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3 years ago

If a layer was deposited but does not appear in the rock record, what has occured

2 answers:

7 0

Erosion. Erosion is the action and movement of a surface process (e.g. water or wind) that removes and transports a material layer from one location to another. Once eroded, the deposited layer will not appear in the rock record.

nalin [4]3 years ago

5 0

I think you forgot to give the choices along with the question. I am answering the question based on my research and knowledge. If a layer was deposited but does not appear in the rock record, the thing that happened is erosion. I hope that this is the ans wer that has actually come to your desired help.

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Which is a standard unit of measurement

ruslelena [56]

There are some more like Time - second (s)
Amount of substance - mole (mole)
Electric current - ampere (A)
Temperature - kelvin (K)
Luminous intensity - candela (cd)

6 0

3 years ago

A ball is pushed into a spring-loaded launcher with a force of 20 N, which compresses the spring 0.08 m.

damaskus [11]

Answer:

0.8J

Explanation:

Given parameters:

Force = 20N

Compression = 0.08m

Unknown:

Spring constant = ?

Elastic potential energy = ?

Solution:

To solve this problem, we use the expression below:

F = k e

F is the force

k is the spring constant

e is the compression

20 = k x 0.08

k = 250N/m

Elastic potential energy;

EPE = $\frac{1}{2}$ k e² = $\frac{1}{2}$ x 250 x 0.08²

Elastic potential energy = 0.8J

3 0

2 years ago

A ball is thrown straight up with a speed of 30 m/s, and air resistance is negligible. How long does it take the ball to reach t

PolarNik [594]

Answer:

3 seconds

Explanation:

Applying,

v = u±gt................ Equation 1

Where v = final velocity, u = initial velocity, t = time, g = acceleration due to gravity.

From the question,

Given: v = 0 m/s ( at the maximum height), u = 30 m/s

Constant: g = -10 m/s

Substitute these values into equation 1

0 = 30-10t

10t = 30

t = 30/10

t = 3 seconds

6 0

3 years ago

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

4 0

3 years ago

While i was jogging, i wad able to run 5 miles in thirty minutes. from this information, we can determine my?

Llana [10]

You can only determine the speed since the only info we know is how much you ran in how long of a time.

7 0

3 years ago

Read 2 more answers