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stiks02 [169]
3 years ago
6

Near the surface of Earth, acceleration due to gravity is 9.8 m/s2. Remember the formula: velocity = acceleration x time. Imagin

e you are in a room with no air, which eliminates fluid friction. Complete the table below for an object that is dropped from rest.

Physics
2 answers:
vagabundo [1.1K]3 years ago
7 0

Answer:

thats hard what grade is this

Explanation:

AveGali [126]3 years ago
5 0

Answer:

10.8

11.8 and all i know.

Explanation:

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The water side of the wall of a 60-m-long dam is a quarter-circle with a radius of 7 m. Determine the hydrostatic force on the d
tamaranim1 [39]

Answer:

26852726.19\ \text{N}

57.52^{\circ}

Explanation:

r = Radius of circle = 7 m

w = Width of dam = 60 m

h = Height of the dam will be half the radius = \dfrac{r}{2}

A = Area = rw

V = Volume = w\dfrac{\pi r^2}{4}

Horizontal force is given by

F_x=\rho ghA\\\Rightarrow F_x=1000\times 9.81\times \dfrac{7}{2}\times  7\times 60\\\Rightarrow F_x=14420700\ \text{N}

Vertical force is given by

F_y=\rho gV\\\Rightarrow F_y=1000\times 9.81\times 60\times \dfrac{\pi 7^2}{4}\\\Rightarrow F_y=22651982.59\ \text{N}

Resultant force is

F=\sqrt{F_x^2+F_y^2}\\\Rightarrow F=\sqrt{14420700^2+22651982.59^2}\\\Rightarrow F=26852726.19\ \text{N}

The hydrostatic force on the dam is 26852726.19\ \text{N}.

The direction is given by

\theta=\tan^{-1}\dfrac{F_y}{F_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{22651982.59}{14420700}\\\Rightarrow \theta=57.52^{\circ}

The line of action is 57.52^{\circ}.

7 0
3 years ago
PLEASE HELP WITH ONE QUESTION!
nata0808 [166]
I think it is -3.99 x 102 j
6 0
3 years ago
Please help on this one?
Sati [7]
I’m pretty sure it’s A
4 0
4 years ago
A 1000 kg car moving at 108 km/h jams on its brakes and comes to a stop. How much work was done by friction?
Nostrana [21]

Answer:

The work done by friction was -4.5\times10^{5}\ J

Explanation:

Given that,

Mass of car = 1000 kg

Initial speed of car =108 km/h =30 m/s

When the car is stop by brakes.

Then, final speed of car will be zero.

We need to calculate the work done by friction

Using formula of work done

W=\Delta KE

W=K.E_{f}-K.E_{i}

W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2

Put the value of m and v

W=0-\dfrac{1}{2}\times1000\times(30)^2

W=-450000
\ J

W=-4.5\times10^{5}\ J

Hence, The work done by friction was -4.5\times10^{5}\ J

3 0
3 years ago
The flow of electrons through a circuit is measured in which of the following units? A. electrical pressure B. amperes C. volts
damaskus [11]

The total quantity of electrons that have flowed through a circuit is a
quantity of charge, measured in Coulombs, or in Ampere-seconds.

The <em><u>rate</u></em> of flow of electrons, or more accurately the rate of flow of
the charge on them, is electrical current.  Its unit is the Ampere. 
1 Ampere is 1 Coulomb of charge per second.


8 0
3 years ago
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