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3 years ago

Convert the following in to SI units without changing its values a. 4000 g b. 2.4 km

Physics

2 answers:

Harrizon [31]3 years ago

5 0

hi :D

i believe i explained this answer properly in my last answer but it would be 4kg and 2400m as these are the SI units for these values.

hope this helps :)

lana66690 [7]3 years ago

3 0

hi <3

i believe i explained this answer properly in my last answer but it would be 4kg and 2400m as these are the SI units for these values.

hope this helps :)

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A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.7 m/s rel

mafiozo [28]

Here it is given that speed of migrating Robin is 12 m/s relative to air

so we can say that

$\vec v_{ra} = 12 m/s$ North

so it will be

Let North direction is along Y axis and East direction is along X axis

$\vec v_{ra} = 12\hat j$

also it is given that speed of air is 6.7 m/s relative to ground

$\vec v_a = 6.7 \hat i$

now as we know by the concept of relative motion

$\vec v_{ab} = \vec v_a - \vec v_b$

$\vec v_{ra} = \vec v_r - \vec v_a$

now by rearranging the terms

$\vec v_r = \vec v_{ra} + \vec v_a$

$\vec v_r = 12 \hat j + 6.7 \hat i$

now we need to find the speed of Robin which means we need to find the magnitude of its velocity which we found above

So here we will say

$v_r = \sqrt{12^2 + 6.7^2}$

$v_r = 13.7 m/s$

so the net speed of Robin with respect to ground will be 13.7 m/s

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3 years ago

A 2.0 kg particle moving along the z-axis experiences the

DochEvi [55]

At point x = 0, the particle accelerates. Since there will be change of velocity at that point. The the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.

Given that a 2.0 kg particle moving along the z-axis experiences the force shown in a given figure.

Force is the product of mass and acceleration. While acceleration is the rate of change of velocity. Both the force and acceleration are vector quantities. They have both magnitude and direction.

If the particle's velocity is 3.0 m/s at x = 0 m, that mean that the particle experience change of velocity at point x = 0. Since the the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.

Learn more here: brainly.com/question/20366032

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3 years ago

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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In the story, The Road, the man is looking for food and supplies with his binoculars. There is a boy that is traveling with the man and they are starving. They continue traveling until they see a big house and they enter it to look for food.

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3 years ago

Convert the following in to SI units without changing its values a. 4000 g b. 2.4 km​

Convert the following in to SI units without changing its values a. 4000 g b. 2.4 km