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Zielflug [23.3K]

2 years ago

13

Seth and Diana were asked to be the leaders for a game of tag during recess at school. Seth and Diana can ensure that all intere

sted students get a chance to participate equally regardless of ability to run fast by having the: Tagged players sit out until the next game starts Tagged players perform a fitness activity then return to the game Two youngest students be the taggers Two fastest students be the taggers

1 answer:

grandymaker [24]2 years ago

3 0

Answer:

B

Explanation:

did this yesterday with my friend

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A gyre is a set of currents that form b. a loop. The circulation of gyres are affected by global wind patterns, landmasses, and the planet's rotation. The circulation is also affected by temperature, as warm water goes up and cold water sinks. There are five major gyres in the world: <span>North Atlantic, South Atlantic, Indian, North Pacific, and South Pacific.</span>

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Elect the correct answer. Two identical projectiles, A and B, are launched with the same initial velocity, but the angle of laun

allsm [11]

Answer:

A. The range of A and B are equal.

Explanation:

Let u be the initial speed of both the projectile.

The gravitational force acting in the projectile is in the downward direction, sp the speed of the projectile in the horizontal direction remains constant and equals to the initial horizontal speed.

For projectile, a projectile having initial velocity u at an angle \theta with the horizontal direction,

The speed in the horizontal direction $= u\cos\theta$

and the speed in the vertical direction is $= u\sin\theta$ upward.

For A:

The speed in the horizontal direction $= u\cos75^{\circ}$

and the speed in the vertical direction is $= u\sin75^{\circ}$ upward.

For B:

The speed in the horizontal direction $= u\cos15^{\circ}$

and the speed in the vertical direction is $= u\sin15^{\circ}$ upward.

Let $t_A$ and $t_B$ are the time of flight for projectile A and B respectively.

As the range is the horizontal distance traveled by the projectile, so

The range for the projectile A $= u\cos75^{\circ}\times t_A\cdots(i)$

The range for the projectile B = $u\cos15^{\circ}\times t_B\cdots(ii)$

At the highest point, the vertical velocity is 0.

Bu using the equation of motion $v^2=u^2 +2a s.$

Here, the final velocity v=0, the initial velocity $u = u \sin \theta$ , h= vertical distance up to the highest point, and $a= -g$ (as per sign convention).

So, $s= \frac{u^2\sin^2 \theta}{2g}$

For projectile A: The maximum height attained.

$s_A= \frac{u^2\sin^2 75^{\circ}}{2g}$

For projectile B: The maximum height attained.

$s_B= \frac{u^2\sin^2 15^{\circ}}{2g}$

As $\sin^2 75^{\circ} > \sin^2 15^{\circ}$ , the height of A is attained by A is more than the heigHt attained by B.

Now, the times required to reach the highest point from the ground and again form the highest point to the ground are the same.

So, the total time of flight = 2 x (Time to reach the highest point)

In a similar way, by using the equation of motion v=u+at,

The time to reach the highest point $=\frac {u\sin\theta}{g}$

where g is the acceleration due to gravity.

So, the total time of flight

$= 2 \times \frac {u\sin\theta}{g}$

The total time of flight for A

$=2 \times \frac {u\sin75^{\circ}}{g}$

The total time of flight for A

$=2 \times \frac {u\sin15^{\circ}}{g}$

Now, from equations (i) and (ii),

The range for the projectile A =

$u\cos75^{\circ}\times \frac {2u\sin75^{\circ}}{g}=\frac {u^2 \sin 150^{\circ}}{g}= \frac {u^2 \sin 30^{\circ}}{g}$

The range for the projectile B =

$u\cos15^{\circ}\times \frac {2u\sin15^{\circ}}{g}=\frac {u^2 \sin 30^{\circ}}{g}.$

Both the projectile have the same range.

Hence, option (A) is correct.

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3 years ago