Why are continental rocks much older than oceanic crust?A. Oceanic crust is continually recycled through convection in the earth's mantle
B. Oceanic crust is made out of much less dense material than continental crust
C. Continental crust is continually renewed through convection in the earth's mantle
D. Continental crust eats oceanic crust for breakfast

Answers

Answer 1
Answer:

Answer:

A. Oceanic crust is continually recycled through convection in the earth's mantle

Explanation:

The oceanic plate is constantly being recycled through the forces of convection within the earth's mantle.

New oceanic plate are formed mid-oceanic ridge for example. As the magma cools and solidifies, they are moved away continually.

This is not the case for the continental curst.


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The most interpersonal constructive passion response to relational conflict is..

Answers

Answer:

loyalty

Explanation:

A volumetric flask made of Pyrex is calibrated at 20.0°C. It is filled to the 150-mL mark with 34.5°C acetone. After the flask is filled, the acetone cools and the flask warms so that the combination of acetone and flask reaches a uniform temperature of 32.0°C. The combination is then cooled back to 20.0°C. (The average volume expansion coefficient of acetone is 1.50 10-4(°C)−1.) (a) What is the volume of the acetone when it cools to 20.0°C?

Answers

Answer:149.73 ml

Explanation:

Given

\beta \ of\ acetone=1.50* 10^(-4) ^(\circ)C^(-1)

change in volume is given by

\Delta V=V_(final)-V_(initial)

\Delta V=\nu_(initial)\beta _(acetone)\left [ T_f-T_i\right ]

V_(final)=\nu_(initial)+\nu_(initial)\beta _(acetone)\left [ T_f-T_i\right ]

V_(final)=150+150* 1.50* 10^(-4)\left [ 20-32\right ]

V_(final)=149.73 ml

Final answer:

The volume of the acetone when it cools to 20.0°C is approximately 142.39 mL.

Explanation:

In order to determine the volume of the acetone when it cools to 20.0°C, we can use the equation for the volume change caused by a temperature change at constant pressure, known as Charles's law. Charles's law states that the volume of a gas is directly proportional to its temperature in Kelvin. We can use the formula V2 = V1 * (T2 / T1) to calculate the volume of the acetone at the lower temperature.

Given that the initial volume of the acetone is 150 mL at a temperature of 34.5°C, we need to convert this temperature to Kelvin by adding 273.15. Therefore, T1 = 34.5°C + 273.15 = 307.65 K.

Since the final temperature is 20.0°C, the final temperature in Kelvin will be T2 = 20.0°C + 273.15 = 293.15 K. We can now plug these values into the equation to find the volume of the acetone at the lower temperature: V2 = 150 mL * (293.15 K / 307.65 K) = 142.39 mL.

Learn more about Volume calculation here:

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A 1.5v battery stores 4.5KJ of energy. How long can it light a flashlight bulb that draws 0.60A​

Answers

Answer:

The 1.5V battery can power the flashlight bulb drawing 0.60A for 83.33 minutes before it is depleted.

Explanation:

To determine how long a 1.5V battery can power a flashlight bulb drawing 0.60A, you can use the formula for calculating the energy (in joules) consumed by an electrical device over time:

Energy (Joules) = Power (Watts) × Time (Seconds)

In this case, the power (P) is given by the product of the voltage (V) and current (I):

Power (Watts) = Voltage (Volts) × Current (Amperes)

So, first, calculate the power consumption of the flashlight bulb:

Power (Watts) = 1.5V × 0.60A = 0.90 Watts

Now, you want to find out how long the battery can power the bulb, so rearrange the energy formula to solve for time:

Time (Seconds) = Energy (Joules) / Power (Watts)

Given that the battery stores 4.5 kJ (kilojoules), which is equivalent to 4,500 joules, and the power consumption is 0.90 watts:

Time (Seconds) = 4,500 J / 0.90 W = 5,000 seconds

Now, to express the time in more practical units, convert seconds to minutes:

Time (Minutes) = 5,000 seconds / 60 seconds/minute ≈ 83.33 minutes

So, the 1.5V battery can power the flashlight bulb drawing 0.60A for approximately 83.33 minutes before it is depleted.

Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 162 cm , but its circumference is decreasing at a constant rate of 14.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.500 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop. Find the magnitude of the emf EMF induced in the loop after exactly time 8.00s has passed since the circumference of the loop started to decrease.

Answers

Answer:

0.00124 V

Explanation:

Parameters given:

Initial circumference = 162 cm

Rate of decrease of circumference = 14 cm/s

Magnetic field, B = 0.5 T

Time, t = 8 secs

The magnitude of the EMF induced in the loop is given as:

V = (-NBA) / t

Where N = number of turns = 1

B = magnetic field

A = area of loop

t = time taken

First, we need to find the area of the loop.

To do this, we will find the radius after the loop circumference has decreased for 8 secs.

The rate of decrease of the circumference is 14 cm/s and 8 secs has passed, which means after 8 secs, it has decreased by:

14 * 8 = 112 cm

The new circumference is:

162 - 112 = 50 cm = 0.5 m

To get radius:

C = 2 * pi * r

r = C / (2 * pi)

r = 0.5 / (2 * 3.142)

r = 0.0796 m

The area is:

A = pi * r²

A = 3.142 * 0.0796²

A = 0.0199 m²

Therefore, the EMF induced is:

V = (-1 * 0.5 * 0.0199) / 8

V = -0.00124V

This is the EMF induced in the coil.

