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4 years ago

What information can scientists obtian from tree rings

Physics

1 answer:

Alexandra [31]4 years ago

5 0

What information can scientists obtain from tree rings?

Answer's <u>I chose</u>:

<h3>how narrow the rings are</h3><h3>how the climate changed in the tree’s life</h3><h3>how wide the rings are</h3>

Please <u>correct</u> me if there are <em>more </em>or <em>less</em>

Please give a brainliest and a thanks.

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In the picture below identify the type of wave in the top picture. *

Lerok [7]

Answer; transverse wave
Explanation;

In transverse waves the particles vibrate perpendicular to the wave's motion.

4 0

3 years ago

the velocity of a curling stone is due east. what will be the direction of the momentum of the curling stone as it moves?

kirill [66]

East.
momentum = mass x velocity
the direction of the momentum is the same as that of velocity.

8 0

3 years ago

Read 2 more answers

A machine shop worker reports the mass of an aluminum cube as 176 g. If one side of the cube measures 4 cm, what is the density

Zarrin [17]

-- Since it's a cube, its length, width, and height are all the same 4 cm .

-- Its volume is (length x width x height) = 64 cm³ .

-- Density = (mass) / (volume)

= (176 g) / (64 cm³)

= 2.75 gm/cm³ .

6 0

3 years ago

Consider a metal bar of initial length L and cross-sectional area A. The Young's modulus of the material of the bar is Y. Find t

RUDIKE [14]

Answer:

Spring constant = YA / L

Explanation:

Let F be the force being applied on cross sectional area A of metal bar due to which an extension of ΔL is obtained in the wire of length L then

stress = F / A

strain = ΔL /L

Young's modulus = ( F / A ) / (ΔL /L)

Y = ( F L / A ΔL )

( F / ΔL ) = ( YA / L )

Spring constant = ( F / ΔL )

Spring constant = YA / L

6 0

3 years ago

A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh

irina1246 [14]

(a) Zero

When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.

This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

$K=\frac{1}{2}mv^2$

where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

$U=mgh$

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

$a=\frac{v-u}{t}$

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

$u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s$

and the positive sign means it points upward.

(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

$\Delta v = v -u$

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

$\Delta v = 0 - (+9.8 m/s)=-9.8 m/s$

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

$a=\frac{v-u}{t}$

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

$v = u+at = 0 +(-9.8 m/s^2)(1 s)= -9.8 m/s$

and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

$\Delta v = v -u$

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

$\Delta v = -9.8 m/s - 0=-9.8 m/s$

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

$\Delta v = v -u$

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

$\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s$

(g) -9.8 m/s^2

There is always one force acting on the ball during the motion: the force of gravity, which is given by

$F=mg$

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

$mg = ma$

which means that the acceleration is

$a= g = -9.8 m/s^2$

and the negative sign means it points downward.

7 0

4 years ago