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vredina [299]
3 years ago
13

How does the comets energy change as it moves from point A to point B?

Physics
2 answers:
Finger [1]3 years ago
6 0

Answer:

KE decreases, GPE increases

Explanation:

A Px

katrin [286]3 years ago
4 0

Answer:

a comet would go faster

Explanation:

the comet is being slingshotted by the suns immense gravitational pull

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ExtremeBDS [4]

Answer:

B!!!

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Friction helps your vehicle stop quickly.<br> A. TRUE<br> B. FALSE
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Answer:

true

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Jack is planning an experiment to see how temperature can affect the reproduction rate of a certain species of bacteria. Jack is
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4 years ago
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A dynamite blast propels a heavy rock straight up with a launch velocity of 160ft/sec (about 109 mph). Write a position for the
Flura [38]
A) Using:
2as = v² - u², where v will be 0 at max height
s = -(160)² / 2 x -32.174
s = 397.8 ft

b) Using:
s = ut + 1/2 at²
256 = 160t - 16.1t²
solving for t,
t = 2.0, t = 7.9
Now, v = u + at, for both times:
v(2) = 160 - 32.174(2)
v(2) = 95.7 ft/sec on the way up

v(7.9) = 160 - 32.174(7.9)
v(7.9) = -94.3 ft/sec; 94.3 ft/sec on the way down

c) -32.174 ft/s², which is the acceleration due to gravity.

d) s = 0
0 = 160t - 1/2 x 32.174t²
t = 9.94 seconds
3 0
4 years ago
Point A of the circular disk is at the angular position θ = 0 at time t = 0. The disk has angular velocity ω0 = 0.17 rad/s at t
statuscvo [17]

Given that,

Angular velocity = 0.17 rad/s

Angular acceleration = 1.3 rad/s²

Time = 1.7 s

We need to calculate the angular velocity

Using angular equation of motion

\omega=\omega_{0}+\alpha t

Put the value in the equation

\omega=0.17+1.3\times1.7

\omega=2.38(k)\ m/s

We need to calculate the angular displacement

Using angular equation of motion

\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}

Put the value in the equation

\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}

\theta=2.1675\times\dfrac{180}{\pi}

\theta= 124.18^{\circ}

We need to calculate the velocity at point A

Using equation of motion

v_{A}=v_{0}+\omega\times r

Put the value into the formula

v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))

v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i

v_{A}=(-0.267j-0.393i)\ m/s

We need to calculate the acceleration at point A

Using equation of motion

a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)

Put the value in the equation

a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)

a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)

a_{A}=-0.146j-0.215i−0.636i+0.937j

a_{A}=0.791j-0.851i

a_{A}=-0.851i+0.791j\ m/s^2

Hence, (a). The velocity at point A is (-0.267j-0.393i)\ m/s

(b). The acceleration at point A is (-0.851i+0.791j)\ m/s^2

3 0
4 years ago
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