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2 years ago

12

PLEASE HELP!! Lab: Electromagnetic Induction

2 answers:

grigory [225]2 years ago

8 0

Answer:

I think this is the answer

Explanation:

Advocard [28]2 years ago

5 0

I am not sure what this is

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Choose one of these three people and write a letter to them to express your concerns about the events of 01/06/21

Mashcka [7]

Bro I didn’t put nothing

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3 years ago

It is 2058 and you are taking your grandchildren to Mars. At an elevation of 34.7 km above the surface of Mars, your spacecraft

Paul [167]

Answer: $1.23\ m/s^2$

Explanation:

Given

At an elevation of $y=34.7\ km$ , spacecraft is dropping vertically at a speed of $u=293\ m/s$

Final velocity of the spacecraft is $v=0$

using equation of motion i.e. $v^2-u^2=2as$

Insert the values

$\Rightarrow 0-(293)^2=2\times a\times (34.7\times 10^3)\\\\\Rightarrow a=-\dfrac{293^2}{2\times 34.7\times 10^3}\\\\\Rightarrow a=-1.23\ m/s^2$

Therefore, magnitude of acceleration is $1.23\ m/s^2$ .

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3 years ago

The planet's hawks and block their near each other in the door again system the dworkin's have very advanced technology and a do

Virty [35]

Answer: A,B, and E

Explanation: Just checked I got them right:)

4 0

3 years ago

at which plate boundary would you expect to see volcanoes A.divergent b.all of the above c.convergent D.transform

natulia [17]

A or d would be most likely it

7 0

3 years ago

A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s

Naddika [18.5K]

Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

A)

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is: $w_{c} =m_{c} g$ and maximum tension is $T=2.50 m_{c} g=1.41*10^4N$

total mass and weight is :

$M =m_{c}+ m_{b} =740kg+550kg=1290 kg$

$w_{M} =1.2650*10^4N$

∑ $F_{y} =ma_{y}$

$T-M_{g} =Ma_{y}$

$a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg$

$=0.645m/s^2$

B)

maximum acceleration

$a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0$

using $y-y_{0} =v_{oy} t+1/2(a_{y} )t^2$

to solve for t

$t=\sqrt{2(y-y_{0} )/a_{y} }$

$t=\sqrt{2(119m)/0.645m/s^2} =19.20s$

6 0

3 years ago