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sweet-ann [11.9K]
4 years ago
12

What is the acceleration of a 30.0-kg go kart if the thrust of its engine is 300.0 N?

Physics
2 answers:
VikaD [51]4 years ago
5 0

Answer : a=10\ m/s^2

Explanation :

It is given that,

Mass of the engine, m = 30 kg

Thrust is equivalent to the force acting perpendicularly and it is F = 300 N

According to Newton's second law of motion :

F = m\times a

a is the acceleration of the engine.

a=\dfrac{F}{m}

a=\dfrac{300\ N}{30\ Kg}

a=10\ m/s^2

So, the acceleration of the engine is 10\ m/s^2.

Hence, this is the required solution.

sp2606 [1]4 years ago
5 0

Answer:

Acceleration, a=10\ m/s^2

Explanation:

It is given that,

Mass of the cart, m = 30 kg

Thurst applied to the engine, F = 300 N

Let a is the acceleration of the cart. According to Newton's second law of motion, force is equal to the product of mass and acceleration. It is given by :

F=ma

a=\dfrac{F}{m}

a=\dfrac{300\ N}{30\ kg}

a=10\ m/s^2

So, the acceleration of the cart is 10\ m/s^2. Hence, this is the required solution.

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A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5
harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

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The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

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This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

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3 years ago
Why do astronomers use the word on to describe angles on the sky rather than angles in the sky?
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A professor's office door is 0.91 m wide, 2.0 m high, and 4.0 cm thick; has a mass of 25 kg; and pivots on frictionless hinges.
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Answer:

F= 5.71 N

Explanation:

width of door= 0.91 m

door closer torque on door= 5.2 Nm

In order to hold the door in open position we need to exert an equal and opposite torque, to the door closer torque, on the door.

so wee need to exert 5.2 Nm torque on the door.

If we want to apply minimum force to exert the required torque we need to apply force perpendicularly on the door knob (end of door) so that to to greater moment arm.

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5 0
3 years ago
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