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2 years ago

A helicopter’s velocity increases from 35 m/s to 70 m/s in 10 seconds. What is the acceleration of this helicopter?

1 answer:

3 0

3.5 meters per second second is the acceleration because we know that acceleration is change in velocity over time and the change is velocity here is 35 and the time is 10 so we can simply divide 35 by 10 which is 3.5 m/s squared

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Which of the following statements CANNOT be supported by Kepler's laws of planetary motion?

horsena [70]

Answer:

B) A planet's speed as it moves around the sun will not be the same in six months.

Explanation:

A planet's speed as it moves around the sun will not be the same in six months, is a statement that CANNOT be supported by Kepler's laws of planetary motion.

8 0

2 years ago

Read 2 more answers

A computational model predicts the maximum kinetic energy a roller coaster car can have given its mass, the speed at the highest

Ymorist [56]

Answer:

Explanation:

The decrease in potential energy is equal to the increase in kinetic energy.

mgh

250 x 9.8 x 30

=73, 500

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2 years ago

7. What does the slope of a position-time graph represent?

allochka39001 [22]

Answer:

a. Velocity

Explanation:

The slope of the tangent line on a position-time graph is the instantaneous velocity.

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3 years ago

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3$

$T^2=\sqrt{9.96\times 10^{14}}\ s$

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0

2 years ago

A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle

liq [111]

Answer:

0.0031792338 rad/s

Explanation:

$\theta$ = Angle of elevation

y = Height of balloon

Using trigonometry

$tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta$

Differentiating with respect to t we get

$\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta$

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

$h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft$

Now y = 2500 ft

$cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974$

$\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s$

The angle is changing at 0.0031792338 rad/s

6 0

2 years ago