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2 years ago

7

Kristina works out seven days a week. Lately, she has been tired, and her body aches. If she is overtraining, which training pri

nciple should Kristina consider before continuing her program?
a. overload
b. recovery
c. transfer
d. variation

1 answer:

sertanlavr [38]2 years ago

8 0

Answer: Option (b) is the correct answer.

Explanation:

As Kristina is over training, therefore, there is pain in her muscles because human body also requires rest in order to work or function properly.

That is why, gym trainers suggest to give one day off in a week for your workout sessions so that body muscles should recover.

Thus, we can conclude that if Kristina is over training, then recovery training principle should Kristina consider before continuing her program.

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If I0 is the intensity of the unpolarized light incident on the first polarizer, and I1 and I2 denote the intensity of the light

e-lub [12.9K]

E or C 10 hope this helps

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2 years ago

A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.

Pie

#1

Volume of lead = 100 cm^3

density of lead = 11.34 g/cm^3

mass of the lead piece = density * volume

$m = 100 * 11.34 = 1134 g$

$m = 1.134 kg$

so its weight in air will be given as

$W = mg = 1.134* 9.8 = 11.11 N$

now the buoyant force on the lead is given by

$F_B = W - F_{net}$

$F_B = 11.11 - 11 = 0.11 N$

now as we know that

$F_B = \rho V g$

$0.11 = 1000* V * 9.8$

so by solving it we got

V = 11.22 cm^3

(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N

(iii) Buoyant force = 0.11 N

(iv)since the density of lead block is more than density of water so it will sink inside the water

#2

buoyant force on the lead block is balancing the weight of it

$F_B = W$

$\rho V g = W$

$13* 10^3 * V * 9.8 = 11.11$

$V = 87.2 cm^3$

(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight = 11.11 N

(iii) Buoyant force = 11.11 N

(iv) since the density of lead is less than the density of mercury so it will float inside mercury

#3

Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid

3 0

2 years ago

A car moving with constant acceleration covers the distance between two points 60 m apart in 6.0 s. Its speed as it passes the s

BlackZzzverrR [31]

Answer:

The speed in the first point is: 4.98m/s

The acceleration is: 1.67m/s^2

The prior distance from the first point is: 7.42m

Explanation:

For part a and b:

We have a system with two equations and two variables.

We have these data:

X = distance = 60m

t = time = 6.0s

Sf = Final speed = 15m/s

And We need to find:

So = Inicial speed

a = aceleration

We are going to use these equation:

$Sf^2=So^2+(2*a*x)$

$Sf=So+(a*t)$

We are going to put our data:

$(15m/s)^2=So^2+(2*a*60m)$

$15m/s=So+(a*6s)$

With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.

$\sqrt{(15m/s)^2-(2*a*60m)}=So$

$15m/s-(a*6s)=So$

$\sqrt{(15m/s)^2-(2*a*60m)}=15m/s-(a*6s)$

$[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}$

$(15m/s)^2-(2*a*60m)}=(15m/s)^{2}-2*(a*6s)*(15m/s)+(a*6s)^{2}$

$-120m*a=-180m*a+36s^{2}*a^{2}$

$0=120m*a-180m*a+36s^{2}*a^{2}$

$0=-60m*a+36s^{2}*a^{2}$

$0=(-60m+36s^{2}*a)*a$

$0=a1$

$\frac{60m}{36s^{2}} = a2$

$1.67m/s^{2}=a2$

If we analyze the situation, we need to have an aceleretarion greater than cero. We are going to choose a = 1.67m/s^2

After, we are going to determine the speed in the first point:

$Sf=So+(a*t)$

$15m/s=So+1.67m/s^2*6s$

$15m/s-(1.67m/s^2*6s)=So$

$4.98m/s=So$

For part c:

We are going to use:

$Sf^2=So^2+(2*a*x)$

$(4.98m/s)^2=0^2+(2*(1.67m/s^2)*x)$

$\frac{24.80m^2/s^2}{3.34m/s^2}=x$

$7.42m=x$

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2 years ago

1. A 930-kg car traveling 56 km/h comes to a complete stop in 2.0 s. What is the

Juli2301 [7.4K]

The force exerted on the car during this stop is 6975N

<u>Explanation:</u>

Given-

Mass, m = 930kg

Speed, s = 56km/hr = 56 X 5/18 m/s = 15m/s

Time, t = 2s

Force, F = ?

F = m X a

F = m X s/t

F = 930 X 15/2

F = 6975N

Therefore, the force exerted on the car during this stop is 6975N

6 0

2 years ago

Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence

Lynna [10]

Answer:

1) Recollapsing universe

2) critical universe

3) Coasting universe

Explanation:

According to the smallest ration (ratio actual mass density to current density) to largest ration, rank of models for expansion of universe are

1) Recollapsing universe -in this, metric expansion of space is reverse and universe recollapses.

2) critical universe - in this, expansion of universe is very low.

3) Coasting universe - in this, expansion of universe is steady and uniform

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2 years ago