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4 years ago

How can you use fossils and geologic features to interpret the relative ages of rock layers

Physics

1 answer:

Salsk061 [2.6K]4 years ago

4 0

Answer:

Explanation:

Geologists follow three main simple rules in order to determine the relative age of rock layers. First and foremost, they use the law of superposition to determine the relative ages of sedimentary rock layers. According to the law of superposition, in horizontal sedimentary rock layers the oldest is at the bottom.

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What is the weight of a 225kg space probe on the moon and the acceleration of gravity on the moon is 1.62

olga nikolaevna [1]

<span>364N should be your answer.. hope this helps

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7 0

3 years ago

Read 2 more answers

Damping is negligible for a 0.135-kg object hanging from a light, 6.30-N/m spring. A sinusoidal force with an amplitude of 1.70

IceJOKER [234]

Answer:

0.72 Hz minimum frequency

Explanation:

When the damping is negligible,Amplitude is given as

$A = (F/m)/[\sqrt{(\omega ^{^{2}}-\omega _{o}^{2}})^2$

here $\omega _{o}^{2}= k/m$ = (6.30)/(0.135) = 46.67 N/m kg

$F / mA$ = 1.70/(0.135)(0.480) = 26.2 N/m kg

From the above equation , rearranging for ω,

$\omega ^{2}= \omega _{o}^{2}\pm F/m$

⇒ ω² =46.67 ± 26.2 = 72.87 or 20.47

⇒ ω = 8.53 or 4.52 rad/s

Frequency = f

ω=2 π f

⇒ f = ω / 2π = 8.53 /6.28 or 4.52 / 6.28 = 1.36 Hz or 0.72 Hz

The lower frequency is 0.72 Hz and higher is 1.36 Hz

8 0

3 years ago

The energy consumed by a home during a month is 90 kWh, how many Joules are we talking about? a good explanation please is for t

andreev551 [17]

Answer:

3.24×10⁸ J, or 324 MJ

Explanation:

"kWh" is a kilowatt-hour. It's the energy used by 1 kilowatt of power after one hour.

A kilowatt is a kilojoule per second.

90 kWh

= 90 kW × 1 hr

= 90 kJ/s × 1 hr

= 90 kJ/s × 3600 s

= 324,000 kJ

= 324,000,000 J

The energy is 3.24×10⁸ J, or 324 megajoules.

3 0

3 years ago

I need help ASAP. This is for 15 points

kramer

Answer:

Latitude :

runs: east to west

measures : distances north and south of the equator

Longitude :

runs : north to south

measures : the distance east or west of the Prime Meridian

7 0

3 years ago

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

$I=\frac{1}{3}mL^2$

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0

3 years ago