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3 years ago

12

Why is it important to include variety in your workouts

1 answer:

grigory [225]3 years ago

4 0

Explanation:

Variety in to an exercise program allows the body to adapt to many demands, from high intensity exercise to slow, steady state exercise. Variety is important in a fitness program because it allows for your body to be challenged on a consistent basis and to overcome a plateau.

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Potential energy

AnnZ [28]

The answer would be B.

7 0

3 years ago

A ball whirls around in a vertical circle at the end of a string. The other end of the string is fixed at the center of the circ

never [62]

Answer:

$T_b=m(g+\frac{v_b^2}{R})$

Explanation:

At the bottom the tension would be upwards and the weight downwards, their difference being the centripetal force. Taking the upwards direction as positive we then have:

$T_b-mg=F_{cp}=ma_{cp}=m\frac{v_b^2}{R}$

where we have used the equation for centripetal acceleration. Thus we have:

$T_b=m(g+\frac{v_b^2}{R})$

4 0

3 years ago

Which items below describes specific kinds of producers.

Katen [24]

Answer:

grocery store owner

medical doctor

trial lawyer

3 0

3 years ago

True False Suppose I have a resistor of some resistance R. If I were to double the length and double the cross-sectional area of

skad [1K]

Explanation:

The resistance of a wire is given by :

$R=\rho\dfrac{l}{A}$

Where

$\rho$ is the resistivity of the wire

l = initial length of the wire

A = initial area of cross section

If length and the area of cross section of the wire is doubled then new length is l' and A', l' = 2 l and A' = 2 A

So, new resistance of the wire is given by :

$R'=\rho\dfrac{l'}{A'}$

$R'=\rho\dfrac{l}{A}$

R' = R

So, the resistance of the wire remains the same on doubling the length and the area of wire.

4 0

3 years ago

A voltaic cell is constructed with two Zn2+-Zn electrodes, where the half-reaction is Zn2+ + 2e− → Zn (s) E° = -0.763 V The conc

wariber [46]

Answer : The cell emf for this cell is 0.077 V

Solution :

The balanced cell reaction will be,

Oxidation half reaction (anode): $Zn(s)\rightarrow Zn^{2+}+2e^-$

Reduction half reaction (cathode): $Zn^{2+}+2e^-\rightarrow Zn(s)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}{diluted}}{[Zn^{2+}{concentrated}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

$[Zn^{2+}{diluted}]$ = 0.0111 M

$[Zn^{2+}{concentrated}]$ = 4.50 M

Now put all the given values in the above equation, we get:

$E_{cell}=0-\frac{0.0592}{2}\log \frac{0.0111M}{4.50M}$

$E_{cell}=0.077V$

Therefore, the cell emf for this cell is 0.077 V

7 0

4 years ago