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Zina [86]
3 years ago
9

Visualize five horizontal sedimentary strata exposed in a cliff or canyon wall identified by consecutive numbers, 1 being the lo

west bed and 5 being the highest. Which of the following statements concerning the strata are true?
a. Bed 4 is older than bed 2.
b. Bed 3 is older than beds 2 and 4.
c. Bed 5 is the oldest.
d. Beds l and 3 are older than bed 4.
Physics
2 answers:
Grace [21]3 years ago
6 0

Answer:

Only the 4 th statement is true that is bed 1 and 3 are older than 4.

Explanation:

The 5 beds are numbered from 1 to 5 , 1 being the lowest and 5 being the topmost bed.

We are given 4 statements and we have to find out which all are true.

(<u>a)Bed 4 is older than bed 2 </u>

This is wrong because the lower beds are older than beds that are higher.

<u>(b)Bed 3 is older than beds 2 and 4</u>

This is also wrong because the 2 is older than 3

(<u>c)Bed 5 is the oldest </u>

This is wrong because bed 1 is the oldest

(<u>d)Beds 1 and 3 are older than 4</u>

This is true as lower beds are older

<u />

Vlad1618 [11]3 years ago
4 0

Answer:

bed 2 was deposited before bed 3

Explanation:

Which of the following conclusions can be made about the sedimentary layers?

bed 2 was deposited after bed 3

bed 1 is the youngest

bed 3 was deposited before bed 1

bed 2 was deposited before bed 3

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The sound intensity level from one solo flute is 70 dB. If 10 flutists standing close together play in unison, what will the sou
Masja [62]

Answer:

80 dB

Explanation:

I_0 = Threshold intensity = 10^{-12}\ W/m^2

I = Intensity of sound

\beta = Intensity level of sound = 70 dB

Intensity level of sound is given by

\beta=10log\dfrac{I}{I_0}\\\Rightarrow 70=10log\dfrac{I}{I_0}\\\Rightarrow \dfrac{70}{10}=log\dfrac{I}{10^{-12}}\\\Rightarrow 10^{\dfrac{70}{10}}=\dfrac{I}{10^{-12}}\\\Rightarrow \dfrac{I}{10^{-12}}=10^{\dfrac{70}{10}}\\\Rightarrow I=10^{7}\times 10^{-12}\\\Rightarrow I=10^{-5}\ W/m^2

If there are 10 flutes I=10\times 10^{-5}\ W/m^2

\beta=10log\dfrac{10\times 10^{-5}}{10^{-12}}\\\Rightarrow \beta=10log10^8\\\Rightarrow \beta=10\times 8\\\Rightarrow \beta=80\ dB

The sound intensity level is 80 dB

5 0
3 years ago
A 37.5 kg box initially at rest is pushed 4.05 m along a rough, horizontal floor with a constant applied horizontal force of 150
vodomira [7]

Answer:

a) 607.5 J

b) 160.531875 J

c)  0 J

d)  0 J

e) 2.925 m\s

Explanation:

The given data :-

  • Mass of the box ( m ) = 37.5 kg.
  • Displacement made by box ( x ) = 4.05 m.
  • Horizontal force ( F ) = 150 N.
  • The co-efficient of friction between box and floor ( μ ) = 0.3
  • Gravitational force ( N ) = m × g = 37.5 × 9.81 = 367.875

Solution:-

a) The work done by applied force ( W )

W = force applied × displacement = 150 × 4.05 = 607.5 J

b)  The increase in internal energy in the box-floor system due to friction.

Frictional force ( f ) = μ × N = 0.3 × 367.875 = 110.3625 N

Change in internal energy = change in kinetic energy.

ΔU = ( K.E )₂ - ( K.E )₁

Since the initial velocity is zero so the  ( K.E )₁ = 0  

ΔU = ( K.E )₂ = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

c) The work done by the normal force .

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

d)  The work done by the gravitational force.

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

e) The change in kinetic energy of the box

( K.E )₂ - ( K.E )₁ = ( K.E )₂ - 0 = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

f) The final speed of the box

( K.E )₂ = 160.531875 J = 0.5 × 37.5 × v²

v² = 8.56

v = 2.925 m\s.

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3 years ago
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zysi [14]

Answer:

The answer is C ..

EXPLANATION:-

The amount of energy transferred between objects as a result of a difference in temperature.

4 0
3 years ago
A bag of sugar has a mass of 2.2kg what is its weight on earth
NARA [144]
The earth's gravitational acceleration is 9.8 m/s^2, and force = mass * acceleration. Therefore, a bag of sugar weighing 2.2 kg on earth would weigh 2.2 * 9.8 = 21.56 Newtons
8 0
3 years ago
Read 2 more answers
In a skate park, you are trying to determine how to get the most speed at the bottom of the ramp. If the ramp is 4.5 m high, and
Oksanka [162]

Answer:

If you have an initial velocity of 1.5m/s and the ramp height is 4.5, and an extreme fart is initiated at the bottom of the ramp exerting an didditional upward force, and the smell of the fart causes you to begin flailing, and everyone running from you causes an additional wind force, then you would be rolling at a speed of 3 m/s given there is no friction. :)

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2 years ago
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