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3 years ago

3m/s for 12 seconds how dar would he walk

Physics

1 answer:

Sunny_sXe [5.5K]3 years ago

4 0

I think 36m/12s because 3×12 =36

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The pictures represent three different states of matter.

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4 years ago

A negatively charged object has?

mixas84 [53]

More, fewer electrons than object is neutral

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3 years ago

(a) Suppose that your measured weight at the equator is one-half your measured weight at the pole on a planet whose mass and dia

stealth61 [152]

Answer:

7160.2812 s or 1.988 hours

Explanation:

m = Mass of person

R = Radius of Earth = $6.37\times 10^{6}\ m$

g = Acceleration due to gravity = 9.81 m/s²

$\omega$ = Angular speed

Force at equator would be

$F_e=m(g-\omega^2R)$

Force at pole

$F_p=mg$

From the question

$F_e=\dfrac{1}{2}F_p\\\Rightarrow m(g-\omega^2R)=\dfrac{1}{2}F_p\\\Rightarrow \omega=\sqrt{\dfrac{g}{2R}}$

Time period is given by

$T=\dfrac{2\pi}{\omega}\\\Rightarrow T=2\pi\sqrt{\dfrac{2R}{g}}\\\Rightarrow T=2\pi\sqrt{\dfrac{2\times 6.37\times 10^6}{9.81}}\\\Rightarrow T=7160.2812\ s=1.988\ hours$

The rotational period of the planet is 7160.2812 s or 1.988 hours

5 0

3 years ago

Define torque qualitatively and quantitatively

andrezito [222]

The rotational effect of a force is called torque.it is the cause of rotation or angular deceleration

τ=rXF
where
τ = F r sin @
hope it helps

4 0

3 years ago

Please show all the steps and work. The answer is -3.9 m/s2

konstantin123 [22]

Given that,

Initial velocity of the airplane = 66 m/s

Distance = 560 m

Final velocity of the airplane = 0 (stops)

To find,

Acceleration of the airplane.

Solution,

Let us consider that a is the acceleration of the airplane. We can find it using third equation of motion as follows :

$v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{0^2-66^2}{2(560)}\\\\a=-3.9\ m/s^2$

Therefore, the acceleration of the airplane is $-3.9\ m/s^2$

8 0

3 years ago