2 Points

miskamm [114]

B . I hope this is right

4 0

3 years ago

A plane traveling horizontally at 120 m/s over flat ground at an elevation of 3610 m must drop an emergency packet on a target

allsm [11]

Answer:

Explanation:

Horizontal displacement

x = 120 t

Vertical position

y = 3610 - 4.9 t²

y = 0 for the ground

0 = 3610 - 4.9 t²

t = 27.14 s

This is the time it will take to reach the ground .

During this period , horizontal displacement

x = 120 x 27.14 m

= 3256.8 m

So packet should be released 3256.8 m before the target.

3 0

3 years ago

Read 2 more answers

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.

7 0

3 years ago

Select the correct answer. The Cretaceous-Paleogene was a mass extinction event in which nearly every single large, land-dwellin

navik [9.2K]

Answer:

d

Explanation:

because

4 0

2 years ago

Why are energy pyramids usually no more than 4 trophic levels?

Sav [38]

There are usually no more than four trophic levels, because of the huge loss of energy at each trophic levels. Alomost 10 percent of the energy is lost at each level, as compared to the previous level, which occurs due to wastage of materials like bones and leftover tissues. So, till the fourth trophic level, very less energy is left to be continued to another level, and thus the process is like that.

4 0

3 years ago