411}{281}

  Acceleration = 1.46m/s²

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

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Answer:

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Explanation:

all this following are indicators use to check the end point of a reaction

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4 years ago
A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s
Paraphin [41]

Answer:

\omega = 0.016\,\frac{rad}{s}

Explanation:

The rotation rate of the man is:

\omega = \frac{v}{R}

\omega = \frac{0.80\,\frac{m}{s} }{5\,m}

\omega = 0.16\,\frac{rad}{s}

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega

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3 0
3 years ago
An electron in a vacuum is first accelerated by a voltage of 51400 V and then enters a region in which there is a uniform magnet
zimovet [89]

Answer:

       F = 8.6 10⁻¹² N

Explanation:

For this exercise we use the law of conservation of energy

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Final. Electron with velocity, just out of the electric field

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          v =√ 2 e ΔV / m

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Now we can use the equation of the magnetic force

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Since the speed and the magnetic field are perpendicular the force that

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       For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

         Emo = U = q DV

Final. Electron with velocity, just out of the electric field

         Emf = K = ½ m v2

          Emo = Emf

          .e DV = ½ m v2

          .v = RA 2 e DV / m

          .v = RA (2 1.6 10-19 51400 / 9.1 10-31)

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