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3 years ago

What best describes the dropping height of a ball that bounced back up to a height of 45 centimeters?

1 answer:

6 0

Answer:C:Less than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy

Less than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy.

Although the initial energy (potential energy is preserved), the energy of deformation as the ball strikes a surface creates energy dissipation in the form of frictional heat and audible sound energy.

Every time the ball bounces, its height will be less than its previous height.

Explanation:

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calcula la potencia por hora de un radiador, sabiendo que esta conectado a un contacto común 110 v. y requiere 20 Amp.

Ira Lisetskai [31]

calculate the power per hour of a radiator, knowing that it is connected to a common 110 v contact. and requires 20 Amp.

Answer:

2.2kWh

Explanation:

Given parameters:

Potential difference = 110v

Current = 20A

Unknown:

Power = ?

Solution:

To solve this problem, we use the expression below:

Power = IV

Power = 110 x 20 = 2200W

This is therefore 2.2kW

Power per hour = 2.2kWh

8 0

3 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

2 years ago

An injured monkey sits perched on a tree branch 9 m above the ground, while a wildlife veterinarian is kneeling down in the bush

Yakvenalex [24]

Answer:

The hunter should aim directly at the perched monkey because the tranquilizer dart will fall away from the line sight at the same rate that the monkey falls from its perch.

Tan theta = 9 / 90 = .1 so theta = 5.71 deg

The time for the monkey to reach the ground is

t = (2 h / g)^1/2 = (18 / 9.8)^1/2 = 1.36 sec

So the horizontal speed of the dart must be at least

Vx = 90 m / 1.36 sec = 66.4 m/s

Vx = V cos theta

V = 66.4 m/s / cos 5.71 = 66.7 m/s

7 0

3 years ago

Explain how Pascal's principle can be used to design a fluid power system and describe how a fluid power system works.

zhenek [66]

Pascal's law of fluid transfer states that when there is an increase in fluid pressure, the rest of the extrinsic variables also increases. For example, in a flow of liquid in an orifice, there is a contraction of diameter in the orifice part. The fluid that will go in there increases in pressure and thereby an increase in velocity as well.

4 0

3 years ago

In 0.60 seconds, a projectile goes from 0 to 610 m/s. What is the acceleration of the projectile?

IceJOKER [234]

Answer: $a=1016.66 m/s^{2}$