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3 years ago

A body weighs less inside water

Physics

1 answer:

Ipatiy [6.2K]3 years ago

7 0

Answer:

Body weighs lesser in water because of the upward force (buoyant force) which acts on our body thereby reducing our actual weight. This is our apparent weight.

Explanation:

hope it helps

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A shuttle bus slows down with an average acceleration of -2.4 m/s2. How long does it

olganol [36]

Answer:

$\boxed {\boxed {\sf 3.75 \ seconds }}$

Explanation:

Average acceleration is found by dividing the change in acceleration by the time.

$a=\frac{ v_f-v_i}{t}$

The shuttle bus has an acceleration of -2.4 meters per square second. It slows from 9.0 meters per second to rest, or 0 meters per second. Therefore:

$a= -2.4 \ m/s^2 \\v_f= 0 \ m/s \\v_i= 9 \ m/s$

Substitute the values into the formula.

$-2.4 \ m/s^2=\frac{0 \ m/s - 9 \ m/s}{t }$

Solve the numerator.

$-2.4 \ m/s^2 = \frac{-9 \ m/s}{t}$

We want to solve for t, the time. We have to isolate the variable. Let's cross multiply.

$\frac{-2.4 \ m/s^2}{1} = \frac{-9 \ m/s}{t}$

$-9 \ m/s *1= -2.4 \ m/s^2 *t$

$-9 \ m/s=-2.4 m/s^2*t$

t is being multiplied by -2.4. The inverse of multiplication is division, so divide both sides by -2.4

$\frac{-9 \ m/s }{-2.4 \ m/s^2} =\frac{ -2.4 \ m/s^2*t}{-2.4 \ m/s^2}$

$\frac{-9 \ m/s }{-2.4 \ m/s^2} =t$

$3.75 \ s=t$

It takes <u>3.75 seconds.</u>

4 0

3 years ago

An airplane is traveling at a fixed altitude with an outside wind factor. The airplane is headed N 40° W at a speed of 600 miles

bija089 [108]

Answer: $\vec{v_R}=\hat{i}[-329.11]+\hat{j}[516.18]$

Explanation:

Given

Plane is initially flying with velocity of magnitude $v=600\ mph$

at angle of $40^{\circ}$ with North towards west

Velocity of plane airplane can be written as

$v_a=600(-\sin 40\hat{i}+\cos 40\hat{j})$

Now wind is encountered with speed of $v=80\ mph$ at angle of $N45^{\circ}E$

$v_w=80(\cos 45\hat{i}+\sin 45\hat{j})$

resultant velocity

$\vec{v_R}=\vec{v_a} +\vec{v_w}$

$\vec{v_R}=600(-\sin 40\hat{i}+\cos 40\hat{j})+ 80(\cos 45\hat{i}+\sin 45\hat{j})$

$\vec{v_R}=\hat{i}[-385.67+56.56]+\hat{j}[459.62+56.56]$

$\vec{v_R}=\hat{i}[-329.11]+\hat{j}[516.18]$

for direction $\tan \theta =\frac{516.18}{329.11}$

$\tan \theta =1.568$

$\theta =57.47^{\circ}$ west of North

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zepelin [54]

True