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2 years ago

An inclined plane is 30 meters long and 2 meters high. What is the mechanical advantage of the inclined plane? MA =

Physics

2 answers:

zalisa [80]2 years ago

6 0

Answer:

15=ma

Explanation:

Bond [772]2 years ago

4 0

In a second class lever, the LF is between the fulcrum load.
Like the first class lever, the MA is an increase in fource and is calculated by dividing the EA by the LA ( EA divided LA = IMA )

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Which statements below are true?

Ilya [14]

The correct statements are that the speed decreases as the distance decreases and speed increases as the distance increases for the same time.

Answer:

Option A and Option B.

Explanation:

Speed is defined as the ratio of distance covered to the time taken to cover that distance. So Speed = Distance/Time. In other words, we can also state that speed is directly proportional to the distance for a constant time. Thus, the speed will be decreasing as there is decrease in distance for the same time. As well as there will be increase in speed as the distance increases for the same time. So option A and option B are the true options. So if there is decrease in the distance due to direct proportionality the speed will also be decreasing. Similarly, if the distance increases, the speed will also be increasing.

7 0

2 years ago

A planet orbits a star, in a year of length 3.37 x 107 s, in a nearly circular orbit of radius 1.04 x 1011 m. With respect to th

PIT_PIT [208]

Answer

Given,

Time period of star,T = 3.37 x 10⁷ s

Radius of circular orbit,R = 1.04 x 10¹¹ m

a) Angular speed of the planet

$\omega = \dfrac{2\pi}{T}=\dfrac{2\pi}{3.37\times 10^{7}}$

$\omega = 1.864\times 10^{-7}\ rad/s$

b) tangential speed

$v = r \omega = 1.04\times 10^{11}\times 1.864 \times 10^{-7}$

v = 1.94 x 10⁴ m/s

c) centripetal acceleration magnitude

$a = \dfrac{v^2}{r}= \dfrac{(1.94\times 10^4)^2}{1.04\times 10^{11}}$

a = 3.62 x 10⁻³ m/s²

8 0

2 years ago

David lifts a book 1.2 meters from the floor to the top of his desk. If the book weighed 0.50 kg, what is the gravitational pote

Anika [276]

Answer:

5.886 J

Explanation:

Given:

The mass of the book is, $m=0.5$ kg

Height of lift is, $\Delta h=1.2$ m

Acceleration due to gravity is, $g=9.81$ m/s²

Now, gain in gravitational potential energy is a function of change in position and is given as:

$\Delta PE=mg\Delta h$

Here, $\Delta PE$ is the change in gravitational potential energy.

Plug in 0.5 kg for $m$ , 9.81 for $g$ and 1.2 for $\Delta h$ . Solve for $\Delta PE$

$\Delta PE=0.50\times 9.81\times 1.2=5.886\ J$

Therefore, the gain in gravitational energy of the book is 5.886 J.

7 0

2 years ago

True or false according to newtons 1st law an unbalanced force can change the motion of an object

lisov135 [29]

I think this is true

6 0

2 years ago

A 75kg Tibetan is trekking along flat, but icy ledge with his 450kg yak when he slips over the edge. Luckily, he is holding the

Tcecarenko [31]

Answer:

minimal coefficient of static friction: $\mu_s=0.1667$

Explanation:

Once the Tibetan is hanging from the strap, he is exerting a horizontal force on the yak equal to his weight which is the product of his mass times the acceleration of gravity (g) as written below:

$w = m\,*\,g= 75\,kg\,*\,g$

The other forces acting on the yak are (see attached diagram):

* the force of gravity on the yak (identified in blue color in the image as $F_g$ ,

* the normal force (indicated in green in the image and identified by the letter "n") of the ledge on the yak as reaction to the yak's weight

* the force of static friction between the yak's hooves and the ledge (pictured in red in the image and identified with $f_s$ )

Since the normal force and the force of gravity on the yak cancel each other (balance - the yak is not moving vertically), the only forces we need to analyse are the force of the Tibetan's weight via the strap, and the force of static friction which should at least be equal in magnitude so the Tibetan doesn't fall. We assume these two forces are acting horizontally (one to the right: the Tibetan's weight, and one to the left: the static friction).

As we said, we want them to be at least equal so thy are in balance.

We recall that the force of static friction is the product of the normal force (n) times the coefficient of static friction ( $\mu_s$ ), such that: $f_s=\mu_s\,*\,n$

In our case these are the forces at play:

$F_g= M\,*\,g=450\, kg \,*\,g\\n=F_g=450\,kg\,*\,g\\f_s=\mu_s\,*\,n=\mu_s\,*450\,kg\,*\,g\\w=m\,*\,g=75\,kg\,*\,g$

So we need to find what is the minimum coefficient of static friction that precludes the Tibetan from falling. We therefore proceed to make an equality between the force of static friction on the yak and the weight of the Tibetan:

$f_s=w\\\mu_s\,*450\,kg\,*g=75\,kg\,*\,g$

and proceed to solve for the coefficient of friction by dividing both sides by "g" (which by the way cancels out), and by the yak's mass:

$\mu_s\,*450\,kg\,*g=75\,kg\,*\,g\\\mu_s=\frac{75}{450} \\\mu_s=0.1667$

where we have rounded to four decimal places the periodic number that the quotient generates. Notice that as expected, the coefficient of friction has no units (they all cancelled out in the division).

5 0

2 years ago