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3 years ago

Free use of uniforms and socialization with classmates are possible advantages of ? A.)

Physics

1 answer:

Galina-37 [17]3 years ago

8 0

The answer is B community based sports program

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A bowling ball has a mass of 7.2 kg and a weight of 70.6 N. It moves down the bowling alley at 1 m/s and strikes a pin with a fo

Inga [223]

The answer is 15.0N its explained in newtons third law hope this helps:)

4 0

3 years ago

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Rick shoots a basketball at an angle of 35' from the horizontal. It leaves his hands 7 feet from the ground with a velocity of 2

Korvikt [17]

Given:

The angle of projection of the basketball, θ=35°

The height at which the ball leaves the hand, h=7 ft

The initial velocity of the basketball, v=20 ft/s

To find:

The parametric equations describing the shot.

Explanation:

The range, x of the basketball is given by,

$x=v\cos\theta t$

On substituting the known values,

$\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}$

The change in the height, y of the basketball is given by,

$y=-v\sin\theta t+\frac{1}{2}gt^2$

Where g is the acceleration due to gravity.

On substituting the known values,

$\begin{gathered} y=-20\times\sin35\degree\times t+\frac{1}{2}\times32\times t^2 \\ \implies y=-11.5t+16t^2 \end{gathered}$

Final answer:

The parametric equations describing the shot are

$\begin{gathered} \begin{equation*} x=16.4t \end{equation*} \\ \begin{equation*} y=-11.5t+16t^2 \end{equation*} \end{gathered}$

8 0

1 year ago

Force has _____. magnitude direction gravity weight

Contact [7]

Answer:

magnitude

Explanation:

7 0

3 years ago

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A particle travels clockwise on a circular path of diameter R, monitored by a sensor on the circle at point P; the other endpo

kotykmax [81]

We make a graphic of this problem to define the angle.

The angle we can calculate through triangle relation, that is,

$sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}$

With this function we should only calculate the derivate in function of c

$\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}$

That is the rate of change of $\theta$ .

b) At this point we need only make a substitution of 0 for c in the equation previously found.

$\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}$

Hence we have finally the rate of change when c=0.

6 0

2 years ago

The brick wall (of thermal conductivity 0.35 W/m ·◦ C) of a building has dimensions of 2.7 m by 6 m and is 16 cm thick. How much

Gwar [14]

Answer:

$Q = 54.577\,MJ$

Explanation:

The heat transfer through brick wall is:

$\dot Q = \frac{k\cdot A}{L}\cdot \Delta T$

$\dot Q = \frac{\left(0.35\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (2.7\,m)\cdot (6\,m)}{0.16\,m} \cdot (31^{\circ}C - 8^{\circ}C)$

$\dot Q = 815.063\,W$

The heat flow in a 18.6-h period is:

$Q = \dot Q \cdot \Delta t$

$Q = (815.063\,W)\cdot (18.6\,h)\cdot \left(\frac{3600\,s}{1\,h} \right)$

$Q = 54576618.48\,J$

$Q = 54.577\,MJ$

8 0

3 years ago

Read 2 more answers