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3 years ago

What is the difference between muscular strength and muscular endurance

2 answers:

6 0

Mark Brainliest please

Answer

While endurance is all about how long a muscle can perform, muscular strength is how hard it can perform.

Doing less repetitions with more weight will help you increase your strength. Doing more repetitions with lighter weights will help you build up endurance.

Muscle strength is the ability to exert a maximal amount of force for a short period of time.
Muscle endurance is the ability to do something over and over for an extended period of time without getting tired.

Summary:

1.Muscular strength is one’s ability to carry heavy weights or wield force as opposed to resistance.
2.Muscular endurance is one’s ability to perform muscular tasks for prolonged periods of time.
3.Muscular strength is linked to fast-twitch muscle fibers while endurance is linked to slow-twitch muscle fibers.
4.Muscular strength must be the focus of strength-based sports like rugby while endurance is the priority of long-lasting sports like running.

MAXImum [283]3 years ago

4 0

Muscular strength is a measure of how much force you can exert in one repetition. Muscular endurance refers to the ability to perform a specific muscular action for a prolonged period of time.

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Trong thí nghiệm giao thoa trên mặt nước, hai nguồn kết hợp A và B cách nhau 32 cm dao động cùng

antiseptic1488 [7]

Answer:

48cm by the minus of difference of 32 CM length 20 CM because of the length of the building

5 0

2 years ago

If an organism starts from only one cell,how do we get different body parts and organs

Vladimir [108]

Every cell has chromosomes in its nucleus. In a human cell,
there are 26 of them. AND ... lined up on every chromosome
are thousands and thousands of GENES. Those carry the
instructions for what kind of organ or bone or tissue or hair
this cell will be used to build, how the structure will work, how
tall you will be, what color your eyes will be, how deep your
voice will be, and what illnesses you will have.

Every cell has the complete set of instructions in it. In the past
several years, we have just started to be able to read them, and
extraordinary progress has been made. There's an awful long way
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8 0

3 years ago

b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul

Nat2105 [25]

Answer:

b) Betelgeuse would be $\approx 1.43 \cdot 10^{6}$ times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is $\approx 1.37 \cdot 10^{-5} }$ the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

$B = \displaystyle{\frac{L}{4\pi d^2}}$ (1)

where $B$ is the brightness, $L$ is the star luminosity and $d$ , the distance from the star to the point where the brightness is calculated (measured). Thus:

b) $B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}}$ and $B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}}$ where $L_{Sun}$ is the Sun luminosity ( $3.9 x 10^{26} W$ ) but we don't need to know this value for solving the problem. $ly$ is light years.

Finding the ratio between the two brightness we get:

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }$

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is $1.581 \cdot 10^{-5}\ ly$ . Then

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }$

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0

3 years ago

What are some of the physical benefits to be derived from aerobic

VladimirAG [237]

Some of the benefits are increased heart muscles, increase in blood flow, and reduced body fat

6 0

3 years ago

A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri

exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

$U_i +K_i = U_f + K_f$

where

$U_i$ is the initial potential energy, at the top

$K_i$ is the initial kinetic energy, at the top

$U_f$ is the final potential energy, at halfway

$K_f$ is the final kinetic energy, at halfway

The equation can be rewritten as

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m = 2.7 kg is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 0.54$ is the initial height

u = 0 is the initial speed

$h_f = 0.25 m$ is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

$v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s$

The total mechanical energy of the block can be calculated from the initial conditions, and it is

$E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J$

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

$E=E_e = \frac{1}{2}kx^2$

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

$x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m$

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

$E=E_e = 14.3 J$