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taurus [48]
3 years ago
13

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:
Physics
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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The answer would be C because there is six electrons and so there will be six protons because the amount of protons and electrons have to be the same otherwise it would be an unbalanced particle and you wouldn't be able to touch the object it is in without worrying about something happening
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Five different forces act on an object. Is it possible for the net force on the object to be zero?
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4 0
3 years ago
When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin
mario62 [17]

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}

where n=1,2,3 or ...n

L=Length of string

T=Tension

\mu =Mass per unit length

When string is clamped at mid-point

Effecting length becomes L'=0.5 L

Thus new Frequency becomes

f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz      

4 0
3 years ago
Some lenses are shaped with one flat side and one spherically-shaped side. This shape is designed to focus parallel light rays o
Komok [63]

Answer:

Some lenses are used to focus light to a pre-defined point based on the amount of curvature of their surfaces.

In a piano design convex, some surfaces are flat while others has positive lenses (biconvex)

Explanation:

Solution

These lenses are applied to pay attention to light in a  point pre-defined  based on the amount of curvature of their surfaces.

For that of a plano-convex design, one surface has a positive curve and for biconvex lenses, both surfaces are positively curved while the other  remains flat.

when used practically, plano-convex lenses are most commonly used where the object being imaged is far apart from lens.

7 0
3 years ago
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\,\%

\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%

\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%

\%K_{loss} = 92.889\,\%

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

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

\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%

\%v_{loss} = 73.333\,\%

The speed of the toboggan is reduced in 73.333 %.

5 0
3 years ago
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