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3 years ago

Newton began his academic career in 1667 for how long was he a working scientists was he a very productive scientist

1 answer:

6 0

Newton was given the title "scientist" when he was given the Merit Badge at the age of 15. He continued to work in the field of science, and was considered a scientist until his death at the age of 84. That is a total of 69 years of work. Hope this helps!

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A quarterback claims that he can throw the football a horizontal distance of 167 m. Furthermore, he claims that he can do this b

pychu [463]

Answer:u=42.29 m/s

Explanation:

Given

Horizontal distance=167 m

launch angle $=33.1^{\circ}$

Let u be the initial speed of ball

Range $=\frac{u^2\sin 2\theta }{g}$

$167=\frac{u^2\sin (66.2)}{9.8}$

$u^2=1788.71$

$u=\sqrt{1788.71}$

$u=42.29 m/s$

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3 years ago

The skateboarder weighs 75 kilogram. Calculate the potential energy of the skateboarder sliding on the track when his height abo

ArbitrLikvidat [17]

Answer:F(of gravity) = MA

F(normal force) = MA * cos(angle)

F = 72 * 9.81 * cos28

Don't have a calculator, so can't really do all the math right there. So just plug that in

Explanation:

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3 years ago

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To measure specific heat, the student flows air with a velocity of 20 m/s and a temperature of 25C perpendicular to the length

san4es73 [151]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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3 years ago

Charging a ballon and rubbing it on wool is an example of ___?

Travka [436]

Charging a balloon and rubbing it on wool is an example of static electricity. :)

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3 years ago

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A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the a

Whitepunk [10]

Answer:

C) one-half as great

Explanation:

We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

$\Delta x=v_0t+\frac{gt^2}{2}$

In this case, the sphere starts from rest, so $v_0=0$ . Replacing the given values and solving for g':

$g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}$

The acceleration due to gravity near Earth's surface is $g=9.8\frac{m}{s^2}$ . So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.

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3 years ago