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3 years ago

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A crate slides down a ramp that makes a 20∘ angle with the ground. to keep the crate from sliding too fast, paige pushes back on

it with a 64 n horizontal force. part a how much work does paige do on the crate as it slides 3.6 m down the ramp?

1 answer:

shutvik [7]3 years ago

8 0

The solution is:
Paige's force is (somewhat) against the direction of motion: Work = F * d Where F is the force; andd is the distance
Our f is 64 N and our distance is 20 and -3.6Plugging that in our equation will give us:
= 64N * cos20º * -3.6m = -217 J

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Answer:

Explanation:

In this case, law of conservation of energy will be implemented. It states that "the energy of the system remains conserved until or unless some external force act on it. Energy of the system may went through the conversion process like kinetic energy into potential and potential into kinetic energy.But their total always remain the same in conserved systems."

Given data:

Height of tower = 10.0 m

Depth of the pool = 3.00 cm

Mass of person = 61.0 kg

Solution:

Initial energy = Final energy

$U_{i} = (K.E) + U_{f}$

As the person was at height initially so it has the potential energy only.

$mg(h_{1} +h_{2}) = K.E + mgh_{2}$

$K.E = mgh_{1}$

$K.E = (61.0)(9.8)(10)\\K.E = 5978 J$

Lets find out the magnitude of the force that the water is exerting on the diver.

W =ΔK.E

$F.h_{2} = 5978\\$

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__ is the apparent in the position of an object when viewed from two locations

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3 years ago

At the local swimming hole, a favorite trick is to run horizontally off a cliff that is 8.0 m above the water. One diver runs of

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Answer:

Number of revolutions=1.532 revolutions

Explanation:

Given data

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To find

Number of revolutions

Solution

From the equation of simple motion we not that

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3 years ago

What is the velocity of the object?

dmitriy555 [2]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2> $\huge\boxed{\text{V = 9.5 m/s}}$ </h2><h2>_____________________________________</h2>

<h2>DATA:</h2>

mass = m = 2kg

Distance = x = 6m

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TO FIND:

Work = W = ?

Velocity = V = ?

<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

${\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)$

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Height is the y-axis of the graph which is Force i.e. 30N

So,

$W\ =\ \frac{1}{2}\:6\:x\:30$

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The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

${\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2$

$\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}$

$\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}$

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

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3 years ago

Structures on a bird feather act like a diffraction grating having 8000 8000 lines per centimeter. What is the angle of the firs

sammy [17]

Answer:

Angle of first order maximum, $\theta=21.19^{\circ}$

Explanation:

Given that,

Wavelength of the light, $\lambda=452\ nm=452\times 10^{-9}\ m$

Number of lines, N = 8000 per cm

The relation between the number of lines and the slit width is given by :

$d=\dfrac{1}{N}$

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3 years ago