The stomach lining is made up of deep muscular grooves.How might these structures help the stomach to break down food?

Two metal disks are welded together and are mounted on a frictionless axis through their common centers. One disk has a radius R

Airida [17]

The Inertia is 22. 488 kg. m² and the speed just before it hits the ground is 6. 4 m/s

<h3>How to determine the inertia</h3>

Using the formula:

I = 1/2 M₁R₁² + 1/2 M₂R₂²

Where I = Inertia

I = 1/2 * 0.810* (2. 60)² + 1/2 * 1. 58 * (5)²

I = 1/2 * 5. 476 + 1/2 * 39. 5

I = 2. 738 + 19. 75

I = 22. 488 kg. m²

To determine the block's speed, use the formula

v = $\sqrt{2gh}$

v = $\sqrt{2* 10 * 2. 10}$

v = $\sqrt{42}$

v = 6. 4 m/s

Therefore, the Inertia is 22. 488 kg. m² and the speed just before it hits the ground is 6. 4 m/s

Learn more about law of inertia here:

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7 0

2 years ago

During takeoff, an airplane goes from 0 to 56 m/s in 9 s. How fast is it going after 4 s?

meriva

24 miles per second
56/9=6
6+6+6+6=24

8 0

3 years ago

A boxer can hit a heavy bag with great force. Why can't he hit a piece of tissue paper in midair with the same amount of force?

12345 [234]

Answer:

This is due to impulse

Explanation:

Impulse equal to mΔv and FΔt

You can set these equal as mΔv = FΔt

When a boxer punches a tissue, it is like punching a cushion or a pillow. The time that the hit takes is much grater than if they were to hit something solid. In addition, the change in velocity of the boxer's arm would be much greater when they hit a punching bag. In this equation, the greater the time, the less force that is needed.

6 0

3 years ago

Can a goalkeeper at his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will

zmey [24]

The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

Explanation:

Consider the vertical motion of ball,

We have equation of motion v = u + at

Initial velocity, u = u sin θ

Final velocity, v = 0 m/s

Acceleration = -g

Substituting

v = u + at

0 = u sin θ - g t

$t=\frac{usin\theta }{g}$

This is the time of flight.

Consider the horizontal motion of ball,

Initial velocity, u = u cos θ

Acceleration, a =0 m/s²

Time, $t=\frac{usin\theta }{g}$

Substituting

s = ut + 0.5 at²

$s=ucos\theta \times \frac{usin\theta }{g}+0.5\times 0\times (\frac{usin\theta }{g})^2\\\\s=\frac{u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{2g}$

This is the range.

In this problem

u = 30 m/s

g = 9.81 m/s²

θ = 45° - For maximum range

Substituting

$s=\frac{30^2\times sin(2\times 45)}{2\times 9.81}=45.87m$

Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.

So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

6 0

3 years ago

A loaded barge has a mass of 1 500 000 kg and is traveling at 3 m/s. If a tugboat applies an opposing force of 12 000 N for 10 s

yan [13]

Answer:

Explanation:

Initial momentum is 1.5e6(3) = 4.5e6 kg•m/s

An impulse results in a change of momentum

The tug applied impulse is 12000(10) = 120000 N•s or 0.12e6 kg•m/s

The remaining momentum is 4.5e6 - 0.12e6 = 4.38e6 kg•m/s

The barge velocity is now 4.38e6 / 1.5e6 = 2.92 m/s

The tug applies 0.012e6 N•s of impulse each second.

The initial barge momentum will be zero in

t = 4.5e6 / 0.012e6 = 375 s or 6 minutes and 15 seconds

To stop the barge in one minute(60 s), the tug would have to apply

4.5e6 / 60 = 75000 N•s /s or 75 000 N

5 0

2 years ago