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guajiro [1.7K]
2 years ago
9

Once a metamorphic rock, always a metamorphic rock. Rocks cannot change from one type of a rock to another.

Physics
1 answer:
Gennadij [26K]2 years ago
5 0

Answer:

False

Explanation:

Metamorphic rock can melt into magma or weathers down (erosion) into sediments.

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The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are l
marshall27 [118]

Answer:

C)the right cord pulls on the pulley with greater force than the left cord

Explanation:

As we can see the figure that B is connected to the right string while A is connected to the left string

Now force equation for B as it will move downwards is given as

(2m)g - T_B = (2m)a

similarly for block A which will move upwards we can write the equation as

T_A - mg = ma

now we know that pulley is also rotating so the tangential acceleration of the rope at the contact point with pulley must be same as that of acceleration of the blocks

so here pulley will rotate clockwise direction

So tension in the right string must be more than the left string

So correct answer will be

C)the right cord pulls on the pulley with greater force than the left cord

3 0
2 years ago
1. Which letter represents the location of the battery in this diagram?
My name is Ann [436]

Answer:

location of battery in this diagram is at A and location of switch is at B.

4 0
3 years ago
Read 2 more answers
A rocket rises vertically, from rest, with an acceleration of 3.2 m/s2 until it runs out of fuel at an altitude of 1300 m. After
svetlana [45]

Explanation:

Below is an attachment containing the solution.

6 0
3 years ago
Two liquids, A and B, have equal masses and equal initial temperatures. Each is heated for the same length of time over identica
DochEvi [55]

Answer:

So the specific heat of the liquid B is greater than that of A.

Explanation:

Liquid A is hotter than the liquid B after both the liquids are heated identically for the same duration of time from the same initial temperature then according to heat equation,

Q=m.c.\Delta T

where:

m = mass of the body

c = specific heat of the body

\Delta T= change in temperature of the body

The identical heat source supplies the heat for the same amount of time then the quantity of heat supplied is also equal.

So for constant heat, constant mass the temperature change is inversely proportional to the specific of heat of the liquid.

\Delta T=\frac{Q}{m} \times \frac{1}{c}

\Delta T\propto\frac{1}{c}

So the specific heat of the liquid B is greater than that of A.

5 0
2 years ago
Derive the formula for the moment of inertia of a uniform, flat, rectangular plate of dimensions l and w, about an axis through
Ad libitum [116K]

Answer:

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in (Figure). We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

λ = m/l (orm) = λl

If we take the differential of each side of this equation, we find

d m = d ( λ l ) = λ ( d l )

since  

λ

is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact,  

d l = d x

in this situation. We can therefore write  

d m = λ ( d x )

, giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain

I=∫r2dm=∫x2dm=∫x2λdx.

The last step is to be careful about our limits of integration. The rod extends from x=−L/2x=−L/2 to x=L/2x=L/2, since the axis is in the middle of the rod at x=0x=0. This gives us

I=L/2∫−L/2x2λdx=λx33|L/2−L/2=λ(13)[(L2)3−(−L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.

4 0
2 years ago
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