Any value giving up from the best alternative is called

What is the ability to contract a muscle, or group of muscles, repeatedly without getting tired?

Wewaii [24]

This is known as Muscular Endurance.
Hope this helped, and please mark brainliest!

3 0

3 years ago

Read 2 more answers

In 1913 Niels Bohr formulated a method of calculating the different energy levels of the hydrogen atom.

Svetlanka [38]

They are right the answer is A true

4 0

3 years ago

The velocity of a 1.3 kg block sliding down a frictionless inclined plane is found to be 1.26 m/s. 1.10 s later, it has a veloci

nasty-shy [4]

Answer:

$\theta = 25.3^\circ$

Explanation:

The acceleration of the block can be found by the kinematics equations:

$v = v_0 + at\\5.88 = 1.26 + a(1.1)\\a = 4.2~m/s^2$

Since the plane is frictionless, the only force acting on the block along the motion of the block is its weight.

$F = mg\sin(\theta) = ma\\g\sin(\theta) = a\\(9.8)\sin(\theta) = 4.2\\\theta = 25.3^\circ$

7 0

3 years ago

Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha

mixer [17]

The density of seawater at a depth where the pressure is 500 atm is $1124kg/m^3$

Explanation:

The relationship between bulk modulus and pressure is the following:

$B=\rho_0 \frac{\Delta p}{\Delta \rho}$

where

B is the bulk modulus

$\rho_0$ is the density at surface

$\Delta p$ is the variation of pressure

$\Delta \rho$ is the variation of density

In this problem, we have:

$B=2.3\cdot 10^9 N/m^2$ is the bulk modulus

$\rho_0 =1100 kg/m^3$

$\Delta p = p-p_0 = 500 atm - 1 atm = 499 atm = 5.05\cdot 10^7 Pa$ is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)

Therefore, we can find the density of the water where the pressure is 500 atm as follows:

$\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3$