Which is a chemical property of iron?

Describe what would happen if you rubbed a mineral with a Mohs hardness value of 7 against a mineral with a value of 5?

chubhunter [2.5K]

The mineral with Mohs hardness would be scratched because the mineral with Mohs 7 hardness is stronger than the Mohs 5 mineral. Eventually, that mineral would turn into dust if you kept rubbing it.

8 0

3 years ago

The half-life of Co-55 is 175 hours. How much of a 4000 g of Cobalt-55 sample would be left after 525 hours?

Harrizon [31]

We know that whatever amount we start with, half of it decays and forms atoms of other elements in 175 hours. So in order to figure out how much is left after 525 hours, we'll need to know how many half-lifes pass in that amount of time.

Well, (525 divided by 175) is exactly 3 half-lifes. So this will be easy.

-- After 1 half-life . . .

. . . . . 50% decays, 50% is still there.

-- After the 2nd half-life . . .

. . . . . (half of the leftover 50%) = another 25% decays, 25% is left.

-- After the 3rd half-life . . .

. . . . . (half of the leftover 25%) = another 12.5% decays, 12.5% is left.

12.5% of 4,000g = (0.125 x 4,000g) = 500 g .

============================================

Another way:

After 1 half-life, 1/2 is left.

After 2 half-lifes, 1/4 is left.

After 3 half-lifes, 1/8 is left.

1/8 of 4,000g = (4,000g/8) = 500 g .

5 0

3 years ago

An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and

Natasha_Volkova [10]

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

$\Delta P.E=mg(h_{2}-h_{1})$

$\Delta P.E=10000\times9.8\times(10000-0)$

$\Delta P.E=10000\times9.8\times10000$

$\Delta P.E=980000000\ J$

$\Delta P.E=980\ MJ$

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

$\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)$

$\Delta K.E=148298642\ J$

$\Delta K.E=148.3\ MJ$

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

7 0

3 years ago

In a coiled spring, the particles of the medium vibrate to and fro about their

Kitty [74]

Answer:

In a coiled spring, the particles of the medium vibrate to and fro about their mean positions at an angle of

A. 0° to the direction of propagation of wave

Explanation:

The waveform of a coiled spring is a longitudinal wave, which is made up of vibrations of the spring which are in the same direction as the direction of the wave's advancement

As the coiled spring experiences a compression force and is then released, it experiences a sequential movement of the wave of the compression that extends the length of the coiled spring which is then followed by a stretched section of the coiled spring in a repeatedly such that the direction of vibration of particles of the coiled is parallel to direction of motion of the wave

From which we have that the angle between the direction of vibration of the particles of the coiled spring and the direction of propagation of the wave is 0°.

8 0

3 years ago

A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of th

Yuri [45]

Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

$E=-\frac{d}{dt}(BACOS\alpha )\\$