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2 years ago

A series of processes on Earth’s surface and in the crust and mantle that slowly change rocks from one kind to another is called

1 answer:

6 0

A series of processes on Earth's surface and in the crust and in the mantle that slowly change rocks from one kind to another is called the rock cycle. This cycle is a group of changes. We all know that the rock has 3 kinds which are igneous rocks, the sedimentary rocks, and the metamorphic rocks. The igneous rocks in the rock cycle can change to the sedimentary rocks or to the metamorphic rocks. The sedimentary rocks can also change to the igneous rocks or to the metamorphic rocks. The metamorphic rocks can change to the igneous rocks or to the sedimentary rocks. All depends on the process it undergoes and the factors present to affect the change it undergoes.

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The 45-g arrow is launched so that it hits and embeds in a 1.40 kg block. The block hangs from strings. After the arrow joins th

worty [1.4K]

Question: How fast was the arrow moving before it joined the block?

Answer:

The arrow was moving at 15.9 m/s.

Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

$\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg$

where $m_a$ is the mass of the arrow, $m_b$ is the mass of the block, $\Delta H$ of the change in height of the block after the collision, and $v$ is the velocity of the arrow before it hit the block.

Solving for the velocity $v$ , we get:

$v = \sqrt{\frac{2(m_b+m_a)\Delta Hg}{m_a} }$

and we put in the numerical values

$m_a = 0.045kg$ ,

$m_b = 1.40kg,$

$\Delta H = 0.4m,$

$g= 9.8m/s^2$

and simplify to get:

$\boxed{ v= 15.9m/s}$

The arrow was moving at 15.9 m/s

6 0

3 years ago

The _______ is the temperature to which air must be cooled at constant pressure to reach saturation.

Dafna1 [17]

The answer is the dew point

8 0

3 years ago

Read 2 more answers

An unstrained horizontal spring has a length of 0.39 m and a spring constant of 350 N/m. Two small charged objects are attached

Vera_Pavlovna [14]

Answer:

A) The possible algebraic signs will either be both positive (+) or both negative (-) charged since the 2 objects are repelling each other to stretch the string.

B) Magnitude of charges = 1.206 × 10^(-6) C

Explanation:

We are given;

Spring constant;k = 350 N/m

Spring length;L = 0.39 m

Stretched length of spring;x = 0.022 m

A) The spring stretches by 0.022m. Therefore, the total force is (350 × 0.022) N = 7.7N. The charged objects will either be both positive (+) or both negative (-) charged since they are repelling each other to stretch the string.

B) Force (F) required to stretch spring is given by the formula;

F = kx

Thus:

F = (350 × 0.022)

F = 7.7 N

Now, if we assume point charges, then the distance (r) between them will be given as:

r = (0.39 + 0.022) = 0.412 m

Coulomb's Law has a formula:

F = k(q1×q2)/r²

where k is coulomb's constant = 8.99 × 10^(9) Nm²/C²

Making q1 × q2 the subject, we have;

(q1 × q2) = Fr²/k = 7.7 × 0.412²/(8.99 × 10^(9))

(q1 × q2) = 14.54 × 10^(-11) C

We are told that both charges are equal, thus; |q1| = |q2|

So;

q = √(14.54 × 10^(-11)) = 1.206 × 10^(-6) C

6 0

3 years ago

A speed-time graph is shown below:

VladimirAG [237]

Answer: $0.5 m/s^{2}$

Explanation:

Average acceleration $a_{ave}$ is the variation of velocity $\Delta V$ over a specified period of time $\Delta t$ :

$a_{ave}=\frac{\Delta V}{\Delta t}}$

Where:

$\Delta V=V_{f}-V_{o}$ being $V_{o}=0 cm/s$ the initial velocity and $V_{f}=4 cm/s$ the final velocity (according to the information given from the described graph)