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Mekhanik [1.2K]

2 years ago

12

A skier of mass 60 kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him 75 m up a 30° slop

e (assumed frictionless) at a constant speed of 2.4 m/s? kJ (b) What power (expressed in hp) must a motor have to perform this task?

1 answer:

postnew [5]2 years ago

5 0

Answer:

Explanation:

Given

mass of skier=60 kg

distance traveled by skier=75 m

inclination $(\theta )=30^{\circ}$

speed (v)=2.4 m/s

as the skier is moving up with a constant velocity therefore net force is zero

$F_{net}=0$

Force applied by cable $=mg\sin \theta$

$F=60\times 9.8\times \sin (30)=294 N$

work done $=F\cdot x$

$W=294\cdot 75=22.125 J$

(b)Power $=F\cdot v$

$P=294\cdot 2.4=705.6 W\approx 0.946 hp$

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Calculate the lowest energy (in ev) for an electron in an infinite well having a width of 0.050 mm.

MA_775_DIABLO [31]

The lowest energy of electron in an infinite well is 1.2*10^-33J.

To find the answer, we have to know more about the infinite well.

<h3>What is the lowest energy of electron in an infinite well?</h3>

It is given that, the infinite well having a width of 0.050 mm.
We have the expression for energy of electron in an infinite well as,

$E_n=\frac{n^2h^2}{8mL^2}$

where;

$m=9.1*10^{-31}kg\\L=0.050*10^{-3}m\\h=6.63*10^{-34}Js\\n=1$

Thus, the lowest energy of electron in an infinite well is,

$E_1=\frac{(6.63*10^{-34})^2}{8*9.1*10^{-31}*(0.050*10^{-3})}=1.2*10^{-33}J$

Thus, we can conclude that, the lowest energy of electron in an infinite well is 1.2*10^-33J.

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11 months ago

A transformer has a secondary voltage of 140 volts and a secondary current of 3.5 amps. if the primary current is 10 amps, what

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For an ideal transformer power loss is assumed to be zero

i.e. the power in primary coil due to input voltage must be equal to power in secondary coil due to output voltage

this can be written in form of equation

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here we know that

$V_2 = 140 volts$

$i_2 = 3.5 A$

$i_1 = 10 A{/tex]now we will use above equation[tex]140*3.5 = 10 * V_1$

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A 52 kg and a 95 kg skydiver jump from an airplane at an altitude of 4750 m, both falling in the pike position. Assume all value

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95 kg skydiver: 12.3 m/s and 386.2 s

Explanation: <u>Drag</u> <u>Force</u> is an opposite force when an object is moving in a fluid.

For skydivers, when falling through the air, the forces acting on it are gravitational and drag forces. At a certain point, drag force equals gravitational force, which is constant on any part of the planet, producing a net force that is zero. Since there is no net force, there is no acceleration and, consequently, velocity is constant. When that happens, the person reached the <u>Terminal</u> <u>Velocity</u>.

Drag Force and Velocity are proportional to the squared speed. So, terminal velocity is given by:

$F_{G}=F_{D}$

$mg=\frac{1}{2}C \rho Av_{T}^{2}$

$v_{T}=\sqrt{\frac{2mg}{\rho CA} }$

where

m is mass in kg

g is acceleration due to gravitational force in m/s²

ρ is density of the fluid in kg/m³

C is drag coefficient

A is area of the object in the fluid in m²

Calculating:

The 52kg skydiver has terminal velocity of:

$v_{T}=\sqrt{\frac{2(52)(9.8)}{(1.21)(0.7)(0.14)} }$

$v_{T}=$ 9.09

The 95kg skydiver's terminal velocity is

$v_{T}=\sqrt{\frac{2(95)(9.8)}{(1.21)(0.7)(0.14)} }$

$v_{T}=$ 12.3

The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.

The time each one will reach the floor will be:

52 kg at 9.09 m/s:

$t=\frac{4750}{9.09}$

t = 522.5

95 kg at 12.3 m/s:

$t=\frac{4750}{12.3}$

t = 386.2

The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.

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Answer:

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Explanation:

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m is the mass

v is the velocity

From the question we have

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We have the final answer as

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