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Digiron [165]
4 years ago
14

MATHPHYS please help thank you in advance

Physics
1 answer:
masya89 [10]4 years ago
6 0

Answer:

1. -6.43 J

2. 6.43 J

3. -12.25 J

4. 196 J

Explanation:

For the box on the ramp, there are two forces:

Weight force mg pulling down

Normal force N pushing perpendicular to the ramp

Sum of the forces in the parallel direction:

∑F = ma

-mg sin θ = ma

a = -g sin θ

a = -9.8 m/s² sin 15°

a = -2.54 m/s²

1. The distance the box moves is:

Δx = vt − ½ at²

Δx = (0 m/s) (1 s) − ½ (-2.54 m/s²) (1 s)²

Δx = 1.27 m

The work done by gravity is:

W = Fd

W = (-mg sin θ) (Δx)

W = -(2 kg) (9.8 m/s²) (sin 15°) (1.27 m)

W = -6.43 J

Notice the answer is negative.  That's because the force vector and displacement vectors are in opposite directions

2. The distance the box moves is:

Δx = v₀t + ½ at²

Δx = (0 m/s) (1 s) + ½ (-2.54 m/s²) (1 s)²

Δx = -1.27 m

The work done by gravity is:

W = Fd

W = (-mg sin θ) (Δx)

W = -(2 kg) (9.8 m/s²) (sin 15°) (-1.27 m)

W = 6.43 J

3. Final KE = initial KE + work by gravity

KE = KE₀ + W

0 = ½ mv² + W

W = -½ mv²

W = -½ (2 kg) (3.50 m/s)²

W = -12.25 J

4. Initial PE = final PE + work by gravity

PE₀ = PE + W

mgh = 0 + W

W = mgh

W = (2 kg) (9.8 m/s²) (10 m)

W = 196 J

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Answer:

2917.4 m/s

Explanation:

From the question given above, the following data were:

Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth

Radius (r) of the Moon = 1737 Km

Escape velocity (v) =?

Next, we shall determine the gravitational acceleration of the Moon. This can be obtained as follow:

Gravitational acceleration of the earth = 9.8 m/s²

Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth

= 0.25 × 9.8 = 2.45 m/s²

Next, we shall convert 1737 Km to metres (m). This can be obtained as follow:

1 Km = 1000 m

Therefore,

1737 Km = 1737 Km × 1000 m / 1 Km

1737 Km = 1737000 m

Thus, 1737 Km is equivalent to 1737000 m

Finally, we shall determine the escape velocity of the rocket as shown below:

Gravitational acceleration of the Moon (g) = 2.45 m/s²

Radius (r) of the moon = 1737000 m

Escape velocity (v) =?

v = √2gr

v = √(2 × 2.45 × 1737000)

v = √8511300

v = 2917.4 m/s

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A 16.9 kg monkey is swinging on a 5.32 m long vine. It starts at rest, with the vine at a 43.0° angle. How fast is the monkey mo
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Answer:

v = 5.7554 m/s

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As the monkey is dropping from the innitial point which is the suspension point, is also dropping from 5.32. Then the actual height of the monkey will be:

Δh = 5.32 - 3.63 = 1.69 m

In order to calculate the speed of the monkey we need to understand that the monkey has a potential energy. This energy, because of the gravity, is converted in kinetic energy, and the value will be the same. Therefore we can say that:

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From here, we can calculate the speed of the monkey.

Ep = mgΔH

Ek = 1/2 mv²

The potential energy is:

Ep = 16.9 * 9.8 * 1.69 = 279.9

Now with the kinetic energy:

1/2 * (16.9) * v² = 279.9

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3 0
3 years ago
A transverse wave on a string is described by the wave functiony(x, t) = 0.350 sin (1.25x + 99.6t)where x and y are in meters an
ella [17]

The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

<h3>How to solve for the time interval</h3>

We have y = 0.175

y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.5

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s = vt

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Read more on waves here

brainly.com/question/25699025

#SPJ4

complete question

A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?

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