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3 years ago

Điện tích Q = 8. 10-6C đặt cố định trong

Physics

1 answer:

ankoles [38]3 years ago

8 0

Answer:

0.72J

Explanation:

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A helicopter blade spins at exactly 110 revolutions per minute. Its tip is 4.50 m from the center of rotation. (a) Calculate th

NeX [460]

Answer:

(a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

Explanation:

Given that,

Angular velocity = 110 rev/m

Radius = 4.50 m

(a). We need to calculate the average speed

Using formula of average speed

$v=r\omega$

$v = 4.50\times110\times\dfrac{2\pi}{60}$

$v=51.83\ m/s$

(b). The average velocity over one revolution is zero because the net displacement is zero in one revolution.

Hence, (a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

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3 years ago

Which statement best describes the law of conservation of energy?

Phantasy [73]

<span>The correct option is C. Energy cannot be created or destroyed. This statement is known as law of conservation of energy, and it implies that whenever a certain form of energy does change, the loss of this form of energy must have converted into an another type of energy. A typical example is an object falling to the ground: initially, the object has gravitational potential energy. As the object falls down, it loses potential energy (since its altitude from the grounf decreases), but it acquires kinetic energy (because its velocity increases). In this example, potential energy has converted into kinetic energy, but the total energy of the object has remained constant.</span>

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3 years ago

Read 2 more answers

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$