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2 years ago

Gradually increasing resistance, increasing repetition, increasing sets are examples of which principle?

Physics

1 answer:

Vikki [24]2 years ago

6 0

Answer:

Resistance training.

Explanation:

Resistance training is based on the principle of gradually increasing resistance, increasing repetition and increasing sets to acquire resistance. Resistance training refers to those physical exercises which are designed and done to improve strength and endurance of the body. Resistance training uses the principle of repetition of a particular exercise in order to attain strength as well as resistance against diseases.

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A muon is a short-lived particle. It is found experimentally that muonsmoving at speed 0.9ccan travel about 1500 meters between

Monica [59]

Answer:

a) $t=5.5\mu s$

b) $\Delta t'=12.5\mu s$

Explanation:

a) Let's use the constant velocity equation:

$v=\frac{\Delta x}{\Delta t}$

v is the speed of the muon. 0.9*c
c is the speed of light 3*10⁸ m/s

$t=\frac{\Delta x}{v}=\frac{1500}{0.9*(3*10^{8})}$

$t=5.5\mu s$

b) Here we need to use Lorentz factor because the speed of the muon is relativistic. Hence the time in the rest frame is the product of the Lorentz factor times the time in the inertial frame.

$\Delta t'=\gamma\Delta t$

$\gamma =\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

v is the speed of muon (0.9c)

Therefore the time in the rest frame will be:

$\Delta t'=\frac{1}{\sqrt{1-\frac{(0.9c)^{2}}{c^{2}}}}\Delta t$

$\Delta t'=\frac{1}{\sqrt{1-0.9^{2}}}\Delta t$

$\Delta t'=\frac{1}{0.44}\Delta t$

No we use the value of Δt calculated in a)

$\Delta t'=2.27*5.5=12.5\mu s$

I hope it helps you!

8 0

3 years ago

The main energy source during exercise of low to moderate intensity is

zhannawk [14.2K]

Answer:

THUS,THE MAJOR SOURCES OF ENERGY DURING EXERCISE ARE CARBOHYDRATES AND FATS.

Explanation:

6 0

3 years ago

g Light that is incident upon the eye is refracted several times before it reaches the retina. As light passes through the eye,

sertanlavr [38]

Answer

Explanation

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7 0

3 years ago

A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em

Sergeeva-Olga [200]

Answer:

a. 0.000002 m

b. 0.00000182 m

Explanation:

36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

$P = \ro g h = 1000 * 10 * 20 = 200000 Pa$

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

$P_1V_1 = P_2V_2$

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

$V_1 = Ah$

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

$P_1Ah_1 = P_2Ah_2$

$h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m$

b) If the temperatures changes, we can still reuse the ideal gas equation above:

$\frac{P_1Ah_1}{T_1} = \frac{P_2Ah_2}{T_2}$

$h_2 = h_1\frac{P_1T_2}{P_2T_1} = 0.36\frac{1.105 * 275}{200000*300} =0.00000182 m$

3 0

3 years ago

A 41 kg girl and a 5.0 kg sled are on the frictionless ice of a frozen lake, 15 m apart but connected by a rope of negligible ma

goldfiish [28.3K]

(a) 0.8 m/s^2

The force exerted on the sled is F = 4.0 N. We can calculate the acceleration of the sled by using Newton's second law:

$F=ma$

where

m = 5.0 kg is the mass of the sled

a is the acceleration of the sled

Solving the equation for a, we find:

$a=\frac{F}{m}=\frac{4.0 N}{5.0 kg}=0.8 m/s^2$

(b) 0.098 m/s^2

According to Newton's third law (action-reaction law), since the girl exerts a force on the sled, then the sled exerts an equal and opposite force on the girl as well. This means that the force exerted on the girl is also F = 4.0 N. As before, we can calculate the acceleration of the girl by using Newton's second law:

$F=ma$

where

m = 41 kg is the mass of the girl

a is the acceleration of the girl

Solving the equation for a, we find:

$a=\frac{F}{m}=\frac{4.0 N}{41 kg}=0.098 m/s^2$

(c) 5.8 s

Taking the initial position of the girl as x = 0, the position at time t of the girl is given by:

$x(t)=\frac{1}{2}a_g t^2$

where $a_g = 0.098 m/s^2$ is the acceleration of the girl.

The sled starts instead its motion from x = 15 m, so its position at time t is given by

$x'(t)=15-\frac{1}{2}a_s t^2$

where $a_s=0.8 m/s^2$ is the acceleration of the sled, and the negative sign is due to the fact that the sled accelerates in opposite direction to the girl's acceleration.

The girl and the sled meet when x(t) = x'(t). So, we find:

$\frac{1}{2}a_g t^2=15-\frac{1}{2}a_s t^2\\(a_g+a_s) t^2=30 m\\t=\sqrt{\frac{30 m}{a_g+a_s}}=\sqrt{\frac{30 m}{0.8 m/s^2+0.098 m/s^2}}=5.8 s$

4 0

3 years ago