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3 years ago

13

The acceleration of gravity is a constant equal to _______ meters per second squared.

2 answers:

OLEGan [10]3 years ago

7 0

9.8 is the answer jus took a test on it

VLD [36.1K]3 years ago

5 0

B. 9.8.
is the Answer

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Imagine you have been asked to create an experimental design to test the hypothesis that talking on a cell phone impairs driving

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3 years ago

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A ball rolls along the floor with a constant velocity of 2.3 m/s. How far will it travel after 15 seconds.

marissa [1.9K]

Answer:

0. 153m

Explanation:

displacement= velocity/ time

S = 2.3 m/s / 15s

S = 0. 153m

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4 years ago

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Bumek [7]

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3 years ago

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A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat

Natali [406]

centripetal acceleration is given by formula

$a_c = \omega^2*R$

given that

$a_c = 34.1 m/s^2$

$R = 5.91 m$

now we have

$\omega^2 R = 34.1$

$\omega^2 * 5.91 = 34.1$

$\omega^2 = 5.77$

$\omega = 2.4 rad/s$

so the ratationa frequency is given by

$\omega = 2 \pi f$

$2.4 = 2 \pi f$

$f = \frac{2.4}{2\pi}$

$f = 0.38 Hz$

7 0

3 years ago

Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug

Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

$F_D=\dfrac{1}{2}\rho C_DA v^2$

$C_D$ =Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

Now by putting the values

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

$\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}$

$\dfrac{F_D_1}{F_D_2}=0.444$

Percentage Change in the drag force

$\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100$

$\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100$

$\Delta F=\dfrac{1-0.444}{0.444}\times 100$

ΔF=125.22 %

Therefore force will increase by 125.22 %.

3 0

3 years ago