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3 years ago

As best you can, describe how we can tell the age of the Earth and rocks found on it.

Physics

1 answer:

Viktor [21]3 years ago

4 0

Answer:

We can determine the age of the Earth and the rocks found on Earth with techniques such as measuring ice cores and digging and analysing fossils.

Explanation:

Hopefully this helped!

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You are designing a delivery ramp for crates containing exercise equipment. The 1470-N crates will move at 1.8 m/s at the top of

alina1380 [7]

Answer:

The force constant ,I = 2394N/m

Explanation:

Given:

Weight of crate,Wg = 1470N

Theta = 22.0°

Kinetic friction,Fk= Fs(max) = 550N

Total length of ramp =8.0m

If y =0 at the bottom of the ramp

y1 = d Sin theta

y1 = 8 × Sin 22°

y1 = 3.0m

y2 = 0

V1=1.8m/s

V2 = 0

The equation combining the gravitational and elastic potential energy, and the work energy theorem is given by:

K1 + Ugrav1 + Uel1 + W = K2 + Uel1 ..eq1

Where KE is given by: 1/2mv^2

Gravitational potential energy is given by: Ugrav = mgy ...eq2

Elastic potential energy = Uel = 1/2Kx^2 ..eq3

The restoring force constant of a spring compressed by a distance x is given by:

Fx = Kx ..eq4

Workdone by a constant force is given by W = Fscostheta ...eq5

Where s = displacement

Theta = the angle between the force and the displacement

Work,W = Wf = 550 ×8 × cos 180°

W = -4400J

From the weight of the crate.mass is:

Wg/g = m

1470/9.8 = 150kg

K1 = 1/2 ×150×1.8^2 = 243m/s

The crate comes to rest at K2=0

Ugrav1 = 150 × 9.8 × 3 = 4410J

Ugrav2 = 150 × 9.8 x 0 = 0J

Uel1= 0 Spring at equilibrium

Substituting the values of the energies and work

253 + 4410 + 0 - 4400 = 0 + 0 + Uel1

Uel1 = 253J

Substituting into eq3

253 = 1/2 Kx^2 = 1/2 Kx(x)

Kx = 506/x

Since crate remains at rest,we use Newton's 2nd law

Fx = fs + Wsin theta

Fx = 550 + 1470 sin 22

Fx = 1100.7N

Substituting into eq4

Kx = 1100.7

X = 506/1100.7 = 0.46m

Kx = 1100.7

K = 1100.7/0.47

K = 2394N/m

5 0

3 years ago

The position-time graph of an object is found to be a straight line passing through the origin. What information about the motio

Nostrana [21]

-- The object either left or crossed the starting line exactly at time=0 .

-- The object has been traveling at constant speed for all time that
we know about.

5 0

3 years ago

TRANSFORMA: 765 mm Hg a atm

-Dominant- [34]

Presión

765

ATMÓSFERA

1,00658

Answer:

Explanation:

5 0

3 years ago

You attach a 2.90 kg mass to a horizontal spring that is fixed at one end. You pull the mass until the spring is stretched by 0.

Murljashka [212]

Answer:

Explanation:

The spring is stretched by .5 m and then released that means its amplitude of oscillation A is 0.5 m .

A = 0.5 m

After the release at one extreme point , the mass comes to rest again at another extreme point after half the time period ie

T / 2 = .3 s

T = 0.6 s

Angular velocity

ω = $\frac{2\times \pi}{T}$

ω = $\frac{2\times \pi}{0.6}$

ω = 10.45

Maximum velocity = ω A

ω and A are angular velocity and amplitude of oscillation.

Maximum velocity = 10.45 x .5

= 5.23 m /s

7 0

3 years ago

For years, the tallest tower in the United States was the Phoenix Shot Tower in Baltimore, Maryland. The shot tower was used fro

Mariulka [41]

Answer:

The velocity of the droplet right before it hits the ground is 40.08 m/s.

Explanation:

To determine the velocity of the droplet right before it hits the ground,

From one of the equations of kinematic for free fall motions,

v = u + gt

Where v is the final velocity

u is the initial velocity

g is acceleration due to gravity (take g = 9.8 m/s²)

and t is time

For the question, v is the velocity of the droplet right before it hits the ground.

u = 0 m/s (Since the molten lead was dropped from rest)

Therefore,

v = gt

First, we will determine the time t

Also, from one of the equations of kinematic for free fall motions,

h = ut + 1/2(gt²)

u = 0 m/s

From the question, the molten lead was dropped from the top of the 82.15 m tall tower, therefore

h = 82.15

Hence,

82.15 = 0×t + 1/2 (9.8 × t²)

82.15 = 1/2 (9.8 × t²)

82.15 = 4.9 t²

t² = 82.15/4.9

∴ t = 4.09 secs

Now, for the velocity v, of the droplet right before it hits the ground,

Recall

v = gt

Then,

v = 9.8 × 4.09

v = 40.08 m/s

Hence, the velocity of the droplet right before it hits the ground is 40.08 m/s.

5 0

3 years ago