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3 years ago

How much work is done by a forklift raising a 125 kg object a distance of 10 meters?

Physics

1 answer:

lozanna [386]3 years ago

6 0

W = Fd

W = 1225 N x 10 m = 12250

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What does a Battery and a human have in common?

Elena L [17]

Both are objects and contain matter and mass. both are made of atoms, and produce reactions.

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2 years ago

An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 16.0 m if her initial speed is 3.60 m/

Airida [17]

The maximum height that is reached by an object thrown or jumped at a certain initial velocity can be calculated through the equation,

2ad = Vf² - Vi²

Vf is zero (0) in this equation because this is the point at the velocity at the maximum height of the object.

Substituting the known values,
2(a)(-16) = 0 - (3.6)²

The value of a from the equation is 0.405 m/s².

<em>Answer: 0.405 m/s²</em>

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3 years ago

If the roller coaster car in the above problem were moving with twice the speed(20m/s), then what would be its new kinetic energ

Diano4ka-milaya [45]

Answer:

KE = 1.05 x105 Joules

Explanation:

KE = 4 * (1.04653 x 105 J) = 4.19 x 105 Joules.

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2 years ago

A constant force of 2.5 N to the right acts on a 4.5 kg mass for 0.90 s.

Alborosie

Answer:

(a) $v_f=0.5\frac{m}{s}$

(b) $v_f=-11\frac{m}{s}$

Explanation:

(a) Since a constant external force is applied to the body, it is under an uniformly accelerated motion. Using the following kinematic equation, we calculate the final velocity of the mass if it is initially at rest( $v_0=0$ ):

$v_f=v_0+at\\v_f=at(1)$

According to Newton's second law:

$F=ma\\a=\frac{F}{m}(2)$

Replacing (2) in (1):

$v_f=\frac{F}{m}t\\v_f=\frac{2.5N}{4.5kg}(0.9s)\\v_f=0.5\frac{m}{s}$

(b) In this case we have $v_0=-11.5\frac{m}{s}$ . So, we use the final velocity equation:

$v_f=v_0+at\\v_f=v_0+\frac{F}{m}t\\v_f=-11.5\frac{m}{s}+\frac{2.5N}{4.5kg}(0.9s)\\v_f=-11\frac{m}{s}$

8 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