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uranmaximum [27]
3 years ago
12

What processes were involved in the formation of the model? How is the model similar to the real sedimentary sample? How is the

model different from the real sedimentary sample?

Physics
1 answer:
Stels [109]3 years ago
8 0

Answer:

1. sediment layering and compacting on top of each other and solidifying

2. sediment layering and compacting (not as much) on top of each other and solidifying, just not as much

Explanation:

hope this helps! :))

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A 20-cm-long spring is attached to a wall. When pulled horizontally with a force of 100 N, the spring stretches to a length of 2
Delvig [45]

Answer:

the length of stretched spring  in cm is 22

Explanation:

given information:

spring length, x1 = 20 cm = 0.2 m

force, F = 100 N

the length of spring streches, x2 = 22 cm = 0.22 m

According to Hooke's law

F = - kΔx

k = F/*=(x2-x1)

  = 100/(0.22 - 0.20)

  = 5000 N/m

if the spring is now suspended from a hook and a 10.2-kg block is attached to the bottom end

m = 10.2 kg

W = m g

    = 10.2 x 9.8

    = 99.96 N

F = - k Δx

Δx = F / k

     = 99.96 / 5000

     = 0.02

Δx = x2- x1

x2 = Δx + x1

    = 0.20 + 0.02

    = 0.22 m

     = 22 cm

7 0
3 years ago
What is the mass of a cannonball if the force a force of 2500 N gives the cannonball an acceleration of 200 m/s^2??
vampirchik [111]

The answer is a.12.5kg because i just did the test and it was correct.

hope this helps


5 0
3 years ago
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g
7nadin3 [17]

Answer:

(a) r = 1.062·R_E = \frac{531}{500} R_E

(b) r = \frac{33}{25} R_E

(c) Zero

Explanation:

Here we have escape velocity v_e given by

v_e =\sqrt{\frac{2GM}{R_E} } and the maximum height given by

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}

Therefore, when the initial speed is 0.241v_e we have

v = 0.241\times \sqrt{\frac{2GM}{R_E} } so that;

v² = 0.058081\times {\frac{2GM}{R_E} }

v² = {\frac{0.116162\times GM}{R_E} }

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r} is then

\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}

Which gives

-\frac{0.941919}{R_E} = -\frac{1}{r} or

r = 1.062·R_E

(b) Here we have

K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m  \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}

Therefore we put  \frac{0.241GM}{R_E} in the maximum height equation to get

\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}

From which we get

r = 1.32·R_E

(c) The we have the least initial mechanical energy, ME given by

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

3 0
3 years ago
Read 2 more answers
You have a spring-loaded air rifle. When it is loaded, the spring is compressed 0.3 m and has a spring constant of 150 N/m. In j
Feliz [49]

The potential energy of the spring is 6.75 J

The elastic potential energy stored in the spring is given by the equation:

E= \frac{1}{2} kx^2

where;

k is the spring constant

x is the compression/stretching of the string

In this problem, we have the spring as follows:

k = 150 N/m is the spring constant

x = 0.3 m is the compression

Substituting in the equation, we get

E=\frac{1}{2} (150) (0.3)^2

E=6.75J

Therefore. the elastic potential energy stored in the spring is 6.75J .

Learn more about potential energy here:

brainly.com/question/10770261

#SPJ4

6 0
1 year ago
Standing at a crosswalk, you hear a frequency of 550 Hz from the siren of an approaching ambulance. After the ambulance passes,
FromTheMoon [43]

There are six steps to this process , I uploaded step one and as you can see you can get all six on Quizlet:). Good luck

6 0
3 years ago
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