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3 years ago

True or False: Unlike a fixed pulley, a moveable pulley multiplies the input force

Physics

1 answer:

IceJOKER [234]3 years ago

6 0

The answer would be Yes True.

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Hi gir_ls join nkd-mbja-nuj

GREYUIT [131]

Answer:

never lol

studying is your work

but why all are doing I don't know=_=

3 0

3 years ago

For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro

viktelen [127]

Answer:

$V_d$ = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, $A = \frac{\pi d^2}{4}$ = $A = \frac{\pi 0.625^2}{4}$

Now,

The current density, J is given as

$J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}$ = 2444619.925 A/mm²

now, the electron density, $n = \frac{\rho}{M}N_A$

where,

$N_A$ =Avogadro's Number

$n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3$

Now,

the drift velocity, $V_d$

$V_d=\frac{J}{ne}$

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

$V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e}$ = 1.75 × 10⁻⁴ m/s

4 0

3 years ago

Read 2 more answers

The wind increases to 14 mph from the north. Now what is your airspeed and what direction are you flying? If your destination is

ladessa [460]

Answer:

I-

Explanation:

...

7 0

3 years ago

A chart labeled table A : effect of height on temperature with initial temperature as 25 degrees Celsius, mass w is 1.0 kilogram

o-na [289]

Answer:

100m: 4.9 kJ

200m: 9.8 kJ

1000m: 49.1 kJ

Explanation:

edge2020

7 0

3 years ago

Read 2 more answers

What would be the escape speed for a craft launched from a space elevator at a height of 54,000 km?

Natasha_Volkova [10]

Answer: 3.63 km/s

Explanation:

The escape velocity equation for a craft launched from the Earth surface is:

$V_{e}=\sqrt{\frac{2GM}{R}}$

Where:

$V_{e}$ is the escape velocity

$G=6.67(10)^{-11} Nm^{2}/kg^{2}$ is the Universal Gravitational constant

$M=5.976(10)^{24}kg$ is the mass of the Earth

$R=6371 km=6371000 m$ is the Earth's radius

However, in this situation the craft would be launched at a height $h=54000 km=54000000 m$ over the Eart's surface with a space elevator. Hence, we have to add this height to the equation:

$V_{e}=\sqrt{\frac{2GM}{R+h}}$

$V_{e}=\sqrt{\frac{2(6.67(10)^{-11} Nm^{2}/kg^{2})(5.976(10)^{24}kg)}{6371000 m+54000000 m}}$

Finally:

$V_{e}=3633.86 m/s \approx 3.63 km/s$

7 0

3 years ago