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emmainna [20.7K]

3 years ago

6

In your discussion entry, create and describe at least four ways to include someone of a diverse culture or ability level in a s

pecific physical activity or sport. Be specific about the type of person you are trying to include. Possible modifications can be made to the:
• Methods of communication during play
• Rules of the physical activity or sport
• Equipment
• Playing area

1 answer:

artcher [175]3 years ago

8 0

The answer is: playing area

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At its limit the history of science touches on

myrzilka [38]

Answer:

Explanation:

the development of the history of science, the histories of the individual scientific disciplines have played an enormously significant role. The goals and functions of these have recently received considerable attention, both because of the influence that such histories have had on the legitimacy and self-image of the disciplines and also because of the adaptability that they have shown when faced with the conceptual and methodological changes that they have undergone.

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3 years ago

A bicyclist of mass 60kg supplies 340W of power while riding into a 5 m/s head wind. The frontal area of the cyclist and bicycle

NikAS [45]

Answer:

87.1 mph

Explanation:

We are given that

Mass,m=60 kg

Power,P=340 W

Speed,v=5 m/s

Area, $A=0.344 m^2$

Drag coefficient, $C_d=0.88$

Coefficient of rolling resistance, $\mu_r=0.007$

Friction force, $f=\mu_rmg=0.007\times 60\times 9.8=4.1 N$

Where $g=9.8 m/s^2$

Let speed of cyclist=v'

Drag force, $F_d=\frac{1}{2}\rho_{air}AC_dv^2$

Density of air, $\rho_{air}=1.225 kg/m^3$

$F_d=\frac{1}{2}\times 1.225\times 0.344\times 0.88(5)^2=4.635N$

Power,P= $(F_d+f)\times v'$

$340=(4.1+4.635) v'=8.735v'$

$v'=\frac{340}{8.735}=38.9m/s$

$v'=87.1 mph$

1 m=0.00062137 miles

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7 0

4 years ago

Now open the simulation. Activate ""grid"" and ""show numbers"" to read values. Place a 1 nC positive (red color) charge on the

zmey [24]

Answer:

The divergence on the sensor shows the magnitude of the charges

Explanation:

This will increase as length increases since it is said to be proportional to the length. note that test charge is always positive and charge on the grid is positive as indicated (1 nC)

8 0

3 years ago

two point charges of 5*10^-19 C and 20*10^-19C are separated by a distance of 2m. at which point on the line joining them will h

Aneli [31]

Answer:

On that line segment between the two charges, at approximately $0.7\; \rm m$ away from the smaller charge (the one with a magnitude of $5 \times 10^{-19}\; \rm C$ ,) and approximately $1.3\; \rm m$ from the larger charge (the one with a magnitude of $20 \times 10^{-19}\; \rm C$ .)

Explanation:

Each of the two point charges generate an electric field. These two fields overlap at all points in the space around the two point charges. At each point in that region, the actual electric field will be the sum of the field vectors of these two electric fields.

Let $k$ denote the Coulomb constant, and let $q$ denote the size of a point charge. At a distance of $r$ away from the charge, the electric field due to this point charge will be:

$\displaystyle E = \frac{k\, q}{r^2}$ .

At the point (or points) where the electric field is zero, the size of the net electrostatic force on any test charge should also be zero.

Consider a positive test charge placed on the line joining the two point charges in this question. Both of the two point charges here are positive. They will both repel the positive test charge regardless of the position of this test charge.

When the test charge is on the same side of both point charges, both point charges will push the test charge in the same direction. As a result, the two electric forces (due to the two point charges) will not balance each other, and the net electric force on the test charge will be non-zero.

On the other hand, when the test charge is between the two point charges, the electric forces due to the two point charges will counteract each other. This force should be zero at some point in that region.

Keep in mind that the electric field at a point is zero only if the electric force on any test charge at that position is zero. Therefore, among the three sections, the line segment between the two point charges is the only place where the electric field could be zero.

Let $q_1 = 5\times 10^{-19}\; \rm C$ and $q_2 = 20 \times 10^{-19}\; \rm C$ . Assume that the electric field is zero at $r$ meters to the right of the $5\times 10^{-19}\; \rm C$ point charge. That would be $(2 - r)$ meters to the left of the $20 \times 10^{-19}\; \rm C$ point charge. (Since this point should be between the two point charges, $0 < r < 2$ .)

The electric field due to $q_1 = 5\times 10^{-19}\; \rm C$ would have a magnitude of:

$\displaystyle | E_1 | = \frac{k\cdot q_1}{r^2}$ .

The electric field due to $q_2 = 20 \times 10^{-19}\; \rm C$ would have a magnitude of:

$\displaystyle | E_2 | = \frac{k\cdot q_2}{(2 - r)^2}$ .

Note that at all point in this section, the two electric fields $E_1$ and $E_2$ will be acting in opposite directions. At the point where the two electric fields balance each other precisely, $| E_1 | = | E_2 |$ . That's where the actual electric field is zero.

$| E_1 | = | E_2 |$ means that $\displaystyle \frac{k\cdot q_1}{r^2} = \frac{k\cdot q_2}{(2 - r)^2}$ .

Simplify this expression and solve for $r$ :

$\displaystyle q_1\, (2 - r)^2 - q_2 \, r^2 = 0$ .

$\displaystyle 5\times (2 - r)^2 - 20\, r^2 = 0$ .

Either $r = -2$ or $\displaystyle r = \frac{2}{3}\approx 0.67$ will satisfy this equation. However, since this point (the point where the actual electric field is zero) should be between the two point charges, $0 < r < 2$ . Therefore, $(-2)$ isn't a valid value for $r$ in this context.

As a result, the electric field is zero at the point approximately $0.67\; \rm m$ away the $5\times 10^{-19}\; \rm C$ charge, and approximately $2 - 0.67 \approx 1.3\; \rm m$ away from the $20 \times 10^{-19}\; \rm C$ charge.

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3 years ago

A tortoise and a hare are having a 1,000-meter race. The tortolse runs the

lukranit [14]

Answer:2 hours

Explanation:

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3 years ago