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k0ka [10]
3 years ago
5

How do intensity, frequency, and time affect physical fitness?

Physics
1 answer:
Y_Kistochka [10]3 years ago
5 0
Important bc it makes it more effective, the specific rate makes or breaks the fitness. Frequency is important to allow your body to rebuild and repair the damage from working out, it allows the body to adapt and time for rest/ healing. Intensity depends on how much your body breaks so the recover time and frequency must be adjusted. Time effects because of the distance between frequencies which plays a role.
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The following nuclear reaction is balanced.<br><br><br><br> True<br> False
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I think its true for tht one!!

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3 years ago
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Question 7 of 25
tiny-mole [99]

Answer:

2PBr₃ + 3Cl₂ → 2PCl₃ + 3Br₂

2Na + MgCl₂ → 2NaCl + Mg

Explanation:

A balanced chemical equation is a chemical equation that have an equal number of elements of each type on both sides of the equation

Among the given chemical reactions, we have;

2PBr₃ + 3Cl₂ → 2PCl₃ + 3Br₂

In the above reaction;

The number of phosphorus, P, on either side of the equation = 2

The number of bromine atoms, Br, on either side of the equation = 6

The number of chlorine atoms, Cl, on either side of the equation = 6

Therefore, the number of elements in the reactant side and products side of the reaction are equal and the reaction is balanced

The second balanced chemical reaction is 2Na + MgCl₂ → 2NaCl + Mg

In the above reaction, there are two sodium atoms, Na,  one magnesium atom and two chlorine atoms on both sides of the reaction, therefore, the reaction is balanced

6 0
3 years ago
Who wants to do my missing physics work
kotegsom [21]

Answer:

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3 years ago
A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500 m/s. It is loaded with 600
3241004551 [841]

Answer:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

Explanation:

The speed of the rocket is given the Tsiolkovsky's differential equation, whose solution is:

v (t) = v_{o} - v_{ex}\cdot \ln \frac{m}{m_{o}}

Where:

v_{o} - Initial speed of the rocket, in m/s.

v_{ex} - Exhaust gas speed, in m/s.

m_{o} - Initial total mass of the rocket, in kg.

m - Current total mass of the rocket, in kg.

Let assume that fuel is burned linearly. So that,

m(t) = m_{o} + r\cdot t

The initial total mass of the rocket is:

m_{o} = 750\,kg

The fuel consumption rate is:

r = -\frac{600\,kg}{30\,s}

r = -20\,\frac{kg}{s}

The function for the current total mass of the rocket is:

m(t) = 750\,kg - (20\,\frac{kg}{s} )\cdot t

The speed function of the rocket is:

v(t) = - 2500\,\frac{m}{s}\cdot \ln \frac{750\,kg -(20\,\frac{kg}{s} )\cdot t}{750\,kg}

The speed of the rocket at given instants are:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

7 0
3 years ago
Interactive Solution 9.1 presents a model for solving this problem. The wheel of a car has a radius of 0.300 m. The engine of th
Murrr4er [49]

Answer:

740 N

Explanation:

We are given that

Radius,r=0.3 m

Torque,\tau=222 Nm

We have to find the magnitude of the static frictional force.

According to question

Torque by engine=Torque by static friction

222=f\times r

f=\frac{222}{r}

f=\frac{222}{0.3}

f=740 N

Hence, the magnitude of static frictional force=740 N

8 0
3 years ago
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