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3 years ago

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A machine takes 0.5 seconds to move a brick 1 meter and put 100 Joules of energy into it. (hint: Power is the amount of energy t

ransferred or converted per unit time. Or we say power is work done per unit time. LaTeX: P\:=\:\frac{W}{t}P = W t) This machine would expend more power in this action if it:

1 answer:

Flauer [41]3 years ago

8 0

Answer:

200 Watts.

Explanation:

Power is defines as the amount of work expended per unit time. Mathematically, it is expressed as Power = Workdone/Time

Given parameters

Energy used up 100Joules

Distance moved by brick = 1 meters

Time taken by the machine = 0.5 secs

Power can also be written as Energy/Time

Required

We need to calculate the amount of power used up.

Power = 100J/0.5s

Power = 100/(1/2)

Power = 100 * 2/1

Power = 200Watts.

This shows that the machine would expend 200Watts of power

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A mass is oscillating with amplitude A at the end of a spring.

Dmitry_Shevchenko [17]

A) $x=\pm \frac{A}{2\sqrt{2}}$

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

$E=\frac{1}{2}kA^2$ (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

$U=\frac{1}{3}K$

Using (2) we can rewrite this as

$U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}$

And using (1), we find

$U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2$

Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

$\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}$

B) $x=\pm \frac{3}{\sqrt{10}}A$

In this case, the kinetic energy is 1/10 of the total energy:

$K=\frac{1}{10}E$

Since we have

$K=E-U$

we can write

$E-U=\frac{1}{10}E\\U=\frac{9}{10}E$

And so we find:

$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

3 0

3 years ago

OLga [1]

Answer:

Doppler Theory

Explanation:

it's a theory regarding the change in wave frequency during the relative motion between a wave source and its observer.

6 0

2 years ago

The statements below are all true. Some of them represent important reasons why the giant impact hypothesis for the Moon’s forma

Molodets [167]

Answer:

the order of importance must be b e a f c

Explanation:

Modern theories indicate that the moon was formed by the collision of a bad plant with the Earth during its initial cooling period, due to which part of the earth's material was volatilized and as a ring of remains that eventually consolidated in Moon.

Based on the aforementioned, let's analyze the statements in order of importance

b) True. Since the moon is material evaporated from Earth, its compassion is similar

e) True. If the moon is material volatilized from the earth it must train a finite receding speed

a) True. The solar system was full of small bodies in erratic orbits that wander between and with larger bodies

f) False. The moon's rotation and translation are equal has no relation to its formation phase

c) false. The amount of vaporized material on the moon is large

Therefore, the order of importance must be

b e a f c

5 0

4 years ago

Lucia kicks a ball on a level playing field with an initial velocity of 11.3 m/s at an angle of 35° above the horizontal. Find:

grandymaker [24]

Explanation:

Given that,

Initial velocity, u = 11.3 m/s

Angle above the horizontal, $\theta=35^{\circ}$

Time of flight :

$t=\dfrac{2u\sin\theta}{g}\\\\t=\dfrac{2\times 11.3\times \sin(35)}{9.8}\\\\t=1.32\ s$

Horizontal distance traveled is given by :

x = ut

x = 11.3 m/s × 1.32 s

x = 14.916 m

Maximum height is given by :

$H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(11.3)^2\times \sin^2(35)}{2\times 9.8}\\\\H=2.14\ m$

Hence, time of flight is 1.32 s, horizontal distance is 14.916 m and maximum height is 2.14 m.

6 0

2 years ago

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

3 0

4 years ago