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3 years ago

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A 10.0 kg mass is at the origin. A 20.0 kg mass is 0.500 m to the left of it, and a 30.0 kg mass is 1.25 m to the right of it, w

hat is the net gravitational force on the 10.o kg mass?
? x 10^-8 n

1 answer:

IRINA_888 [86]3 years ago

4 0

Explanation:

100N please mark me brainliest

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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level.

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Answer:

There is an interval of 24.28s in which the rocket is above the ground.

Explanation:

$y_{i}=0m$

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$y_{e}=1000m$

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From Kinematics, the position $y$ as a function of time when the engine still works will be:

$y(t)=v_it+\frac{1}{2}at^2$

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$v_it+\frac{1}{2}at^2=y_{e}$ ⇒ $\frac{1}{2}at^2+v_it-y_{e}=0$

Using the quadratic formula: $t_1=10s$ .

How much time does it take for the rocket to touch the ground? No the function of position is:

$y(t)=y_{e}+v_et-\frac{1}{2}gt^2$

Where our new initial position is $y_{max}$ , the velocity when the engine breaks is $v_e=v_i+at=120\frac{m}{s}$ and the only acceleration comes from gravity (which points down).

Now, when the rocket tounches the ground:

$y_{e}+v_et-\frac{1}{2}gt^2=0$

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Question on Resistance. WORTH 20 POINT

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Answer:

t = 1964636.542 sec

Explanation:

Given data:

sphere diameter is 10 mm

Density is 1150 kg/m^3

viscosity 105 N s/m^2

We knwo that time taken by sphere can be calculated by following procedure

$\tau = \mu \frac{du}{dy}$

$\frac{F}{A} = \mu \frac{du}{r}$

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