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DanielleElmas [232]

3 years ago

13

Forms when heat, pressure, or fluids act on igneous, sedimentary, or other metamorphic rock to change its form or composition, o

r both

1 answer:

BARSIC [14]3 years ago

4 0

I think its a metamorphic rock because i know 100% that a metamophic rocks are created by heat and pressure

You might be interested in

Hull (1943) had rats push a lever that required 21 grams of force to budge. After they had learned to push the lever in order to

Zina [86]

Years of research have demonstrated that rats are intelligent creatures who experience pain and pleasure, care about one another, are able to read the emotions of others, and would assist other rats, even at their own expense.

<h3>Experiments:</h3>

In trials carried out at Brown University in the 1950s, rats were trained to press a lever for food, but they stopped pressing the lever when they noticed that with each press, a rat in an adjacent cage would scream in pain (after experiencing an electric shock).

Rats were trained to press a lever to lower a block that was hanging from a hoist by electric shocks administered by experimenters. A rat was subsequently hoisted into a harness by the experimenters, and according to their notes, "This animal normally shrieked and wriggled sufficiently while dangling, and if it did not, it was jabbed with a sharp pencil until it exhibited indications of discomfort." Even if it wasn't in danger of receiving a shock, a rat watching the scenario from the floor would pull a lever to lower the hapless rodent to safety.

Learn more about experiments on rats here:

brainly.com/question/13625715

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6 0

2 years ago

A 0.46-kg cord is stretched between two supports, 7.2 m apart. When one support is struck by a hammer, a transverse wave travels

Kryger [21]

Answer:

T = 6.0 N

Explanation:

given,

mass of the cord = 0.46 Kg

length of the supports = 7.2 m

time taken to travel = 0.74 s

tension in the chord = ?

using formula for tension calculation

$T = \dfrac{v^2.m}{l}$

$v = \dfrac{l}{s}$

$v = \dfrac{7.2}{0.74}$

v = 9.73 m/s

now, calculation of tension

$T = \dfrac{9.73^2\times 0.46}{7.2}$

T = 6.0 N

The tension in the cord is equal to 6.0 N.

7 0

3 years ago

The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the st

Aleonysh [2.5K]

Answer:

$a = 5.83 \times 10^{-4} m/s^2$

Explanation:

Since the system is in international space station

so here we can say that net force on the system is zero here

so Force by the astronaut on the space station = Force due to space station on boy

so here we know that

mass of boy = 70 kg

acceleration of boy = $1.50 m/s^2$

now we know that

$F = ma$

$F = 70(1.50) = 105 N$

now for the space station will be same as above force

$F = ma$

$105 = 1.8 \times 10^5 (a)$

$a = \frac{105}{1.8 \times 10^5}$

$a = 5.83 \times 10^{-4} m/s^2$

3 0

3 years ago

A point charge q1=2.0μC is located on the positive y axis at y=0.30m, and an identical charge q2 is at the origin. Find the magn

dalvyx [7]

Answer:

(A) 0.279N at angle 38.02°

(B) 0.701N

(C) 14.19°

Explanation:

(A) The net force on q3 is given as:

F = Fxi + Fyj

Fx is the x component of the force

Fy is the y component of the force

Fx = -F(1, 3)cos(90 - x) + F(2, 3)cos0

Fy = -F(2, 3)cosx - F(2, 3)cos90 = -F(2, 3)cosx

First let us find y and angle x from the diagram.

Using Pythagoras theorem,

y² = 0.3² + 0.4²

y² = 0.25

y = 0.5m

Using SOHCAHTOA to find x,

sinx = 0.4/0.5

x = 53.13°

Electrostatic force, F is given as:

F = kqQ/r²

Where k = Coulumbs constant

F(1,3) = (k*q1*q3) / r²

F(1, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.5²)

F(1, 3) = 0.288N

F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

F(2, 3) = 0.45N

Therefore,

Fx = -0.288cos36.87 + 0.45

Fx = 0.22N

Fy = 0.288cos53.13

Fy = 0.172N

=> F = 0.22i + 0.172j

The magnitude of the force will be

F(mag) = √(0.22² + 0.172²)

F(mag) = 0.279N

The direction of the force makes will be

tanθ = Fy/Fx

tanθ = 0.172/0.22 = 0.781

θ = 38.02° to the x axis.

(B) q2 = - 2.0 * 10^(-6)

This implies that:

F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * -2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

F(2, 3) = -0.45N

Therefore,

Fx = -0.288cos36.87 - 0.45

Fx = -0.68N

Fy = 0.172N

=> F = - 0.68i + 0.172j

The magnitude of the force will be

F(mag) = √((-0.68)² + 0.172²)

F(mag) = 0.701N

(C) The direction of the force makes will be

tanθ = 0.172/0.68

θ = 14.19° to the x axis

4 0

3 years ago

Read 2 more answers

which is not part of maturity? excising self control, using good judgement, making impulsive decisions, being dependable

IgorLugansk [536]

Answer:

using good judgement

Explanation:

7 0

2 years ago

Read 2 more answers