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3 years ago

5

A cheap effective and simple way to help body recover is to get adequate amounts of. ?

2 answers:

skelet666 [1.2K]3 years ago

3 0

Get adequate amounts of liquid

lara31 [8.8K]3 years ago

3 0

Answer:

liquid and water is wrong try sleep

Explanation:

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What is the fundamental unit for measuring mass in the metric system ?

goldenfox [79]

Kilogram(kg)
It's not the SI unit of mass in the metric system however.

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2 years ago

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A bag of cement weighing 325 N hangs in equilibrium from three wires. Two of the wires make angles of theta1=60.0 degrees and th

Murljashka [212]

The sketch of the system is: two strings, 1 and 2, are attached to the ceiling and to a third string, 3.The third string holds the bag of cement.

The free body diagram of the weight with the string 3, drives to the tension T3 = weihgt => T3 = 325 N

The other free body diagram is around the joint of the three strings.

In this case, you can do the horizontal forces equilibrium equation as:

T1* cos(60) - T2*cos(40) = 0

And the vertical forces equilibrium equation:

Ti sin(60) + T2 sin(40) = T3 = 325 N

Then you have two equations with two unknown variables, T1 and T2

0.5 T1 - 0.766 T2 = 0

0.866 T1 + 0.643T2 = 325

When you solve it you get, T1 = 252.8 N and T2 = 165 N

Answer: T1 = 252.8 N, T2 = 165N, and T3 = 325N

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3 years ago

Help !!! Pick all the apply

Sergio [31]

The answer is the Second one

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2 years ago

A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

Rate of change of magnetic field is $3.466\times 10^3T/sec$

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop $r=\frac{0.13}{2}=0.065m$

Length of the circular loop $L=2\pi r=2\times 3.14\times 0.065=0.4082m$

Wire is made up of diameter of 2.6 mm

So radius $r=\frac{2.6}{2}=1.3mm=0.0013m$

Cross sectional area of wire $A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2$

Resistivity of wire $\rho =2.18\times 10^{-8}m$

Resistance of wire $R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm$

Current is given i = 11 A

So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

Emf induced in the coil is $e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}$

$0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=3.466\times 10^3=T/sec$

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3 years ago

Which diagram represents deposition?

Vera_Pavlovna [14]

Your answer will be gas

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3 years ago