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2 years ago

6

What is being described in the following definition: An opening in the floor, platform, or pavement that measures 12 inches or m

ore, and through which persons may fall.
A.
Floor Hole

B.
Floor Opening

C.
Wall Opening

D.
Railing

1 answer:

azamat2 years ago

3 0

Answer is Floor Opening

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Calculate the displacement of the volleyball in sample problem 2f when teh volleyballs final velocity is 1.1 m/s upward

stich3 [128]

We have: v i (initial velocity) = 6 m/sv = 1.1 m/sa = - 9.8 m/s²v = v i + a · t1.1 m/s = 6 m/s - 9.8 m/s² t9.8 t = 6 - 1.19.8 t = 4.9t = 4.9 : 9.8t = 0.5 sThen the replacement:x = xi + vi · t + a t² / 2( xi = 0 )x = 6 · 0.5 - 9.8 · 0.25 / 2x = 3 - 1.225Answer:
x = 1.775 m

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3 years ago

Two balls of mass 0.09 kg hang on strings attached to the same point on the ceiling. The balls are given charges Q that cause th

telo118 [61]

Answer:

Q = 6.33μC

Explanation:

To find the value of the charge Q you take into account both gravitational force and electric force over each ball. By symmetry you can use the fact that both balls experiences the same forces. Hence you only take into account the forces for one ball for the x component and y component:

$-Mg+Tcos\theta=0\\\\F_e-Tsin\theta=0$

M: mass of the ball = 0.09kg

T: tension of the string

F_e: electric force between charges

angle = 45°

The electric force is given by:

$F_e=k\frac{Q^2}{r^2}$

Q: charge of the balls

r: distance between balls = 2m

You divide both equation in order to eliminate the tension T:

$tan\theta=\frac{F_e}{Mg}=k\frac{Q^2}{Mgr^2}$

By doing Q the subject of the formula and replacing you obtain:

$Q=\sqrt{\frac{tan\theta Mgr^2}{k}}=\sqrt{\frac{tan45\°(0.09kg)(2m)^2}{(8.89*10^{9}Nm^2/C^2)}}=6.33*10^{-6}C=6.33\mu C$

hence, the charge of the balls is 6.33μC

4 0

3 years ago

Can anyone solve this?

love history [14]

Answer:

F = 39.2 N

Explanation:

Since, the object is in uniform motion. Therefore, the frictional force on object will be:

Frictional Force = μk N = μk mg

where,

μk = coefficient of kinetic friction = 0.2

m = mass of crate = 10 kg

g = 9.8 m/s²

Therefore,

Frictional Force = (0.2)(10 kg)(9.8 m/s²)

Frictional Force = 19.6 N

The horizontal component of force must be equal to this frictional force to continue the uniform motion:

F Sin 30° = 19.6 N

F = 19.6 N/Sin 30°

<u>F = 39.2 N</u>

3 0

2 years ago

Find the value of F1 + F2 + F3.<br>

Dovator [93]

Answer:

F = 0.78[N]

Explanation:

The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.

<u>For F₁</u>

<u /> $F_{y}=2[N]$ <u />

<u>For F₂</u>

$F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N]$

<u>For F₃</u>

<u /> $F_{x}=-1*sin(60)\\F_{x}=-0.866[N]\\F_{y}=1*cos(60)\\F_{y}=0.5 [N]$ <u />

Now we can sum each one of the forces in the given axes:

$F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N]$

Now using the Pythagorean theorem we can find the total force.

$F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N]$

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2 years ago

Who can do my worksheet for me

timurjin [86]

Answer:

what is it on? like name one of the questions

Explanation:

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3 years ago