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antoniya [11.8K]
3 years ago
15

A 20 kg object in a 2 kg object falls toward earth with a acceleration of 9.8m/sWhat is true about the force of gravity on the t

wo objects
Physics
1 answer:
Vitek1552 [10]3 years ago
5 0
Look it up BYEEEEEEEEEEEEEEEEEEEEEE
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A wheelchair moves upward on a 7.1 degree ramp at a speed of 20km/h what is the horizontal velocity
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Answer:

If the wheelchair is up 7.1 ft. In hight the time of flight should be 0.664 seconds and the distance should be 12.108 ft.

Explanation: I divided the displacement by the time and I used the equation Vx = 20 km/m

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How often does a 16-year-old need a medical evaluation from a doctor?
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Explanation: A

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student saw a science fiction movie in which a spaceship was zooming through space. The sound of the engine was loud. However, h
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Sound waves move through molecules. When there are no molecules in a certain space, sound cannot travel - whether in a vacuum or in space. 
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Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r
Butoxors [25]

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current y_{1} = 102 mA = 0.102 A

According to ohm's law, V = IR

R = V/I

Here, I = y_{1}

R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms

For the second measurement, current y_{2} = 97 mA = 0.097 A

R = \frac{V}{y_{2} }

R = \frac{10}{0.097} \\R = 103 .09 ohms

b) y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}

y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]

y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3}  \end{array}\right]

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

x = \left[\begin{array}{ccc}100\end{array}\right]

\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3}  \end{array}\right] =  \left[\begin{array}{ccc}G_{1} \\G_{2}  \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2}  \end{array}\right] =  \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5}  \end{array}\right]

3 0
3 years ago
During a baseball game, a batter hits a high .
Harman [31]

Answer: Due that we don't know the initial speed after hitting the ball, we are going to accept that the ball goes up for half of the time and then falls during other half part, that is 3.0 seconds each. Then we know that ball's movement is ruled by the acceleration of gravity formula, as follows: H = Vi * T + 1/2 * g * T^2 V = Vi + g * T where: H is height, Vi initial speed, g gravity acceleration and T time When we only consider the second half of the trajectory, we have that initial speed at the top of that movement is zero, because ball goes up till top, where stops and starts to go down, so : H = 0 * 3 + 1/2 * 32 * 3^2 = 144 ft. So the height of the pop-up is 144 feet.

7 0
3 years ago
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