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2 years ago

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Arm abcd is pinned at b and undergoes reciprocating motion such that θ=(0.3 sin 4t) rad, where t is measured in seconds and the

argument for the sine is in radiaus. determine the largest speed of point a during the motion and the magnitude of the acceleration of point d at this instant.

1 answer:

storchak [24]2 years ago

8 0

<span>θ=0.3sin(4t)
w=0.3cost(4t)(4)=1.2cost(4t)
a=-4.8sin(4t)

cos4t max will always be 1 (refer to cos graph), for same reason, sin4t will always be 0

therefore, wmax=1.2rad/s

vAmax=r*w=250*1.2=300mm/s
(may be different if your picture/radius is from a different picture)

adt=a*r=200*-4.8sin(4t)=0 (sin(4t)=0)

adn=r*w^2=200*1.2^2=288

ad= square root of adt^2+adn^2 = 288mm/s^2</span>

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A solar heated house loses about 5.4 × 107 cal through its outer surfaces on a typical 24-h winter day.

mojhsa [17]

Answer:

C

Explanation:

Q=mcΔθ

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Δθ= temperature change.

m= Q/(cΔθ)

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Δθ=62°C-20°C

=42°C

c=0.21cal/g.°C

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2 years ago

An object is located 25.0 cm from a convex mirror. The image distance is -50.0 cm. What is the magnification?

Lynna [10]

Answer:

$\boxed{\sf Magnification \ (m) = 2}$

Given:

Object distance (u) = 25.0 cm

Image distance (v) = -50.0 cm

To Find:

Magnification (m)

Explanation:

$\boxed{\bold{\sf Magnification \: (m) = - \frac{Image \: distance \: (v)}{Object \: distance \: (u)}}}$

Substituting values of Image distance(v) & Object distance (u) in the equation:

$\sf \implies m = - \frac{( - 50)}{25}$

-(-50) = 50:

$\sf \implies m = \frac{50}{25}$

$\sf \implies m = \frac{2 \times \cancel{25}}{ \cancel{25}}$

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2 years ago

The image shows an example of white light entering a prism and coming out as colors of the rainbow. How does a prism a produce t

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The answer is C because when white light enters a prism, its gets separated into component colors which are red, orange, yellow, green, blue, indigo, and violet. This can also be referred to as <span>dispersion.</span>

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2 years ago

Read 2 more answers

Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball's trajectory and notices that it lands,

Elenna [48]

Let $u$ be the initial velocity of the soccer ball at an angle of inclination of $\theta_0$ with the positive x-axis.

Given that:

$\theta_0=45^{\circ}$

The horizontal distance covered by the projectile=20 m

Time of flight, $t_f=2$ seconds

Acceleration due to gravity, $g= 10 m/s^2$ downward.

As "north" and "up" as the positive x ‑ and y ‑directions, respectively.

So, $g= -10 m/s^2$

As the acceleration due to gravity is in the vertical direction, so the horizontal component of the initial velocity remains unchanged.

The x-component of the initial velocity, $u_x=u\cos\theta_0$ .

The horizontal distance covered by the projectile $= u_x\times t_f$

$\Rightarrow u_x\times t_f=20$

$\Rightarrow u_x\times 2=20$

$\Rightarrow u_x=10$ m/s

So, the horizontal component of the velocity is 10 m/s which is constant and the graph has been shown in the figure (i).

Now, $u\cos(45^{\circ})=10$ [as $u_x=u\cos\theta_0$ ]

$\Rightarrow u=10\sqrt{2}$ m/s.

The vertical component of the initial velocity,

$u_y= u\sin\theta_0$

$\Rightarrow u_y=10\sqrt{2}\sin(45^{\circ})$

$\Rightarrow u_y=10$ m/s

Let v be the vertical component of the velocity at any time instant t.

From the equation of motion,

$v=u+at$

where u: initial velocity, v: final velocity, a: constant acceleration, and t: time taken to change the velocity from u to v.

In this case, we have $u=u_y, a= -10 m/s^2$ .

So at any time instant, t.

$v=u_y+(-10)t$

$\Rightarrow v=10-10t$

The vertical component of the velocity, v, is the function of time and related as $v=10-10t$ .

This is a linear equation.

At 2 second, the vertical component of the velocity

v=10-10x2=-10 m/s.

The graph has been shown in figure (ii).

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2 years ago

You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph from due north.

Afina-wow [57]

Answer:

<em>a) 17.05 mph</em>

<em>b) 54.7° northeast direction</em>

<em>c) 10.71 mph</em>

<em>The direction is -22.58° relative to the east.</em>

<em></em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

Explanation:

The question is a little confusing but, I guess the correct question should be;

You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph due north.

a) What is your airspeed?

b) What angle (direction) are you flying?

c) The wind increases to 14 mph from north. Now what is your airspeed and what direction are you flying? If your destination is to the northeast, how would you change your speed or direction so you might make it there?

NB: The difference in the question and my suggestion is highlighted boldly.

Your speed = 14 mph

direction is 45° northeast

Th wind speed = 4 mph

direction is north

We resolve the your speed and the wind speed into the horizontal and vertical components

For vertical the component component

$V_{y}$ = 14(sin 45) + 4 = 9.89 + 4 = 13.89 mph

For the horizontal speed component

$V_{x}$ = 14(cos 45) + 0 = 9.89 + 0 = 9.89 mph

Resultant speed = $\sqrt{V^{2} _{y}+V^{2} _{x} }$

==> $\sqrt{13.89^{2} +9.89^{2} }$ = <em>17.05 mph This is your airspeed</em>

b) To get your direction, we use

tan ∅ = $V_{y}$ / $V_{x}$

tan ∅ = 13.89/9.89 = 1.413

∅ = $tan^{-1}$ (1.413) = <em>54.7° northeast direction</em>

c) If the wind increases to 14 mph from the north, then it means the wind blows due south. As before, only the vertical component is affected .

In this case,

$V_{y}$ = 14(sin 45) - 14 = 9.89 - 14 = -4.11 mph

Resultant speed = $\sqrt{V^{2} _{y}+V^{2} _{x} }$

==> $\sqrt{4.11^{2} +9.89^{2} }$ = <em>10.71 mph This is your airspeed</em>

Your direction will be,

tan ∅ = $V_{y}$ / $V_{x}$

tan ∅ = -4.11/9.89 = -0.416

∅ = $tan^{-1}$ (-0.416) =<em> -22.58° this is the angle you'll travel relative to the east.</em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

5 0

2 years ago