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3 years ago

The terms intrusive and extrusive are used to describe which one of the three rock groups

Physics

1 answer:

tino4ka555 [31]3 years ago

8 0

Answer:

IGNEOUS ROCKS

Explanation: Igneous rocks are those rocks that solidify from magma.

Igneous rock is divided into two ,they are:

1. Intrusive

Igneous rocks crystallized belowearth"s crust. Its cooling material is called lava.

2 Extrusive igneous rock is also known as known as volcanic rocks

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A 1.0-kg ball has a velocity of 12 m/s downward just before it strikes the ground and bounces up with a velocity of 12 m/s upwar

Nezavi [6.7K]

Answer:

The change in momentum of the ball is 24 kg-m/s

Explanation:

It is given that,

Mass of the ball, m = 1 kg

Initial velocity of the ball, u = -12 m/s (in downwards)

Final velocity of the ball, v = +12 m/s (in upward)

We need to find the change in momentum of the ball.

Initial momentum of the ball, $p_i=mu=1\ kg\times (-12\ m/s)=-12\ kg-m/s$

Final momentum of the ball, $p_f=mv=1\ kg\times (12\ m/s)=12\ kg-m/s$

Change in momentum of the ball, $\Delta p=p_f-p_i$

$\Delta p=12-(-12)=24\ kg-m/s$

So, the change in momentum of the ball is 24 kg-m/s. Hence, this is the required solution.

3 0

3 years ago

The density of a certain type of plastic is 0.75 g/cm3 . If a sheet of this plastic is 10.0 m long, 1.0 m wide, and 1 cm thick,

Oksanka [162]

Answer:

7.5 g

Explanation:

8 0

2 years ago

Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water

Ksenya-84 [330]

To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

$v_2^2 = v_1^2 + 2gh$

Where,

$V_i =$ Velocity in each state

g= Gravity

h = Height

Our values are given as,

$A_1 = 2.4*10^{-4} m^2$

$v_1 = 0.8 m/s$

$h = 0.11m$

Replacing at the kinetic equation to find $V_2$ we have,

$v_2 = \sqrt{v_1^2 + 2gh}$

$v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}$

$v_2= 1.67 m/s$

Applying the concepts of continuity,

$A_1v_1 = A_2v_2$

We need to find A_2 then,

$A_2= \frac{A_1v_1 }{v_2}$

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

$A_2= \frac{A_1v_1 }{v_2}$

$A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}$

$A_2= 1.14*10^{-4} m2$

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is $1.14*10^{-4} m2$

8 0

3 years ago

PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (47pts)

Reptile [31]

Answer:

can I have points, i don't have any money

5 0

3 years ago

Read 2 more answers

3. Jeremy has a solution with a mass of 228 g. He knows that the mass of the solvent was 224 g. How much is the mass of the solu

mariarad [96]

228 - 224 = 4

there is 4g of solute in the solution.

3 0

2 years ago