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3 years ago

What idea about work and force does the diagram describe? Explain your answer.

Physics

1 answer:

Misha Larkins [42]3 years ago

5 0

Answer:

W = F*d, where displacement is always in the same direction as the force. The work done by a constant force is proportional to the force applied times the displacement of the object.

Explanation:

F and W are directly proportional to each other, while F and d are indirectly propotional.

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A golfer hits her tee-shot due north towards the fairway. Her shot has an initial velocity of 60 m/s. A 15 m/s wind is blowing i

vovangra [49]

Answer:

$c=71.4m/s$

$\theta=8.54\textdegree$

Explanation:

From the question we are told that

Initial velocity of 60 m/s

Wind speed $V_w= 15 m/s \angle 45 \textdegree$

Generally Resolving vector mathematically

$sin(45\textdegree)15=10.6\\cos(45\textdegree)15=10.6$

Generally the equation Pythagoras theorem is given mathematically by

$c^2=a^2+b^2$

$c^2=10.6^2 +(10.6+60)^2$

$c=\sqrt{10.6^2 +(10.6+60)^2}$

Therefore Resultant velocity (m/s)

$c=71.4m/s$

b)Resultant direction

Generally the equation for solving Resultant direction

$\theta=tan^-1(\frac{y}{x})$

Therefore

$\theta=tan^-1(\frac{10.6}{70.6})$

$\theta=8.54\textdegree$

7 0

2 years ago

In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a

matrenka [14]

Answer:

Approximately $\displaystyle\rm \left[ \begin{array}{c}\rm191\; m\\\rm-191\; m\end{array}\right]$ .

Explanation:

Consider this $45^{\circ}$ slope and the trajectory of the skier in a cartesian plane. Since the problem is asking for the displacement vector relative to the point of "lift off", let that particular point be the origin $(0, 0)$ .

Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at $45^{\circ}$ to the x-axis. Since this slope is connected to the ramp, it should also go through the origin. Based on these conditions, this line should be represented as $y = -x$ .

Convert the initial speed of this diver to SI units:

$\displaystyle v = \rm 110\; km\cdot h^{-1} = 110 \times \frac{1}{3.6} = 30.556\; m\cdot s^{-1}$ .

The question assumes that the skier is in a free-fall motion. In other words, the skier travels with a constant horizontal velocity and accelerates downwards at $g$ ( $g \approx \rm -9.81\; m\cdot s^{-2}$ near the surface of the earth.) At $t$ seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:

$x$ -coordinate: $30.556t$ meters (constant velocity;)
$y$ -coordinate: $\displaystyle -\frac{1}{2}g\cdot t^{2} = -\frac{9.81}{2}\cdot t^{2}$ meters (constant acceleration with an initial vertical velocity of zero.)

To eliminate $t$ from this expression, solve the equation between $t$ and $x$ for $t$ . That is: express $t$ as a function of $x$ .

$x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}$ .

Replace the $t$ in the equation of $y$ with this expression:

$\begin{aligned} y = &-\frac{9.81}{2}\cdot t^{2}\\ &= -\frac{9.81}{2} \cdot \left(\frac{x}{30.556}\right)^{2}\\&= -0.0052535\;x^{2}\end{aligned}$ .

Plot the two functions:

$y = -x$ ,
$\displaystyle y= -0.0052535\;x^{2}$ ,

and look for their intersection. Refer to the diagram attached.

Alternatively, equate the two expressions of $y$ (right-hand side of the equation, the part where $y$ is expressed as a function of $x$ .)

$-0.0052535\;x^{2} = -x$ ,

$\implies x = 190.35$ .

The value of $y$ can be found by evaluating either equation at this particular $x$ -value: $x = 190.35$ .

$y = -190.35$ .

The position vector of a point $(x, y)$ on a cartesian plane is $\displaystyle \left[\begin{array}{l}x \\ y\end{array}\right]$ . The coordinates of this skier is approximately $(190.35, -190.35)$ . The position vector of this skier will be $\displaystyle\rm \left[ \begin{array}{c}\rm191\\\rm-191\end{array}\right]$ . Keep in mind that both numbers in this vectors are in meters.

4 0

3 years ago

A space station in the shape of a 100 m-diameter (50m radius) wheel is spinning so as to impart a linear tangential speed of 22.

zalisa [80]

Answer:

correct option is b. 31.3 m/s

Explanation:

given data

artificial gravity a1 = 1 g

artificial gravity a2 = 2 g

diameter = 100 m

radius r= 50 m

speed v1 = 22.1 m/s

solution

As acceleration is ∝ v²

so we can say

$\frac{a2}{a1} = \frac{v2}{v1}$ .....................1

put here value

$\frac{2}{1} = \frac{v2}{22.1}$

solve it

v2 = $\sqrt{2 }$ × 22.1

v2 = 31.25 m/s

so correct option is b. 31.3 m/s

4 0

3 years ago

The metal ions in iron are held together by

Sidana [21]

Metals are giant structures of atoms held together by metallic bonds. “Giant” implies that large but variable numbers of Atoms are involved - depending on the size of the bits of metal. most metals are close packed - that is, they fit as many items as possible into the available volume.

5 0

3 years ago

A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave

Oliga [24]

Answer:3.51

Explanation:

Given

Coefficient of Friction $\mu =0.4$