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4 years ago

If Susan exerted a force of 5 newtons and the output force of the machine is 15

Physics

1 answer:

Anon25 [30]4 years ago

4 0

Answer:

Mechanical advantage = 3

Explanation:

Mechanical advantage, MA = Output force/Input force

Output force = 15 N. Input force = 5 N

MA = Output force/Input force = 15 N/ 5 N = 3

Read more about doppler shift here:

brainly.com/question/4052291

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7 0

2 years ago

when a temparature of a coin is 75°C, the coin's diameter increases. if the original diameter of a coin is 1.8*10^-2 m and its c

andrey2020 [161]

Answer:

ΔD = 2.29 10⁻⁵ m

Explanation:

This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation

ΔA = 2α A ΔT

the area is

A = π r² = π D² / 4

we substitute

ΔA = 2α π D² ΔT/4

as they do not indicate the initial temperature, we assume that ΔT = 75ºC

α = 1.7 10⁻⁵ ºC⁻¹

we calculate

ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4

ΔA = 6.49 10⁻⁷ m²

by definition

ΔA = A_f- A₀

A_f = ΔA + A₀

A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4

A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴

A_f = 2,551 10⁻⁴ m²

the area is

A_f = π D_f² / 4

A_f = $\sqrt{4 A_f /\pi }$

D_f = $\sqrt{4 \ 2.551 10^{-4} /\pi }$

D_f = 1.80229 10⁻² m

the change in diameter is

ΔD = D_f - D₀

ΔD = (1.80229 - 1.8) 10⁻² m

ΔD = 0.00229 10⁻² m

ΔD = 2.29 10⁻⁵ m

5 0

3 years ago

1) A spring, which has a spring constant k=7.50 N/m, has been stretched 0.40 m from ts equilibrium position . What the potential

cupoosta [38]

<h3>Answer:</h3>

$\displaystyle U_s = 0.6 \ J$

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Physics</u>

<u>Energy</u>

Elastic Potential Energy: $\displaystyle U_s = \frac{1}{2} k \triangle x^2$

U is energy (in J)
k is spring constant (in N/m)
Δx is displacement from equilibrium (in m)

<h3>Explanation:</h3>

<u>Step 1: Define</u>

k = 7.50 N/m

Δx = 0.40 m

<u>Step 2: Find Potential Energy</u>

Substitute in variables [Elastic Potential Energy]: $\displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.40 \ m)^2$
Evaluate exponents: $\displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.16 \ m^2)$
Multiply: $\displaystyle U_s = (3.75 \ N/m) (0.16 \ m^2)$
Multiply: $\displaystyle U_s = 0.6 \ J$

6 0

3 years ago

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

3 years ago

help, please! Do you think that plate tectonics play a major or minor role in the rock cycle? please explain

Dominik [7]

Answer:

Major.

Explanation:

Because if we didn't have plate tectonics, we would not know anything about the inside of the Earth.

: ) hope this helps.

8 0

3 years ago