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2 years ago

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

Physics

1 answer:

tino4ka555 [31]2 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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Billiard ball A of mass mA = 0.125 kg moving with speed vA = 2.80 m/s strikes ball B, initially at rest, of mass mB = 0.140 kg .

lianna [129]

Answer:

V=1.309

β= -41.997

Explanation:

Law Newton's conservation motion

Axis x

$m_{1}*v_{x1}+m_{2}*v_{x2}=m_{1}*v_{fx1}+m_{2}*v_{fx2}\\v_{x1}=2.8\frac{m}{s}\\v_{x2}=0 \frac{m}{s}\\m_{1}=0.125kg\frac{m}{s}\\m_{2}=0.140kg\frac{m}{s}\\0.125kg*2.8\frac{m}{s}+0.14kg*0=0.125kg*2.10\frac{m}{s}*cos(30) +0.14kg*v_{fx2}\\0.35 \frac{kg*m}{s} =0.125kg*1.81\frac{m}{s}+0.14kg*v_{fx2}\\v_{fx2}=\frac{0.12\frac{kg*m}{s} }{0.14kg} \\v_{fx2}=0.876 \frac{m}{s}$

Axis y

$m_{1}*v_{y1}+m_{2}*v_{y2}=m_{1}*v_{fy1}+m_{2}*v_{fy2}$

$v_{y1} =0\\v_{y2} =0$

$0=m_{1}*v_{fy1} +m_{2}*v_{fy2} \\v_{fy2}=-\frac{m_{1}*v_{fy1} }{m_{2}}\\ v_{fy2}=-\frac{0.125kg*2.10*sen(30)\frac{m}{s}}{0.14kg}\\v_{fy2}= -0.937\frac{m}{s}$

So the velocity $v_{f2}$

$v_{f2}=\sqrt{v_{fx2}^{2} +v_{fy2}^{2} } \\v_{f2}=\sqrt{0.876^{2} +0.983^{2} } \\v_{f2}=1.309\frac{m}{s}$

The angle can be find using both velocity factors

$\alpha =tanx^{-1}*\frac{v_{fx2}}{v_{fy2}} \\\alpha =tanx^{-1}*\frac{0.876}{-0.973}\\ \alpha =tanx^{-1}*\\ \alpha =-41.997$

Check:

$m_{1}*v_{x1}+m_{2}*v_{x2}=m_{1}*v_{fx1}+m_{2}*v_{fx2}$

$0.125*2.80=0.125*2.1*cos(30)+0.14*1.03*cos(-41.997)\\0.35=0.227+0.107\\$

0.35≅0.3489

4 0

3 years ago

A block of density pb = 9.50 times 10^2 kg/m^3 floats face down in a fluid of density pt = 1.30 times 10^3 kg/m^3. The block has

Nookie1986 [14]

Answer:

The depth and acceleration are 0.1919291 ft and 3.61 m/s².

Explanation:

Given that,

Density of block $\rho_{b} =9.50\times10^2\ kg/m^3$

Density of fluid $\rho_{t} =1.30\times10^3\ kg/m^3$

We need to calculate the depth

Using balance equation

$mg=\rho g V$ ....(I)

We know that,

The density is

$\rho=\dfrac{m}{V}$

$m=\rh0\times V$

Put the value of m in equation (I)

$\rho_{b}\times V\times g=\rho_{t}\times g\times V$

$\rho_{b}\times A\times H\times g=\rho_{t}\times g\times A\times h$

$h=\dfrac{\rho_{b}H}{\rho_{t}}$

Put the value into the formula

$h=\dfrac{9.50\times10^2\times8.00\times10^{-2}}{1.30\times10^3}$

$h= 5.85\ cm$

$h=0.1919291\ ft$

We need to calculate the acceleration

Using formula of net force

$F_{net}=\rho_{t}\times g\times V- \rho_{b}\times g\times V$

$ma=\rho_{t}\times g\times V- \rho_{b}\times g\times V$

$\rho_{b}\times V\times a=\rho_{t}\times g\times V- \rho_{b}\times g\times V$

$a=(\dfrac{\rho_{t}}{\rho_{b}}-1)g$ ....(II)

Put the value in the equation (II)

$a=(\dfrac{1.30\times10^3}{9.50\times10^2}-1)\times9.8$

$a=3.61\ m/s^2$

Hence, The depth and acceleration are 0.1919291 ft and 3.61 m/s².

4 0

2 years ago

Which statement correctly describes gravity?

Advocard [28]

Answer:

a force that attracts matter to the earth

Explanation:

depends on where you are the gravity can be different in space there is no gravity on Earth there is , that's why when you jump you come back down

5 0

2 years ago

A moped has a mass of 120kg . It accelerates at a rate of 2 m/s^2 . What is the size of the resultant force acting on it ?

stiv31 [10]

Answer:

240N

Explanation:

The formula for force is F=mass×acceleration.

Therefore, Force= 120×2= 240N

7 0

3 years ago

A jogger hears a car alarm and decides to investigate. While running toward the car, she hears an alarm frequency of 872.10 Hz.

Nata [24]

Answer:

v = 4.18 m/s

Explanation:

given,

frequency of the alarm = 872.10 Hz

after passing car frequency she hear = 851.10 Hz

Speed of sound = 343 m/s

speed of the jogger = ?

speed of the

$v_f = \dfrac{872.10-851.10}{2}$

$v_f =10.5\ Hz$

v_o = 872.1 - 10.5

$V_0 = 861.6\ Hz$

The speed of jogger

$v = \dfrac{v_1 \times 343}{v_0}-343$

$v = \dfrac{872.1 \times 343}{861.6}-343$

v = 4.18 m/s

5 0

2 years ago