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vodomira [7]
3 years ago
8

Sedimentary rocks were initially deposited as? will give brainliest

Physics
2 answers:
viva [34]3 years ago
5 0

Answer:

B. horizontal layers

Explanation:

nlexa [21]3 years ago
3 0

Based on my notes the answer must be B. horizontal layers

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A grandfather clock is controlled by a swinging brass pendulum that is 1.2 m long at a temperature of 27°C. (a) What is the leng
stealth61 [152]

Answer:

L2 = 1.1994 m

the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m

Explanation:

Given;

Initial length L1 = 1.2m

Initial temperature T1 = 27°C

Final temperature T2 = 0.0°C

Linear expansion coefficient of brass x = 1.9 × 10^-5 /°C

The change i length ∆L;

∆L = L2 - L1

L2 = L1 + ∆L ...........1

∆L = xL1(∆T)

∆L = xL1(T2 - T1) ......2

Substituting the given values into equation 2;

∆L = 1.9 × 10^-5 /°C × 1.2m × (0 - 27)

∆L = 1.9 × 10^-5 /°C × 1.2m × (- 27)

∆L = -6.156 × 10^-4 m

From equation 1;

L2 = L1 + ∆L

Substituting the values;

L2 = 1.2 m + (- 6.156 × 10^-4 m)

L2 = 1.2 m - 6.156 × 10^-4 m

L2 = 1.1993844 m

L2 = 1.1994 m

the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m

3 0
3 years ago
As the distance between two objects decreases, what happens to the force of gravity between the two objects
KonstantinChe [14]
The force of gravity between them increases.
3 0
4 years ago
A body travels a distance of 20m in the 7th second and 24m in the 9th second. How much distance shall it travel in the 15th seco
DaniilM [7]

Answer:

<u>36 m</u>

Explanation:

We can consider this to be an AP.

Then,

  • a₇ = 20
  • a₉ = 24

<u>Subtract a₇ from a₉.</u>

  • a + 8d - a + 6d = 24 - 20
  • 2d = 4
  • d = 2

  • a + 6(2) = 20
  • a = 8

<u>Finding a₁₅</u>

  • a₁₅ = a + 14d
  • a₁₅ = 8 + 14(2)
  • a₁₅ = 8 + 28
  • a₁₅ = <u>36 m</u>
4 0
2 years ago
Read 2 more answers
4. Johnny exerts a 3.55 N rightward force on a 0.200-kg box to accelerate it across a low-friction track. If the total resistanc
muminat

a) 15.2 m/s^2

b) 1.96 N

c) 1.96 N

Explanation:

a)

To find the acceleration of the box, we apply Newton's second law of motion:

\sum F=ma

where

\sum F is the net force on the box

m is the mass of the box

a is its acceleration

Here we have to consider the horizontal direction, which is the one in which the box is moving. The net force is given by:

\sum F=3.55 N - 0.52 N=3.03 N

which is the difference between the forward force and the resistive force. Then we have

m = 0.200 kg (mass of the box)

Therefore, the acceleration of the box is:

a=\frac{\sum F}{m}=\frac{3.03}{0.200}=15.2 m/s^2

b)

The gravitational force (also called weight) is the force with which an object is pulled by the Earth towards the Earth's centre.

It is given by

F_g = mg

where

m is the mass of the object

g is the acceleration due to gravity

In this problem, we have:

m = 0.200 kg is the mass of the box

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

Therefore, the gravitational force on the box is:

F_g=(0.200)(9.8)=1.96 N

c)

The normal force is the force with which a surface pushes back on an object.

For an object lying on a flat surface, there are two forces acting along the vertical direction:

- The gravitational force, F_g, which pushes the object downward

- The normal force, N, which pushes the object upward

As the object is at rest, the vertical acceleration of the object is zero, therefore according to Newton's second law of motion, the net force must be zero:

\sum F=F_g-N=0

Which means that the normal force is equal to the gravitational force:

N=F_g

And so, for the box in this problem, the normal force is

N=1.96 N

6 0
4 years ago
Two children are pulling and pushing a 30.0 kg sled. The child pulling the sled is exerting a force of 12.0 N at a 45o angle. Th
irinina [24]
I'm not quite sure what happens to Fay so I didn't finish but hope it helps

5 0
3 years ago
Read 2 more answers
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