The magnitude is |-0.00124| V = 0.00124 V.

A 75 kg man starts climbing a ladder that leans against a wall. If the weight of the ladder is negligible, determine how far up the ladder the man can climb before the ladders starts to slip. The coefficient of friction on both surfaces is μS=0.25

Answers

The man can climb \bold { X (max) = 0.25* L* tan \alpha }, before  the ladders starts to slip.  

   

A - point at the top of the ladder  

B - point at the bottom of the ladder  

C - point where the man is positioned in the ladder  

L- the length of the ladder  

α - angle between ladder and ground  

x - distance between C and B  

 

The forces act on the ladder,  

Horizontal reaction force (T) of the wall against the ladder  

Vertical (upward) reaction force (R) of ground against the ladder.  

Frictionalhorizontal ( to the left ) force (F)  

Vertical( downwards) of the man,

mg = 75 Kg x 9.8 m/s² = 735 N  

in static conditions,  

∑Fx = T - F = 0                   Since,  T = F  

∑Fy = mg - R = 0                Since,  735 - R = 0, R = 735  

∑ Torques(b) = 0  

In point B the torque produced by forces R and F is Zero  

Then:  

∑Torques(b) = 0        

And the arm lever for each force,  

mg = 735  

   

Since, ∑Torques(b) = 0    

 \bold {735* x* cos\alpha  = F* L* sin\alpha    }     Since,T = F  

 

\bold {F = \frac {735* x* cos\apha }{L* sin\alpha }}      \bold {  \frac {cos \alpha } { sin\alpha }=  cot\alpha =\frac 1{tan\alpha}}  

\bold {F = \frac {735* x* cos\apha }{L }}    

\bold {F =  735* x* tan\alpha }}  

F < μR the ladder will starts slipping over the ground  

μ(s) = 0.25    

 

\bold { X (max) = 0.25* L* tan \alpha }

Therefore, the man can climb \bold { X (max) = 0.25* L* tan \alpha }, before  the ladders starts to slip. \

To know more about  Torque,

brainly.com/question/6855614  

Answer:

x (max) = 0,25*L*tanα

Explanation:

Letá call  

A: point at the top of the ladder

B: the point at the bottom of the ladder

C: point where the man is up the ladder

L the length of the ladder

α angle between ladder and ground

"x" distance between C and B

Description

The following forces act on the ladder

Point A: horizontal (to the right)  reaction (T) of the wall against the     ladder

Point B : Vertical (upwards) reaction (R)  of ground against the ladder

               frictional horizontal ( to the left ) force (F)

Point C : Weight (vertical downwards)) of the man mg

mg = 75 Kg * 9,8 m/s²       mg = 735 [N]

Then in static conditions:

∑Fx = T - F = 0    ⇒   T = F

∑Fy = mg - R = 0       ⇒   735 - R = 0     ⇒  R = 735

∑Torques(b) = 0

Note: In point B the torque produced by forces R and F are equal to 0

Then:

∑Torques(b) = 0      

And the arm lever for each force is:

mg = 735

d₁ for mg     and d₂  for T

cos α = d₁/x     then    d₁ = x*cosα

sin α  = d₂ / L   then    d₂ = L*sinα

Then:

∑Torques(b) = 0     ⇒   735*x*cosα  - T*L*sinα = 0

735*x*cosα =  T*L*sinα

T = F then       735*x*cosα = F*L*sinα

F = (735)*x*cosα/L*sinα         cos α / sinα = cotgα = 1/tanα

F = (735)*x*cotanα/L     or   F = (735)*x/L*tanα

When F < μ* R  the ladder will stars slippering over the ground

μ(s) = 0,25           0,25*R = 735*x/L*tanα

x   = 0,25*R*tanα*L/735

But R = mg = 735 then

0,25*L*tanα = x

Then  x (max) = 0,25*L*tanα

The radius of a typical human eardrum is about 4.15 mm. Calculate the energy per second received by an eardrum when it listens to sound that is at the threshold of hearing, assumed to be 1.20E-12 W/m2

Answers

The energy per second received by an eardrum is 6.4884 * 10^(-17) watt

Calculation of the energy per second;

The area should be

= \pi r^2\n\n= 3.14 * 0.00415m\n\n= 5.407 * 10^(-5)m^2

Now

The power should be

= 1.2 * 10^(-12) * 5.407 * 10^(-5)\n\n= 6.4884 * 10^(-17) watt

Learn more about the energy here: brainly.com/question/14338287

Answer:

Power energy per second will be equal to 6.4884* 10^(-17)watt

Explanation:

We have given radius of human eardrum r = 4.15 mm = 0.00415 m

Intensity at threshold of hearing I=1.2* 10^(-12)w/m^2

Area is given by A=\pi r^2=3.14* 0.00415^2=5.407* 10^(-5)m^2

We know that power is given by P=I* A=1.2* 10^(-12)* 5.407* 10^(-5)=6.4884* 10^(-17)watt

So power energy per second will be equal to 6.4884* 10^(-17)watt