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3 years ago

A power boat pulls a water skier 2.6 km maintaining a

Physics

1 answer:

Luba_88 [7]3 years ago

7 0

Answer:

W = 650 [kJ]

Explanation:

The definition of work is denoted by the product of force by the distance traveled by the body, this distance traveled corresponds to the direction of the force.

In this case we have:

d = distance = 2.6[km] = 2600 [m]

F = force = 250 [N]

W = F*d = 250 * 2600 = 650000 [J] or 650 [kJ]

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A particular HeNe laser beam at 633 nm has a spot size of 0.8 mm. Assuming a

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Answer:

Explanation:

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3 years ago

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Ksenya-84 [330]

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3 years ago

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

belka [17]

Answer:

$F_a=5.67\times 10^{-5}\ N$

$F_b=3.49\times 10^{-5}\ N$

$F_c=9.16\times 10^{-5}\ N$

Explanation:

Given:

mass of particle A, $m_a=363\ kg$

mass of particle B, $m_b=517\ kg$

mass of particle C, $m_c=154\ kg$

All the three particles lie on a straight line.

Distance between particle A and B, $x_{ab}=0.5\ m$

Distance between particle B and C, $x_{bc}=0.25\ m$

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

Force on particle A due to particles B & C:

$F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}$

$F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})$

$F_a=5.67\times 10^{-5}\ N$

Force on particle C due to particles B & A:

$F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}$

$F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )$

$F_c=9.16\times 10^{-5}\ N$

Force on particle B due to particles C & A:

$F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}$

$F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2} )$

$F_b=3.49\times 10^{-5}\ N$

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3 years ago

what are three activities in your everyday life that involve electromagnetic energy? and explain what type of electromagnetic en

mihalych1998 [28]

1). While I'm driving the car, I keep my eyes pointed out the front window,

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them. Using this process, I'm able to SEE what's in front of me, so that I don't

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2). Also while I'm driving the car, I usually have the radio on, and I tune it

to a frequency where the RADIO WAVES that it responds to are being

modulated by people who are either talking or else playing classical music.

3). Before I leave the house on most days, I take a small container of

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above the stove, where MICROWAVE ENERGY thaws it, and warms

it to a nice eating temperature.

3A). When I arrive at my job, I usually do things that are also associated

in some way with MICROWAVE energy, since I work as a microwave

system design engineer. Most of the time it's just paperwork, but there

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3 years ago

Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.

Ksivusya [100]

$|v| =\sqrt{ G \cdot M / r}$ , where

$M$ the mass of the planet, and
$G$ the universal gravitation constant.

Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

Equation 1 (see below) relates net force the object experiences, $\Sigma F$ to its orbit velocity $v$ and its mass $m$ required for it to stay in orbit :

$\Sigma F = m \cdot v^{2} / r$ (equation 1)

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force $\Sigma F$ acting on the soccer ball shall equal to its weight, $W = m \cdot g$ where $g$ the gravitational acceleration constant. Thus

$\Sigma F = W = m \cdot g$ (equation 2)

Substitute equation 2 to the left hand side of equation 1 and solve for $v$ ; note how the mass of the soccer ball, $m$ , cancels out:

$m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0)$ (equation 3)

Equation 4 gives the value of gravitational acceleration, $g$ , a point of negligible mass experiences at a distance $r$ from a planet of mass $M$ (assuming no other stellar object were present)

$g = G \cdot M/ r^{2}$ (equation 4)

where the universal gravitation constant $G = 6.67408 \times 10^{-11} \cdot \text{m}^{3} \cdot \text{kg}^{-1} \cdot \text{s}^{-2}$

Thus

$\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}$

3 0

3 years ago