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2 years ago

What are the building blocks of minerals? (1 point)?

1 answer:

8 0

Minerals are composed mainly of "elements". Elements are mainly the basic building blocks of nearly everything around us.
Minerals are defined as substances that are found naturally in Earth.
Minerals have definite structures and compositions.
Currently, there are over 3000 known minerals.

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A stationary 6-kg shell explodes into three pieces. One 4.0 kg piece moves horizontally along the negative x-axis. The other two

natta225 [31]

Answer:

-15 m/s

Explanation:

The computation of the velocity of the 4.0 kg fragment is shown below:

For this question, we use the correlation of the momentum along with horizontal x axis

Given that

Weight of stationary shell = 6 kg

Other two fragments each = 1.0 kg

Angle = 60

Speed = 60 m/s

Based on the above information, the velocity = v is

$1\times 60 \times cos\ 60 + 1\times 60 \times cos\ 60 - 4\ v = 0$

$\frac{60}{2} + \frac{60}{2} - 4\ v = 0$

$v = \frac{60}{4}$

= -15 m/s

4 0

2 years ago

The engine oil at 150 degree Celsius is cooled to 80 degree Celsius in a parallel flow heat exchanger by water entering at 25 de

Setler [38]

Answer:

Explanation:

Attach is the solution

6 0

2 years ago

A 20-turn coil of area 0.32 m2 is placed in a uniform magnetic field of 0.055 T so that the perpendicular to the plane of the co

makvit [3.9K]

Answer:

1.5 * 10^-2 Tm^2

Explanation:

Electric Flux = B.A cos(theta)

B = 0.055 T

A = 0.32 m^2

theta = 30

Electric Flux = (0.055 T).(0.32 m^2).Cos(30) = 0.0152 = 1.5 * 10^-2 Tm^2

5 0

1 year ago

An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra

slega [8]

Answer:

Part a)

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

$E = 4.4 \times 10^{-13} J$

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

$m_1v_1 + m_2v_2 + m_3v_3 = 0$

$(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0$

$(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0$

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

By equation of kinetic energy we have

$E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$

$E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}$

$E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}$

$E = 4.4 \times 10^{-13} J$

8 0

2 years ago

What conditions must be satisfied for a body to be in equilibrium <br>

dem82 [27]

For an object to be in equilibrium, it must be experiencing no acceleration. This means that both the net force and the net torque on the object must be zero.

5 0

2 years ago