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4 years ago

Compared to a 1-kg block of solid iron, a 2-kg block of solid iron has twice as much

Physics

1 answer:

Tcecarenko [31]4 years ago

3 0

Answer:

D. All of the above.

Explanation:

Iron has a constant density, which means 2-kg block will have twice as much volume as 1-kg block; therefore, choice A is correct.

Inertia is defined by the equation F = ma: it measures how hard it is to change the motion of an object. The inertia of the the 1-kg solid iron is

F = 1a,

And the inertia of the 2-kg solid iron is

F = 2a,

which is twice as much that of the 1-kg block; therefore, choice B is correct.

The mass of the 2-kg block is twice as much as that of the 1-kg block; therefore, choice C is also correct.

Thus, all of the choices are correct (D).

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If you're swimming underwater and knock two rocks together, you will hear a very loud noise. But if your friend above the water

Svetradugi [14.3K]

Answer:

The air-water interface is an example of boundary. The transmitted portion of the initial wave energy is way smaller than the reflected portion. This makes the boundary wave hard to hear.

When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can travel directly to your ear.

Explanation:

The air-to-water sound wave transmission is inhibited because more of reflection than transmission of the wave occurs at the boundary. In the end, only about 30% of the sound wave eventually reaches underwater. For sound generated underwater, all the wave energy is transmitted directly to the observer. Sound wave travel faster in water than in air because, the molecules of water are more densely packed together, and hence can easily transmit their vibration to their neighboring molecules, when compared to air.

4 0

4 years ago

Are eggs a element, a compound, or a mixture

enyata [817]

Answer:

I'm gonna go with compund

5 0

3 years ago

Read 2 more answers

A ball is thrown upward and reaches a maximum height of 100m and then returns back to where it was launched. The displacement of

Natalka [10]

The correct answer is
A) 0 m

In fact, the displacement is a vectorial quantity that corresponds to the difference between the initial and the final position of an object in motion. In this problem, the final position of the ball is equal to its initial position (because the ball returns back to where it was launched), so the displacement of the ball is zero.

6 0

3 years ago

Read 2 more answers

A giant chorus of 1000 vocalists is singing the same note. Suddenly, 999 vocalists stop,leaving one soloist. By how many decibel

stiks02 [169]

Answer:

The decrease in decibels is 0.1 dB.

Explanation:

Let the intensity of one chorus is Io.

let the intensity of 1000 vocalist is dB.

The intensity of 1000 vocalist is 1000 Io.

$dB = 10log\frac{1000Io}{Io}=30$ ..... (1)

let the intensity of 999 vocalist is dB'.

$dB' = 10log\frac{999Io}{Io}=29.9$ ..... (2)

So, the change is

= dB - dB' = 30 - 29.9 = 0.1 dB

6 0

3 years ago

A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly

Mnenie [13.5K]

Answer:

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The acceleration of the tortoise is 0.28 m/s²

Explanation:

The equations that describe the position and velocity of the hare and the tortoise are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

To find the acceleration of the hare once it begins to slow down, we have to find how much time the hare traveled during the deceleration and what was its initial speed.

First, the hare moves with constant acceleration for 4.7 s. Then, its velocity at t = 4.7 s will be:

v = v0 + a · t (v0 = 0 because the hare starts form rest)

v = a · t = 0.9 m/s² · 4.7 s = 4.2 m/s

The distance traveled by the hare while accelerating can be calculated using the equation of the position:

x = x0 + v0 · t + 1/2 · a · t² (x0 = 0 and v0 = 0)

x = 1/2 · a · t² = 1/2 · 0.9 m/s² · (4.7)² = 9.9 m

Then, the hare runs at a constant speed of 4.2 m/s for 11.7 s. The distance traveled at constant speed will be:

x = v · t

x = 4.2 m/s · 11.7 s = 49.1 m

Then, the distance traveled by the hare while slowing down was:

Distance traveled while slowing down = 72 m - 49.1 m - 9.9 m = 13 m

Let´s find how much time it took the hare to come to stop, so we can calculate the acceleration. We know that when the position is 13 m, the velocity is 0.

v = v0 + a · t

0 = 4.2 m/s + a · t

-4.2 m/s / t = a

Replacing in the equation of the position:

x = v0 · t + 1/2 · a · t² (considering x0 as the point at which the hare started to slow down)

13 m = 4.2 m/s · t - 1/2 · 4.2 m/s / t · t²

13 m = 4.2 m/s · t - 2.1 m/s · t

13 m = 2.1 m/s · t

t = 13 m / 2.1 m/s

t = 6.2 s

Then, the acceleration of the hare while slowing down will be:

-v0/t = a

-4.2 m/s / 6.2 s = a

a = -0.68 m/s²

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The hare traveled 72 m in (6.2 s + 11.7 s + 4.7 s) 22.6 s. The tortoise reaches the final position of the hare at the same time, so, using the equation of the position we can calculate the acceleration of the tortoise:

x = x0 + v0 · t + 1/2 · a · t² (x0 = 0 and v0 = 0)

x = 1/2 · a · t²

72 m = 1/2 · a · (22.6 s)²

144 m / (22.6 s)² = a

a = 0.28 m/s²

The acceleration of the tortoise is 0.28 m/s²

6 0

3 years ago