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4 years ago

when a person uses an iron to remove the wrinkles from a shirt, why does heat travel from the iron to the shirt?

1 answer:

3 0

I think the answer to your question is that:
There is chemical energy being produced (by the water) and that causes the base of iron to absorb the heat. Then when the heated base of the iron is placed onto the shirt, the heat flattens the wrinkles on it.

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platform diving in the olympic games takes place at two heights: 5 meters and 10 meters. What is the velocity of a diver enterin

posledela

1) Velocity: 9.9 m/s and 14 m/s

The motion of the diver is a free-fall motion, so it is a uniform accelerated motion. Choosing downward as positive direction, the final velocity can be found by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u = 0 is the initial velocity (the diver starts from rest)

$a=g=9.8 m/s^2$ is the acceleration of gravity

s is the displacement

For the diver jumping from 5 m, s = 5 m, so

$v=\sqrt{2as}=\sqrt{2(9.8)(5)}=9.9 m/s$

For the diver jumping from 10 m, s = 10 m, so

$v=\sqrt{2as}=\sqrt{2(9.8)(10)}=14 m/s$

2) Time: 1.01 s and 1.43 s

The time of flight of each diver can be found by using the other suvat equation

$s=ut+\frac{1}{2}at^2$

And since u = 0, it can be reduced to

$s=\frac{1}{2}at^2$

For the diver jumping from 5 m, s = 5 m, so we find

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(5)}{9.8}}=1.01 s$

For the diver jumping from 10 m, s = 10 m, so we find

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(10)}{9.8}}=1.43 s$

5 0

3 years ago

Draw a distance over time graph of a dog that is tied to a 4 foot rope and travels in 1 complete circle

Soloha48 [4]

I think this is the answer. I hope you can understand.

6 0

3 years ago

The gravitational force between two objects is 2400 N. What will be the gravitational force between the objects if the mass of o

motikmotik

Answer:

4800N

Explanation:

Lets assume,

Mass of first object = m₁

Mass of second object = m₂

Distance between the two objects = r

Thus the force between the two objects will be

$F = \frac{G\times m_{1}\times m_{2}}{r^{2}}$

where, G = Universal gravitational constant

Given, F = 2400N

New mass of second object = 2m₂

Now, the force will be

$F_{2} = \frac{G\times m_{1}\times 2m_{2}}{r^{2}}$

$F_{2}= 2\frac{G\times m_{1}\times m_{2}}{r^{2}}$

$F_{2}= 2F$

$F_{2}= 2\times2400$

Thus, F₂ = 4800N

6 0

3 years ago

A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).

Ede4ka [16]

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c. The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω² + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh = 1/2Iω² + 1/2mv² = 1/2Iω² + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω² + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω² + 1/2mr²ω²

mgh = mr²ω²/5 + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

6 0

3 years ago

If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, what

charle [14.2K]

Answer:

She will make the mass of the smiley face twice as massive in order to keep the mobile in perfect balance.

Explanation:

mass of an object is directly proportional to the cube of its length. In this case the length is constant, the mass will also be constant for the smiley face, so that the mobile will be kept in perfect balance.

Therefore, If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, she will make the mass of the smiley face twice as massive in order to keep the mobile in perfect balance.

4 0

3 years ago