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Fudgin [204]
3 years ago
10

Why is the gravitational potential energy of an object 1 meter above the moon’s surface less than its potential energy 1 meter a

bove Earth’s surface?
A. The object’s mass is less on the moon.
B. The object’s weight is more on the moon.
C. The object’s acceleration due to gravity is less on the moon.
D. both a and c
Physics
1 answer:
dsp733 years ago
3 0

Potential energy relative to the surface is

         (mass) x (<span>acceleration due to gravity) x (height above the surface).

At 1.0 meter above the surface, that is

</span>          (mass) x (<span>acceleration due to gravity) x (1.0 meter) .

The object's mass doesn't change, so the only thing that has any effect
on its potential energy at 1 meter above the surface is the acceleration
of gravity or, in other words, the surface of <em><u>what</u></em> ?

</span>
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Given the wave function: Y(x,t) = 5sin27(0.2x - 3t); (x = meters, t = sec.): What are the amplitude, frequency, wavelength, angu
MariettaO [177]

Answer:

Explanation:

Y = 5 Sin27( .2x-3t)

= 5 Sin(5.4x - 81 t )

Amplitude = 5 m

Angular frequency ω = 81

frequency = ω / 2π

= 81 / (2 x 3.14 )

=12.89

Wave length λ = 2π / k ,

k = 5.4

λ = 2π / 5.4

= 1.163 m

Phase velocity =ω / k

= 81 / 5.4

15 m / s.

The wave is travelling in + ve x - direction.

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3 years ago
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When you whisper you produce a 10-dB sound.
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4 0
3 years ago
while test flying the new f-22 raptor fighter jet, test pilot bob put the plane in a horizontal loop while flying at the speed o
sveticcg [70]

Given

v = 343 m/s

ac = 5g

ac = 5*9.8 m/s^2

ac = 49 m/s^2

where,

v: velocity

ac = centripetal aceleration

Procedure

We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration ac; centripetal means “toward the center” or “center seeking”.

Formula

\begin{gathered} a_c=\frac{v^2}{r} \\ r=\frac{v^2}{a_c} \\ r=\frac{(343m/s)^2}{49m/s^2} \\ r=2401\text{ m} \end{gathered}

The minimum radius not to exceed the centripetal acceleration is 2401 m.

8 0
1 year ago
A truck travels up a hill with a 5.7° incline. The truck has a constant speed of 22 m/s. What is the horizontal component of the
tester [92]

Answer:

The horizontal component of the velocity is 21.9 m/s.

Explanation:

Please see the attached figure for a better understanding of the problem.

Notice that the vector v and its x and y-components (vx and vy) form a right triangle. Then, we can use trigonometry to find the magnitude of vx, the horizontal component of the velocity.

To find vx, let´s use the following trigonometric rule of right triangles:

cos α = adjacent / hypotenuse

cos 5.7° = vx / 22 m/s

22 m/s · cos 5.7° = vx

vx = 21.9 m/s

The horizontal component of the velocity is 21.9 m/s.

8 0
3 years ago
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