All categories

2 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]2 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.

You might be interested in

An electrical circuit is powered by a 1.5-volt battery and contains a light bulb producing 5 Ohms of resistance. There is also a

Fed [463]

Answer:

if the resistor is fitted in series with the bulb , then the current flowing will be 0.15 A.....

if the resistor if fitted in parallel with the bulb ,

the current flowing will be 0.6 A

Explanation:

total potential difference = 1.5V

when resistors in series ,

total equivalent resistance is = 5 + 5 = 10 ohm

so current = 1.5 ÷ 10 = 0.15

when resistors in parallel ,

total equivalent resistance is = (5 × 5)÷(5 + 5) =2.5 ohm

so current = 1.5÷2.5 = 0.6 A

8 0

2 years ago

Which of the following units would need to be converted before being used for a calculation

Andreyy89

The units which would need to be converted before being used for a calculation are cm and 0 ks.

This is referred to a standard which is used to make comparisons in the aspect of measurement.

The units cm and 0 ks aren't in their standard form which is m and s respectively.

Read more about Unit here brainly.com/question/4895463

#SPJ1

6 0

1 year ago

a boy runs from his house to school at speed 5m/s on a straight road and returns with speed of 10m/s.The distance between the ho

CaHeK987 [17]

Explanation:

time spent to run from house to school=100/5=20s

time spent to return from school=100/10=10s

average velocity=200m/(10+20)

=6.66m/s

3 0

2 years ago

Câu 1. Con lắc lò xo treo thẳng đứng, dao động điều hòa với biên độ 2cm và tần số góc 20 rad/s. Chiều dài tự nhiên của lò xo là

natka813 [3]

Answer:

28 cm and 32 cm

Explanation:

1. The spring pendulum hangs vertically, oscillates harmonic with amplitude 2cm and angular frequency 20 rad/s. The natural length of

a spring is 30cm. What is the minimum and maximum length of the spring during the oscillation? Take g = 10m/s2.

As the amplitude is 2 cm and the natural length is 30 cm. So, it oscillates between 30 -2 = 28 cm to 30 + 2 = 32 cm.

So, the minimum length is 28 cm and the maximum length is 32 cm.

8 0

2 years ago

Two electrons are initially at rest separated by a distance of 2nm. At time t=0, they start to move apart due to Coulombic repul

Gnom [1K]

Answer:

$t=2.5\times 10^{-14}\ s$

Explanation:

We know that charge on electron

$q=1.6\times 10^{-19}\ C$

r= 2 nm

We know that force between two charge given

$F=K\dfrac{Q_1Q_2}{r^2}$

Now by putting the value

$F=9\times10^9\dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(2\times 10^{-9})^2}$

$F=5.67\times 10^{-11}\ N$

We know that mass of electron

The mass of electron

$m=9.1\times 10^{-31}\ kg$

F= m a

a= Acceleration of electron

a= F/m

$a=\dfrac{5.67\times 10^{-11}}{9.1\times 10^{-31}}\ m/s^2$

$a=6.2\times 10^{19} m/s^2$

$S=ut+\dfrac{1}{2}at^2$

initial velocity given that zero ,u=0

$20\times 10^{-9}=\dfrac{1}{2}\times 6.2\times 10^{19} t^2$

$t=\sqrt {\dfrac{40\times 10^{-9}}{6.2\times 10^{19}}}$

$t=2.5\times 10^{-14}\ s$

3 0

2 years ago