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DaniilM [7]
3 years ago
14

One important relationship between chemical and mechanical weathering is ________. 1. mechanical weathering forms more brittle m

aterials that are easily broken down by chemical weathering.2. mechanical weathering can produce smaller pieces of rock that have more surface area for chemical weathering to work on. 3. chemical weathering can produce smaller pieces of rock that can then be modified by mechanical weathering 4. chemical weathering can form brittle materials that are easily broken into smaller pieces by mechanical weathering
Physics
1 answer:
Paha777 [63]3 years ago
5 0

Answer:

2. mechanical weathering can produce smaller pieces of rock that have more surface area for chemical weathering to work Explanation:

Mechanical weathering involves activities of living organisms or some geological processes. The bigger rocks are usually reduced to smaller rocks and further reduction might be limited or not posibble mechanically. This reduced rocks now increases the surface area available for chemical weathering; which further reduces the sizes of the rocks below the size range of mechanical weathering. one will recall that the rate of chemical reaction increases with exposed surface area.

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12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L
Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

6 0
3 years ago
The energy delivered to the resistive coil is dissipated as heat at a rate equal to the power input of the circuit. However, not
Nuetrik [128]

Answer:

P(bat) = V²r/(R+r)²

Explanation:

Let the resistance of the coil be R

Internal resistance of the battery be r

Emf of the battery = V

Power dissipated in the internal resistance of the battery is normally given as P = I²r

where I is the current flowing in the circuit.

From Ohm's law,

V = I R(eq)

R(eq) = (R + r)

I = V/(R+r)

P = I²r

P = [V/(R+r)]²r

P = V²r/(R+r)²

Hope this Helps!!!

6 0
3 years ago
Please help I have no idea
DochEvi [55]

Answer:

b

Explanation:

bbbbbbbbbbbbbbvgh c tyvftj xf

3 0
3 years ago
A coin released at rest from the top of a tower hits the ground after falling 5.7 s. What is the speed of the coin as it hits th
Bad White [126]

Given parameters:

Initial velocity of Coin = 0m/s

Time taken before coin hits ground  = 5.7s

Unknown:

Final velocity of the coin  = ?

Velocity is displacement with time. To solve this problem, we have to apply one of the equations of motion.

The fitting one of them here is shown below;

             V = U + gt

where;

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

t is the time taken

Here we use positive value of acceleration due to gravity because the coin is falling with the effect of acceleration and not against it.

Now input the parameters and solve;

               V  = 0 + 9.81 x 5.7

               V = 55.917m/s

Therefore, the final velocity is 55.917m/s.

8 0
3 years ago
A block of mass m = 4.8 kg slides from left to right across a frictionless surface with a speed Vi=7.3m/s. It collides with a bl
nirvana33 [79]

Answer:

The speed of the 11.5kg block after the collision is V≅4.1 m/s

Explanation:

ma= 4.8 kg

va1= 7.3 m/s

va2= - 2.5 m/s

mb= 11.5 kg

vb1= 0 m/s

vb2= ?

vb2= ( ma*va1 - ma*va2) / mb

vb2= 4.09 m/s ≅ 4.1 m/s

7 0
3 years ago
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